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MasterKill
2003-Nov-19, 01:55 AM
What is the difference between the special theory of realtivity and the general theory of relativity?

Also, did I spell everything in the title right? It looks sort of funny.

Humphrey
2003-Nov-19, 02:13 AM
This is really werid. This was discussed just today in my history of science class! This is how it was described to us: (I am not a physics majoy, I am just repeating what my professor said)

1.Special relativity: the laws of physics are the same for all observers in uniform motion.

2. General relativity:
Adds saying that it allows different frames of reference due to acceleration.

This is all with the assumtion of a constant veliocity of the speed of light.

So basically with the equation v=d/t than with a constant v, distance and time have to change/distort. So if everyone was moving at the speed of light, nobody will notice a difference. But if one person was moving slower, his perception of time or distance would change relative to the other person.



Yah i know its confusing. I barely understand it and i am very sure i got it wrong. But this is what i have down in my note (well I edited out some other stuff). If i am wrong can someone correct me please?

Normandy6644
2003-Nov-19, 02:51 AM
It gets much more complicated on both a physics and mathematical level (I'm reading myself into it now).

Essentially, special relativity has two main postulates:
1) the speed of light is constant in all inertial reference frames
2) the laws of physics will remain the same no matter which reference frame they are applied.

General relativity has 3 main ideas:
1) spacetime is curved and can be described by mathematical "things" called pseudo-Riemannian manifolds
2) there are locally inertial reference frames (flat coordinates) in which the physics in GR are the same as in SR (known as the principle of equivalence - there is no experement that can distinguish an accelerating reference frame from a gravitational field)
3) mass curves spacetime (think of a bowling ball on a stretched out blanket; the ball makes a "dent" in the fabric, curving the are around it)

I think that's about it. Hope that helps :D

Humphrey
2003-Nov-19, 02:58 AM
WhooHoo! i had the basics right. :-D

wtgmatt
2003-Nov-19, 08:13 AM
Everyone always like to go over relativity in terms of a train traveling at the speed of light. Without getting scientific at all, the basic difference is:

Special relativity explains what happens when you are on a train traveling at the speed of light.
General relativity explains what happens when a train traveling at the speed of light goes by you.

With my luck I've probably got that backa**wards, but if I remember the math and physics (which I hopefully remember more of than I do of trains), I think that makes sense.

swansont
2003-Nov-19, 03:58 PM
Everyone always like to go over relativity in terms of a train traveling at the speed of light. Without getting scientific at all, the basic difference is:

Special relativity explains what happens when you are on a train traveling at the speed of light.
General relativity explains what happens when a train traveling at the speed of light goes by you.

With my luck I've probably got that backa**wards, but if I remember the math and physics (which I hopefully remember more of than I do of trains), I think that makes sense.

Relativity explains why your train can't go the speed of light.

wtgmatt
2003-Nov-19, 04:09 PM
Relativity explains why your train can't go the speed of light.

Oops... make that very close to the speed of light. I knew there was a reason why I shouldn't be up early in the morning, much less up early in the morning trying to make sense. Coffee time!

Glom
2003-Nov-19, 04:28 PM
Special relativity applies the experimental result that the speed of light is the same in all inertial frames to extend Newtonian physics to relativistic speeds.

General relativity explains why a person standing on a platform appears to accelerate backwards when you're in a departing train by introducing the idea of equivalence between gravity and acceleration.

Wally
2003-Nov-20, 03:38 PM
The train analogy doesn't work for me. Wouldn't both the person on the train as well as the person standing at the station platform simply be 2 different inertial frames of reference, hence both fit snuggly into the realm of SR? I guess I always thought of GR simply as SR, but with accelleration taken into account (accel. being any "curve" in the path of an object). A simplistic view, granted, but I'm a simplistic kind of guy!

Glom
2003-Nov-20, 04:47 PM
The train analogy doesn't work for me. Wouldn't both the person on the train as well as the person standing at the station platform simply be 2 different inertial frames of reference, hence both fit snuggly into the realm of SR? I guess I always thought of GR simply as SR, but with accelleration taken into account (accel. being any "curve" in the path of an object). A simplistic view, granted, but I'm a simplistic kind of guy!

No. The passenger on the platform observes you. He sees you accelerate relative to him with the train out of the station because of the ballet of forces with train. Everything fits. You are accelerating because there is a force applied to you.

Now, you observe the passenger on the platform. You seem him accelerate relative to you. But there is no force causing him to do that. Acceleration without force? Defies Newton's Second Law. You're in a non-inertial frame because the frame is accelerating with respect to an inertial one. That's where Newtonian mechanics fails. It applies only to inertial frames. It cannot be applied to non-inertial frames. An inertial frame is one where all accelerations observed are due to forces according to Newton's Second Law. An non-inertial frame is one that is accelerating relative to an inertial one.

The way to account for this is to presume a gravitational field acting in the opposite direction to your acceleration relative to the inertial. That explains why you feel like you're being pushed back into your seat and why the passenger on the platform appears to accelerate. It's this gravity equivalence.

General relativity found a way to consider the equivalence to be not only apparent, but actual.

swansont
2003-Nov-20, 06:19 PM
I think if you're fixated on using a train analogy, you have to have the train accelerating for the general relativity. SR will work for either observer for a train moving at constant speed.

Glom
2003-Nov-20, 06:20 PM
Yes. The train is accelerating.

Cougar
2003-Nov-21, 01:26 AM
What is the difference between the special theory of realtivity and the general theory of relativity?
This may not be the difference you're looking for, but....

Special theory: easy.
General theory: hard.

The special theory is actually fairly easy to understand. And a high school student who's fair in math can handle the mathematics involved.

The general theory... well, the general idea isn't so tremendously strange. For the general theory Einstein adds the (very relevant) ideas of gravitation and acceleration into the mix, which were not treated in the special theory. And the math is considerably harder (but I guess something always seems harder when you don't have a good handle on it).
http://www.xmission.com/~dcc/cougar.gif

ljbrs
2003-Nov-21, 01:44 AM
General relativity explains why a person standing on a platform appears to accelerate backwards when you're in a departing train by introducing the idea of equivalence between gravity and acceleration.


glom:

That is part of Special Relativity, too.

ljbrs :wink:

kilopi
2003-Nov-21, 05:26 AM
General relativity explains why a person standing on a platform appears to accelerate backwards when you're in a departing train by introducing the idea of equivalence between gravity and acceleration.

That is part of Special Relativity, too.
Which part? The equivalence of gravity and acceleration is not a part of special relativity.

Glom
2003-Nov-21, 01:21 PM
Special relativity covers inertial frames only.

CincySpaceGeek
2003-Nov-21, 03:23 PM
#-o I think I need a very, VERY big Tylenol and a practice nap. :)

kilopi
2003-Nov-21, 04:12 PM
Special relativity covers inertial frames only.
But even Einstein's original 1905 paper dealt with non-inertial effects (the "twin paradox"). Perhaps that is what she is referring to.

Eta C
2003-Nov-21, 04:20 PM
The twin paradox comes about because both see the other's clock as moving slow while the flight is in progress. The paradox is that on the return the space faring twin is younger. Since SR is limited to inertial (non-accelerated) referance frames, it cannot explain why there is an age difference. Hence the paradox.

In GR, which does include accelerated frames, there is no paradox. The space faring twin undergoes two major accelerations the ground-based one does not. The first is to launch the ship in the first place. The second is to turn the ship around at its destination and return it to earth. General relativity makes the correct predictions for the age difference of the twins and the reasons why are well understood and no longer paradoxical.

SeanF
2003-Nov-21, 05:26 PM
The twin paradox is strictly SR. It doesn't need to involve GR.

The fact is that in the twin paradox there are three inertial frames - the original, planet-bound frame, the frame in which the twin is moving away, and the frame in which the twin is moving back.

In order to get back to the planet, the twin needs to switch inertial frames, and that breaks the reciprocity. The acceleration, the means by which the twin changes frames, isn't really relevent, so neither is GR.

Eroica
2003-Nov-21, 05:32 PM
In GR, which does include accelerated frames, there is no paradox. The space faring twin undergoes two major accelerations the ground-based one does not. The first is to launch the ship in the first place. The second is to turn the ship around at its destination and return it to earth.
There is still a paradox in the GR version, but not quite the one you're talking about. Which of the twins is younger at the moment the space-faring twin turns for home?

Let's assume that they are both at rest at the start relative to a third inertial frame. Then the space-faring twin makes his trip to his destination at a higher velocity than the stay-at-home, and so is younger when he turns for home.

But imagine that at the start both twins were flying through space away from the destination at a constant velocity relative to a fourth frame of reference. Now the space-faring twin has to slow down to get to the destination. So he's the older twin when he turns for home!

Identical situation, but the space-farer is the older twin in one frame, and the younger twin in another!

Of course, by the time he gets home, he'll be the younger twin in all frames (because the journey home requires that he accelerate to catch up with the stay-at-home).

SeanF
2003-Nov-21, 05:47 PM
That's not right, Eroica - there's no difference between "speeding up" and "slowing down." They're both acceleration, and the GR effects would be the same.

The two twins would disagree on who's older at the time of turn-around, though - in fact, the space-faring twin immediately before the turn-around would disagree with the space-faring twin immediately after the turn-around! :o

Eta C
2003-Nov-21, 06:15 PM
GR is relevant since that's the theory you need to use to handle accelerated reference frames (or gravitational fields). SR is "special" because it can only deal with the "special" case of intertial reference frames. The twin paradox arises because SR cannot explain why there would be an age difference upon return. Simply stated, the paradox asks why this should occur since each twin sees the other's clock as running slow. The answer is that the space twin undergoes accelerations at launch and turn-around that the earth twin does not. At this point SR breaks down since we no longer have two inertial frames. When the calculations are done with general relativity (which is "general" because it deals with all possible accelerations and gravitational fields) the correct answers are found and there is no paradox.

Think about all of the train and flashlight gedanken experiments that are use to illustrate SR. In all of them, the train has already reached a steady speed and does not accelerate or turn. That makes it an inertial frame. We don't worry about how it got there, just that it is.

SeanF
2003-Nov-21, 07:38 PM
You need GR to deal with accelerating reference frames, but you don't need to deal with accelerating frames in the twin paradox - they're there, but can be ignored. Consider this:

Put both the twins on spaceships sitting next to each other pointing in the same direction (they're in an inertial frame). Both ships begin accelerating uniformly and simultaneously, and reach .6c simultaneously, at which point they both cut acceleration and coast (in a new inertial frame). After a certain amount of time, one ship decelerates uniformly until it comes to a stop (relative to the starting point - back in the first inertial frame). The second ship continues to coast along for a while long, then goes through the same uniform deceleration and also stops.

The second ship (the one that went farther) then goes through a uniform acceleration back to .6c in the other direction. The first ship begins its uniform acceleration back in the return direction as the second ship approaches so that it reaches .6c as the second ship comes alongside. They they continue to coast back towards the starting point, where they simultaneously decelerate and stop at their original starting point.

Any and all GR / acceleration related effects can be discounted because they will be identical for both ships. However, the twin in the second ship will be younger. Why? Because of the SR effects of the different amounts of time spent in the different inertial frames.

In the standard twin paradox, of course, only one twin experiences acceleration (and the related GR effects), which will compound the age difference, but that twin is still younger as a result of simple SR effects - and the longer he coasts between accelerating and decelerating, the more difference in age there'll be.

There is no actual "paradox", even in SR, because the situation is not reciprocal. One twin remains in the same inertial frame through-out the experiment, while the other spends time in two additional inertial frames.

daver
2003-Nov-21, 07:38 PM
GR is relevant since that's the theory you need to use to handle accelerated reference frames (or gravitational fields). SR is "special" because it can only deal with the "special" case of intertial reference frames. The twin paradox arises because SR cannot explain why there would be an age difference upon return.
No, you can deal with the problem in SR as well. As SeanF pointed out, there are three reference frames (the stationary twin, the inbound twin, and the outbound twin). If you formulate the problem to make the reference frames explicit then the paradox goes away. No need for GR.

Eroica
2003-Nov-21, 08:41 PM
That's not right, Eroica - there's no difference between "speeding up" and "slowing down." They're both acceleration, and the GR effects would be the same.
I guess that's right, but there is a difference in the SR effects of travelling faster and travelling slower (after the accelerations are over). I still contend that I described the paradox correctly, but it's really an SR effect. At the turning point, one twin is older in one frame of reference and younger in another. :-k

SeanF
2003-Nov-21, 08:55 PM
That's not right, Eroica - there's no difference between "speeding up" and "slowing down." They're both acceleration, and the GR effects would be the same.
I guess that's right, but there is a difference in the SR effects of travelling faster and travelling slower (after the accelerations are over). I still contend that I described the paradox correctly, but it's really an SR effect. At the turning point, one twin is older in one frame of reference and younger in another. :-k

Actually, at any point while the twin is moving away from Earth, he's older in his own reference frame and younger in the Earth's reference frame.

After he's turned around and is heading back, he's younger in both reference frames, but there's disagreement as to how much younger - until he gets back to Earth and stops, at which point everybody agrees on everything! :D

Eta C
2003-Nov-21, 09:15 PM
I'm in the midst of checking up on this one some more before making another declarative statement. But here's a preliminary observation. While one can resolve the twin paradox (calculate the age difference) using just the SR time dilation relations, you cannot explain why that time difference occurs without invoking the perception of the astronaut. Preferably, one (especially Einstein) wants to get a purely physical description that doesn't require an observer. One of the texts I'm looking at calls this "one of the loose ends that drove Einstein to go beyond the simple version of relativity (SR) we have studied so far."

The answer has to do with the effect gravitational fields (or accelerated frames, they are equivalent) have on clocks as a function of height. It's a well known (and experimentally proven) consequence of GR that clocks at high altitude run faster than those at ground level. The effect is proportional to the acceleration and the height. For the spaceship changing direction at the distant star both the "height" (in this case the distance to earth) and the acceleration are large. The overall impact is that during the turnaround period of acceleration, the clock on earth is running much faster than the one on the spaceship. Thus most of the time difference occurrs during the turn-around. To quote the same text "Note this is not the case of two observers each believing the other's clock is slow. Everyone agrees that in an accelerated reference frame the speed of a clock depends on its position and clocks higher "up" run faster."

Anyway, I'm still checking the math on this one, so I reserve the right to change my opinion. It's been a while since I've had to deal with this sort of qual problem question, so I'm rusty. I did recollect this argument though, which is why I was so insistent on why GR was required to fully resolve the twin paradox. More to come.

Glom
2003-Nov-21, 09:21 PM
I think that the time dilation under GR is related to the gravitional field. At higher altitude, the G-field is weaker and so there is less time dilation. In an accelerating frame though, the time dilation is uniform throughout.

Eta C
2003-Nov-21, 09:26 PM
That is right, as the acceleration due to gravity decreases with distance, down the time dilation effect will also drop off. For a constant acceleration, though that won't be the case. Anyway, I'm still working on this one, so I'll withold further comment for a while. 8)

daver
2003-Nov-21, 10:53 PM
I'm in the midst of checking up on this one some more before making another declarative statement. But here's a preliminary observation. While one can resolve the twin paradox (calculate the age difference) using just the SR time dilation relations, you cannot explain why that time difference occurs without invoking the perception of the astronaut. Preferably, one (especially Einstein) wants to get a purely physical description that doesn't require an observer. One of the texts I'm looking at calls this "one of the loose ends that drove Einstein to go beyond the simple version of relativity (SR) we have studied so far."

This doesn't seem at all right to me. I don't see how the perception of the astronaut has anything to do with the problem.


For the spaceship changing direction at the distant star both the "height" (in this case the distance to earth) and the acceleration are large. The overall impact is that during the turnaround period of acceleration, the clock on earth is running much faster than the one on the spaceship. Thus most of the time difference occurrs during the turn-around. To quote the same text "Note this is not the case of two observers each believing the other's clock is slow. Everyone agrees that in an accelerated reference frame the speed of a clock depends on its position and clocks higher "up" run faster."

I don't believe you're applying the right principle here--simultaneity is i think more the issue.

Cougar
2003-Nov-21, 11:32 PM
After he's turned around and is heading back, he's younger in both reference frames, but there's disagreement as to how much younger - until he gets back to Earth and stops, at which point everybody agrees on everything! :D
That's a relief!

kilopi
2003-Nov-22, 12:12 AM
While one can resolve the twin paradox (calculate the age difference) using just the SR time dilation relations, you cannot explain why that time difference occurs without invoking the perception of the astronaut.
Einstein did, in his original 1905 paper on SR, see the third of the following links:

http://mentock.home.mindspring.com/twins.htm

http://mentock.home.mindspring.com/twin2.htm

http://mentock.home.mindspring.com/twinrdux.htm

Eroica
2003-Nov-22, 10:32 AM
All this talk about the Earth's gravitational field and the height above the Earth is just a distraction. While it is true that clocks run more slowly on the surface of the Earth (where gravity is stronger) than in deep space, the difference is minuscule - we're talking nanoseconds - compared to the difference due to high velocities close to c. The Twin Paradox is best considered out in space, far from all gravitational fields.

Eroica
2003-Nov-22, 10:48 AM
The overall impact is that during the turnaround period of acceleration, the clock on earth is running much faster than the one on the spaceship. Thus most of the time difference occurs during the turn-around.
Hmmm ... that's not how I see it. The accelerations are what "boost" the space-faring twin into a different time frame, one in which his clock is running more slowly that the other twin's clock. But how much younger he ends up depends on how long he remains in that time frame (even if he's no longer accelerating).

Try to imagine the paradox in which the space-faring twin is able to accelerate and decelerate instantaneously without splattering himself against the bulkheads. (Inertial dampeners should do the trick! :D ) He doesn't actually gain any time during the accelerations, but they are still the key to the paradox.

Glom
2003-Nov-22, 06:53 PM
With regard to the Twins Paradox, I asked about it here (http://www.badastronomy.com/phpBB/viewtopic.php?t=5814). Daver gave a pretty good response, which I still don't fully comprehend. :oops:

daver
2003-Nov-24, 10:00 PM
With regard to the Twins Paradox, I asked about it here (http://www.badastronomy.com/phpBB/viewtopic.php?t=5814). Daver gave a pretty good response, which I still don't fully comprehend. :oops:

For me, i like to see the numbers before i think i understand something, but in this case the numbers might be getting in the way. I could try giving another (probably equally confusing) explanation if you'd like, or it might be more helpful to find some victim and try to explain the apparent paradox to them.

Glom
2003-Nov-24, 10:46 PM
I vote for finding a victim.

Wally
2003-Nov-26, 02:26 PM
I think if you're fixated on using a train analogy, you have to have the train accelerating for the general relativity. SR will work for either observer for a train moving at constant speed.

Bingo, I was under the impression the train was zooming by the station at a constant velocity rather than accelorating as it pulls out.

Wally

Wally
2003-Nov-26, 03:08 PM
GR is relevant since that's the theory you need to use to handle accelerated reference frames (or gravitational fields). SR is "special" because it can only deal with the "special" case of intertial reference frames. The twin paradox arises because SR cannot explain why there would be an age difference upon return. Simply stated, the paradox asks why this should occur since each twin sees the other's clock as running slow. The answer is that the space twin undergoes accelerations at launch and turn-around that the earth twin does not. At this point SR breaks down since we no longer have two inertial frames. When the calculations are done with general relativity (which is "general" because it deals with all possible accelerations and gravitational fields) the correct answers are found and there is no paradox.



First off, I stopped reading this thread after the above quote to post this response, so pardon me if this was covered already.

I disagree wth the above statement that SR cannot explain the twin paradox. It can, and does. The "paradox" has nothing to do with acceleration at launch or at turn around (well, at least not acceleration in a GR sense of the word). It exists (such as it is) even when you consider "instanteous acceleration" at both points. The key, as others have posted, is that you MUST take into account the 3rd inertial frame of reference that exists as soon as twin B instantaneously reverses course and heads back to earth. It is whichever twin inhabits this 3rd frame that determines who will be the older and who will be the younger.

If, rather than the "space bound" twin reversing course and heading back, you had the earth-bound twin suddenly take off in pursuit and eventually catch up with the other one, it'd be him that "entered" the 3rd FoR, and hence he would be now be the younger of the 2.

Eta C
2003-Nov-26, 03:43 PM
I don't disagree that you can compute the outcomes of the twin paradox without reference to GR. I've done it myself back during my student days. My point is that the spaceborne twin undergoes massive accelerations to get close to the speed of light and to reverse that motion. Thus he is in a decidedly non-inertial reference frame. To understand what is happening to him (or her) during those periods requires GR (note this is not the same as saying you need GR to predict the outcome). As he accelerates to change direction at the far point he perceives the Earthbound clock as running fast due to that acceleration. Every text I've looked at agrees with this point of view.

(edited once to correct myself. He perceives the earth clock as running fast, not slow.) :oops:

Sam5
2003-Nov-29, 01:40 AM
The twin paradox is strictly SR. It doesn't need to involve GR.

The fact is that in the twin paradox there are three inertial frames - the original, planet-bound frame, the frame in which the twin is moving away, and the frame in which the twin is moving back.

In order to get back to the planet, the twin needs to switch inertial frames, and that breaks the reciprocity. The acceleration, the means by which the twin changes frames, isn't really relevent, so neither is GR.

Hi.

Excuse me, but this is not correct.

The number of “inertial frames” is irrelevant. Einstein’s first clock paradox thought experiment had only two inertial frames, one for the A clock and one for the B clock. These two clocks move toward each other with “relative motion”, no acceleration, and no “absolute motion”.

The fundamental error of his basic 1905 theory is clearly revealed in his first thought experiment in which the clock paradox appears:

“From this there ensues the following peculiar consequence. If at the points A and B of K there are stationary clocks which, viewed in the stationary system, are synchronous; and if the clock at A is moved with the velocity v along the line AB to B, then on its arrival at B the two clocks no longer synchronize, but the clock moved from A to B lags behind the other which has remained at B by 1/2tv^2/c^2 (up to magnitudes of fourth and higher order), t being the time occupied in the journey from A to B.”

But based on the terms of the first 3 sections of the theory and the symmetry principle, this is incorrect, since both the A observer and the B observer are supposed to “see” each other’s clock “slow down” exactly the same amount. So, during the “relative motion”, B “sees” A move toward B and “time dilate”, while A “sees” B move toward A and “time dilate”.

The error Einstein missed in this thought experiment was that he neglected to report what the A observer “sees” during the “relative motion”. He only reports what the B observer “sees”, and that’s how he wound up with the paradox, which is unresolvable, and it points out the basic flaw and error in his 1905 SR theory. He later corrected the error in his GR theory.

The reason this original thought experiment has been converted into the so-called “twins paradox”, is so that the errors in SR theory can be covered up, usually by two common methods: 1) by adding “acceleration” to the “traveling twin’s” frame, or 2) as you have done, by claiming that only one “twin” travels and that twin "changes frames", and somehow the changing of the frames clears up all the errors of the 1905 SR theory. But it doesn’t clear them up, it only hides them.

Sam5
2003-Nov-29, 02:06 AM
But here's a preliminary observation. While one can resolve the twin paradox (calculate the age difference) using just the SR time dilation relations, you cannot explain why that time difference occurs without invoking the perception of the astronaut. Preferably, one (especially Einstein) wants to get a purely physical description that doesn't require an observer. One of the texts I'm looking at calls this "one of the loose ends that drove Einstein to go beyond the simple version of relativity (SR) we have studied so far."

One cannot resolve the clock paradox of SR by just using SR. That is because the SR theory is in error. In the fist place, “relative motion” can not possibly cause any clock to change its rate.

What happened was, after the SR theory was published, the errors of the theory were pointed out to Einstein and they became obvious. He later got rid of the clock paradox in the GR theory, by having only ONE of the clocks actually slow down, and generally speaking, it is an atomic clock that does this, due to acceleration. This is a physical effect that acceleration has on atomic clocks.

Other clocks can speed up due to acceleration or increased temperature, such as pendulum clocks and thermodynamic clocks. So it is not all of “time” in an accelerating frame that slows down or speeds up, it is the clock, and the direction and rate of the speed-up or slow-down depends on the type of clock used. Generally speaking, human beings are not like atomic clocks. They are more like thermodynamic clocks. In traveling spacecraft, the local thermodynamic effects must be considered in order to minimize clock drift due to temperature changes inside the spacecraft, which are totally unrelated to either “relative motion” or “acceleration”.

SeanF
2003-Nov-29, 03:08 AM
The twin paradox is strictly SR. It doesn't need to involve GR.

The fact is that in the twin paradox there are three inertial frames - the original, planet-bound frame, the frame in which the twin is moving away, and the frame in which the twin is moving back.

In order to get back to the planet, the twin needs to switch inertial frames, and that breaks the reciprocity. The acceleration, the means by which the twin changes frames, isn't really relevent, so neither is GR.

Hi.

Hi! :)

Excuse me, but this is not correct.
Yes, it is.

The number of “inertial frames” is irrelevant. Einstein’s first clock paradox thought experiment had only two inertial frames, one for the A clock and one for the B clock. These two clocks move toward each other with “relative motion”, no acceleration, and no “absolute motion”.

The fundamental error of his basic 1905 theory is clearly revealed in his first thought experiment in which the clock paradox appears:

“From this there ensues the following peculiar consequence. If at the points A and B of K there are stationary clocks which, viewed in the stationary system, are synchronous; and if the clock at A is moved with the velocity v along the line AB to B, then on its arrival at B the two clocks no longer synchronize, but the clock moved from A to B lags behind the other which has remained at B by 1/2tv^2/c^2 (up to magnitudes of fourth and higher order), t being the time occupied in the journey from A to B.”

You're right that there's only two inertial frames in that description. However, it is still true that Clock B changes inertial frames while Clock A does not, hence there is no paradox.

But based on the terms of the first 3 sections of the theory and the symmetry principle, this is incorrect, since both the A observer and the B observer are supposed to “see” each other’s clock “slow down” exactly the same amount. So, during the “relative motion”, B “sees” A move toward B and “time dilate”, while A “sees” B move toward A and “time dilate”.

Yes, that is what happens.

The error Einstein missed in this thought experiment was that he neglected to report what the A observer “sees” during the “relative motion”. He only reports what the B observer “sees”, and that’s how he wound up with the paradox, which is unresolvable, and it points out the basic flaw and error in his 1905 SR theory. He later corrected the error in his GR theory.

The reason this original thought experiment has been converted into the so-called “twins paradox”, is so that the errors in SR theory can be covered up, usually by two common methods: 1) by adding “acceleration” to the “traveling twin’s” frame, or 2) as you have done, by claiming that only one “twin” travels and that twin "changes frames", and somehow the changing of the frames clears up all the errors of the 1905 SR theory. But it doesn’t clear them up, it only hides them.
Sorry, but you're wrong. SR is not flawed in the way you claim. GR is an extension to SR, not a correction of it. The Twin Paradox is entirely SR-related and SR-driven, but it is not really a Paradox.


One cannot resolve the clock paradox of SR by just using SR. That is because the SR theory is in error. In the fist place, “relative motion” can not possibly cause any clock to change its rate.

Relative motion doesn't cause the clock to change its rate. Relative motion causes a difference in the perception of time itself. Since a clock measures time, it runs at a different rate.

Sam5
2003-Nov-29, 03:31 AM
You're right that there's only two inertial frames in that description. However, it is still true that Clock B changes inertial frames while Clock A does not, hence there is no paradox.

No, sorry, but you are incorrect again.

In “On the Electrodynamics of Moving Bodies”, published in 1905, in the Einstein thought experiment I quoted from the original paper, there is no change in inertial frames for either clock after the initial “relative motion” begins between clock A and clock B. There is no “blast off” and there is no “turn-around”.

What you are talking about is a high-school level, popular science book, “Mary and Bill”, “human twins”, thought experiment, with only one twin “traveling”, and “blasting off” from the earth, and then “turning around” and “returning to earth”.

This type of thought experiment was a modification of Einstein’s original, and it was modified so that the excuse of “acceleration” and “frame change” could be used to cover up the errors of the 1905 SR theory.

SeanF
2003-Nov-29, 04:33 AM
You're right that there's only two inertial frames in that description. However, it is still true that Clock B changes inertial frames while Clock A does not, hence there is no paradox.

No, sorry, but you are incorrect again.

In “On the Electrodynamics of Moving Bodies”, published in 1905, in the Einstein thought experiment I quoted from the original paper, there is no change in inertial frames for either clock after the initial “relative motion” begins between clock A and clock B. There is no “blast off” and there is no “turn-around”.


Mea culpa. As you originally quoted it:

"If at the points A and B of K there are stationary clocks which, viewed in the stationary system, are synchronous; and if the clock at A is moved with the velocity v along the line AB to B, then on its arrival..."

So the key here is that the clocks are synchronized when "viewed in the stationary system." So, an observer with Clock A says the clocks are synchronized at the start of the experiment, but an observer with Clock B would not agree - the Clock B observer would say that Clock B is significantly behind Clock A. This is why they both see the other clock as ticking more slowly, but agree that B is still behind A when they meet.


What you are talking about is a high-school level, popular science book, “Mary and Bill”, “human twins”, thought experiment, with only one twin “traveling”, and “blasting off” from the earth, and then “turning around” and “returning to earth”.

This type of thought experiment was a modification of Einstein’s original, and it was modified so that the excuse of “acceleration” and “frame change” could be used to cover up the errors of the 1905 SR theory.

Again, GR is an extension of SR, it is not a correction. And believe me, you're not the first person to claim SR is "flawed." You're not even the first to do it on this BB. It's not.

Sam5
2003-Nov-29, 05:05 AM
So the key here is that the clocks are synchronized when "viewed in the stationary system." So, an observer with Clock A says the clocks are synchronized at the start of the experiment, but an observer with Clock B would not agree - the Clock B observer would say that Clock B is significantly behind Clock A. This is why they both see the other clock as ticking more slowly, but agree that B is still behind A when they meet.

No, sorry. Read it again carefully.

He said:


“If at the points A and B of K there are stationary clocks which, viewed in the stationary system, are synchronous;”

He is favoring the “K” system in which B always resides. But in the thought experiment we also have the K1 system, in which A always resides.

See his symmetry principle a few paragraphs up, which states:

“It is clear that the same results hold good of bodies at rest in the “stationary'' system, viewed from a system in uniform motion.”

He put the word “stationary” in quotes because he knows that either system is viewed as the “stationary system” by observers that are fixed within that system. This means that in his thought experiment, observer A is fixed and stationary with clock A in the K1 system, and observer B is fixed and stationary with clock B in the K system. Before the motion begins, both K and K1 are the same system, and they share points A and B on the x axis. After the motion begins. K and K1 split. A stays with K1 and B stays with K. While the relative motion between the two is along the x axes of K and K1.

And remember, the motion is only “relative”.

In both the K and K1 system, before the relative motion begins, both A and B see their own and each other’s clocks as being both “stationary” and “synchronous”.

He trips up his readers a bit (and himself) by referring only to points A and B in the K system, but these are also points A and B in the K1 system, before the motion begins, since the relative motion hasn’t yet begun in the first part of his first sentence. Then, when the relative motion begins. A stays with the K1 system, while B stays with the K system.

And it’s very important to remember that the motion is only “relative”. So, points A and B are “moving” only “relative” to one another. It is only observer B that “sees” A “move”, while A “sees” B “move”.

By his keeping his mind focused only on the “K” system, he forgets all about the “K1” system and the A observer.

So, when you say, “So the key here is that the clocks are synchronized when ‘viewed in the stationary system’”, read the thought experiment again, carefully. He says, “If at the points A and B of K there are stationary clocks.” Ok? So, points A and B are stationary in K and K1 before the relative motion begins, and at A and B there are stationary clocks, that are “synchronous”.

Before the motion begins, both K and K1 are “stationary systems”, as seen by observers in BOTH systems. And then.... the relative motion begins, and observer A in K1 “sees” B in K “move”, while observer B in K “sees” A in K1 “move”.

You said, “And believe me, you're not the first person to claim SR is "flawed."”

I realize that. The information is all over the internet now. And you are not the first person who has said it is “not flawed”.

Sam5
2003-Nov-29, 05:20 AM
So, an observer with Clock A says the clocks are synchronized at the start of the experiment, but an observer with Clock B would not agree - the Clock B observer would say that Clock B is significantly behind Clock A. This is why they both see the other clock as ticking more slowly, but agree that B is still behind A when they meet.


You said:

“So, an observer with Clock A says the clocks are synchronized at the start of the experiment, but an observer with Clock B would not agree”

No, this is not true. The thought experiment itself says that in the beginning, points A and B are “stationary”, ie “not moving”. So there is no reason for B to disagree with A at this point in the thought experiment.

And he didn’t use the word “synchronized”, he used the word “synchronous”, which means they are running at the same rate.

Then you said:

“the Clock B observer would say that Clock B is significantly behind Clock A.”

Why would he say that, if, as Einstein said, “at the points A and B of K there are stationary clocks which, viewed in the stationary system, are synchronous;”?

Before the clocks move, neither is “significantly behind” the other, since they are “synchronous”.

Then you said:

“This is why they both see the other clock as ticking more slowly,”

No, in the SR theory, they only “see” the other clock as “ticking more slowly” by means of light signals when the relative motion is taking place, and A “sees” the B clock “tick more slowly”, while B “sees” the A clock “tick more slowly,” and only during the relative motion.

Then you said:

”but agree that B is still behind A when they meet.”

No.

Notice that you said “B is still behind A when they meet”, but Einstein said, “but the clock moved from A to B lags behind the other which has remained at B”, which is just the opposite of what you said.

And there you have the eternal paradox.

SeanF
2003-Nov-29, 07:48 AM
Maybe I shouldn't post after midnight. I was describing as if Clock B moved, not Clock A. I've probably got everybody confused now.

At one point you said:

In “On the Electrodynamics of Moving Bodies”, published in 1905, in the Einstein thought experiment I quoted from the original paper, there is no change in inertial frames for either clock after the initial “relative motion” begins between clock A and clock B. There is no “blast off” and there is no “turn-around”.


I bolded that phrase. There is a change in inertial frames when the relative motion begins. That is the "blast off". You acknowledge that in later posts with K and K1. If Clock A starts out in K and then moves to K1, it's changing inertial frames. It's not written that way, of course - A is "always in K1" and B is "always in K", and K1 somehow changes from being the same as K to being different. The effect is the same, though - an object at rest will remain at rest and an object in motion will remain in motion unless a force acts upon it. If A and B were motionless relative to each other and are now moving relative to each other, there was a force somewhere and at least one of them changed frames.

The "motion is relative" only means that any definition of K or K1 as "at rest" is arbitrary. We can consider both Clocks A and B to be at rest at the beginning and A starts moving towards B. We can consider that they were both already moving and A stopped, allowing B to catch up to it. We can also consider that they were already moving in the opposite direction and A sped up to catch B. But if we're establishing that B always remains in K and A does not, we cannot pretend that B changed its velocity.

The thought experiment can be looked at two ways. Either

1) The clocks are motionless relative to each other at the beginning of the experiment and then begin moving towards each other; or

2) The clocks are already in motion relative to each other when the experiment starts.

If we're dealing with situation 1, then we have a change in inertial frames - at least one of the clocks has to change from the original "at rest" frame to a new one. Which clock changes frames will determine which clock is behind when they meet.

If we're dealing with situation 2, then we can have no agreement on their status of synchronization. There will be agreement on what times they are displaying at the moment they pass each other, but not on what times they are displaying when the experiment starts (and therefore not on which clock runs "slow").

Either way, there's no paradox.

Now, I'm going to bed. I'll check in again tomorrow. :)

Eroica
2003-Nov-29, 10:04 AM
Einstein’s first clock paradox thought experiment had only two inertial frames, one for the A clock and one for the B clock. These two clocks move toward each other with “relative motion”, no acceleration, and no “absolute motion”.

“From this there ensues the following peculiar consequence. If at the points A and B of K there are stationary clocks which, viewed in the stationary system, are synchronous; and if the clock at A is moved with the velocity v along the line AB to B, then on its arrival at B the two clocks no longer synchronize, but the clock moved from A to B lags behind the other which has remained at B ...."

You're right that Einstein's thought-experiment has only two inertial frames, but you're wrong when you assert that it has no acceleration. At the start of the experiment, both clocks are stationary. Then the clock at A is accelerated to velocity v, thus entering a new inertial frame.

That, at least, is my reading of the passage.

If Clock A is already moving towards B at velocity v at the start of the experiment, then the experiment breaks down - because we don't know which clock has shifted from a stationary frame to a moving frame.

PS: Someone once said that if you want to learn about relativity, don't read Einstein! :D

Glom
2003-Nov-29, 11:58 AM
I think the reason for the problem in solving the twins paradox in SR is the assumption of symmetry. We think if one twin is experiencing time dilation relative to the other, then the other should be experiencing the same thing relative to the first. The time dilation is symmetrical.

However, we forget about length contraction. Length contraction is not symmetrical. The length contraction of the traveller observed by the homebound twin affects the spacecraft, making it squashy. But it doesn't really matter all that much. However, the length contraction observed by the traveller is of the whole region of space, including the distance he has to travel. So he has less distance to travel and gets back sooner.

Normandy6644
2003-Nov-29, 04:02 PM
The twin paradox is all about who is moving, the brother one the spaceship or the brother on the earth. If the ship is moving, one would expect that the brother on the ship would not have aged so much, whereas the brother on earth would have aged many years. On the other hand, if it is the earth that is moving and not the ship, the oppositie effect occurs. SR attempts to resolve this, as does GR. I can't remember off the top of my head what SR says about it, but GR would say there is not absolute motion in either reference frame.

Sam5
2003-Nov-29, 05:30 PM
No, sorry.

K and K1 can be thought of as “the same” or as “separate”, it doesn’t matter. So there is no “change of frames” for either A or B.

You know enough to know that in SR theory, Einstein was always talking about two inertial frames that were both either “stationary” or “relatively moving”. When he said “A moves to B”, he was only thinking of what the B observer saw. But of course, the A observer would see B move to A.

He left out acceleration because he wanted to describe what he thought the effect “relative motion” would have on the clocks. So A and B never “change frames” in his thought experiment. Their two frames are always inertial. They are either moving relatively or they are not. And, anyway, Einstein never mentioned what you keep bringing up, the “changing” of frames, since it wasn’t important. The only thing that was important was the “relative motion”, and what each observer “saw” during the motion.

You said, “The effect is the same, though - an object at rest will remain at rest and an object in motion will remain in motion unless a force acts upon it.” Right, that’s Newton’s First Law.

You said, “If A and B were motionless relative to each other and are now moving relative to each other, there was a force somewhere and at least one of them changed frames.”

No, he leaves out “force”. There is no “force” in the Kinematical part of the SR theory. He says, “Now to the origin of one of the two systems (k) let a constant velocity v be imparted in the direction of the increasing x of the other stationary system (K)....” So you are supposed to imagine the motion being “imparted” with no “force” or “acceleration” involved. You keep thinking in terms of first “reality” and then “SR theory”. But you can’t do that and understand the theory. You must think only in terms of SR theory, and then you will begin to see that it doesn’t match reality.

You said, “The "motion is relative" only means that any definition of K or K1 as "at rest" is arbitrary.”

Ahh, that is a common error. What relative motion actually means in the theory is that K “sees” himself as “stationary” while he “sees” K1 as “moving”, and K1 “sees” himself as “stationary” while he “sees” K as moving, and neither is moving “absolutely”. We can tell that because neither felt “acceleration”.

You said, “We can consider both Clocks A and B to be at rest at the beginning and A starts moving towards B.”

You can’t take his thought experiment in isolation. You’ve got to follow the rules of the first three sections of his paper. Since this motion is only “relative”, then A sees B “move”, while B sees A “move”.

He didn’t clear up this error until about 10 years later when he added acceleration to the General Theory and we can ALL agree on which one really “moved”, because that one felt acceleration.

It is quite common for people to try to add acceleration to the SR theory, but he said many times in different papers that the SR theory purposely leaves out acceleration, and he made a mistake in doing that. A mistake which he corrected later with GR theory.

You said, “The clocks are motionless relative to each other at the beginning of the experiment and then begin moving towards each other...” Yes, you are correct about that. And according to the theory, A sees B’s clock “dilate”, while B sees A’s clock “dilate”. So, just one of the clocks can’t “really” dilate.

The “change in inertial frames” doesn’t matter. An inertial frame is an inertial frame is an inertial frame. The errors of SR are very subtle and not easy to detect. That’s why so many people usually just give up trying to understand the 1905 paper, and they just settle for reading modern (and incorrect) thought experiments to which “acceleration” or “frame changing” are added.

Sam5
2003-Nov-29, 05:38 PM
However, we forget about length contraction. Length contraction is not symmetrical. The length contraction of the traveller observed by the homebound twin affects the spacecraft, making it squashy. But it doesn't really matter all that much. However, the length contraction observed by the traveller is of the whole region of space, including the distance he has to travel. So he has less distance to travel and gets back sooner.

No, sorry.

Einstein specifically said in the 1905 theory:

“It is clear that the same results hold good of bodies at rest in the "stationary'' system, viewed from a system in uniform motion.”

In his 1907 paper on Special Relativity, he said that the actual shape of a body is called its “geometrical shape”. And he said, “The latter obviously does not depend on the state of motion of a reference system.”

This means that the so-called “length contraction” of SR is not “real”, and the bodies really do NOT change shape during the relative motion.

Sam5
2003-Nov-29, 05:47 PM
You're right that Einstein's thought-experiment has only two inertial frames, but you're wrong when you assert that it has no acceleration. At the start of the experiment, both clocks are stationary. Then the clock at A is accelerated to velocity v, thus entering a new inertial frame.

No.

He stated in the SR theory and in other books and papers that he left any consideration of the “acceleration” of frames out of the SR theory.

What you are thinking about is “reality”, not the SR theory. In reality, there is “acceleration”, but in SR theory there is not.

Sam5
2003-Nov-29, 05:54 PM
The twin paradox is all about who is moving, the brother one the spaceship or the brother on the earth. If the ship is moving, one would expect that the brother on the ship would not have aged so much, whereas the brother on earth would have aged many years. On the other hand, if it is the earth that is moving and not the ship, the oppositie effect occurs. SR attempts to resolve this, as does GR. I can't remember off the top of my head what SR says about it, but GR would say there is not absolute motion in either reference frame.


You are getting SR mixed up with GR. In SR, there is only “relative motion” between two observers. In GR, there is “absolute motion” of one observer, and when he moves absolutely, he feels acceleration and that slows down his atomic clocks.

In GR the accelerated one sees an unaccelerated atomic clock as speeding up, while he sees his own clock as not changing rates. While the unaccelerated observer sees the accelerated clock as slowing down, while he sees his own clock as not changing rates. But here we have no paradox, since one atomic clock really is ticking more slowly than the other.

But in SR theory, there is a paradox, because each of two observers “see” the other clock as “slowing down”. Neither “sees” either clock “speeding up”, while both see each other’s clock as “slowing down”. Thus, the paradox.

Glom
2003-Nov-29, 06:16 PM
You are getting SR mixed up with GR. In SR, there is only “relative motion” between two observers. In GR, there is “absolute motion” of one observer, and when he moves absolutely, he feels acceleration and that slows down his atomic clocks.

That's a contradiction in terms. There is no such thing as absolute motion! That's what relativity is all about.

Sam5
2003-Nov-29, 06:41 PM
You are getting SR mixed up with GR. In SR, there is only “relative motion” between two observers. In GR, there is “absolute motion” of one observer, and when he moves absolutely, he feels acceleration and that slows down his atomic clocks.

That's a contradiction in terms. There is no such thing as absolute motion! That's what relativity is all about.

In these thought experiments, what I mean by “absolute motion” can be determined by the moving observer feeling acceleration. By this I mean, a “change in the rate or direction of motion”, an “absolute motion” relative to the previous “unaccelerated motion state” of the object that experiences the absolute motion. In this sense, the “absolute motion” of GR theory takes place during the change of velocity, during the acceleration, but when the acceleration stops, then the motion becomes “relative” to all observers.

In the SR theory, the “absolute motion” of one or the other or of both observers can’t be determined, since neither feel acceleration and neither changes velocity, since both frames are always “inertial frames”, as per Newton’s First Law. What Einstein does in the SR theory is imagine two frames at first not moving “relatively”, and then he imagines them “moving relatively”.

In the “Mary and John” twins thought experiment, both Mary and John start off on the earth, and then John does experience an “absolute motion” relative to the earth and Mary, while Mary does not experience an “absolute motion” relative to the earth or to John. But that’s GR theory, not SR theory. And of course, in this thought experiment, a real “force” is impressed upon John, but not on Mary.

Sam5
2003-Nov-29, 06:52 PM
That's a contradiction in terms. There is no such thing as absolute motion! That's what relativity is all about.


Here. See this quote in this link.

“The term 'absolute motion' has no meaning in the following sense; if you were moving in a straight line, with constant velocity, and there were no windows to see the outside, there is no way you can tell what speed you are moving at (or, for that matter, whether you are moving at all) with any measurement. Thus, speed has only meaning relative to something else.

However, the term 'absolute acceleration' _does_ have a meaning. If you were on a roller coaster, even on one which has closed cars with no windows, you would still be able to tell you were moving -- you'd be thrown every which way, and you would even be able to feel the motion in your guts, given you were securely fastened in your seat. Now, when riding a roller-coaster, you definitely know it is YOU that is moving, and not your friend standing on the ground waving to you.”

Absolute Motion Link (http://www.physlink.com/Education/AskExperts/ae118.cfm)

Eroica
2003-Nov-29, 08:27 PM
However, the term 'absolute acceleration' _does_ have a meaning. If you were on a roller coaster, even on one which has closed cars with no windows, you would still be able to tell you were moving -- you'd be thrown every which way, and you would even be able to feel the motion in your guts, given you were securely fastened in your seat.
No, you wouldn't. For all you know, all that tossing about might be caused by gravity. That's what GR's principle of equivalence is all about. The effects of gravity and acceleration cannot be distinguished.

Sam5
2003-Nov-29, 10:36 PM
No, you wouldn't. For all you know, all that tossing about might be caused by gravity. That's what GR's principle of equivalence is all about. The effects of gravity and acceleration cannot be distinguished.


At the surface of the earth, we know that gravity is not going to change quickly and toss people around. But we do know that if you and I get into two different rail cars, and we close the curtains, and if you start to be tossed around, that means you are moving and accelerating. It doesn’t mean that the gravity of the earth suddenly is changing rapidly or that you are moving around because I’m moving relative to you. My motion is not going to have any affect on what you feel in your car. And my relative motion is not going to change the rate of your atomic clock. Only your change in motion and acceleration will do that.

SeanF
2003-Nov-30, 12:23 AM
All righty, I'm back! Anybody miss me? :)

First things first. The thought experiments used in Einstein's original papers and elsewhere are not the Theory of Relativity. They are merely demonstrations of it. SR itself can be summed up in less than 20 words:

"The speed of light in a vacuum (c) is the same for all observers in inertial frames."

Everything else, all the Lorentz contraction and time dilation, and the "Twin Paradox," flows logically from that.

Consider SR's predictions in the following experiment:

You are observer A. You have a clock with you. I am observer B. I have a clock with me. There is a third clock, call it C, sitting some distance away from you, motionless relative to you. You and I are moving relative to each other at 0.6c.

OBSERVER A (YOU) SEES: You are motionless. Clock C is also motionless, one light-minute distant from you. Clock C is synchronized with your clock (they are not only ticking at the same rate, but they actually show the same time at the same time). I am moving towards you at 0.6c. My clock is running at 80% the rate of you and Clock C. At the moment I pass Clock C, all three clocks read exactly 12:00:00. At the moment I pass you, both Clock A and Clock C read 12:01:40 while Clock B, my Clock, reads 12:01:20. Therefore, you say that Clocks A and C both ticked 100 seconds while Clock B only ticked 80 seconds.

OBSERVER B (ME) SEES: I am motionless. Clock C and Clock A are both moving towards me at 0.6c. They are 0.8 light-minutes apart. Clock A and Clock C are both running at 80% the rate of my clock. They are therefore synchronous. However, they are not synchronized - Clock A is 36 seconds ahead of Clock C. At the moment Clock C passes me, both Clock B (my clock) and Clock C read 12:00:00 while Clock A reads 12:00:36. At the moment Clock A passes me, Clock A reads 12:01:40, Clock B (my clock) reads 12:01:20, and Clock C reads 12:01:04. Therefore, I say that Clocks A and C both ticked 64 seconds while Clock B ticked 80 seconds.

Accepted or not?

Glom
2003-Nov-30, 12:37 AM
Alright, campers. Let's start from scratch.

Fred resides on Earth and observes as his twin George hijacks the C-ship Lorentz and flies it away at v. We'll ignore the powerful and brief acceleration period. So Fred resides in inertial frame S while George resides in inertial frame S', which is travelling at speed v relative to S.

As Fred observes from frame S, he sees the Lorentz travel a distance x and then come to a stop relative to him. In S, a time t has elapsed. But what time, t', has elapsed for George? Use the Lorentz transformation.

t' = t sqrt(1-v²/c²)

Since 1-v²/c² must always be less than 1 for real v, this means that t' < t and hence George hasn't experienced as much time as Fred. Hence George is younger.

The Lorentz turns around, quickly gets up to speed v again, heading home and again travels x in time t relative to S and hence the duration of the return leg is the same as the outbound leg for the respective frames. You get the idea, George arrives home younger than Fred.

Now! What if we observe this from George's perspective? :-s

George observes Fred along with Earth, and the rest of the local area in fact, heading away from him at speed v. Because of this, everything starts squashing up according to the length contraction.

x' = x sqrt(1-v²/c²)

Clearly, x' < x.

Earth is observed to move away at speed v. Hence, the time of travel observed by George is

t' = x'/v = x sqrt(1-v²/c²)

Since x' < x, this means that t' < t, (and ditto for the return journey) which was also implied when observing from Fred's perspective.

Hence, the paradox is explained without the need for GR.

Sam5
2003-Nov-30, 12:43 AM
All righty, I'm back! Anybody miss me? :)

Yes, we missed you very much.


First things first. The thought experiments used in Einstein's original papers and elsewhere are not the Theory of Relativity. They are merely demonstrations of it. SR itself can be summed up in less than 20 words:

"The speed of light in a vacuum (c) is the same for all observers in inertial frames."

Surely you don’t believe that. Light slows down when it passes near the sun. Light leaves a distant high speed galaxy traveling at about c relative to the galaxy but less than c relative to us. The light speeds up on its way to us and arrives at us traveling at about c relative to us. Light gets sucked into a “black hole” and can’t leave a “black hole”. There’s no evidence that light always has the same speed in a vacuum in deep space, and there is plenty of evidence that it does not.


OBSERVER B (ME) SEES: I am motionless. Clock C and Clock A are both moving towards me at 0.6c. They are 0.8 light-minutes apart. Clock A and Clock C are both running at 80% the rate of my clock. They are therefore synchronous. However, they are not synchronized - Clock A is 36 seconds ahead of Clock C. At the moment Clock C passes me, both Clock B (my clock) and Clock C read 12:00:00 while Clock A reads 12:00:36. At the moment Clock A passes me, Clock A reads 12:01:40, Clock B (my clock) reads 12:01:20, and Clock C reads 12:01:04. Therefore, I say that Clocks A and C both ticked 64 seconds while Clock B ticked 80 seconds.

Accepted or not?

Lol, you are just trying to complicate the thought experiment, hoping I'll get mixe up.

Glom
2003-Nov-30, 12:48 AM
Surely you don’t believe that. Light slows down when it passes near the sun. Light leaves a distant high speed galaxy traveling at about c relative to the galaxy but less than c relative to us. The light speeds up on its way to us and arrives at us traveling at about c relative to us. Light gets sucked into a “black hole” and can’t leave a “black hole”. There’s no evidence that light always has the same speed in a vacuum in deep space, and there is plenty of evidence that it does not.

Ah, no wonder we can't agree. You've just said that Einstein was talking bartsibrel!

No. A photon always travels at the same speed relative to all observers, but it's path can be changed by gravity or equivalent effects.

Sam5
2003-Nov-30, 12:55 AM
OBSERVER A (YOU) SEES: You are motionless. Clock C is also motionless, one light-minute distant from you. Clock C is synchronized with your clock (they are not only ticking at the same rate, but they actually show the same time at the same time). I am moving towards you at 0.6c. My clock is running at 80% the rate of you and Clock C. At the moment I pass Clock C, all three clocks read exactly 12:00:00. At the moment I pass you, both Clock A and Clock C read 12:01:40 while Clock B, my Clock, reads 12:01:20.


You are forgetting something. In the SR theory, since you are stationary with your own clock, you never see your own clock running slowly. You only see the other clocks running slowly. So you would disagree with A and C about what your and their clocks read.

You would say that your clock reads the normal time, while you would say that the A and C clocks lag behind yours. So you would disagree with the A and C observers, and they would disagree with you.

This technique is known as the “majority opinion” attempt to reconcile the paradox, but you can’t get rid of that pesky ol’ paradox.

Sam5
2003-Nov-30, 01:12 AM
At the moment I pass Clock C, all three clocks read exactly 12:00:00. At the moment I pass you, both Clock A and Clock C read 12:01:40 while Clock B, my Clock, reads 12:01:20.

You don’t see 12:01:20 on your clock just because you see your clock as “time dilated”. You see it because you see both A and C as “moving”, and therefore you see the distance AC as “length contracted”. So you can’t say they are “motionless” in your thought experiment. And you don’t see a greater lapsed time on their clocks than on yours, because you see their clocks “dilating”, slowing down, so you would see their clocks with a lesser lapsed time than yours.

russ_watters
2003-Nov-30, 01:19 AM
Light slows down when it passes near the sun. Light leaves a distant high speed galaxy traveling at about c relative to the galaxy but less than c relative to us. The light speeds up on its way to us and arrives at us traveling at about c relative to us. Light gets sucked into a “black hole” and can’t leave a “black hole”. There’s no evidence that light always has the same speed in a vacuum in deep space, and there is plenty of evidence that it does not. This quite simply is not what is observed to occur in reality. You're wrong about relativity being wrong.

It seems your error is simply that you don't accept or can't get your arms around the concept of light having no mass.

Sam5
2003-Nov-30, 01:30 AM
At the moment Clock C passes me, both Clock B (my clock) and Clock C read 12:00:00 while Clock A reads 12:00:36

Now you are winding up with too many different times on just three clocks. You already said that A and C read the same time, 12:00:00, when B is at C. But now you have A lagging behind C by 36 seconds when B is at C, even though A and C are in the same inertial frame. You’ve already said A and C “are not only ticking at the same rate, but they actually show the same time at the same time.”

So which is it? A and C both read 12:00:00 when B and C pass, or C reads 12:00:00 while A reads 12:00:36 when B and C pass? When you reset all the clocks at 12:00:00 in the first thought experiment when B and C pass, why didn’t you do it in the second thought experiment? Why did you set A at 12:00:36 when you set B and C at 12:00:00?

You are getting yourself confused.

The Supreme Canuck
2003-Nov-30, 01:32 AM
Because you are a different person with a different frame of reference in the second experiment. The times are all relative to your frame of reference.

Sam5
2003-Nov-30, 01:34 AM
You've just said that Einstein was talking bartsibrel!

Sorry, I don't know what "bartsibrel" means. I think he wrote the 1905 paper in the German language.

SeanF
2003-Nov-30, 01:56 AM
At the moment Clock C passes me, both Clock B (my clock) and Clock C read 12:00:00 while Clock A reads 12:00:36

Now you are winding up with too many different times on just three clocks. You already said that A and C read the same time, 12:00:00, when B is at C. But now you have A lagging behind C by 36 seconds when B is at C, even though A and C are in the same inertial frame. You’ve already said A and C “are not only ticking at the same rate, but they actually show the same time at the same time.”

So which is it? A and C both read 12:00:00 when B and C pass, or C reads 12:00:00 while A reads 12:00:36 when B and C pass? When you reset all the clocks at 12:00:00 in the first thought experiment when B and C pass, why didn’t you do it in the second thought experiment? Why did you set A at 12:00:36 when you set B and C at 12:00:00?

You are getting yourself confused.

The last paragraph of Section 2 of Einstein's 1905 paper:


So we see that we cannot attach any absolute signification to the concept of simultaneity, but that two events which, viewed from a system of co-ordinates, are simultaneous, can no longer be looked upon as simultaneous events when envisaged from a system which is in motion relatively to that system.

In my experiment, Clock A reading 12:00:00 and Clock C reading 12:00:00 are the two pertinent events. In the Observer-A reference frame (or "system"), those two events are simultaneous. In the Observer-B system, "which is in motion relatively to" Observer-A's, they are not.

This aspect of SR, the issue of simultaneity, is often forgotten in trying to deal with the ramifications of it, but it can't be ignored. If you're going to try to knock SR, you have to make sure you fully understand it.

Also, what I described in my last post is a single thought experiment, not two. You made mention that Einstein's error in his experiment was that he never described what the other observer sees, so I did in mine. I described both what Observer A ("you") would see and what Observer B ("me") would see.



"The speed of light in a vacuum (c) is the same for all observers in inertial frames."
Surely you don’t believe that. Light slows down when it passes near the sun. Light leaves a distant high speed galaxy traveling at about c relative to the galaxy but less than c relative to us. The light speeds up on its way to us and arrives at us traveling at about c relative to us. Light gets sucked into a “black hole” and can’t leave a “black hole”. There’s no evidence that light always has the same speed in a vacuum in deep space, and there is plenty of evidence that it does not.
Again from Einstein's 1905 paper:

Any ray of light moves in the "stationary" system of co-ordinates with the determined velocity c, whether the ray be emitted by a stationary or by a moving body.
As Glom pointed out, gravity can bend light (and traveling through a medium, in which the photons are absorbed and re-emitted, can seem to slow light down), but in a vacuum it always travels at c.

If Clock C in my experiment were sending out light pulses, both Observer A (who sees Clock C as motionless) and Observer B (who sees Clock C as moving) would see those light pulses as traveling at c relative to themselves.

Sam5
2003-Nov-30, 02:16 AM
Any ray of light moves in the "stationary" system of co-ordinates with the determined velocity c, whether the ray be emitted by a stationary or by a moving body.

He said that in 1905. But in 1911 he said:

“If we call the velocity of light at the origin of co-ordinates c1, then the velocity of light c at a place with the gravitational potential “x” will be given by the relation

[equation]

The principle of the constancy of the velocity of light holds good according to this theory in a different form from that which usually underlies the ordinary theory of relativity.”

And in the 1952 5th Appendix to his book, he wrote:

“The principle of inertia and the principle of the constancy of the velocity of light are valid only with respect to an inertial system.”

So, after 1905, he understood that light slows down in a gravitational field.

SeanF
2003-Nov-30, 02:31 AM
Yes, light measured from an accelerating frame or deep in a gravitational field (which are indistinguishable from each other) will measure differently. But that's really irrelevant to the matter of SR, which only deals with inertial frames. It certainly doesn't change my last statement, that Clock C's light pulses would be measure at c by both observers in my experiment.

And it's got nothing to do with the question of whether Clocks A and C are synchronized, or whether the "Twin Paradox" is strictly SR...

Sam5
2003-Nov-30, 02:59 AM
You specifically said that clocks A, C, and B were all reset to 12:00:00 when C and B pass, and you certainly implied that observers A, C, and B agree with this. But later you said that C and B are reset to 12:00:00, while A is reset to 12:00:36 when C and B pass.

You specifically said, “At the moment I pass Clock C, all three clocks read exactly 12:00:00.”

You said that from a presumed “objective” point of view. You didn’t mention anything about a delay in the light signals going from A to B. You said what all three clocks would see, including B, and then later you changed it.

If you’ve already reset all clocks to 12:00:00 in your first thought experiment when C and B meet, and all observers agree, then what causes B to disagree with his earlier opinion and see something different in your second thought experiment?

In your second thought experiment, you said, “However, they are not synchronized - Clock A is 36 seconds ahead of Clock C,” but you didn’t say why. In SR theory, how could either C or B see A ticking faster than C or in advance of C?

And you said, “At the moment Clock C passes me, both Clock B (my clock) and Clock C read 12:00:00 while Clock A reads 12:00:36.” And at the start of this paragraph, you said, “OBSERVER B (ME) SEES”. So how are you, Observer B, going to “see” the A clock in advance of the C clock, when you are at C, if the light signal from the A clock takes far longer to reach you than the light signal from the C clock? If A and C are in the same frame, how can “OBSERVER B [YOU] SEE” the A clock ticking more rapidly than C, when you, B, are at C?

And even if you do set the A and C clocks at 12:00:00 in the AC frame, what would cause B to see A in advance of C, when B is at C, when B is still a long way from A?

Even assuming you don’t see its signal yet, what causes you to “see” A running faster than C, since you see A and C in the same frame? When you reset A, C, and B to 12:00:000 in the first thought experiment, why didn’t you reset A, C, and B to 12:00:00 in the second thought experiment? What caused you to reset A to 12:00:36 when C and B pass?

If A and C aren’t moving relative to each other, how does A get ahead of C by 36 seconds, at the moment B is at C? And how can B “see” this, when he’s at C, but the light from A emitted at that same moment he is at C, hasn’t reached him yet?

Nope, in trying to complicate the thought experiments, you cause even more paradoxes than the single paradox of the one in the original SR theory.

Sam5
2003-Nov-30, 04:19 AM
Ok, I’ve taken the time to go over your thought experiment numbers, in an attempt to find out why you artificially advanced the A clock by 36 seconds in your second thought experiment. If you see that I have made any math calculation errors, please point them out to me so I can correct them.

According to your specifications of distances and relative velocities, this is what would happen in SR theory:

First, no motion:

A------C/B

With relative motion at .6 c, frame AC would “see”:

A------C/B

And then:

A/B------C

Observers in frame AC would “see” the B clock time dilate by 20%, and they would “see” B take 100 seconds, by their own A and C clocks, to travel 1 light minute (11160000 miles) from C to A.

If all timers are set at 12:00:00 when B and C pass, what observers A and C will see on their own clocks when B reaches A will be 12:01:40 (we of course are not taking into consideration the travel time for the light signals to go from A to C or C to A), and observer’s A and C would “see” on the B clock readout, when B reaches A, the reading of 12:01:20, since they would see the B clock time dilated by 20% (100 seconds – 20% = 80 seconds).

Notice in this thought experiment that there is no clock “D” in the same frame as B. Why not? I think you left it out so you could make the overall situation appear to be asymmetrical.

Ok, let’s continue to leave out clock “D”.

So what would B “see”?

A---C/B

And then:

A/B---C

B would see the distance AC as “length contracted” by 20% due to the “relative motion”. So, B would see himself, by his own undilated clock, travel the contracted distance of .8 x 11160000 miles = 8928000 miles in 80 undilated seconds. So, if all the clocks were set at 12:00:00 when B passed C, then when B passes A, B would see 12:01:20 on his own clock, and he would see 12:01:04 on the A clock, because he would see the time on the A clock as “time dilated” by 20% (80 seconds – 20% = 64 seconds).

So, observers A and C would disagree with observer B about what they “see”, and we still have the paradox.

The only reason that you reset the timer A to 12:00:36 when C and B pass in your second thought experiment (or the 2nd half of your first thought experiment) is to try to force an artificial reconciliation of the paradox. I see that you got the 36 number by subtracting 64 from 100. 100 seconds is the undilated time that A and C would “see” (on their own clocks) B travel the uncontracted distance AC in the AC frame, as “seen” by A and C.

By artificially advancing clock A by 36 seconds when you reset all the clocks in your second thought experiment, you made sure that the A clock would read 12:01:40 when B reaches A. Based on SR theory, what B would actually “see” is 12:01:04 on the A clock, when B reaches A, so you had to advance the A clock by 36 seconds at the start of your second thought experiment, so you could force the A clock to read 12:01:40 when B reached A.

Sam5
2003-Nov-30, 04:42 AM
Here’s all you have to do. Einstein made it very simple in SR theory.

During the relative motion, George would “see” Fred “move”, while Fred would “see” George “move”.

George “sees” Fred’s length contact by the ratio of

1 : √1- (v^2/c^2)

And Fred sees George’s length contract by the same ratio.

George “sees” Fred’s time dilate by the amount of:

1 - √1- (v^2/c^2)

And Fred sees George’s time dilate by the same amount.

During the relative-motion travel (in both directions) both would see each other age more slowly. And when they reunite on earth, they would get into a big argument about which one actually aged more slowly, because a “time dilation” paradox would exist.

But, if you play a trick on yourself and pretend that only George “moves”, then you will think you have resolved the paradox, but you haven’t.

SeanF
2003-Nov-30, 07:45 PM
Okay, let's talk about the simultaneity issue. We'll use the same basic set-up as I described in my experiment above.

First question is, how can Observer A know that Clock A and Clock C are synchronized? He's at Clock A, so he can measure it directly. Clock C is a fair distance away, however, and he can't read it directly, so how could he determine synchronicity?

As I mentioned, the basic premise of Relativity is that the speed of light (in a vacuum) is a constant, c. So let's say that Clock A is configured to send signals to Clock C (via light waves) with its readout. Clock C responds to each signal with its own readout.

When Clock A reads 11:59:00 it sends a signal to Clock C. Exactly two minutes later (Clock A = 12:01:00) it receives the response from Clock C. The response essentially says "Clock C received the 11:59:00 signal when Clock C read 12:00:00" (For the sake of this experiment, we'll assume that Clock C's response time is instantaneous - rather than concerning ourselves with how much time it spends processing A's signal and formulating a response, we'll just assume the response is immediate, like A's signal simply bounces off C and comes right back).

Now, Observer A knows that the speed of light is constant (and he's not moving), so he knows that the light signal must've taken the same amount of time going out to C as it took coming back from C. Hence, the signal "hit" C exactly in the middle of the round-trip time: A=12:00:00. Since he knows that Clock C said "12:00:00" at this same time, he can determine that A=12:00:00 and C=12:00:00 were simultaneous. He also knows that Clock C is exactly one light-minute distant - or at least he knows that it was exactly one light-minute distant at 12:00:00. From repeatedly bouncing these signals off, he can determine that Clock C is remaining exactly one light-minute away, and that it's remaining synchronized.

Now, Observer B (who perceives A and C as moving) could bounce signals off both A and C to determine their distance from him. He could thus determine that they are, in fact, moving at 0.6c towards him, and he could determine the distance between them (note: I'm going to ignore the contraction of distance and time dilation that SR predicts here for the moment - I just want to talk about simultaneity).

But what can B determine about the times on A and C? He can also receive the information that A sent a signal at A=11:59:00 and received the response at A=12:01:00, as well as the information that that signal hit C at C=12:00:00.

However, Observer B sees A and C as moving - and he still measures the speed of light as c relative to himself. That means that his perception of those signals would be that Clock C was moving away from A's original signal at 0.6c and that A was moving into C's response signal at 0.6c. This means that he has to conclude the A-to-C trip takes longer than the C-to-A trip.

Mathematically, he can determine that 80% of the total round-trip was spent on A-C and only 20% was spent on C-A. This is simple math given that the velocity of A (and C), at 0.6c, is 60% the velocity of the signals.

But this means that of that entire two-minute round-trip, 1:36 of it was going out and only 0:24 was coming back. This means that the moment the signal hit Clock C, which we know happened when C=12:00:00, must've happened when A=12:00:36.

This is why Observer A says A=12:00:00 and C=12:00:00 were simultaneous, but Observer B says A=12:00:36 and C=12:00:00 were simultaneous.

Add the time dilation and length contraction to this, and it all works out in the end.

Sam5
2003-Nov-30, 08:48 PM
Okay, let's talk about the simultaneity issue. We'll use the same basic set-up as I described in my experiment above.

We don’t need to do any other stuff, and we don't need to think up any new thought experiments. You can keep making the thought experiments more and more complicated, but it still won’t work out. The theory just doesn’t work, and it leaves the clocks and observers with an unresolvable paradox.

You can have all three clocks and observers, A, C, and B (with B at C) stationary at the very beginning, and you can do whatever you need to synchronize their clocks and make them synchronous (all run at the same rate), and then you can have B and A move toward each other, relatively, (with the start-up acceleration disregarded), and you’re going to get the same result as I got.

A and C will “see” B’s clock running slow while B “sees” the A and C clocks running slow, and when B meets A, observers A and C will disagree with what B “sees”, and B will disagree with what A and C “see”.

You’ve given me several different thought experiments already, and I’ve shown you that it never works out with just one frame’s clock “really” time dilated, so there is no need to try to invent more complicated thought experiments.

Sam5
2003-Nov-30, 08:56 PM
Now, Observer B (who perceives A and C as moving) could bounce signals off both A and C to determine their distance from him

You can synchronize all three clocks at 12:00:00 before any motion begins, with B stationed at C. Then you move B and A toward each other and stop B and A when they are together, disregarding all acceleration during the start and stop, and you’ll get the same result I got earlier. B “sees” A and C as “moving” and time dilated, and he sees the line AC as “contracted”. A and C “see” B as “moving” and time dilated. And you end up with the same paradox.

SeanF
2003-Nov-30, 09:16 PM
Sam5, it does work out, but only if you look at all of it. If you only look at part of SR - if you just look at the length contraction and time dilation but ignore the simultaneity - then of course it's not going to work out. But that's not because SR is wrong, it's because you're doing it wrong.

If you start out with A, B, and C all motionless relative to each other and synchronize them and then move A and B together, it's not enough to simply say "we moved A & B together." Moving A towards B is different than moving B towards A - and if you think SR says they're not different, you're not looking at the whole thing.

The whole "motion is relative" thing means that we can consider A, B, and C to all be moving in the beginning and then we stop B or we can consider A, B, and C to be at rest in the beginning and then we move B - we'll get the same results. Even though in the first case we have A and C moving, and in the second we have B moving. In both cases, we changed B.

It does not mean we can consider A, B, and C to all be at rest and then we move B or we can consider A, B, and C to all be at rest and then we move A and C. Those will yield different results, and SR predicts those different results. In the first case, we changed B. In the second, we changed A and C.

A light signal that goes from A to C and then back to A will take the exact same amount of time A-C as it takes C-A if A is not moving relative to you (C's motion doesn't matter).

That same light signal will take different amounts of time A-C and C-A if A is moving relative to you.

If we have two observers, one moving relative to A and one not, watching the same light signals travel from A-C-A, they will disagree on whether the trips took the same amount of time. In other words, they will disagree on what time A's clock said when the signal hit C.

If you don't understand that, then you don't understand SR. And if you don't understand SR, you can't reach a valid conclusion on whether or not it's flawed.

Sam5
2003-Nov-30, 09:43 PM
But that's not because SR is wrong, it's because you're doing it wrong.


No, you’re doing it wrong, trying to get it to work out. You are trying all sorts of asymmetrical thought experiments trying to get it to work out, but it doesn’t work out.

Sam5
2003-Nov-30, 09:58 PM
If you start out with A, B, and C all motionless relative to each other and synchronize them and then move A and B together, it's not enough to simply say "we moved A & B together."

Lol, that’s because if you do that, I mean if you use real “relative motion” like the theory requires, it won’t work out because the theory is wrong. That’s why people have to add things like “acceleration”, “instantaneous acceleration”, “a frame change”, and all sorts of other asymmetrical stuff in an attempt to make it seem like the theory works.

Remember, in a 1907 paper Einstein admitted that the so-called “length contraction” is not real, and it doesn’t actually happen with just “relative motion”. He called the real shape of a “stationary” object its “geometrical” shape, and then he said:

“The latter obviously does not depend on the state of motion of a reference system.”

And then in his 1911 paper he admitted that the speed of light slows down when a light beam passes the sun, so out goes the “constancy of the speed of light” postulate.

Then in GR he added acceleration to one of two atomic clocks, and that one clock slowed down, and he finally did away with the “clock paradox”.

But “relative motion” can not possibly cause any lengths to really “contract” or any clocks to really “slow down”.


The whole "motion is relative" thing means that we can consider A, B, and C to all be moving in the beginning and then we stop B or we can consider A, B, and C to be at rest in the beginning and then we move B - we'll get the same results.

No, that’s not correct. You are trying to get favorable results by saying only one is “really” moving while the other is “stationary”, then you switch them around, and you think the theory works out. But what you’ve got to do is get it to work out with both moving “relatively”, and you won't ever be able to do that, because the theory is wrong.

Sam5
2003-Nov-30, 10:08 PM
If we have two observers, one moving relative to A and one not, watching the same light signals travel from A-C-A, they will disagree on whether the trips took the same amount of time. In other words, they will disagree on what time A's clock said when the signal hit C.

I know that. That’s exactly what the theory says, and that's what we've been discussing. That’s because of his “mutually observed time dilation” idea. The same would happen (in the theory) if you had A-C and B-D moving relatively. A and C would both disagree with B and D, and B and D would both disagree with A and C. It’s an irreconcilable paradox. He built it into the theory by leaving out “acceleration” and by trying to make the speed of light “constant”.

SeanF
2003-Nov-30, 10:32 PM
No, it's not because of the time dilation. Here are a couple diagrams. Clocks A and C, sending a light signal back and forth. If A and C are not moving, it will look something like this:



1 A* C
2 A * C T
3 A * C I
4 A * C M
5 A *C E
6 A * C |
7 A * C |
8 A * C V
9 A* C


However, if A and C are moving, it will look something like this:



1 A* C
2 A * C T
3 A * C I
4 A * C M
5 A * C E
6 A * C |
7 A *C |
8 A * C V
9 A* C


The * characters, representing the light pulses, move at the same rate in both cases, that being c. A and C, however, are moving at 1/2c. In the first case, the light's trip from A to C (4 lines) takes the same amount of time as the trip from C to A (4 lines). In the second case, however, the light takes longer (6 lines) going out than coming back (2 lines).

This has got nothing to do with time dilation. It is the simple fact, as I quoted Einstein before, "Any ray of light moves in the 'stationary' system of co-ordinates with the determined velocity c, whether the ray be emitted by a stationary or by a moving body." In the first drawing we are looking from a system "stationary" relative to A and C (the ray-emitting bodies are stationary). In the second, we are looking from a system that is not "stationary" relative to A and C (the ray-emitting bodies are moving).

We can clearly see that the point along A which is simultaneous with the light hitting C is different in the two drawings. This is why Einstein concluded, "So we see that we cannot attach any absolute signification to the concept of simultaneity, but that two events which, viewed from a system of co-ordinates, are simultaneous, can no longer be looked upon as simultaneous events when envisaged from a system which is in motion relatively to that system."

Agreed?

Sam5
2003-Nov-30, 10:38 PM
1 A* C
2 A * C T
3 A * C I
4 A * C M
5 A * C E
6 A * C |
7 A *C |
8 A * C V
9 A* C




Lol, looks like you’ve got a pretty strong ether wind blowing in that bottom diagram.

Sam5
2003-Nov-30, 10:51 PM
In the second, we are looking from a system that is not "stationary" relative to A and C (the ray-emitting bodies are moving).

Moving relative to what? Shouldn’t you call them B and D?

I hate to tell you this, but what you show in the second illustration is A and C moving to the right, relative to a stationary light propagating medium. The beam of light moves slowly, relative to C, to get from A to C, because C is moving away from the source, then the beam from C to A travels faster relative to A, since A is moving toward the source. You’ve set up a c – v and a c + v situation. I don’t know if that’s what you intended, but that’s what you’ve done.

kilopi
2003-Nov-30, 11:15 PM
No, you’re doing it wrong, trying to get it to work out. You are trying all sorts of asymmetrical thought experiments trying to get it to work out, but it doesn’t work out.
O yes it does.

It’s an irreconcilable paradox. He built it into the theory by leaving out “acceleration” and by trying to make the speed of light “constant”.
Fred25, is that you?

SeanF
2003-Nov-30, 11:15 PM
In the second, we are looking from a system that is not "stationary" relative to A and C (the ray-emitting bodies are moving).

Moving relative to what? Shouldn’t you call them B and D?

I hate to tell you this, but what you show in the second illustration is A and C moving to the right, relative to a stationary light propagating medium. The beam of light moves slowly, relative to C, to get from A to C, because C is moving away from the source, then the beam from C to A travels faster relative to A, since A is moving toward the source. You’ve set up a c – v and a c + v situation. I don’t know if that’s what you intended, but that’s what you’ve done.

A and C are moving relative to the Observer who's watching them. There's no light-propagating medium. "Any ray of light moves in the 'stationary' system of co-ordinates with the determined velocity c, whether the ray be emitted by a stationary or by a moving body." So our observer is sitting in his "'stationary' system of co-ordinates" watching A and C move by him, emitting light rays as they go.

And yes, this observer sees the first light pulse moving at a velocity of c-v relative to A and C and the second pulse moving at a velocity of c+v relative to A and C (in this case, v is 0.5c). However, this observer sees the light pulses moving at exactly c relative to himself, which is what SR requires.

Do you think I've done something here that is at odds with SR?

SeanF
2003-Nov-30, 11:16 PM
No, you’re doing it wrong, trying to get it to work out. You are trying all sorts of asymmetrical thought experiments trying to get it to work out, but it doesn’t work out.
O yes it does.

It’s an irreconcilable paradox. He built it into the theory by leaving out “acceleration” and by trying to make the speed of light “constant”.
Fred25, is that you?

Hey, K-pi, I was wondering if you were going to chime in on this!

Fred25? I have to keep fighting from typing, "JW . . ." :)

Sam5
2003-Nov-30, 11:45 PM
And yes, this observer sees the first light pulse moving at a velocity of c-v relative to A and C and the second pulse moving at a velocity of c+v relative to A and C (in this case, v is 0.5c). However, this observer sees the light pulses moving at exactly c relative to himself, which is what SR requires.


Right, the “stationary” observer sees the light travel at c – v and c + v relative to A and C. That’s why the light is slow to get to C in your second diagram and fast to get to A. The “stationary” observer sees the light speed as “c” relative to himself, because he’s not moving through the light-speed-regulating medium, which is, essentially represented by the white background of your diagrams. In SR theory, the light speed regulator is “stationary” with the observer that is deemed to be “stationary”.

Figure out what that "regulator" is, and you might receive a Nobel Prize.

Sam5
2003-Nov-30, 11:48 PM
SeanF, back on subject... So, how about showing me a diagram of two clocks that move relatively, but just one of them really slows down?

Sam5
2003-Nov-30, 11:54 PM
SeanF, and also, while you’re at it, see if you can reconcile this from the 1905 paper, with us learning the opinion of the A observer who never moves relative to the A clock. So far you haven’t been able to do it. If anyone can, you can, but I think you can’t. But if you can, I’ll change my mind.

“From this there ensues the following peculiar consequence. If at the points A and B of K there are stationary clocks which, viewed in the stationary system, are synchronous; and if the clock at A is moved with the velocity v along the line AB to B, then on its arrival at B the two clocks no longer synchronize, but the clock moved from A to B lags behind the other which has remained at B by 1/2tv^2/c^2 (up to magnitudes of fourth and higher order), t being the time occupied in the journey from A to B.”

SeanF
2003-Dec-01, 12:32 AM
In SR theory, the light speed regulator is “stationary” with the observer that is deemed to be “stationary”.

Figure out what that "regulator" is, and you might receive a Nobel Prize.

That's not quite accurate. Everything that happens can be deemed an "event." An event is basically a single moment in time at a single point in space. A light source "turning on" and sending out a pulse of light is an event. That pulse of light impacting on a light detector some distance away is another event. Speed is, of course, distance divided by time. So, the speed of that light pulse would be the distance between the source and the detector divided by the amount of time that passed between the emitting of the pulse and the receiving of it.

In SR, two observers who are in motion relative to each other perceive space and time differently. One of them may perceive the two events as occuring closer together (in both time and space) than the other. For example, one might perceive the two events as occuring 186000 miles apart in space and one second apart in time. The other might perceiving them as occurring 372000 miles apart in space and two seconds apart in time. The speed of the light will always be the same.

So, basically, space and time are connected in such a way that it is entirely accurate to say 186000 miles = one second. When relative motion distorts perception of space and time, they are always distorted equally so that 186000 miles always equals one second. Since light is massless, it travels at that specific "natural" velocity, 186000 miles per second.

Einstein already figured this out, but he didn't get a Nobel prize for it . . . :(

SeanF
2003-Dec-01, 12:33 AM
SeanF, back on subject... So, how about showing me a diagram of two clocks that move relatively, but just one of them really slows down?

I can certainly do that. Well, I don't know if I can do it using a text-based system like this one, but I can draw one up. I've done it many a time.

We're not there yet, though - the diagram would be absolutely meaningless without a firmer grounding of SR itself.

SeanF
2003-Dec-01, 12:38 AM
SeanF, and also, while you’re at it, see if you can reconcile this from the 1905 paper, with us learning the opinion of the A observer who never moves relative to the A clock. So far you haven’t been able to do it. If anyone can, you can, but I think you can’t. But if you can, I’ll change my mind.

“From this there ensues the following peculiar consequence. If at the points A and B of K there are stationary clocks which, viewed in the stationary system, are synchronous; and if the clock at A is moved with the velocity v along the line AB to B, then on its arrival at B the two clocks no longer synchronize, but the clock moved from A to B lags behind the other which has remained at B by 1/2tv^2/c^2 (up to magnitudes of fourth and higher order), t being the time occupied in the journey from A to B.”

Simultaneity is an issue in SR. If the two clocks are synchronous "viewed in the stationary system," then they're not when viewed from a system in relative motion to them. So, as soon as clock A begins "mov[ing] with the velocity v along the line AB to B," it can no longer consider B to be synchronous with itself. Basically, it would seem to A that clock B must have suddenly "jumped ahead" when A started moving, then proceeded to tick more slowly than A while A is approaching, but not enough to drop it back behind A by the time A arrives.

Sam5
2003-Dec-01, 01:00 AM
The speed of the light will always be the same.


No. Sorry. There apparently is a fairly steady “speed of light” through a local medium, but there is also a “relative speed of light” relative to an observer that is moving through the medium or that is in an other area of space, in a different area which I think is now called his own area of “comoving” space.

The distant galaxies that appear to be moving away from us at speeds faster than light are apparently carrying their own medium with them. They are not “traveling through a medium”, so they are not limited by the “c” limit (which apparently applies only inside and relative to a local medium) but they are carrying their own local medium with them.

And, as I said, this “medium” seems to presently be called simply their “comoving space”, the space that immediately surrounds the galaxies. This in itself is an “ether theory”, and a new one at that. You might want to read up on the latest theory about how light from a 3c galaxy reaches us. It’s interesting.

Let me ask your opinion about something. How do you suppose we see a blueshift in the starlight coming from a distant star that is fixed relative to the sun, when the earth moves toward that star every six months? Don’t you suppose it’s because of the c + v phenomenon, relative to the earth?

Sam5
2003-Dec-01, 01:04 AM
I can certainly do that. Well, I don't know if I can do it using a text-based system like this one, but I can draw one up. I've done it many a time.

Uh Oh. I feel a lopsided Minkowski diagram coming my way.


We're not there yet, though - the diagram would be absolutely meaningless without a firmer grounding of SR itself.

We?

Sam5
2003-Dec-01, 01:12 AM
So, as soon as clock A begins "mov[ing] with the velocity v along the line AB to B,"

So you really believe that only A “moves”, while B does not? How can we tell this with just two clocks and no acceleration?

What would the A observer say about it?


Basically, it would seem to A that clock B must have suddenly "jumped ahead" when A started moving,

No. You know there is no "jumping ahead" in SR theory. The observers only see a "slowdown". They never see a "jump ahead".

Glom
2003-Dec-01, 01:54 AM
Who would have thought we'd be getting into flame wars over SR?

Sam5
2003-Dec-01, 02:03 AM
Who would have thought we'd be getting into flame wars over SR?

Oh, this isn’t a flame war at all. It’s an interesting discussion. And to think, all of this started with the Michelson-Morley experiment of 1886, and it hasn’t been resolved yet.

SeanF
2003-Dec-01, 02:22 AM
The speed of the light will always be the same.


No. Sorry. There apparently is a fairly steady “speed of light” through a local medium, but there is also a “relative speed of light” relative to an observer that is moving through the medium or that is in an other area of space, in a different area which I think is now called his own area of “comoving” space.

The distant galaxies that appear to be moving away from us at speeds faster than light are apparently carrying their own medium with them. They are not “traveling through a medium”, so they are not limited by the “c” limit (which apparently applies only inside and relative to a local medium) but they are carrying their own local medium with them.

And, as I said, this “medium” seems to presently be called simply their “comoving space”, the space that immediately surrounds the galaxies. This in itself is an “ether theory”, and a new one at that. You might want to read up on the latest theory about how light from a 3c galaxy reaches us. It’s interesting.

My understanding of the latest theory was that it deals with the expansion of space. The space between us and those distant galaxies is expanding, and the distances are so great that that expansion becomes greater than c over the entire distance. Unless you're aware of something that I haven't read, there's nothing that overrides SR in any of it.


Let me ask your opinion about something. How do you suppose we see a blueshift in the starlight coming from a distant star that is fixed relative to the sun, when the earth moves toward that star every six months? Don’t you suppose it’s because of the c + v phenomenon, relative to the earth?
Doppler effect is unrelated to SR as well. Blueshifts and reshifts do not occur because the speed of light is different, just the frequency.

Uh Oh. I feel a lopsided Minkowski diagram coming my way.
:D

So you really believe that only A “moves”, while B does not? How can we tell this with just two clocks and no acceleration?

What would the A observer say about it?
Motion is relative. Only A changes, however.

And didn't I just describe what the A observer says about it?


You know there is no "jumping ahead" in SR theory. The observers only see a "slowdown". They never see a "jump ahead".
Oh-ho? Seems to me that until a few posts ago, you didn't even know there was a "simultaneity issue" in SR, and it's the simultaneity issue that leads to the "jumping ahead." Thought experiments are not SR - just because Einstein never pointed it out in his papers doesn't mean it isn't there.

All that's necessary is to take the supposition that the speed of light is c for all observers in all inertial reference frames and work out what happens. Do the math, and voila - time dilation, length contraction, simultaneity discrepencies! All rolled up in one neat little package with no paradoxes. :)

Sam5
2003-Dec-01, 02:33 AM
My understanding of the latest theory was that it deals with the expansion of space. The space between us and those distant galaxies is expanding, and the distances are so great that that expansion becomes greater than c over the entire distance. Unless you're aware of something that I haven't read, there's nothing that overrides SR in any of it.


Sure there is. The so-called “constancy of the velocity of light, and also the so-called “limiting speed”, such as in the paper, “For velocities greater than that of light our deliberations become meaningless; we shall, however, find in what follows, that the velocity of light in our theory plays the part, physically, of an infinitely great velocity.” So, if the galaxies are moving at 3 c relative to the earth, then c isn’t a “limiting velocity” or an “infinitely great velocity”.


Doppler effect is unrelated to SR as well. Blueshifts and reshifts do not occur because the speed of light is different, just the frequency.

Well why do the frequencies change? The waves see us coming and so they change frequency? No, we and the light change speeds relative to one another.

Sam5
2003-Dec-01, 02:47 AM
Oh-ho? Seems to me that until a few posts ago, you didn't even know there was a "simultaneity issue" in SR, and it's the simultaneity issue that leads to the "jumping ahead." Thought experiments are not SR - just because Einstein never pointed it out in his papers doesn't mean it isn't there.

Oh, come on, the B clock doesn’t “jump head”. You NEED it to “jump ahead” so you say it “jumps ahead”, just like you needed those 36 seconds on one particular clock at the end of one of your examples, so you put them on that clock at the beginning. LOL. But in the theory no clock “jumps ahead” and then “slows down” during the relative motion.

So, so far you’ve got “frame changes”, an extra 36 seconds that came from out of nowhere, and a clock that “jumps ahead” and suddenly starts ticking slowly, but not TOO slowly so that you make sure it doesn’t lose more time than it gained when it “jumped ahead”. Lol, I love this! I’m adding these to my collection. Have you brought up “instantaneous acceleration” yet? I can’t remember.

Sam5
2003-Dec-01, 02:56 AM
All that's necessary is to take the supposition that the speed of light is c for all observers in all inertial reference frames and work out what happens. Do the math, and voila - time dilation, length contraction, simultaneity discrepencies! All rolled up in one neat little package...........


Right, and you wind up with such a big clock paradox, and a big disagreement between what observers A and B “see”, hundreds of guys in books (and now on the internet) have been trying to get rid of the paradox for the past 98 years, and they ALL have to add stuff that’s not in the theory, such as “jumping clocks”, “frame changes”, “36 seconds”, “acceleration”, “instantaneous acceleration”, and lopsided Minkowski diagrams, none of which are in the 1905 SR theory. It seems to me that sooner or later, with all of these problems and with reconcilability being impossible without adding something that isn’t in the theory, that you would eventually begin to realize something is wrong with the theory.

Kaptain K
2003-Dec-01, 05:45 AM
Fred25? I have to keep fighting from typing, "JW . . ."
Iwas thinking "oriel36"! :roll:

SeanF
2003-Dec-01, 12:29 PM
Well why do the frequencies change? The waves see us coming and so they change frequency? No, we and the light change speeds relative to one another.

When the emitting object is relatively moving away from us, we perceive the light as moving relative to the object at c+v, hence the frequency is shifted at the source. When the emitting object is relatively moving towards us, we perceive the light as moving relative to the object at c-v, and so again, the frequency is shifted at the source.

An observer motionless with respect to the emitting object, would, of course, see no Doppler shift - but he would see the light moving at c+v or c-v relative to us, so would expect us to detect a Doppler shift anyway.

See, it's not too difficult if you think about it.

SeanF
2003-Dec-01, 12:30 PM
Oh-ho? Seems to me that until a few posts ago, you didn't even know there was a "simultaneity issue" in SR, and it's the simultaneity issue that leads to the "jumping ahead." Thought experiments are not SR - just because Einstein never pointed it out in his papers doesn't mean it isn't there.

Oh, come on, the B clock doesn’t “jump head”. You NEED it to “jump ahead” so you say it “jumps ahead”, just like you needed those 36 seconds on one particular clock at the end of one of your examples, so you put them on that clock at the beginning. LOL. But in the theory no clock “jumps ahead” and then “slows down” during the relative motion.

So, so far you’ve got “frame changes”, an extra 36 seconds that came from out of nowhere, and a clock that “jumps ahead” and suddenly starts ticking slowly, but not TOO slowly so that you make sure it doesn’t lose more time than it gained when it “jumped ahead”. Lol, I love this! I’m adding these to my collection. Have you brought up “instantaneous acceleration” yet? I can’t remember.

Well, I am glad you find it all amusing. I wish you'd think about it instead of just laughing it off, but I guess if you're not interested in learning you're not interested in learning.

"Instantaneous acceleration" is just another way of saying "change in inertial frames" ("we impart a velocity v," which is how Einstein worded it).

SeanF
2003-Dec-01, 12:37 PM
All that's necessary is to take the supposition that the speed of light is c for all observers in all inertial reference frames and work out what happens. Do the math, and voila - time dilation, length contraction, simultaneity discrepencies! All rolled up in one neat little package...........


Right, and you wind up with such a big clock paradox, and a big disagreement between what observers A and B “see”, hundreds of guys in books (and now on the internet) have been trying to get rid of the paradox for the past 98 years, and they ALL have to add stuff that’s not in the theory, such as “jumping clocks”, “frame changes”, “36 seconds”, “acceleration”, “instantaneous acceleration”, and lopsided Minkowski diagrams, none of which are in the 1905 SR theory. It seems to me that sooner or later, with all of these problems and with reconcilability being impossible without adding something that isn’t in the theory, that you would eventually begin to realize something is wrong with the theory.

Okay, let's start at the very beginning.

1. Light is always propagated in empty space (disregarding gravity wells and acceleration) with a definite velocity c which is independent of the state of motion of the emitting body.

THEREFORE:

2. An observer monitoring a light emitter motionless relative to himself would measure the light pulse as moving at c relative to himself and as moving at c relative to the emitter.

3. An observer monitoring a light emitter moving relative to himself would measure the light pulse as moving at c relative to himself but at c-v relative to the emitter (assuming the emitter is moving with velocity v in the same direction as the emitted light).

AND:

4. Conclusions 2 and 3 are both valid, even if the two observers are monitoring the same emitter and the same light pulse at the same time (the two observers would be in relative motion to each other).

Now, where, from 1 to 4, do you begin to disagree with these statements?

Glom
2003-Dec-01, 01:20 PM
Okay, we had this discussion in tutorial and these were the answers we got.

The twins paradox is a paradox in SR and can only be resolved in GR.

GR is the extension of SR to cover non-inertial frames. It does not describe absolute motion as there is no such thing.

Gravity and acceleration can change the speed of light.

SeanF
2003-Dec-01, 02:11 PM
Okay, we had this discussion in tutorial and these were the answers we got.

The twins paradox is a paradox in SR and can only be resolved in GR.


Uh, who told you that? The twins paradox is not a paradox in SR. :-?

You know, I think the original reason it was referred to as the "Twins Paradox" is because you end up with two people who used to be the same age but aren't any longer. If you consider that to be an unacceptable paradox, then I guess there is one - but that's going to exist in GR as well. There isn't really a "paradox" in the vein of "both twins should be older than the other," not even in SR.

Eta C
2003-Dec-01, 02:40 PM
You know, I think the original reason it was referred to as the "Twins Paradox" is because you end up with two people who used to be the same age but aren't any longer. If you consider that to be an unacceptable paradox, then I guess there is one - but that's going to exist in GR as well. There isn't really a "paradox" in the vein of "both twins should be older than the other," not even in SR.

The "Paradox" is that in SR both twins observe the other's clock as running slowly. Why then is the spaceborne twin younger upon return? The answer; he is not in an inertial reference frame. He undergoes accelerations to get up to speed and to change direction.

Now one can compute the age difference without resorting to GR by using the SR time dilation relations from the earthbound twin's point of view (the space twin's clock is slow) given the time it takes the spacecraft to fly out and back from the earth's point of view. This is incomplete, however, since it doesn't cover what happens during accelerations. The complete description requires a theory that can handle non-inertial (i.e. accererated) reference frames and that theory is GR. In GR there is no paradox as both twins will agree on who's clock is fast and who's clock is slow.

Check out some physics texts Sean, and you'll find that this is the correct interpretation. "Elementary Modern Physics" by Weidner & Seils is the one I checked, but there are many others. This is the interpretation Glom's instructor was presenting as well.

Normandy6644
2003-Dec-01, 03:48 PM
That's the way I've always understood it. Though I don't remember about the twins having an "agreement." The paradox is under the heading of SR (I just checked my Fenyman Lectures, and it's at least how he tought it), and GR resolves it by introducing non-inertial reference frames.

SeanF
2003-Dec-01, 03:51 PM
You know, I think the original reason it was referred to as the "Twins Paradox" is because you end up with two people who used to be the same age but aren't any longer. If you consider that to be an unacceptable paradox, then I guess there is one - but that's going to exist in GR as well. There isn't really a "paradox" in the vein of "both twins should be older than the other," not even in SR.

The "Paradox" is that in SR both twins observe the other's clock as running slowly. Why then is the spaceborne twin younger upon return? The answer; he is not in an inertial reference frame. He undergoes accelerations to get up to speed and to change direction.

Now one can compute the age difference without resorting to GR by using the SR time dilation relations from the earthbound twin's point of view (the space twin's clock is slow) given the time it takes the spacecraft to fly out and back from the earth's point of view. This is incomplete, however, since it doesn't cover what happens during accelerations. The complete description requires a theory that can handle non-inertial (i.e. accererated) reference frames and that theory is GR. In GR there is no paradox as both twins will agree on who's clock is fast and who's clock is slow.

Check out some physics texts Sean, and you'll find that this is the correct interpretation. "Elementary Modern Physics" by Weidner & Seils is the one I checked, but there are many others. This is the interpretation Glom's instructor was presenting as well.

The spaceborne twin is younger because he changes inertial frames, but not because of anything that happens while he's changing inertial frames. That may sound like I'm picking nits, but it is important. GR deals with what happens during the acceleration, but that doesn't lead to the twin paradox.

Use triplets, with two of them in two different spaceships. Have both spaceships undergo equal acceleration, cruise at a relativistic speed for a while, then decelerate equally, turn around, re-accelerate, cruise again, decelerate again, and stop back where they started. The two triplets who travelled will be the same age as each other, but they'll be younger than the one who stayed behind, right?

Now, what if we have one of the triplets cruise at the relativistic speed for longer before stopping and turning around? All the accelerations and decelerations will be identical for the two traveling triplets, one just cruised farther. Because the accelerations are all the same, GR would predict the same results for both of those two travelling triplets, but the one who travelled farther will be younger, simply because of the amount of time spent in the different inertial frames - straightforward SR.

Sam5
2003-Dec-01, 05:06 PM
When the emitting object is relatively moving away from us, we perceive the light as moving relative to the object at c+v, hence the frequency is shifted at the source. When the emitting object is relatively moving towards us, we perceive the light as moving relative to the object at c-v, and so again, the frequency is shifted at the source.



No. Sorry. The frequency of light from a star that is FIXED relative to the sun does not shift “at the source” (ie at the star) just because the little ol’ earth is orbiting the sun! That is a geocentric theory that says astronomical objects in deep space do things “at the source” because the earth is moving and because humans are looking at them, and that is wrong.

Anyway, if we orbit the sun once a year, how can that cause a star that’s 10 light-years away to give out higher and lower frequencies of light “at the source”, while we’re 10 light-years away from it? That would mean the star would have to “blink” every six months and gain energy and lose energy and increase its light frequency and then decrease its light frequency, just because the earth is orbiting the sun and humans are looking at the star from a distance of 10 light-years. Surely you don’t believe that?

What you need to accept that starlight passes through a light-speed-regulating medium while on its way from the star to the solar system. That light is not blueshifted or redshifted during that travel, just because the earth is orbiting the sun. When the earth moves toward the oncoming light waves, then the earth-relative speed of the light waves as they travels in deep space is c + v and when the earth is moving away from them, their earth-relative speed is c – v.

Since light speed is regulated to c at the surface of the earth, by some local light-speed-regulating medium, when the starlight waves approach and begin to enter that medium, THAT is where the shift occurs. If the earth is moving toward the star, the waves slow down a little in the earth’s local medium and they “bunch up” and we see a blueshift when those waves finally reach the earth. If the earth is moving away from the star, the waves speed up a little when they reach the earth’s local medium and they “stretch out” and we see a redshift when the waves finally arrive at the earth. So, the frequency and wavelength shifting occur near and at the earth, NOT at the source.

Sam5
2003-Dec-01, 05:10 PM
The twins paradox is a paradox in SR and can only be resolved in GR.

Gravity and acceleration can change the speed of light.

You are right.

Sam5
2003-Dec-01, 05:12 PM
There isn't really a "paradox" in the vein of "both twins should be older than the other," not even in SR.

You are wrong.

Sam5
2003-Dec-01, 05:16 PM
In GR there is no paradox as both twins will agree on who's clock is fast and who's clock is slow.

Check out some physics texts Sean, and you'll find that this is the correct interpretation. "Elementary Modern Physics" by Weidner & Seils is the one I checked, but there are many others. This is the interpretation Glom's instructor was presenting as well.

You are correct also.

Sam5
2003-Dec-01, 05:25 PM
The spaceborne twin is younger because he changes inertial frames, but not because of anything that happens while he's changing inertial frames.

Nope. There is no “frame change” in the SR theory or in the 1905 clock paradox thought experiment we’ve been debating.

First A and B are NOT moving relatively, and then they ARE moving relatively. However, they always remain inside their same Newtonian “inertial frame” whether they are moving or not moving, since acceleration was left out of the theory.

As I mentioned to you before, several times, the earth “twins” story is a GR story, not an SR story. The twin that “blasts off” and “turns around” experiences “acceleration”, and of course that acceleration would slow down his atomic clocks. But SR theory has no “acceleration”, no “blast off”, and no “turn around”. And SR has the paradox, while GR does not.

SeanF
2003-Dec-01, 05:52 PM
The spaceborne twin is younger because he changes inertial frames, but not because of anything that happens while he's changing inertial frames.

Nope. There is no “frame change” in the SR theory or in the 1905 clock paradox thought experiment we’ve been debating.

First A and B are NOT moving relatively, and then they ARE moving relatively. However, they always remain inside their same Newtonian “inertial frame” whether they are moving or not moving, since acceleration was left out of the theory.

That doesn't really even make sense. When they're not moving relatively, they're both in the same inertial frame. When they are moving relatively, they are in two different inertial frames. Therefore, there had to be a change in inertial frames. Acceleration was left out to the extent that SR doesn't deal with what happens during the change, but the whole basis of SR is "this inertial frame is different than that inertial frame." Without allowing changes in inertial frame, it'd be meaningless!

Inertial frames are not dependent on objects or observers being in them. When we start out with two clocks sitting motionless relative to each other, there is still an inertial frame that is moving at 0.6c relative to those clocks even if there are no physical objects in that other inertial frame - even if there are no other physical objects in the universe! Actually, there are a lot of different inertial frames moving relatively at 0.6c, because the direction of motion is important.

We can deduce how the clocks in the current inertial frame would appear to an observer viewing from one of the 0.6c inertial frames - and when we start moving one of the clocks (meaning we put it into that new inertial frame), what it then "sees" is exactly what we deduced he would see from that new frame.

Eta C
2003-Dec-01, 05:55 PM
The spaceborne twin is younger because he changes inertial frames, but not because of anything that happens while he's changing inertial frames. That may sound like I'm picking nits, but it is important. GR deals with what happens during the acceleration, but that doesn't lead to the twin paradox.


Wrong. Perhaps we have different interpretations of what "paradox" means. As I understand it the twin paradox comes about because both twins see the other's clock as moving slowly during the periods between accelerations. Why then is the space-faring twin younger? The difference is that he undergoes accelerations the earth-bound twin does not. Contrary to what you're claiming, what happens during those accelerations is crucial to understanding why the spaceborne twin is younger upon return. You can calculate an answer with SR alone, but you don't know why that answer is correct until you invoke GR. Until you can explain what happens during the accelerations your understanding of the situation is incomplete.


Use triplets, with two of them in two different spaceships. Have both spaceships undergo equal acceleration, cruise at a relativistic speed for a while, then decelerate equally, turn around, re-accelerate, cruise again, decelerate again, and stop back where they started. The two triplets who travelled will be the same age as each other, but they'll be younger than the one who stayed behind, right?

Now, what if we have one of the triplets cruise at the relativistic speed for longer before stopping and turning around? All the accelerations and decelerations will be identical for the two traveling triplets, one just cruised farther. Because the accelerations are all the same, GR would predict the same results for both of those two travelling triplets, but the one who travelled farther will be younger, simply because of the amount of time spent in the different inertial frames - straightforward SR.

If you look at the GR you'll find that the greater distance between the clocks also perfectly explains why the one who went farther is still younger on return.

There is a difference between computing the result, which I've already acknowledged can be done with the SR time dilation relations only, and understanding why that result is correct, which requires GR. Simply saying that the acceleration periods are unimportant is not an answer. They are what distinguishes the twins (or triplets) from each other and what happens during them is key to understanding the situation. Again I ask you to look into the textbooks. If you do you'll find that every one of them invokes GR as the ultimate explanation of the twin paradox.

SeanF
2003-Dec-01, 06:01 PM
No. Sorry. The frequency of light from a star that is FIXED relative to the sun does not shift “at the source” (ie at the star) just because the little ol’ earth is orbiting the sun! That is a geocentric theory that says astronomical objects in deep space do things “at the source” because the earth is moving and because humans are looking at them, and that is wrong.

It's not geocentricity, it's Relativity. Relativity says any inertial frame can be considered to be "at rest" - There is no difference between "Earth is moving towards the distant star" and "the distant star is moving towards the Earth." Therefore, Relativity predicts the Doppler Effect would be the same in either case.

Since we are on the Earth, we perceive our own inertial frame as being at rest and the star's inertial frame as being in motion. In that case, we'd see a Doppler Effect for the reasons I described, and so we do.

If we were at the distant star, we'd perceive the star's inertial frame as being at rest and the Earth as moving. In that case, we'd expect Earth to measure a Doppler Effect as I described above, and so it does.

Eroica
2003-Dec-01, 07:23 PM
As I understand it the twin paradox comes about because both twins see the other's clock as moving slowly during the periods between accelerations.
Not entirely true. During the first part of the space-faring twin's journey - when he's moving away from the Earthbound twin - each twin sees the other's clock as moving slowly. But when the space-farer turns for home, he immediately sees the other's clock running fast. However, the Earthbound twin continues to see the spacefarer's clock running slowly. As far as he's aware, the spacefarer hasn't yet turned for home. So the two situations are not symmetrical, even before we start to discuss acceleration.

You're right that acceleration is the key difference which ultimately explains why one twin ages less. It's the acceleration which gets him from one inertial frame into another. But very little of the difference in age occurs during the accelerations. And if the accelerations are instantaneous, then no age difference will accrue during them.

SeanF
2003-Dec-01, 07:44 PM
The spaceborne twin is younger because he changes inertial frames, but not because of anything that happens while he's changing inertial frames. That may sound like I'm picking nits, but it is important. GR deals with what happens during the acceleration, but that doesn't lead to the twin paradox.


Wrong. Perhaps we have different interpretations of what "paradox" means. As I understand it the twin paradox comes about because both twins see the other's clock as moving slowly during the periods between accelerations. Why then is the space-faring twin younger? The difference is that he undergoes accelerations the earth-bound twin does not. Contrary to what you're claiming, what happens during those accelerations is crucial to understanding why the spaceborne twin is younger upon return. You can calculate an answer with SR alone, but you don't know why that answer is correct until you invoke GR. Until you can explain what happens during the accelerations your understanding of the situation is incomplete.

Okay.

Clock A is sitting in space. Clock B is moving relative to A at a rate of 0.6c. Clock C is also moving relative to A at a rate of 0.6c, but in the opposite direction. From A's motionless point of view, events occur like this:



B A C
BA C
B C
AB C
A B C
A B C
A BC
A CB
A C B
A C B
AC B
C B
CA B
C A B


None of the clocks ever accelerate in any form during this experiment.

Observer A says that Clocks A & B both said "12:00:00" when they passed.
Observer A says that Clocks B & C both said "12:04:00" when they passed.
Observer A says that Clock C said "12:08:00" and Clock A said "12:10:00" when they passed.
Observer A says that Clocks B & C were both running slower than Clock A at all times.

Observer B says that Clocks A & B both said "12:00:00" when they passed.
Observer B says that Clocks B & C both said "12:04:00" when they passed.
Observer B says that Clock C said "12:08:00" and Clock A said "12:10:00" when they passed.
Observer B says that Clocks A & C were both running slower than Clock B at all times.

Observer C says that Clocks A & B both said "12:00:00" when they passed.
Observer C says that Clocks B & C both said "12:04:00" when they passed.
Observer C says that Clock C said "12:08:00" and Clock A said "12:10:00" when they passed.
Observer C says that Clocks A & B were both running slower than Clock C at all times.

Everybody agrees on this. There's no paradox. Even though the "other clocks" are always running slower, the reciprocity is broken by the fact that B&C are in different reference frames. You don't need GR for this.

The "Twin Paradox" is this exact same thing with the "stay-at-home" twin as A and the "traveling" twin as B in the first half and C in the second. Since you don't need GR for this, why would you need it for the "Twin Paradox"?

Sam5
2003-Dec-01, 08:50 PM
SeanF, as I said earlier, since none of your earlier simpler thought experiments worked out, you have to keep making more and more complex thought experiments, and you hope that people will just eventually get lost while trying to figure out the more complex ones.

But there is absolutely no legitimate reason why you should deviate from the original simpler thought experiment from the 1905 paper itself, except that it leads to a paradox and you want to try to avoid the paradox.

You said:

“Observer A says that Clocks A & B both said "12:00:00" when they passed.
Observer B says that Clocks A & B both said "12:00:00" when they passed.
Observer C says that Clocks A & B both said "12:00:00" when they passed.

Observer A says that Clocks B & C both said "12:04:00" when they passed.
Observer B says that Clocks B & C both said "12:04:00" when they passed.
Observer C says that Clocks B & C both said "12:04:00" when they passed.”

According to SR theory, your second set of observations is not correct, since observer A would see the B and C clocks time dilating in the ratio of 1 : √ 1 – (v^2/c^2), because observer A would see himself as being “stationary” and he would see the B and C clocks as “moving”. AND, observers B and C would see themselves as being “stationary”, while they would see each other and A as “moving”. So, observer A could not see the same time on the B and C clocks that the B and C observers see on their own clocks when B and C pass. According to SR theory, observer A would see a “time dilated” time on the B and C clocks when B and C pass. B would see dilated time on the C clock but not on his own, and C would see dilated time on the B clock but not on his own. So observers B and C would disagree about what time they see on each other’s and their own clocks.

According to SR theory, B would not see the same time on the C clock as on his own clock since he would see himself as “stationary” and the C clock as “moving”, so B would “see” the C clock as time dilated. And C would not see the same time on his own clock as he sees on the B clock, since C would see himself as “stationary” and the B clock as “moving”, so C would see the B clock as being time dilated.

In fact, in this thought experiment of yours, each observer, A, B, and C, would “see” his own clock as ticking normally, while he would “see” the other two clocks as ticking more slowly.

SeanF
2003-Dec-01, 09:41 PM
Sam5, nobody can disagree as to what B and C said at the moment B and C pass each other (and I do mean nobody - regardless of relative motion of inertial frame, everybody who can see that B and C passed each other will agree that they both said 12:04:00 at the moment it happened). This is a single point in space-time, at which Clocks B and C are co-located; that is, they are at the same point in space and the same point in time. It's a single "event," and thus must be agreed upon by all observers.

Surely you understand that two clocks which are ticking at different rates could still end up showing the same time simultaneously at some point, don't you?

Sam5
2003-Dec-01, 10:14 PM
Sam5, nobody can disagree as to what B and C said at the moment B and C pass each other (and I do mean nobody - regardless of relative motion of inertial frame, everybody who can see that B and C passed each other will agree that they both said 12:04:00 at the moment it happened). This is a single point in space-time, at which Clocks B and C are co-located; that is, they are at the same point in space and the same point in time. It's a single "event," and thus must be agreed upon by all observers.


Yes, but, B and C were already moving relatively, so B “saw” C’s clock time dilating, while C “saw” B’s clock time dilating, while neither saw their own clock time dilating. So they would not have the same readings unless you reset their timers exactly when they meet.

In your thought experiment, you’ve got several errors and misconceptions. In the first place, you say that B and C are moving relative to A at .6c, but you don’t point out that C and B are moving relative to each other at a higher rate.

Notice down at the bottom of your chart that you have C further away from B than A is. So you have A and B separating at one rate, while you have C and B separating at a faster rate. If B were separating from A at the same rate as B separates from C, then C and A would not separate at the bottom of your chart. They would move relative to B, but they would not separate relative to each other.

What you have in your chart is B and C converging and then separating from A at the same rate, but you’ve got B and C converging and separating from each other at a faster rate. This is because you have A as “stationary” in your chart, relative to us, the chart viewers, and also relative to our computer screens.

Another problem with your thought experiment is that you say:

“Observer A says that Clocks B & C both said "12:04:00" when they passed.
Observer B says that Clocks B & C both said "12:04:00" when they passed.”

So what you have here is only A’s “opinion” about what B is “supposed to say” he would “see” on his clock. But you can’t do that in SR theory. You can only say what A “sees” on his own and the B and C clocks. Then, if you want to get B’s opinion, you have to see the thought experiment from B’s perspective and he will tell you what he “sees” on his own and A’s clock, and what he “sees” on the C clock too. And of course, what B will actually “see” is his own clock ticking normally, while he “sees” the A clock as time dilated and the C clock time dilated more than the A clock.

Since the relative velocity of A and B is different from the relative velocity of B and C, according to your chart, with the relative velocity between B and C being greater than the velocity between B and A, then B would “see” the A clock time dilate a little, while he would “see” the C clock time dilate more during the relative motion. So, by the time B and C meet, B would “see” the A clock dilated a little and the C clock dilated more. So B can’t “see” the C clock reading the same as his own clock, unless you did some thought experiment manipulation and had the C clock start out with an advanced reading, or if you reset the B and C clocks when they meet. But even if you did that, the A observer and the B observer would disagree about what time they both “see” on the C clock, since C is moving at one velocity relative to A and a different velocity relative to B.

The more complicated you make these thought experiments, the more time it takes me to figure them out and find your errors, but I’m willing to continue when I have the spare time to work on them.

SeanF
2003-Dec-01, 10:48 PM
Sam5, nobody can disagree as to what B and C said at the moment B and C pass each other (and I do mean nobody - regardless of relative motion of inertial frame, everybody who can see that B and C passed each other will agree that they both said 12:04:00 at the moment it happened). This is a single point in space-time, at which Clocks B and C are co-located; that is, they are at the same point in space and the same point in time. It's a single "event," and thus must be agreed upon by all observers.

Yes, but, B and C were already moving relatively, so B “saw” C’s clock time dilating, while C “saw” B’s clock time dilating, while neither saw their own clock time dilating. So they would not have the same readings unless you reset their timers exactly when they meet.

Sure they can - if their clocks are set right to begin with.


In your thought experiment, you’ve got several errors and misconceptions. In the first place, you say that B and C are moving relative to A at .6c, but you don’t point out that C and B are moving relative to each other at a higher rate.
Neglecting to point out something that is so painfully obvious as that is not an error.


Notice down at the bottom of your chart that you have C further away from B than A is. So you have A and B separating at one rate, while you have C and B separating at a faster rate. If B were separating from A at the same rate as B separates from C, then C and A would not separate at the bottom of your chart. They would move relative to B, but they would not separate relative to each other.

What you have in your chart is B and C converging and then separating from A at the same rate, but you’ve got B and C converging and separating from each other at a faster rate. This is because you have A as “stationary” in your chart, relative to us, the chart viewers, and also relative to our computer screens.

I still don't see what the problem is. B's relative motion to C is greater than his relative motion to A, yes. Of course. If B and C weren't moving relative to each other, they would never pass each other. Why is this a problem?

Another problem with your thought experiment is that you say:

“Observer A says that Clocks B & C both said "12:04:00" when they passed.
Observer B says that Clocks B & C both said "12:04:00" when they passed.”

So what you have here is only A’s “opinion” about what B is “supposed to say” he would “see” on his clock. But you can’t do that in SR theory. You can only say what A “sees” on his own and the B and C clocks. Then, if you want to get B’s opinion, you have to see the thought experiment from B’s perspective and he will tell you what he “sees” on his own and A’s clock, and what he “sees” on the C clock too. And of course, what B will actually “see” is his own clock ticking normally, while he “sees” the A clock as time dilated and the C clock time dilated more than the A clock.


No, no, no. Perhaps that was bad terminology on my point. It's not Observer A's opinion of Observer B's opinion of Clock B, it's Observer A's opinion of Clock B. How about, "Observer A says that Clocks B & C were both at '12:04:00' when they passed." Is that better? Adjust all the other statements accordingly.


Since the relative velocity of A and B is different from the relative velocity of B and C, according to your chart, with the relative velocity between B and C being greater than the velocity between B and A, then B would “see” the A clock time dilate a little, while he would “see” the C clock time dilate more during the relative motion. So, by the time B and C meet, B would “see” the A clock dilated a little and the C clock dilated more. So B can’t “see” the C clock reading the same as his own clock, unless you did some thought experiment manipulation and had the C clock start out with an advanced reading, or if you reset the B and C clocks when they meet. But even if you did that, the A observer and the B observer would disagree about what time they both “see” on the C clock, since C is moving at one velocity relative to A and a different velocity relative to B.

I never said anything about how the clocks were initially set up (again, pardon me for not explicitly stating the obvious). When one clock is running slower than the other, that does not mean the first clock is always behind the other. If the slower clock started out ahead of the faster clock, it will stay ahead (but less and less ahead) until the faster clock "catches" it. At that moment, they'll both show the same time (how about that?). From then on out, the slower clock will be behind the faster, lagging farther and farther behind as we go on.


The more complicated you make these thought experiments, the more time it takes me to figure them out and find your errors, but I’m willing to continue when I have the spare time to work on them.

Hmm. First you complain about things I've left out, then you complain that I'm making things unnecessarily complicated. The "simple" version of this experiment was the first three sentences:


Clock A is sitting in space. Clock B is moving relative to A at a rate of 0.6c. Clock C is also moving relative to A at a rate of 0.6c, but in the opposite direction.

Everything I added after that was just specifics - the actual times displaying on the clocks at specific points, etc.

(I also feel compelled to point out that this particular thought experiment was intended to show Eta C how I see the Twin Paradox as strictly SR. As such, whether it's simpler than others I've described to you never came into my mind.)

At any rate, I keep trying to make the experiments simpler until I can find one you understand, and then move ahead from there. :-?

(Speaking of which, what happened with my post with the four statements and asking you where the logic breaks down in it? You never answered...)

Sam5
2003-Dec-01, 11:16 PM
I never said anything about how the clocks were initially set up

Ah Ha! See? I knew you knew something that you weren’t telling us. Now, why don’t you tell us the start-out time reading on all the clocks, as seen by their own “stationary” observers, at the very start of your thought experiment?

You know what this reminds me of? That little disk shaped plastic toy that has the four buttons around the edges, and a button in the center. So I hit the center button and I get a “beep”. Then I have to hit the button that lit up when I heard the “beep”. Then I hit the center button again and I get a “beep” and a “bop”. Then I have to hit those two buttons in order to keep the game going. But it keeps getting more complicated. By the time I get up to a “beep”, “bop”, “boop”, “beep”, “beep”, “bonk”, “bop”, I’m just about bonkers, and finally I hit the wrong button and I get an offensive “buzzzzzz”.

My analysis of your last thought experiment is correct, but if you add just one more moving observer to your thought experiment, I might go bonkers.

Sam5
2003-Dec-01, 11:20 PM
Hmm. First you complain about things I've left out, then you complain that I'm making things unnecessarily complicated.

No, I’m not complaining at all. I find this very relaxing and enjoyable. I’m just pointing out what you did, so some of the others can understand what you are doing.

Kaptain K
2003-Dec-02, 12:18 AM
Sam5,
You are the only one here who does not understand. The rest uf us are waiting for you to "get it". There may be some quibbles with some finer points, but you are the only one here who is disagreeing with relativity (either SR or GR).

SeanF
2003-Dec-02, 12:33 AM
I never said anything about how the clocks were initially set up

Ah Ha! See? I knew you knew something that you weren’t telling us. Now, why don’t you tell us the start-out time reading on all the clocks, as seen by their own “stationary” observers, at the very start of your thought experiment?

Sam5, I shouldn't have to tell you anything else about the clocks. If you understand SR, you already know (or at least you know how to calculate it). So let's take a little quiz.

Did you notice that for each "passing" (A & B, B & C, and A & C), I told you how all three observers would "read" the two clocks that pass each other, but said nothing about the distant clock? Specific example: When B passes C, I told you what everybody says about Clocks B & C, but I didn't say anything about what anybody says about Clock A. So, please answer these three questions:

What time does Observer A conclude was displaying on Clock A at the moment B passed C?
What time does Observer B conclude was displaying on Clock A at the moment B passed C?
What time does Observer C conclude was displaying on Clock A at the moment B passed C?

You know Special Relativity, right? So what does it predict for those three answers?

“Beep”, “bop”, “boop”, “beep”, “beep”, “bonk”, “bop” :)

Sam5
2003-Dec-02, 12:35 AM
Kaptain K, as far as GR is concerned, there is no irresolvable clock paradox, since only one of two clocks “really” slows down. This is due to an acceleration force being experienced by one of the clocks, the one that slows down. But with SR there is an irresolvable clock paradox, since both of two relatively moving observers “see” each other’s clock slow down at the same rate, by means of light signals, while they do not “see” their own clock slow down at all Therefore, one of the two relatively moving clocks in SR theory can not “really” be slowed down while the other is not, not based solely on “relative motion”.

Sam5
2003-Dec-02, 01:53 AM
So let's take a little quiz.

I’m still waiting for you to answer my earlier questions: Please tell us how you came up with the 36 seconds to advance the A clock at the beginning of one of your earlier thought experiments. In the first half of the experiment you had A and C set at 00:00 at the beginning, and in the second half you had A set at 00:36 while you had C set at 00:00. They weren’t moving relative to each other in either half of the experiment. This advance for the A clock gave a false reading at the end of your experiment and made it appear that observer B saw A’s clock ticking rapidly, whereas, according to SR theory, he actually saw the A clock ticking more slowly.

And also, I asked you to tell us the start times on all your clocks, A, B, and C, as seen by their own observers, at the beginning of your last thought experiment. And I would like to know the speed of B relative to C.

Some of the readers of this thread might not be as familiar with the SR numbers as we are, so when you artificially set your clocks to different times at the beginning of your thought experiments, so that you can get their times to work out the way you want them too later in the thought experiments, I think you need to tell everyone what those different beginning set-times are and why you used those particular numbers. Otherwise you might leave people thinking all the start times are 00:00 or 12:00:00, and I’m sure you wouldn’t want to give anyone a false impression or a misconception about any of your clock start times or your own thought experiments.

And when you have two different relative velocities in your thought experiments, please let us know what both of them are so we can run them through the Lorentz Transformation equation to find out what A, B, and C actually “see” on their own clocks and the other clocks.

SeanF
2003-Dec-02, 03:12 AM
So let's take a little quiz.

I’m still waiting for you to answer my earlier questions: Please tell us how you came up with the 36 seconds to advance the A clock at the beginning of one of your earlier thought experiments. In the first half of the experiment you had A and C set at 00:00 at the beginning, and in the second half you had A set at 00:36 while you had C set at 00:00. They weren’t moving relative to each other in either half of the experiment. This advance for the A clock gave a false reading at the end of your experiment and made it appear that observer B saw A’s clock ticking rapidly, whereas, according to SR theory, he actually saw the A clock ticking more slowly.

It takes all the fun out of it if you don't do it yourself, Sam5, but since you insist...

"Light is always propogated in empty space with a definite velocity c which is independent of the state of motion of the emitting body." - Albert Einstein

We have an emitter/receiver with a mirror. A pulse of light is sent from the emitter, bounces off the mirror, and returns to the receiver. A clock on the emitter/receiver times the duration of the light's round-trip.

Since the velocity of light is c regardless of the motion of the emitting body, the velocity of light relative to the emitter & mirror will be c-v and c+v, where v is the velocity of the emitter assembly relative to the observer.

So, t1 = d/(c-v) and t2 = d/(c+v), where t1 is the time from the emitter to the mirror, t2 is the time from the mirror back to the receiver, and d is the distance between the emitter/receiver and the mirror. t1+t2=t, of course, where t is the total round-trip time.

Now, if v=0, meaning the emitter assembly is motionless relative to the observer, then both c-v and c+v are equal to c. This means t2=d/c and t1=d/c, so t1=t2. Therefore, t=t1+t2 becomes t=t1+t1, then t=t1*2, and finally t1=t*0.5. So, the light pulse spends half the total round-trip time on the way out and half on the way back.

Now, if v=0.6c, then c-v=0.4c and c+v=1.6c. This means t2=d/1.6c and t1=d/0.4c. From that last equation, d=t1*0.4c. Plugging that in for d in the next-to-last equation gives us t2=(t1*0.4c)/1.6c, which simplifies to t2=t1*0.25. Therefore, t=t1+t2 becomes t=t1+t1*0.25, or t=t1*1.25, or t1=t*0.8. So, the light pulse spends four-fifths (80%) the total round-trip time on the way out and only one-fifth (20%) on the way back.

If the round-trip time takes 120 seconds, then an observer stationary to the system says the light took 60 seconds out and 60 seconds back. An observer who sees the system moving at 0.6c relative to himself says the light took 96 seconds out and 24 seconds back. 96-60 = 36 = 60-24. Therefore the "moving" observer concludes the clock is 36 seconds later when the pulse hits the mirror than the "stationary" observer does.

Specifically, if a clock at the emitter reads 11:59:00 when the pulse leaves and 12:01:00 when the pulse returns, the "stationary" observer would conclude the light hit the mirror when that clock read 12:00:00. However, an observer moving relatively at 0.6c says the light would hit the mirror when that clock read 12:00:36.

If a clock located at the mirror says 12:00:00 when the light hits the mirror, the "stationary" observer would conclude the two clocks are synchronized. The "moving" observer, however, would conclude that the emitter clock is 36 seconds ahead of the mirror clock.

So from this we can conclude that two clocks located one light-minute apart (which would be the required distance for the pulse to take two minutes round trip) which are synchronized in the "stationary" frame would be off from each other by 36 seconds in the relatively "moving" frame.

The fact that 36 seconds is exactly what I needed to set the clocks off by to make the final numbers work out is not coincidence or something I "fudged." It is a result of the fact that Special Relativity is internally consistent and so it will all work out in the end.



And also, I asked you to tell us the start times on all your clocks, A, B, and C, as seen by their own observers, at the beginning of your last thought experiment. And I would like to know the speed of B relative to C.

When adding relativistic velocities, v = ( v1 + v2 ) / ( 1 - ( v1 * v2) / ( c * c ) ), where v is the total velocity, v1 and v2 are the separate original velocities, and c, of course, is the speed of light. Plugging 0.6c in for both v1 and v2 results in approximately 0.88c (unless I did my math wrong), which should be the velocity of B relative to C and vice-versa.

You've got to know that equation (or know where to look it up :) ) if you're going to deal with SR...


Some of the readers of this thread might not be as familiar with the SR numbers as we are, so when you artificially set your clocks to different times at the beginning of your thought experiments, so that you can get their times to work out the way you want them too later in the thought experiments, I think you need to tell everyone what those different beginning set-times are and why you used those particular numbers. Otherwise you might leave people thinking all the start times are 00:00 or 12:00:00, and I’m sure you wouldn’t want to give anyone a false impression or a misconception about any of your clock start times or your own thought experiments.

And when you have two different relative velocities in your thought experiments, please let us know what both of them are so we can run them through the Lorentz Transformation equation to find out what A, B, and C actually “see” on their own clocks and the other clocks.

The explanation I gave above for 36 seconds is the basis for the simultaneity issue in SR. At the point when A & B pass each other in this experiment (which we can consider to be the "start of the experiment"), C is quite some distance away, and A, B, and C are all three in different inertial frames. Therefore, while it's easy and universal to say that A & B were both at 12:00:00 when they passed each other, the time displayed on C at that same moment (in other words, what C-clock "tick" event is simultaneous with the A-B-passing event) will depend on which of the three inertial frames you're looking from.

This will also be true for the question of what time is displayed on A when B & C pass each other, and also for the question of what time is displayed on B when A & C pass each other.

Kaptain K
2003-Dec-02, 05:59 AM
Sam5
Some of the readers of this thread might not be as familiar with the SR numbers as we are...
I cannot speak for all readers of this thread, but of all the posters, you are the one who is least familiar with SR. You are arguing about thought experiments, which are just an attempt to put into words the mathematics of SR. The word descriptions are not the math!

Sam5
2003-Dec-02, 06:40 AM
Kaptain K, as far as I know, the math of the theory is fine. It’s some of the concepts that are in error. For example, Einstein has his light-speed c-regulator as being fixed with the frame he calls “stationary”. But when we switch to the other frame and call it “stationary”, then the c-regulator becomes fixed with that frame.

The theory seems to work out as long as we think from the point of view of only one “stationary” frame per thought experiment. But when we realize that both frames are equal, inertial, and moving relatively, then the observer in the so-called “moving” frame will see his frame as the “stationary” one and the other frame as the one that is “moving”. When we switch frame and observer points of view and alternate each frame as being “stationary”, which we must do, since all motion in the theory is relative, then that’s when a major error of the theory is revealed and the clock paradox results.

Jobe
2003-Dec-02, 07:19 AM
Sam5, you're embarrassing yourself.

Kaptain K
2003-Dec-02, 07:23 AM
Please, get this straight:

The math is the theory!
No more, no less. If "the math of the theory is fine", the theory is also (by definition) "fine". Again, you are arguing about the word decription of the theory, not the theory itself. If you understood the theory (the math), you would not be picking nits over the words.

Diamond
2003-Dec-02, 10:39 AM
Sam5:

http://math.ucr.edu/home/baez/RelWWW/wrong.html

Sam5
2003-Dec-02, 01:58 PM
Sorry Kaptain K, but the math is only part of the Special Relativity theory. If you want to say that a separate individual light-propagating medium is stationary with every observer in every inertial frame, and then when there is relative motion between inertial frames, the universe splits up into multiple parallel universes so that all the observers in all the different inertial frame see different sets of things happen, then you still don’t have a valid theory that is based on reality, but you’ve got a fine plot for a science fiction movie.

If you take the theory down to its most fundamental level, in Einstein’s own clock paradox thought experiment he winds up with observer A and observer B disagreeing about what they “see” at the end of the experiment. Observer A “sees” observer B’s clock running slow and lagging behind, while observer B “sees” observer A’s clock running slow and lagging behind. So we wind up with two sets of clock face readings when A and B unite, a total of 4 readings in all for only 2 clocks.

We have what A sees on his own clock, and what A sees on B’s clock, and they are different readings from what B sees on his own clock and what B sees on A’s clock. This does not conform to reality. In General Theory, atomic clocks slow down when they experience acceleration forces. That is based on reality and conforms to observation. So there is no paradox in the GR theory. If you think there is no paradox in the SR theory, then you just don’t understand the SR theory.

And by the way, pendulum clocks speed up when they experience acceleration while atomic clocks slow down, so it is not “all of time that slows down” when clocks experience acceleration, since different clocks react to acceleration in different ways.

SeanF
2003-Dec-02, 02:13 PM
Sorry Kaptain K, but the math is only part of the Special Relativity theory. If you want to say that a separate individual light-propagating medium is stationary with every observer in every inertial frame, and then when there is relative motion between inertial frames, the universe splits up into multiple parallel universes so that all the observers in all the different inertial frame see different sets of things happen, then you still don’t have a valid theory that is based on reality, but you’ve got a fine plot for a science fiction movie.

There is no light-propagating medium. I told you before that one second and 186,000 miles are the same thing, and that's why light always travels 186000 miles in one second regardless of who's looking at it.

{Edited to Add}: K's right, the math is the theory. The "Twin Paradox" and such thought experiments are merely demonstrations or examples.


If you take the theory down to its most fundamental level, in Einstein’s own clock paradox thought experiment he winds up with observer A and observer B disagreeing about what they “see” at the end of the experiment. Observer A “sees” observer B’s clock running slow and lagging behind, while observer B “sees” observer A’s clock running slow and lagging behind. So we wind up with two sets of clock face readings when A and B unite, a total of 4 readings in all for only 2 clocks.

We have what A sees on his own clock, and what A sees on B’s clock, and they are different readings from what B sees on his own clock and what B sees on A’s clock. This does not conform to reality.
No, we don't. Observer's A and B both agree on what Clocks A and B read when they meet. They disagree on what the clocks read while they are separated (SR predicts these disagreements), but when they meet, all is well.

In General Theory, atomic clocks slow down when they experience acceleration forces. That is based on reality and conforms to observation. So there is no paradox in the GR theory. If you think there is no paradox in the SR theory, then you just don’t understand the SR theory.

And by the way, pendulum clocks speed up when they experience acceleration while atomic clocks slow down, so it is not “all of time that slows down” when clocks experience acceleration, since different clocks react to acceleration in different ways.
No, in the General Theory of Relativity time flows differently in gravity wells and during acceleration. It is not a physical response of the clocks, it is a fundamental property of space and time itself.

There are some physical responses - a pendulum clock will change its rate if you grab hold of the pendulum, for pete's sake - but those aren't part of Relativity (Special or General). General Relativity predicts changes in time (as does Special Relativity), and the clocks merely reflect that.

Sam5
2003-Dec-02, 02:15 PM
SeanF, in your previous thought experiment, on page 3 of this thread, you had clocks A and C and they weren’t moving relatively. They were in the same frame. And you said the start times for both clocks was 00:00, then later you said their start times were 00:36 for the A clock and 00:00 for the C clock.

So you wound up with two different start times for the A clock.

What you did was remove the “paradox” from the end of your thought experiment and place it at the beginning of your thought experiment. So you didn’t resolve the paradox, you just moved it from one end of the thought experiment to the other.

SeanF
2003-Dec-02, 02:31 PM
SeanF, in your previous thought experiment, on page 3 of this thread, you had clocks A and C and they weren’t moving relatively. They were in the same frame. And you said the start times for both clocks was 00:00, then later you said their start times were 00:36 for the A clock and 00:00 for the C clock.

So you wound up with two different start times for the A clock.

What you did was remove the “paradox” from the end of your thought experiment and place it at the beginning of your thought experiment. So you didn’t resolve the paradox, you just moved it from one end of the thought experiment to the other.

Nope, it's not a paradox. A and C remain physically separated for that entire experiment, and so it does not make one whit of difference what A reads when C reads "12:00:00".

When B and C pass each other, they tell each other what they think A reads at that moment. B says "A reads 12:00:36 right now, and when it gets to me (because it's moving) it'll read 12:01:40." while C says "No, A reads 12:00:00 right now and when you get to it (because you're moving) it'll read 12:01:40."

A isn't there, it's off in the distance. They're both calculating the "current value" of A based on light that left A some time ago, so it does not cause any kind of "paradox" that they disagree.

Sam5
2003-Dec-02, 02:34 PM
I told you before that one second and 186,000 miles are the same thing, and that's why light always travels 186000 miles in one second regardless of who's looking at it.

Your statement is misleading. There is a light-speed-regulator that travels with every large astronomical body. And because of that, light slows down when it pass near the sun or other massive bodies. The stronger the gravitational field, the slower light travels as it passes through it.

I’m not sure if we should call this a “propagation medium” or what. But this is why we on earth see light traveling at c, here locally at the surface of the earth, while a distant high-c galaxy sees light traveling at c relative to that galaxy.

When the light is emitted by that distant galaxy it is traveling at c relative to the galaxy but less than c relative to the earth. So you can’t say that, “light always travels 186000 miles in one second regardless of who's looking at it.” You can say that light travels at 186000 miles in one second while traveling near the surface of the earth, but you can’t say that light travels inside a distant high-c galaxy at 186,000 miles in one second both through that galaxy and relative to the earth at the same time.

This is one of the most fundamental and basic concepts of nature, and I don’t understand why you can’t understand it.

Now, we can take your wording and make it “work out” by saying this: “people on earth see light traveling locally on earth at c”, and “people in high-c galaxies see light traveling locally in their galaxy at c”, but we can’t say that their light is traveling at c relative to the earth while it is traveling through their galaxy.

Sam5
2003-Dec-02, 02:44 PM
Nope, it's not a paradox. A and C remain physically separated for that entire experiment, and so it does not make one whit of difference what A reads when C reads "12:00:00".

It does make a difference when you’ve already said the A clock starts out at 00:00. So you can’t change that in the same thought experiment. If you do change it, then you’ve got the paradox of A “seeing” his own clock read both 00:00 and 00:36 at the same time, and that is impossible.

Do you understand that in reality a single observer holding a single clock can not see two different readings on the single clock at the same time?

Sam5
2003-Dec-02, 02:54 PM
When B and C pass each other, they tell each other what they think A reads at that moment.


But B doesn’t “see” clock A reading :36 seconds at the beginning. You already said that in the beginning all clocks read 00:00 and you said that you were using instantaneous light signals, so all observers know that all clocks are reading 00:00 in the beginning.

In Einstein’s thought experiment, he has all clocks as being “stationary” in the beginning so that all observers can synchronize all of them and make them synchronous. We can do the very same thing while B is stationary at C. So all observers know that all clocks are reading 00:00 at the beginning. Then the relative motion begins. And when A and B unite, observers A and B will disagree about what their own and each other's clock readings are.

SeanF
2003-Dec-02, 03:23 PM
Nope, it's not a paradox. A and C remain physically separated for that entire experiment, and so it does not make one whit of difference what A reads when C reads "12:00:00".

It does make a difference when you’ve already said the A clock starts out at 00:00. So you can’t change that in the same thought experiment. If you do change it, then you’ve got the paradox of A “seeing” his own clock read both 00:00 and 00:36 at the same time, and that is impossible.

Do you understand that in reality a single observer holding a single clock can not see two different readings on the single clock at the same time?

No, A sees A at 12:00:00 when B passes C. C, in the same inertial frame as A, sees A at 12:00:00 when B passes C. B, in a different inertial frame as A and C, sees A at 12:00:36 when B passes C. No single Observer disagrees with himself at any point.


But B doesn’t “see” clock A reading :36 seconds at the beginning. You already said that in the beginning all clocks read 00:00 and you said that you were using instantaneous light signals, so all observers know that all clocks are reading 00:00 in the beginning.

I never said anything regarding "instantaneous light signals," and I never said that all observers saw the same thing on all three clocks at the same time. You better go back and read that again.

Instantaneous communication is prohibited under the rules of SR (not arbitrarily - the logic flowing from the constant speed of light concludes it), and it's a good thing, too, because there would be paradoxes without it. That's just another part of how SR is internally consistent. The simultaneity difference in relative motion would result in paradoxes if the observers in relative motion could communicate instantaneously. But they can't.


In Einstein’s thought experiment, he has all clocks as being “stationary” in the beginning so that all observers can synchronize all of them and make them synchronous. We can do the very same thing while B is stationary at C. So all observers know that all clocks are reading 00:00 at the beginning. Then the relative motion begins. And when A and B unite, observers A and B will disagree about what their own and each other's clock readings are.

In my thought experiment, B is never stationary at C, so we can't "do the very same thing." In Einstein's thought experiment, the observer who changes his relative motion sees the distant clock differently after the change than before. When comparing Einstein's thought experiment to mine, My C is Einstein's B before the relative motion starts, and my B is Einstein's B after the relative motion starts.

Crap. Does Einstein's experiment say "We move B to A" or "We move A to B"? I'm assuming it's the former. If it's the latter, than my A is Einstein's B; my C is Einstein's A before the relative motion starts; and my B is Einstein's A after the relative motion starts. Either way, the result is the same.

Sam5
2003-Dec-02, 03:59 PM
No, in the General Theory of Relativity time flows differently in gravity wells and during acceleration. It is not a physical response of the clocks, it is a fundamental property of space and time itself.

No, its a physical response of the clocks, depending on the type of clocks used. An atomic clock slows down in a “gravity well”, but a pendulum clock speeds up in the same gravity well. If we set an atomic clock and a pendulum clock to tick out an accurate astronomical “year”, then if we put them in a gravity well, the atomic clock will run slow and the pendulum clock will run fast, while the astronomical year hasn’t changed at all. We can be in a gravity well here on earth and then move out into space in a spacecraft, and the astronomical year won’t change, whereas an atomic clock in space will speed up while the pendulum clock in space stops running altogether.

These kinds of clock-related “time” dilations were discovered centuries ago. The knowledge that gravity wells affected the tick rate of pendulum clocks was so well known in the 19th Century, the gravitational potential at different elevations on the earth’s surface was measured with pendulum clocks.

In 1877 Maxwell wrote about the force of gravity, “Its intensity, as measured by pendulum experiments, decreases as we ascend”. He also wrote this about time, “We thus see the theoretical possibility of comparing intervals of time however distant, though it is hardly necessary to remark that the method cannot be put in practice in the neighborhood of the earth, or any other large mass of gravitating matter.” So he knew very well that gravity wells can alter the tick rates of clocks. By experiment we have learned that gravity wells slow down atomic clocks and they speed up pendulum clocks. These are simply the laws of nature. But relative motion can’t possibly alter the tick rate of any clock, since the clock experiences no “force” as a result of the “relative motion”.

Sam5
2003-Dec-02, 04:11 PM
Crap. Does Einstein's experiment say "We move B to A" or "We move A to B"? I'm assuming it's the former. If it's the latter, than my A is Einstein's B; my C is Einstein's A before the relative motion starts; and my B is Einstein's A after the relative motion starts. Either way, the result is the same.


LOL. “Boop”, “bop”, “beep”, “bip”. Was it a “bip” before a “bop” or a “boop” before a “beep”? I know how you feel. So we won't go bonkers simultaneously, in our own frames, maybe we need to take a break.

parejkoj
2003-Dec-02, 04:26 PM
Umm... Sam5, do you understand what determines the "tick rate" of a pendulum clock? Go and derive the equation for the period of a pendulum clock (Newtonian or Lagrangian, whatever you choose) and then come back. While you're at it, do the same for a horizontal spring (which would be a better analog for an atomic clock) and tell us which one would be better to measure time, "outside of" the affect of a gravitational field/accelerating reference frame.

Here's a hint though: the period is dependent on the acceleration (gravitational or otherwise) that the pendulum experiences. The period of a spring is not. Just like SeanF (I think it was) said: of course a pendulum will move when you yank on the string!


Sorry Kaptain K, but the math is only part of the Special Relativity theory.

Sorry Sam5, but the math is the theory, just like everyone has been saying. The math is what determines what each clock measures. What determines whether the theory is correct is whether the mathematical calculations line up with observation. They do (we wouldn't have GPS otherwise!). If you don't understand -- and by that I mean DO -- the math, you won't understand the theory. It's that simple! :)

There isn't anything I can add to what SeanF has said about SR/GR, except that you (Sam5) seem to be misunderstanding several very fundamental things. In GR, time actually does flow differently in an accelerating reference frame. And in SR, the relative "tick time" of the clocks in two different reference frames is actually different. And light always moves at the same speed through a given medium, no matter where you are observing from. If my own understanding is correct (someone please correct me if I'm wrong), light emitted from within a strong gravitational well is stretched (is that the right analogy?), not slowed down: this accounts for the observed redshift -- right? :-?

SeanF
2003-Dec-02, 04:28 PM
No, in the General Theory of Relativity time flows differently in gravity wells and during acceleration. It is not a physical response of the clocks, it is a fundamental property of space and time itself.

No, its a physical response of the clocks, depending on the type of clocks used. An atomic clock slows down in a “gravity well”, but a pendulum clock speeds up in the same gravity well. If we set an atomic clock and a pendulum clock to tick out an accurate astronomical “year”, then if we put them in a gravity well, the atomic clock will run slow and the pendulum clock will run fast, while the astronomical year hasn’t changed at all. We can be in a gravity well here on earth and then move out into space in a spacecraft, and the astronomical year won’t change, whereas an atomic clock in space will speed up while the pendulum clock in space stops running altogether.

These kinds of clock-related “time” dilations were discovered centuries ago. The knowledge that gravity wells affected the tick rate of pendulum clocks was so well known in the 19th Century, the gravitational potential at different elevations on the earth’s surface was measured with pendulum clocks.

Yes, these kinds of clock dilations were discovered centuries ago. But Einstein, through Relativity, noted that there are also going to be time dilations as a result of gravitational fields or accelerations that are separate from the physical clock changes and so affect any and all clocks (atomic, pendulum, your own "body clock") equally.


In 1877 Maxwell wrote about the force of gravity, “Its intensity, as measured by pendulum experiments, decreases as we ascend”. He also wrote this about time, “We thus see the theoretical possibility of comparing intervals of time however distant, though it is hardly necessary to remark that the method cannot be put in practice in the neighborhood of the earth, or any other large mass of gravitating matter.” So he knew very well that gravity wells can alter the tick rates of clocks. By experiment we have learned that gravity wells slow down atomic clocks and they speed up pendulum clocks. These are simply the laws of nature. But relative motion can’t possibly alter the tick rate of any clock, since the clock experiences no “force” as a result of the “relative motion”.

Actually, relative motion doesn't really "affect the tick rate of the clock," as such. However, it does affect the perception of simultaneity of spatially distant events, which mathematically amounts to the same thing.

After all, if my perception of our clocks is that 12:00:00 on my clock is simultaneous with 12:00:00 on your clock, and then 12:01:00 on my clock is simultaneous with 12:00:48 on your clock, and then 12:02:00 on my clock is simultaneous with 12:01:36 on your clock (which is when we meet up), then that's the same thing as "your clock is running at 80% the rate of mine," isn't it (2:00 on my clock and 1:36 on yours)?

And if your perception of our clocks is that 12:00:00 on my clock is simultaneous with 11:59:06 on your clock, and then 12:01:00 on my clock is simultaneous with 12:00:21 on your clock, and then 12:02:00 on my clock is simultaneous with 12:01:36 on your clock (which is when we meet up), then that's the same thing as "my clock is running at 80% the rate of yours," isn't it (2:00 on my clock and 2:30 on yours)?

And yet when we meet up, my clock's at 12:02:00 and yours is at 12:01:36 for both of us and everybody's happy! :)

Sam5
2003-Dec-02, 04:59 PM
But Einstein, through Relativity, noted that there are also going to be time dilations as a result of gravitational fields or accelerations that are separate from the physical clock changes and so affect any and all clocks (atomic, pendulum, your own "body clock") equally.

No, sorry. Our “body clocks” are mainly a thermodynamic-time phenomenon, not an atomic or a “swinging mass” pendulum phenomenon. Just ask any biologist. They discovered this a long time ago.

This reminds me of the hospital in Russia that performs open heart surgery with no heart-lung machines. They do it by slowing down the deterioration rate (due to lack of oxygen) of the patient’s brain by means of a time dilation technique used on the patient, so the patient can go about a normal “earth hour” without a blood supply to his brain. How do you suppose they do it? By making the patient travel at nearly the speed of light?

No, they do it by packing the patient in ice.

SeanF
2003-Dec-02, 05:14 PM
But Einstein, through Relativity, noted that there are also going to be time dilations as a result of gravitational fields or accelerations that are separate from the physical clock changes and so affect any and all clocks (atomic, pendulum, your own "body clock") equally.

No, sorry. Our “body clocks” are mainly a thermodynamic-time phenomenon, not an atomic or a “swinging mass” pendulum phenomenon. Just ask any biologist. They discovered this a long time ago.

This reminds me of the hospital in Russia that performs open heart surgery with no heart-lung machines. They do it by slowing down the deterioration rate (due to lack of oxygen) of the patient’s brain by means of a time dilation technique used on the patient, so the patient can go about a normal “earth hour” without a blood supply to his brain. How do you suppose they do it? By making the patient travel at nearly the speed of light?

No, they do it by packing the patient in ice.

What? No, seriously, what are you talking about? Do you seriously expect me to respond to this by thinking, "Oh, cold does slow the deterioration of living flesh, so therefore obviously a gravity well can't slow down time"?

Sam5
2003-Dec-02, 05:33 PM
The math is what determines what each clock measures. What determines whether the theory is correct is whether the mathematical calculations line up with observation. They do (we wouldn't have GPS otherwise!). If you don't understand -- and by that I mean DO -- the math, you won't understand the theory.

No, the gravitational potential and other environmental conditions are what determine the rates at which each clock ticks in different places and under different environmental conditions. For example, your electronic wristwatch will tick more slowly when cold than when warm.

Once you understand the physics involved with different kinds of “clock drift”, then you can develop the appropriate math equations for each set of circumstances.

We have accurate GPS because calculations are made as to the rate of clock drift inside the GPS clocks and inside the earth-based atomic clocks. The GPS clocks drift for several reasons, and they can drift inside a satellite if there are temperature changes inside the satellite. You can go to several GPS “clock drift” websites and find out all about the temperature-related clock drift problems of GPS satellites and receivers. For example, try this Google search and look for the PDF file papers posted by universities:

GPS CLOCK DRIFT (http://www.google.com/search?hl=en&lr=&ie=ISO-8859-1&q=gps+clock+drift+temperature&btnG=Google+Search )

parejkoj
2003-Dec-02, 05:39 PM
No, the gravitational potential and other environmental conditions are what determine the rates at which each clock ticks in different places and under different environmental conditions. For example, your electronic wristwatch will tick more slowly when cold than when warm.


I am well aware of that (the temperature dependance of clocks having bitten me many times before when making measurements). But that has ABSOLUTELY NOTHING to do with the changes that SeanF and others have been talking about.

Let me repeat that: temperature, pulling on the pendulum, cooling the body and other such environmental conditions have ABSOLUTELY NOTHING to do with the actual changes in the rate of TIME due to GR, nor do they have anything to do with the actual changes in measured relative (there's that word again!) time due to SR. The GR/SR corrections are a completely separate set of corrections, if you want to be able to compare the times measured on two different clocks. And they are the same no matter what kind of clocks you are using.

Kaptain K
2003-Dec-02, 05:47 PM
http://www.winternet.com/~mikelr/flame63.html

Sam5
2003-Dec-02, 05:51 PM
What? No, seriously, what are you talking about? Do you seriously expect me to respond to this by thinking, "Oh, cold does slow the deterioration of living flesh, so therefore obviously a gravity well can't slow down time"?


A “gravity well” does slow down atomic clocks a very tiny amount. But if you wanted to slow down a person’s lifespan a noticeable amount by using that method, then I think you will discover that you would first kill him with the excessive gravitational potential inside the well. In other words, you would crush him to death while you are trying to slow down his atomic harmonic oscillation rates enough to notice any “lifespan change” in him due to the harmonic oscillation rate changes. So I think it is pretty much useless to try to conduct such a silly experiment.

In the meantime, the Russian hospital is slowing down the patients’ brain deterioration rates by using the thermodynamic technique. And they don’t have to subject the patient to excessive g forces by operating on him inside a whirling centrifuge. Anyway, if they tried that, the “Einstein method” of inducing time dilation in the patient, they’d wind up killing the patient.

But don’t take my word for it, go ask a biologist.

Since you are pretty good at relativity math, why don’t you calculate how many g forces a patient would have to undergo in order to slow down his body clock a ratio of about 6 minutes to 60 minutes. In other words, a patient can sometimes survive a lack of oxygen to the brain for about 6 minutes, but the Russian doctors stop the oxygen flow for about one hour at a time. So, how much g force would it require to stretch that 6 minutes out to one hour, using the “Einstein GR method” of time dilation?

Sam5
2003-Dec-02, 05:56 PM
parejkoj, you seem to think that there is no other cause of “time dilation” other than the “Einstein cause”. Ok, I’ll put the challenge to you: You can calculate how many g forces a patient would have to undergo in order to slow down his body clock a ratio of about 6 minutes to 60 minutes. In other words, a patient can sometimes survive a lack of oxygen to the brain for about 6 minutes, but the Russian doctors stop the oxygen flow for about one hour at a time. So, how much g force would it require to stretch that 6 minutes out to one hour, using the “Einstein GR method” of time dilation?

Emspak
2003-Dec-02, 06:34 PM
Sam5, I am going to try this one out because you seem to be missing something important -- whether deliberately or not I don't know.

I can slow chemical reactions down by cooling the medium they are in. But this has (as a couple of people have tried to point out) nothing to do with time dilation at all. It is a differnt class of phenomena. By analogy, I can kill a person by throwing them off a roof or poisoning them. But the fact that you can die after falling six floors has nothing to do with the fact that poison can also stop your bodily functions entirely.

In fact, were you to be travelling on a fast spaceship or something, and measuring the slowdown in reaction rates (or in your watch or whatever) due to packing them in ice, you would get the same number as someone doing the same experiment in a still spaceship. Relativistic effects are only discernable if you are "outside" a given reference frame.

Another example is the accelerating windowless spaceship -- outside any other forces, were you to be in a ship accelerating at 9.8 m/s/s, or in a ship standing on its tail on Earth, or in a ship facing away from the Sun at the requisite distance to get a 1g measurement of local gravity, there would be no way to tell unless you looked outside. Everything inside the spaceship behaves exactly the same in all three situations.

The reason everything looks the same inside all three spaceships is because when you do the relevant Lorentz transforms and correct for the shortening of spatial dimensions in the direction of travel you find that the numbers all sort of work out so that someone in one reference frame (the accelerating spaceship) gets the same numbers as the guy on the planet in the upended ship. This is also why you never get past lightspeed -- or at least why you never measure anything going faster than that. Dimensions get shorter, seconds get longer, mass increases and requires more and more energy to push you. But the guy in the ship won't notice -- in fact, his time dilation could easily reach a point where he gets to his destination "before" a light beam would by his own clock. If he were going to Sirius, say, and his time dilation made his clocks say it took hiim 2 years to get there as opposed to 8 for a light beam. Has he gone past lightspeed? No. He is moving at about 80% lightspeed or so.

I see alot of people trying to show they have seen some clever flaw in Einstein's reasoning or whatever. If debunking him were that easy it would have been done long ago because none of the things that depend on GR/SR working properly would function correctly. (Like GPS, and before that the time lag of satellite signals). I mean, are you a guy who thinks that people accept SR/GR because they like Einstein's hair? Or because they just want to put one over on everyone? His stuff works.

These theories are hard and complicated to understand, no doubt about it. But that doesn't mean that it can't be done without a little work. Do as SeanF, Kaptain K and others have suggested and do the math. Learn it, if you haven't got calculus (though that isn't completely necessary, it does make it simpler).
-J[/i]

daver
2003-Dec-02, 07:11 PM
Do as SeanF, Kaptain K and others have suggested and do the math. Learn it, if you haven't got calculus (though that isn't completely necessary, it does make it simpler).
-J[/i]
I agree. You can do it with simple algebra, and I caught the tail end of a lecture giving a geometrical rationale, but calculus makes things easier.

SeanF
2003-Dec-02, 07:17 PM
What? No, seriously, what are you talking about? Do you seriously expect me to respond to this by thinking, "Oh, cold does slow the deterioration of living flesh, so therefore obviously a gravity well can't slow down time"?


A “gravity well” does slow down atomic clocks a very tiny amount. But if you wanted to slow down a person’s lifespan a noticeable amount by using that method, then I think you will discover that you would first kill him with the excessive gravitational potential inside the well. In other words, you would crush him to death while you are trying to slow down his atomic harmonic oscillation rates enough to notice any “lifespan change” in him due to the harmonic oscillation rate changes. So I think it is pretty much useless to try to conduct such a silly experiment.

In the meantime, the Russian hospital is slowing down the patients’ brain deterioration rates by using the thermodynamic technique. And they don’t have to subject the patient to excessive g forces by operating on him inside a whirling centrifuge. Anyway, if they tried that, the “Einstein method” of inducing time dilation in the patient, they’d wind up killing the patient.

But don’t take my word for it, go ask a biologist.

Since you are pretty good at relativity math, why don’t you calculate how many g forces a patient would have to undergo in order to slow down his body clock a ratio of about 6 minutes to 60 minutes. In other words, a patient can sometimes survive a lack of oxygen to the brain for about 6 minutes, but the Russian doctors stop the oxygen flow for about one hour at a time. So, how much g force would it require to stretch that 6 minutes out to one hour, using the “Einstein GR method” of time dilation?

First off, a gravity well slows down time. Depending on the mechanics of the particular clock you are using to measure time, there may be additional effects on the clock itself. But time itself is changed in a gravity well.

Now, your question about slowing down the body clock is ridiculous for several reasons, not the least of which is because the doctors would have to be right there with the patient, meaning they'd be experiencing all the same time dilation as the patient, so it wouldn't buy them any more time to work on him.

But it's also irrelevant. The fact that you can't use relativistic time dilation for keeping patients alive longer doesn't mean that relativistic time dilation doesn't happen. The fact that there are physical means of slowing physical clocks (like grabbing the bloody pendulum) also doesn't mean that relativistic time dilation doesn't happen. So why are you talking about these things?

SeanF
2003-Dec-02, 07:25 PM
But the guy in the ship won't notice -- in fact, his time dilation could easily reach a point where he gets to his destination "before" a light beam would by his own clock. If he were going to Sirius, say, and his time dilation made his clocks say it took hiim 2 years to get there as opposed to 8 for a light beam.

Whoops, Emspak, this part isn't right. If your astronaut is traveling at a relativistic velocity, he would see contraction of distance - Sirius would no longer appear to be 8 light-years away, so it wouldn't take 8 years for the light beam to reach it. No matter how fast he goes, he'll still always see the light beam as going faster and reaching Sirius before he does.

[EDIT]I did some quick math.

Assuming 8.0 light-years to Sirius (and keeping to two significant digits), I figure a velocity of 0.97c will get the astronaut there in 2.0 years by his own clock. At that velocity, however, he will see the 8.0 light-year distance to Sirius contracted to 1.9 light-years, so the beam of light will still beat him there by about 0.1 years. :)

Sam5
2003-Dec-02, 08:16 PM
If debunking him were that easy it would have been done long ago because none of the things that depend on GR/SR working properly would function correctly.



Sorry, I’m not debunking “him”. I’m pointing out some of the errors in SR. And I don’t care who wrote the theory. That doesn’t matter.

The time lag of satellite signals doesn’t have anything at all to do with SR. And anyway, Doppler worked out the Doppler shift theory of moving emitters ages ago, back in 1842. A lot of this stuff isn’t “new”, and it doesn’t date just to 1905. A lot of it can be attributed to Lorentz, Poincaré, Maxwell, Hertz, Faraday, Doppler, and many others.

For example, in 1902 Poincaré wrote in his chapter on “Non-Euclidean Geometries”:

“Let us imagine to ourselves a world only peopled with beings of no thickness, and suppose these "infinitely flat" animals are all in one and the same plane, from which they cannot emerge. Let us further admit that this world is sufficiently distant from other worlds to be withdrawn from their influence, and while we are making these hypotheses it will not cost us much to endow these beings with reasoning power, and to believe them capable of making a geometry. In that case they will certainly attribute to space only two dimensions. But now suppose that these imaginary animals, while remaining without thickness, have the form of a spherical, and not of a plane figure, and are all on the same sphere, from which they cannot escape. What kind of a geometry will they construct ? In the first place, it is clear that they will attribute to space only two dimensions. The straight line to them will be the shortest distance from one point on the sphere to another—that is to say, an arc of a great circle. In a word, their geometry will be spherical geometry. What they will call space will be the sphere on which they are confined, and on which take place all the phenomena with which they are acquainted. Their space will therefore be unbounded, since on a sphere one may always walk forward without ever being brought to a stop, and yet it will be finite; the end will never be found, but the complete tour can be made.”

Oh, now, where have we heard that before? And is it usually attributed to Poincaré in 1902? No, not very often.

Sam5
2003-Dec-02, 08:20 PM
But the guy in the ship won't notice -- in fact, his time dilation could easily reach a point where he gets to his destination "before" a light beam would by his own clock. If he were going to Sirius, say, and his time dilation made his clocks say it took hiim 2 years to get there as opposed to 8 for a light beam.

SeanF says you are wrong, so which one of you two geniuses should I believe?

Sam5
2003-Dec-02, 08:22 PM
Emspak and SeanF, I’ve got two of you lecturing me about what’s what about Special Relativity, but both of you disagree, so what’s the problem? Are you two guys in different inertial frames, or am I?

SeanF
2003-Dec-02, 08:31 PM
But the guy in the ship won't notice -- in fact, his time dilation could easily reach a point where he gets to his destination "before" a light beam would by his own clock. If he were going to Sirius, say, and his time dilation made his clocks say it took hiim 2 years to get there as opposed to 8 for a light beam.

SeanF says you are wrong, so which one of you two geniuses should I believe?

Everybody makes mistakes, Sam5. I've made them, and I bet Emspak has, too. I know you have. Let's wait and see what Emspak says about my comments.

At any rate, you shouldn't just "believe" either one of us. You should endeavor to understand it. If you really understand SR, you can tell us which one of us is right.

kilopi
2003-Dec-02, 08:37 PM
Oh, now, where have we heard that before? And is it usually attributed to Poincaré in 1902? No, not very often.
Do you have a place where that quote is not attributed to Poincare, if it is Poincare?

Or do you mean the idea? I would say that's usually attributed to mathematicians who preceded Poincare.

Sam5
2003-Dec-02, 09:10 PM
At any rate, you shouldn't just "believe" either one of us. You should endeavor to understand it. If you really understand SR, you can tell us which one of us is right.


Lol, seems that we are at an impasse. I do understand it, but you don’t seem to.

Let me ask you this, what makes the “stationary” observer in SR “see” a time dilation in the “moving” clock? Doesn’t it have to do with the distorted “light signals” that the stationary observer is “seeing” as they arrive from the distant clock, rather than something that is actually going on physically “at” the distant clock? And remember, Einstein said in 1907 that the “geometrical shape” of a body does not actually change just due to relative motion, so you can’t tell me that something is happening physically at the distant clock, such as "length contraction", that makes it “slow down”.

The way he set up the theory, the “stationary observer” sees DISTORTED light signals coming from the distant frame. It’s the distorted light signals that give the illusion of “length contraction”, but he said in the 1907 paper that the “length contraction” is NOT real. It is the distorted light signals caused by the illusory “length contraction” that makes the distant clock appear to be “time dilated”, but the “length contraction” isn’t real, and therefore the “time dilation” isn’t real either.

The truth is, NOTHING unusual is happening at the distant clock. With relative motion, the distant clock is ticking normally. The distant clock DOES NOT time dilate because of “relative motion”.

The error in the SR theory is due to the “constancy of the speed of light” postulate, and Einstein’s thought-technique of having the local light-speed c-regulator being “stationary” with the “stationary” observer. All the “stationary” observer is “seeing” in the light signals in SR are distortions in the light signals coming from the distant clock. And this distortion is completely reversible, since the so-called “moving observer” becomes the “stationary observer” when we imagine ourselves fixed inside his frame. When we do that, then in SR theory, the c-regulator becomes “stationary” with that frame, and it is the other frame that is “seen” as “moving” and “time dilating”. I don’t understand why you can’t understand that.

If the distant clock itself, either one, “slowed down”, along with the time of its observer, that observer would see the so-called “stationary” clock SPEED UP. You just can’t have BOTH clocks “slowing down” the same amount at the same time, with just ONE of them winding up “really” time dilated.

Sam5
2003-Dec-02, 09:21 PM
kilopi, I’ve never heard the idea attributed to Poincaré. I just happened to find it in his 1902 book a few weeks ago.

His book is on the internet here (http://spartan.ac.brocku.ca/~lward/Poincare/Poincare_1905_toc.html)

SeanF
2003-Dec-02, 10:22 PM
At any rate, you shouldn't just "believe" either one of us. You should endeavor to understand it. If you really understand SR, you can tell us which one of us is right.


Lol, seems that we are at an impasse. I do understand it, but you don’t seem to.

Let me ask you this, what makes the “stationary” observer in SR “see” a time dilation in the “moving” clock? Doesn’t it have to do with the distorted “light signals” that the stationary observer is “seeing” as they arrive from the distant clock, rather than something that is actually going on physically “at” the distant clock?

See, I knew that it was the most basic aspects of SR that gave you trouble.

You put your camera on a tripod and set off the flash. ::poof:: Flash of light. You then wait for the flash to reset and do it again. ::poof::

Did those two flashes of light occur at the same point in space? Are you sure? The Earth is moving, isn't it . . . seems like your camera would've moved to a different point in space by the time you set off the second flash, wouldn't it?

So let's say we take the camera off the Earth, way way out into intergalactic space. We set off the flash ( ::poof:: ), wait a few seconds, and then do it again ( ::poof:: ). Did those two flashes occur at the same point in space. Are you sure?

The truth is that there is no such thing. If you're sitting motionless with respect to the camera, then of course you're going to conclude the two flashes occurred in the same place. But if you're in relative motion, meaning you see the camera as moving relative to yourself, then of course you're going to conclude that they did not occur in the same place, because clearly the camera moved between the two flashes. How far apart depends on how fast the camera is moving. This means that two observers who are both moving relative to the camera but not at the same velocity will not even agree on how far apart the two flashes were.

This doesn't just mean that it's impossible to determine absolutely how far apart the two flashes occurred, it means that there is no absolute answer.

The same is true for time. If you have two cameras located some distance apart, and they both flash, it is not only impossible to determine absolutely whether or not the flashes were simultaneous, but there actually is no absolute answer. It can not be absolutely true to say they were simultaneous, or to say this one happened one second before that one, or that one happened two seconds before this one.

Let's say you're motionless relative to the two cameras, right in between them, with each camera one light-minute away in opposite directions. You have sensors that will detect the light from the flashes. Both sensors detect the flashes at the same time. You conclude that the flashes were simultaneous, right? They both occurred one minute ago.

But what if there's another observer, with his own sensors, in relative motion? He sees the cameras and you moving at 0.6c relative to himself. As luck would have, he happens to just be passing you when the light from the flashes reaches you. Therefore, his sensors both go off simultaneously at the same time yours do. Since he's in the same place you are, the cameras are equidistant from him right now . . . but they weren't in the past, when the flashes actually occurred! If the two flashes had occurred simultaneously, one camera would've been closer to him and the other farther away, so he wouldn't receive the light simultaneously. Since he did receive the light simultaneously, the flashes themselves must have been non-simultaneous.

Just like a camera that flashes twice could flash in the same place for one observer but different places for another, two separate cameras that both flash could flash at the same time for one observer but different times for another.

You want to lay out space-time in a nice four-dimensional grid with right angles everywhere and say that any two chosen points are n meters apart along the x-axis, n meters apart along the y-axis, n meters apart along the z-axis, and n seconds apart along the t-axis. It just doesn't work that way. If it did, we would be able to measure "absolute motion."

So, we have that clock ticking off its seconds. It ticks off "12:00:00" and some time later it ticks off "12:01:00". Were those two ticks, those two points in space-time, located 60 seconds apart along the universe's t-axis? There is no absolute answer to that, because there is no absolute t-axis - it's relative.

An observer motionless relative to the clock says those two events occurred 60 seconds apart in time and 0 miles apart in space.

An observer in motion relative to the clock (say, at 0.6c) says those two events occurred 75 seconds apart in time and just under 14,000,000 miles apart in space.

They're both right. We're not changing the clock, we don't have to physically change anything. The difference is a result of how the two observers measure the universe.

I think the 14 million mile difference in the location measurements is a little easier to comprehend than the 15 second difference in the time measurements, but it's basically the same thing.

Sam5
2003-Dec-02, 10:26 PM
See, I knew that it was the most basic aspects of SR that gave you trouble.

Perhaps you misunderstood the reason for my questions. I'm trying to find out what part of the theory is giving you the most trouble.

Sam5
2003-Dec-02, 10:28 PM
Did those two flashes of light occur at the same point in space? Are you sure? The Earth is moving, isn't it . .

Hey! Give me a chance to answer your questions!

Sam5
2003-Dec-02, 10:31 PM
If you're sitting motionless with respect to the camera, then of course you're going to conclude the two flashes occurred in the same place.

No I'm not. I'm not an idiot.

SeanF
2003-Dec-02, 10:31 PM
Did those two flashes of light occur at the same point in space? Are you sure? The Earth is moving, isn't it . .

Hey! Give me a chance to answer your questions!

Ah, they're obvious questions. And the "Are you sure?" comes up no matter how you answer it. :D

Sam5
2003-Dec-02, 10:39 PM
But what if there's another observer, with his own sensors, in relative motion? He sees the cameras and you moving at 0.6c relative to himself. As luck would have, he happens to just be passing you when the light from the flashes reaches you. Therefore, his sensors both go off simultaneously at the same time yours do. Since he's in the same place you are, the cameras are equidistant from him right now . . . but they weren't in the past, when the flashes actually occurred! If the two flashes had occurred simultaneously, one camera would've been closer to him and the other farther away, so he wouldn't receive the light simultaneously. Since he did receive the light simultaneously, the flashes themselves must have been non-simultaneous.

Ok, we shall all now turn to page 25 of the Crown reprint of “Relativity, the Special and the General Theory”, and we shall study Chapter 9.

Then later I will give you a test about Chapter 9 and ask you some questions about it, particularly about Einstein’s statement:

“Now in reality (considered with reference to the railway embankment) he is hastening towards the beam of light coming from B, whilst he is riding on ahead of the beam of light coming from A. Hence the observer will see the beam of light emitted from B earlier than he will see that emitted from A.”

Sam5
2003-Dec-02, 10:44 PM
Ah, they're obvious questions. And the "Are you sure?" comes up no matter how you answer it. :D

Ok, ok.

Do you know what Einstein means by, “Now in reality...... he is hastening towards the beam of light coming from B, whilst he is riding on ahead of the beam of light coming from A.”

I know the answer, but I want to know if you know the answer.

Sam5
2003-Dec-02, 10:48 PM
SeanF, no fair looking up the answer on Google. I want to know if you already know the answer.

SeanF
2003-Dec-02, 11:08 PM
Google. Cute.

I don't have access to that book, and I don't know how much I want to hazard without reading the entire section in context. Which specific paper is this from?

A couple of notes I will make:

The thought experiments are not the theory, they are examples and/or demonstrations. As such, they are attempts to put the math into words, which can often result in ambiguous results.

A specific example of this can be found in what you've already quoted. The second time, you ellipsed out the qualifying parenthetical phrase, thereby endowing "in reality" with a quality of absoluteness it should not have.

My guess (** I RESERVE THE RIGHT TO ALTER OR APPEND THIS AFTER READING THE ENTIRE SECTION **) is that "he" refers to an observer in the center of a railway car traveling along the tracks. There are light emitters A and B that are located either at the two ends of the rail car or along the tracks. Emitter B is located at the leading edge of the car, with A at the trailing edge. He (Einstein) appears to be saying that when the emitters are lit up simultaneously from the frame of reference which is stationary relative to the tracks and presumably at the moment the observer is located directly in between them, the light from emitter B would reach "him" (the car-bound observer) before the light from emitter A does.

Sam5
2003-Dec-02, 11:22 PM
SeanF, here’s the Chapter. This is the origin of the thought experiment you were just telling me, about the flashes.

I’ll give you some time to study it. He is referring to the moving observer in the statement I quoted, and I would like some detailed information from you about what you think he means. I don’t mean to put you on the spot about this. I just want your opinion. Then I’ll give you mine, and you’ll probably say I’m wrong and don’t understand the theory. But we’ll deal with that when we get to it.

You answered the question partially, but there are some more details contained in the statement that I am interested in knowing your opinion about.

Chapter 9 (http://www.marxists.org/reference/archive/einstein/works/1910s/relative/ch09.htm)

daver
2003-Dec-02, 11:26 PM
Lol, seems that we are at an impasse. I do understand it, but you don?t seem to.

That isn't my impression.


Let me ask you this, what makes the ?stationary? observer in SR ?see? a time dilation in the ?moving? clock? Doesn?t it have to do with the distorted ?light signals? that the stationary observer is ?seeing? as they arrive from the distant clock, rather than something that is actually going on physically ?at? the distant clock?

There are two effects here, the time distortion and the doppler shift. On an object heading away from you, the effects add, making a clock seem to be running really slowly (time distortion slows the frequency, doppler slows the frequency further). On an object heading towards you, the effects are in different directions (doppler increases the frequency, time distortion slows the frequency). You can get rid of the doppler effect by sampling at the proper moment (as the object passes).


The way he set up the theory, the ?stationary observer? sees DISTORTED light signals coming from the distant frame.

No. For one thing, distance has nothing to do with it. The experiment can be set up so that distorted light has no bearing. For instance, assume one observer is in a mile-long spaceship. He heads out to Pluto, turns around, and zooms back, passing directly overhead at 90%c. How long does it take his shadow to pass? (2.6 microseconds, and we aren't relying on any distorted light coming from the ship's frame of reference).


It?s the distorted light signals that give the illusion of ?length contraction?

The length contraction is real, according to one observer. The pilot of the ship notices nothing wrong with his ship, but does see the earth as being seriously oblate.


The truth is, NOTHING unusual is happening at the distant clock. With relative motion, the distant clock is ticking normally. The distant clock DOES NOT time dilate because of ?relative motion?.

The ship's clock appears to inhabitants in the ship as if it were ticking normally. A "stationary" observer behind the ship looking at the ship's clock through a telescope would see the clock ticking more slowly (red shifted); a "stationary" observer ahead of the ship looking through a telescope would see the clock ticking more rapidly (blue shifted), a stationary observer being passed by the ship would see the clock ticking more slowly (time distortion).


When we do that, then in SR theory, the c-regulator becomes ?stationary? with that frame, and it is the other frame that is ?seen? as ?moving? and ?time dilating?. I don?t understand why you can?t understand that.

We do see that. There is no preferred inertial reference frame. What observer A sees about observer B, observer B sees about observer A.


If the distant clock itself, either one, ?slowed down?, along with the time of its observer, that observer would see the so-called ?stationary? clock SPEED UP. You just can?t have BOTH clocks ?slowing down? the same amount at the same time, with just ONE of them winding up ?really? time dilated.
No. Each observer sees the other's clock slow down.

Sam5
2003-Dec-02, 11:44 PM
Daver, well, I’ll take your word for that number about the shadow, and I’ll assume you have the mile-long space ship “length contracted”, right? But Einstein said in 1907 that geometrical length contraction is not a real phenomenon in relative motion. It doesn’t actually occur. Perhaps you missed reading that paper of his. It is titled, “On the Relativity Theory and the Conclusions Drawn From it,” published in 1907.

So all the short little space ships in all your physics books aren’t really short at all, and no spacecraft will be length contracted when it moves relative to the earth, unless of course, as Lorentz suggested, it passes near the earth and feels a resistance put up by some of the earth’s fields, but that is GR theory. But just “relative motion” won’t length-contract a space ship.

Your statement, “and we aren't relying on any distorted light coming from the ship's frame of reference).”

Ah, but you are. That “distorted light” is in the SR theory, and that’s where you got the idea of “length contraction”.

Sam5
2003-Dec-02, 11:49 PM
No. Each observer sees the other's clock slow down.


Ok, great, that’s what I’ve been saying all along that is a requirement in SR theory. Not in “reality”, and not in GR, but in SR theory. So, now, tell me what Observer A, who is stationary relative to Clock A, will “see” on the B clock in the first “clock paradox” thought experiment in “On the Electrodynamics of Moving Bodies”? Will Observer A see the B clock as running slow or running fast?

Sam5
2003-Dec-03, 12:04 AM
Daver, Ok, here we go. I just scanned this text from his 1907 paper:

”General remarks concerning space and time

1. We consider a number of rigid bodies in nonaccelerated motion with
velocities (i.e.. at rest relative to each other). In accordance with
principle of relativity we conclude that the laws according to which
the bodies can be grouped in space relative to each other do not change with
the change of these bodies' common state of motion. From this it follows that
laws of geometry determine the possible arrangements of rigid bodies in
accelerated motion always in the same way, independent of their common
of motion. Assertions about the shape of a body in nonaccelerated
motion therefore have a direct meaning. The shape of a body in the sense
indicated we will call its "geometric shape." The latter obviously does not
depend on the state of motion of a reference system.”

See?? There is NO change in the actual geometrical shape of a spacecraft that is moving “relative” to the earth. The “length contraction” in the 1905 paper is not a real thing. It is only an illusion perceived by the “observer” who “sees” it, and that’s because of the way he set up the theory in the first place.

SeanF
2003-Dec-03, 02:40 AM
SeanF, here’s the Chapter. This is the origin of the thought experiment you were just telling me, about the flashes.

I’ll give you some time to study it. He is referring to the moving observer in the statement I quoted, and I would like some detailed information from you about what you think he means. I don’t mean to put you on the spot about this. I just want your opinion. Then I’ll give you mine, and you’ll probably say I’m wrong and don’t understand the theory. But we’ll deal with that when we get to it.

You answered the question partially, but there are some more details contained in the statement that I am interested in knowing your opinion about.

Chapter 9 (http://www.marxists.org/reference/archive/einstein/works/1910s/relative/ch09.htm)

Okay, the sentence in question is: "Now in reality (considered with reference to the railway embankment) he is hastening towards the beam of light coming from B, whilst he is riding on ahead of the beam of light coming from A." So, let's break it down:

"Now in reality..."
In the sentence previous to the one in question, Einstein said, "If an observer sitting in the position M' in the train did not possess this velocity, then he would remain permanently at M, and . . . " so he was discussing what would happen if the observer were motionless with respect to the embankment. These first three words of this sentence indicate that he is done discussing that fictional observer (who remains at M) and is instead discussing the actual observer (who remains at M').

"...(considered with reference to the railway embankment)..."
We are using the embankment as our motionless reference frame, not the railcar.

"...he is hastening towards the beam of light coming from B,..."
The beam of light from the lightning strike at B is moving in the direction B->A with velocity c. The observer at M' is moving in the direction A->B with velocity v. Therefore, the observer and this light beam are approaching each other at c+v.

"...whilst he is riding on ahead of the beam of light coming from A."
The beam of light from the lightning strike at A is moving in the direction A->B with velocity c. The observer at M' is moving in the direction A->B with velocity v (which is less than c). Therefore, the observer and this light beam are approaching each other at c-v.

So, that's all pretty much what I said before. What more do you believe is in there?

Sam5
2003-Dec-03, 02:54 AM
SeanF, Man, you don’t know how many SR supporters have argued with me and claimed that c + v and c – v are “impossible” in SR. I’m glad you agree with me about something.

Sam5
2003-Dec-03, 03:00 AM
SeanF,


Ok, now, I suppose that you will say that the observer on the moving train will “not” measure the oncoming light at c + v and c – v??

And if I asked you where the “blueshift” and "redshift" originate for the moving observer, you will say “at the source”, rather than “at the observer”??

SeanF
2003-Dec-03, 03:07 AM
SeanF, Man, you don’t know how many SR supporters have argued with me and claimed that c + v and c – v are “impossible” in SR. I’m glad you agree with me about something.




Ok, now, I suppose that you will say that the observer on the moving train will “not” measure the oncoming light at c + v and c – v??


Yes, the observer on the train measures both beams of oncoming light at exactly c. It is "impossible in SR" for an observer to measure light moving at c+v or c-v relative to himself.



And if I asked you where the “blueshift” and "redshift" originate for the moving observer, you will say “at the source”, rather than “at the observer”??

Yes. The observer on the train sees one light source as moving away from him and the other towards him, so the Doppler shift he measures would originate "at the sources."

Sam5
2003-Dec-03, 03:38 AM
Yes. The observer on the train sees one light source as moving away from him and the other towards him, so the Doppler shift he measures would originate "at the sources."

What do you mean “at the sources”? Maybe I’m misunderstanding what you mean, or maybe you are misunderstanding what the moving observer sees and why he sees it.

I tend to think of an analogy with sound. For example, we’ve got a guy on a flatbed car at the end of a moving train. The train whistle sounds, and the sound waves are “redshifted” in the air behind the whistle and at the whistle. They are “stretched out” in the air.

But the guy at the end of the train doesn’t hear a “redshifted” sound (a lower tone). Why? Because he is moving toward the oncoming sound waves at s + v, the speed of sound in air plus his own velocity through the air. So, he encounters the “stretched out” waves at a rate that is faster than normal, and this has a “blueshift” effect. The Doppler “blueshift” cancels out the Doppler “redshift”, and the lower frequency (due to the whistle’s motion through the air) is canceled out by the higher frequency (due to the observer’s motion through the air).

The redshift takes place at the source, but the blueshift takes place at the observer. In Einstein's example, the blueshift and redshift take place "at the observer", not "at the sources", since the sources are not emitting redshifted or blueshifted light or “stretched out” or “shorter” wavelengths.

And Einstein tells why it happens when he says, “Now in reality....” It’s because of the c + v and c – v phenomenon experienced by the moving train observer as he moves through the embankment-stationary c-regulator. The velocity of light in his thought experiment is regulated to “c” only in reference to the embankment, but not in reference to the moving observer

Sam5
2003-Dec-03, 05:17 AM
parejkoj,


Here is an article on Eigen and Thermodynamic time that you mght find interesting:

Eigen and Thermodynamic time (http://www.pubmedcentral.nih.gov/articlerender.fcgi?artid=122860)

And here is one about biological thermodynamic time:

Thermodynamic biological time (http://www.endeav.org/evolut/text/ta/ta.htm)

Celestial Mechanic
2003-Dec-03, 05:18 AM
What is the difference between the special theory of relativity and the general theory of relativity?
In answer to the original post in this thread, I would say that special relativity is the study of physics in a four-dimensional manifold called spacetime that is endowed with a constant metric known as the Minkowski metric. General relativity is the study of physics in a four-dimensional manifold that now has a metric that is a function of space and time coordinates.

SeanF
2003-Dec-03, 12:46 PM
Yes. The observer on the train sees one light source as moving away from him and the other towards him, so the Doppler shift he measures would originate "at the sources."

What do you mean “at the sources”? Maybe I’m misunderstanding what you mean, or maybe you are misunderstanding what the moving observer sees and why he sees it.

I tend to think of an analogy with sound. For example, we’ve got a guy on a flatbed car at the end of a moving train. The train whistle sounds, and the sound waves are “redshifted” in the air behind the whistle and at the whistle. They are “stretched out” in the air.

But the guy at the end of the train doesn’t hear a “redshifted” sound (a lower tone). Why? Because he is moving toward the oncoming sound waves at s + v, the speed of sound in air plus his own velocity through the air. So, he encounters the “stretched out” waves at a rate that is faster than normal, and this has a “blueshift” effect. The Doppler “blueshift” cancels out the Doppler “redshift”, and the lower frequency (due to the whistle’s motion through the air) is canceled out by the higher frequency (due to the observer’s motion through the air).

Good example. Let's also add the idea that there is another train sitting motionless on parallel tracks. This train's whistle would not be Doppler shifted at the source, right? But the guy riding the flatbed of the first train would hear a Doppler shifted whistle from the second train because of his motion relative to the air. Right?

So, in Einstein's experiment, the observer on the embankment says "There's no Doppler shift at the light sources, and I'm not moving, so I should see no Doppler shifts. The guy on the railcar, though, is moving, so he should." To the embankment observer, the railcar observer is in the position of the guy as I him described above - moving while listening to a motionless whistle.

The observer on the railcar says, "There are Doppler shifts at the light sources, and I'm not moving, so I'll see those Doppler shifts. The guy by the embankment, though, is also moving, so he'll be creating his own Doppler shifts which will cancel out the source shifts, so he should not see the shifts." He sees the embankment observer like the guy on the flatbed as you described him above - moving while listening to a co-moving whistle.

Therefore, both observers predict that the railcar observer will see Doppler-shifted light and both observers predict that the embankment observer will not see Doppler-shifted light.



The redshift takes place at the source, but the blueshift takes place at the observer. In Einstein's example, the blueshift and redshift take place "at the observer", not "at the sources", since the sources are not emitting redshifted or blueshifted light or “stretched out” or “shorter” wavelengths.

And Einstein tells why it happens when he says, “Now in reality....” It’s because of the c + v and c – v phenomenon experienced by the moving train observer as he moves through the embankment-stationary c-regulator. The velocity of light in his thought experiment is regulated to “c” only in reference to the embankment, but not in reference to the moving observer

No, I already went over that. "Now in reality" does not mean that the stationary-embankment-and-moving-railcar scenario is "real" and the moving-embankment-and-stationary-railcar scenario is "not real". It means that he's no longer talking about the stationary-embankment-and-stationary-railcar-observer situation he described in the previous sentence, which is not real.

kilopi
2003-Dec-03, 01:08 PM
kilopi, I’ve never heard the idea attributed to Poincaré. I just happened to find it in his 1902 book a few weeks ago.

His book is on the internet here (http://spartan.ac.brocku.ca/~lward/Poincare/Poincare_1905_toc.html)
Why would you expect it to be attributed to Poincare? It'd seem like there are plenty of folk who had preceded him. Or are you claiming that someone has attributed Poincare's ideas to someone who came up with the ideas after him?

Sam5
2003-Dec-03, 02:05 PM
The observer on the railcar says, "There are Doppler shifts at the light sources, and I'm not moving, so I'll see those Doppler shifts. The guy by the embankment, though, is also moving, so he'll be creating his own Doppler shifts which will cancel out the source shifts, so he should not see the shifts." He sees the embankment observer like the guy on the flatbed as you described him above - moving while listening to a co-moving whistle.


No, no, no, no. This moving guy is not stupid. First, the man on the train knows he’s on a train that’s moving across the surface of the earth. He sees the trees and houses go by, the train is rocking back and forth, he bought his ticket at Chicago, he’s on his way to Los Angeles, and so he knows he riding on a train. This guy is not in the same position of our unfortunate friends, observers A and B, who are lost out in space somewhere, still arguing about what time it is.

The guy on the moving train knows the flashes did not take place on the train. Why? Because he does see a redshift and a blueshift, and he knows he is moving relative to the embankment, so he knows the sources are not moving relative to the embankment and not moving relative to the local light-speed regulator, and he also knows that the regulator through which he is moving is the local AIR (not a vacuum), and he knows the air is stationary with the embankment.

Therefore, he knows that he is approaching the front flash at the additive velocity of c + v, whilst he is riding on ahead of the flash coming from behind him that is approaching him at the subtractive velocity of c – v. The guy on the train knows this, the guy on the embankment knows this, the conductor on the train knows this, and little Sammy and Sean (two young physics students watching from the coach of the train) know this. And in this case, the phenomena are totally and completely explained by the Doppler theory.

So, the train observer has enough information to deduce that the Doppler shifts he sees occur at himself, due to his motion through the air, rather than “at the source”, and he has enough information to be able to tell that the sources are stationary and that he will meet the front flash at c + v and the rear flash at c – v.

And Einstein knows this too, and that’s why he said, “Now in reality.......[what will happen is.....]”.

Sam5
2003-Dec-03, 02:10 PM
No, I already went over that. "Now in reality" does not mean that the stationary-embankment-and-moving-railcar scenario is "real" and the moving-embankment-and-stationary-railcar scenario is "not real". It means that he's no longer talking about the stationary-embankment-and-stationary-railcar-observer situation he described in the previous sentence, which is not real.


Trust me, the embankment is not “moving”. The guy on the train felt an acceleration when the train took off from the last station and the guy on the embankment did not. They used their cell phones to communicate and so both knows which is moving and which isn’t. Plus the guy on the train feels an air breeze, while the guy on the embankment does not.

SeanF
2003-Dec-03, 02:12 PM
The observer on the railcar says, "There are Doppler shifts at the light sources, and I'm not moving, so I'll see those Doppler shifts. The guy by the embankment, though, is also moving, so he'll be creating his own Doppler shifts which will cancel out the source shifts, so he should not see the shifts." He sees the embankment observer like the guy on the flatbed as you described him above - moving while listening to a co-moving whistle.


No, no, no, no. This moving guy is not stupid. First, the man on the train knows he’s on a train that’s moving across the surface of the earth. He sees the trees and houses go by, the train is rocking back and forth, he bought his ticket at Chicago, he’s on his way to Los Angeles, and so he knows he riding on a train. This guy is not in the same position of our unfortunate friends, observers A and B, who are lost out in space somewhere, still arguing about what time it is.

The guy on the moving train knows the flashes did not take place on the train. Why? Because he does see a redshift and a blueshift, and he knows he is moving relative to the embankment, so he knows the sources are not moving relative to the embankment and not moving relative to the local light-speed regulator, and he also knows that the regulator through which he is moving is the local AIR (not a vacuum), and he knows the air is stationary with the embankment.

Therefore, he knows that he is approaching the front flash at the additive velocity of c + v, whilst he is riding on ahead of the flash coming from behind him that is approaching him at the subtractive velocity of c – v. The guy on the train knows this, the guy on the embankment knows this, the conductor on the train knows this, and little Sammy and Sean (two young physics students watching from the coach of the train) know this. And in this case, the phenomena are totally and completely explained by the Doppler theory.

So, the train observer has enough information to deduce that the Doppler shifts he sees occur at himself, due to his motion through the air, rather than “at the source”, and he has enough information to be able to tell that the sources are stationary and that he will meet the front flash at c + v and the rear flash at c – v.

And Einstein knows this too, and that’s why he said, “Now in reality.......[what will happen is.....]”.


The thought experiments are not the theory, they are examples and/or demonstrations. As such, they are attempts to put the math into words, which can often result in ambiguous results.

"Light is always propogated in empty space with a definite velocity c which is independent of the state of motion of the emitting body." - Albert Einstein

"Now in reality" does not mean that the stationary-embankment-and-moving-railcar scenario is "real" and the moving-embankment-and-stationary-railcar scenario is "not real". It means that he's no longer talking about the stationary-embankment-and-stationary-railcar-observer situation he described in the previous sentence, which is not real.

Sam5
2003-Dec-03, 02:20 PM
SeanF,
Are you saying that: 1) the guy on the train doesn’t know he’s moving, 2) he doesn’t know he is moving through the local light-propagating medium, 3) he thinks the embankment is moving, 4) he thinks the frequency shifts occurred at the sources, and 5)he is not moving relative to the light signals at c + v and c - v?

Sam5
2003-Dec-03, 02:28 PM
SeanF,

There are no “ambiguous results” in this thought experiment, the guy on the train is not in a vacuum, he’s not in empty space, the term “Now in reality” means “here’s what happens when the guy on the rail car moves relative to the embankment”.

SeanF
2003-Dec-03, 02:30 PM
SeanF,
Are you saying that: 1) the guy on the train doesn’t know he’s moving, 2) he doesn’t know he is moving through the local light-propagating medium, 3) he thinks the embankment is moving, 4) he thinks the frequency shifts occurred at the sources, and 5)he is not moving relative to the light signals at c + v and c - v?

You know, Sam5, those questions actually brought up an interesting point about SR . . . SR doesn't necessarily make predictions about what observers will think. What it really says is that all motion is relative, and thus the two situations

1) The railcar and its observer are stationary and everything else is moving
2) The railcar and its observer are moving and everything else is stationary

will be indistinguishable from each other. In comparing those two situations, it is generally described as if each observer considers himself to be stationary, but the people can "think" whatever they want.

Let me ask you this, though:


...he also knows that the regulator through which he is moving is the local AIR (not a vacuum)...

Do you believe that if a strong wind was blowing and the air outside the railcar and inside the railcar were moving in the same direction at the same velocity relative to the ground, that the embankment observer would see the light as Doppler-shifted?

Emspak
2003-Dec-03, 02:36 PM
Hey SeanF -- thanks for pointing out the weird mistake, I looked at it and discovered that I had garbled it as I was in a hurry.

I was thinking of the fact that when you figure in the shortening of dimensions, the guy in the moving spaceship sees Sirius as closer and his time slows down so he never gets to "beat" a light beam. I got a figure of .8c for that dilation factor because I was doing it in my head and tooka rough guesstimate of

t = t0 / sqrt(1- v^2/c^2) and took t as equal to 4. I think I messed that up tho.

Point is still valid tho. Time dilation - heck, Einstein's theory in general -- has no "errors' in the way Sam5 is describing. There may be some as yet unforseen problem with it, but I doubt any of us would see it since I don't see any experts in tensor math yet here. (Newton's theory doesn't have any "errors" either, in that sense, it just doesn't work at lightspeed because Newton didn't consider that, because he honestly had no reason to).

On another note, I do find it sot of fascinating that Einstein's theory works as "classical" physics. That is, you can leave out quantum effects and equations beave as though stuff moves, accelerates and does all that other stuff continuously.

Anyhow, I think this guy is just one of those who thinks he has hit on something terribly clever and won't do the math. His objections are essentially on philosophical grounds, and the universe doesn't give a hoot about that. Nor does it care what Maxwell or anyone else thinks.

Relativity works. Mass dilation (is that the right word?) is demonstrated every time you fire up a particle accelerator (as one fellow physics student said once, "We fling atoms around real fast and they get heavy. Seems good to me.") No bones about it. Certain particles last longer than they should when zooming through the upper atmosphere as cosmic rays. Heck, that's two out of three. I'll wait on the length contraction but someone someday can rig up a camera and zoom a spaceship by it. Bets, anyone?

Sam5 tried saying the time lag on signals has nothing to do with it but it does -- the gravity well of earth alters (local) time a bit relative to the sattellites, as well as redshifting signals, and altering the strict Newtonian behavior or orbits, so the satellites have to account for it or they'd be farther off than otherwise. Not much, but GPS is accurate within -- what, a few yards? -- because the software on the satellite takes this stuff into account and corrects for it. (This is terribly oversimplified I know).

That bit about "SeanF and Emspak don;t agree" well, not even Nobel-winning physicists agree completely on all of Relativity's ramifications, but that dog still hunts (and I misstated something). In fact, even Einstein was a bit off on some of his predictions -- Mercury's orbit and the sun's light-bending effect during an eclipse -- because the instruments they had in 1918 were just not as accurate as what we have now. Guess what happened when they could measure the aberration of starlight far more accurately? (It was actually a while after Einstein died, as I remember, but someone correct me here if that's wrong).

Just take a look at the the basic equations -- the ones that describe time dilation, contraction and increased mass. Plug some numbers in. You will see that they work, and it all becomes much clearer. Really, it does. (You can even use a cool graphing program -- the simple one that came with my Mac works).

#-o

Sam5
2003-Dec-03, 02:39 PM
SR doesn't necessarily make predictions about what observers will think. What it really says is that all motion is relative, and thus the two situations

1) The railcar and its observer are stationary and everything else is moving
2) The railcar and its observer are moving and everything else is stationary


I’m talking about “in reality” here and I want your opinions about “reality”. I’m not talking about a 98 year old theory. We know the rail car is moving across the surface of the earth because it experienced acceleration and the embankment did not, and we know the train has to keep its engine running to move, while the embankment has no engine. The car is moving through he air while the embankment is not.

Sam5
2003-Dec-03, 02:57 PM
Let me ask you this, though:


...he also knows that the regulator through which he is moving is the local AIR (not a vacuum)...


Do you believe that if a strong wind was blowing and the air outside the railcar and inside the railcar were moving in the same direction at the same velocity relative to the ground, that the embankment observer would see the light as Doppler-shifted?

I’d like for you to answer my questions first and not try to avoid answering them, but I will give you an opinion about your question.

(think, think, think......) Since the air acts as a sort of a "light propagating medium", I think what would happen would be that the embankment observer would see the left flash first, but he would not see the light as Doppler shifted. The light waves would be “stretched out” (redshifted) in the air (the wind), but the wind (the light propagating medium in this case) would cause the waves to travel at c + v relative to the embankment observer, with v being the velocity of the wind, so he would receive the redshifted light as being blueshifted, so the two effects would cancel each other out. However, since the wind is moving left to right, he would see the left flash first, and then the right flash.

But I'm not yet sure what roles the earth's magnetic, electric, and gravitational fields play in the regulating of the speed of light through the moving air at the surface of the earth. It could be that these fields act more as the local light-speed regulator than the air does, and if that’s the case, then my answer would apply to sound, but not necessairly to light. Now you will please answer my questions.

SeanF
2003-Dec-03, 03:09 PM
SR doesn't necessarily make predictions about what observers will think. What it really says is that all motion is relative, and thus the two situations

1) The railcar and its observer are stationary and everything else is moving
2) The railcar and its observer are moving and everything else is stationary


I’m talking about “in reality” here and I want your opinions about “reality”. I’m not talking about a 98 year old theory. We know the rail car is moving across the surface of the earth because it experienced acceleration and the embankment did not, and we know the train has to keep its engine running to move, while the embankment has no engine. The car is moving through he air while the embankment is not.


You keep thinking in terms of first “reality” and then “SR theory”. But you can’t do that and understand the theory. You must think only in terms of SR theory, and then you will begin to see that it doesn’t match reality. (Emphasis mine)

Why the change, Sam5? You don't want to think "only in terms of SR theory" anymore? And you're "not talking about a 98 year old theory" now, either? If you don't want to talk about SR, then you've been wasting a lot of time talking about SR.

Friction and inertia will keep the car and the embankment stationary relative to each other. The car runs its engine in order to overcome that inertia and friction (and it needs to continuously run that engine to continuously overcome the friction) so it will move relative to the embankment. But the idea that this means the embankment is absolutely stationary and the car is absolutely in motion is absolutely false - and that is "reality."

Sam5
2003-Dec-03, 03:10 PM
Relativity works. Mass dilation (is that the right word?) is demonstrated every time you fire up a particle accelerator (as one fellow physics student said once, "We fling atoms around real fast and they get heavy. Seems good to me.")

Lorentz described his mass increase theory and published it in 1904:

“Hence, in phenomena in which there is an acceleration in the direction of motion the electron behaves as if it had a mass m1; in those in which the acceleration is normal to the path, as if the mass were m2. Thee quantities m1 and m2 may therefore properly be called the ‘longitudinal’ and ‘transverse’ electromagnetic masses of the electron. I shall suppose that there is no other, no ‘true’ or ‘material’ mass.”

Sam5
2003-Dec-03, 03:15 PM
Why the change, Sam5?


Because the quote to which you refer had to do with our discussion about the alleged “time dilation” in relatively moving clocks in space.

Now we are talking about the c + v and c – v phenomenon of relative light speed at the surface of the earth. Now will you please try to answer all my questions? Or at least say, "I don't know".

Sam5
2003-Dec-03, 03:18 PM
Friction and inertia will keep the car and the embankment stationary relative to each other. The car runs its engine in order to overcome that inertia and friction (and it needs to continuously run that engine to continuously overcome the friction) so it will move relative to the embankment.

Thanks, but I already know how trains work. I figured that out when I was 5 years old. Now please try, attempt, to answer my questions.

SeanF
2003-Dec-03, 03:23 PM
SeanF,
Are you saying that: 1) the guy on the train doesn’t know he’s moving,

What the guy "knows" is irrelevant, and nobody knows absolutely whether he's moving or not. But whether he is moving or not, the results are the same.

2) he doesn’t know he is moving through the local light-propagating medium,
There is no "local light-propagating medium," so he can't be moving relative to it, so he can't "know" whether or not he's moving relative to it.

3) he thinks the embankment is moving,
What the guy "thinks" is irrelevant, and nobody knows absolutely whether the embankment's moving or not. But whether the embankment is moving or not, the results are the same.

4) he thinks the frequency shifts occurred at the sources, and
What the guy "thinks" is irrelevant, but in the reference frame where he is stationary, the frequency shifts occur at the source. In the reference frame where he is moving and the light sources are stationary, the frequency shifts occur at him. Either way, the results are the same.

5)he is not moving relative to the light signals at c + v and c - v?
In any reference frame, the light signals are moving relative to him at c + v and c - v where v is his velocity relative to that reference frame's definition of "stationary". In the reference frame where he is stationary, that velocity is zero, and so both c + v and c - v are equal to c.

SeanF
2003-Dec-03, 03:35 PM
I’m talking about “in reality” here and I want your opinions about “reality”. I’m not talking about a 98 year old theory.

Sam5, What you're asking of me now is the equivalent of the old, "When did you stop beating your wife?" question.

As far as I'm concerned, SR is reality. If you want me to describe how this thought experiment would play out if the embankment were absolutely at rest and the car were absolutely in motion, I will do that, but not under the guise that that is "reality."

Okay?

Normandy6644
2003-Dec-03, 03:42 PM
Sam5,

Why do you keep mentioning the Galilean transformations c+v and c-v? T he whole point of SR was to show that these can never be true, and that c is a constant no matter how it is observed. I don't why you keep bringing them into the argument.

Glom
2003-Dec-03, 03:46 PM
Okay, following this thread, full of thought experiments and screwed up clocks is difficult. But, from what I can gather, we must first establish two things, the postulates that form the basis of SR:

1) The laws of physics are the same in all inertial frames.
2) The speed of light is the same as observed in all inertial frames.

These two postulates eliminate the concept of absolute motion. There's no such thing. Get over it!

SeanF
2003-Dec-03, 04:01 PM
Okay, following this thread, full of thought experiments and screwed up clocks is difficult. But, from what I can gather, we must first establish two things, the postulates that form the basis of SR:

1) The laws of physics are the same in all inertial frames.
2) The speed of light is the same as observed in all inertial frames.

These two postulates eliminate the concept of absolute motion. There's no such thing. Get over it!

See, this is where Sam5's problem is. He doesn't accept these postulates - at least, he doesn't accept the second one. He's trying to prove that SR is false through thought experiments, but he wants to set up these thought experiments under his definition of "reality," which includes the idea that postulate 2 is false! That's circular reasoning.

SR is just a logical chain of reasoning which starts with those two postulates and ends with several conclusions.

If someone wants to disprove SR, they have two choices:

1) Prove that at least one of the postulates is false through actual, physical experimentation (eg. Michelson-Morely).
2) Prove that the conclusions of SR do not flow logically from the postulates. This requires accepting the postulates as true to begin with, and then reaching different conclusions than SR reaches (or reaching self-contradictory conclusions) through logical connections.

Sam5 seems to be trying the second method (which stands to reason, as the first needs a lab, not an Internet BB), but he doesn't want to accept the postulates as true to begin with.

As a side note, neither Einstein nor I nor anybody on this board has claimed that our thought experiments "prove" SR. As I've said numerous times, they are examples and demonstrations that seek to explain it. The "proof" is in the mathematical logic flowing from the postulates to the conclusions, as well as in the actual, physical experimentation (eg. Michelson-Morely) which proves the postulates.

Sam5
2003-Dec-03, 04:14 PM
What the guy "knows" is irrelevant, and nobody knows absolutely whether he's moving or not. But whether he is moving or not, the results are the same.

Well, maybe I’m just odd, but I generally know when my train is moving or not.

In the case of this thought experiment, everyone knows the train is moving relative to the embankment.



There is no "local light-propagating medium," so he can't be moving relative to it, so he can't "know" whether or not he's moving relative to it.

Something has to control the speed of light, locally, here on earth and relative to the earth. Just as something has to control the speed of light, locally, at the surface of a distant planet in a 3-c galaxy. Physics now accepts the theory that when the light leaves that galaxy, it is traveling toward the earth at c relative to the galaxy but at less than c relative to the earth, and as that light progresses toward the earth, it speeds up relative to the earth, and so, the light gradually moves from one local light-speed regulator to another. I don’t know what this local “regulator” is, but it exists.




What the guy "thinks" is irrelevant, and nobody knows absolutely whether the embankment's moving or not. But whether the embankment is moving or not, the results are the same.


Again I say that in this particular thought experiment, everyone knows the train is moving relative to the light sources, while the embankment is not. The motion of the train is why the guy on the train sees the right flash first and blueshifted.


What the guy "thinks" is irrelevant, but in the reference frame where he is stationary, the frequency shifts occur at the source.

This is just not so, and it’s nonsense. The reference frame in which he is stationary is moving toward the source on the right and away from the source on the left, and, thus, the frequency shifts occur at the observer, and not at the source. The source doesn’t even know the guy is there and moving, so they can’t shift at the source just for him when they are emitted. They travel at c relative to the local propagating medium, not relative to him. When the light signals are emitted, they don’t even know the guy exists. They are obeying the laws of physics, which require them to propagate locally at the speed of c relative to the earth, not relative to him. The light waves are not required to violate the laws of physics just so they can obey an incorrect 98 year old theory postulate. You might want to violate the laws of physics, so you can claim the 98 year old theory postulate is “right”, but the light waves aren’t working under such restrictions. They don’t have to obey theories, the theories have to obey the laws of physics



In any reference frame, the light signals are moving relative to him at c + v and c - v where v is his velocity relative to that reference frame's definition of "stationary". In the reference frame where he is stationary, that velocity is zero, and so both c + v and c - v are equal to c.

Not so, because in this thought experiment he sees two reference frames, the one he is moving with and the one he is moving relative to, and he knows his reference frame is moving through the local propagating medium toward the source of light on the right, while it is moving away from the source on the left. It is his frame’s motion, relative to the two sources and the propagating medium, that causes the frequency shifts and the c + v and c – v phenomena at the moving observer, and he should know this because the light waves are obeying the laws of physics. You are trying to have him and the light waves obey an obsolete theory postulate and not obey the laws of physics.

Glom
2003-Dec-03, 04:16 PM
If that is true, then I have to agree that Sam5's method is fallicious. You cannot make an assumption and then use it to prove the conclusion upon which the assumption is based.

The postulates are experimentally verified. Unless you have experimental evidence to dispute the postulates, you cannot assume they are false.

kilopi
2003-Dec-03, 04:37 PM
The light waves are not required to violate the laws of physics just so they can obey an incorrect 98 year old theory postulate. You might want to violate the laws of physics, so you can claim the 98 year old theory postulate is “right”, but the light waves aren’t working under such restrictions. They don’t have to obey theories, the theories have to obey the laws of physics
Right now, that 98 year old theory is a law of physics. :)

Sam5
2003-Dec-03, 04:38 PM
1) The laws of physics are the same in all inertial frames.


Glom & SeanF,

1) The laws of physics are the same in all inertial frames.

That quoted statement is correct, and the laws of physics – the laws of nature – state that the frequency shift in this thought experiment occurs at the moving observer and not at the source.

If SeanF tries to have the frequency shift occur at the source for the moving observer, then the stationary observer would also observe a frequency shift. In this thought experiment, the laws of physics require no shift at the source, they require no shift at the stationary observer, but they do require a shift at the moving observer, and that’s where the shift takes place.

The laws of physics in this case require no shift at the source, whether there are or aren’t any observers present, and if different observers are present, that does not change the basic laws of physics and nature. What the laws of nature require is that any shift observed takes place at the moving observer, not at the source, and this occurs in both frames.

We could look at it this way: Both observers are stationary. The lights flash. Both observers are still stationary. Thus, NO shifts occur “at the source”. Then one observer begins to move, and then he sees the first flash. And we all know that the shift occurs “at the observer”, not “at the source”.

If you say that in this case the shift occurs “at the source”, after the light has already been emitted, unshifted at the source, then you are violating one of the most basic laws of physics and you are trying to go back in time and change what happened at the source.

Sam5
2003-Dec-03, 04:43 PM
Right now, that 98 year old theory is a law of physics. :)

It is my understanding that the “laws of physics” should match the “laws of nature”. Please correct me if I am wrong.

And let me ask you this, in the train thought experiment we have been discussing, do you think the shift that the moving observer sees occurs “at the source” or “at the moving observer”?

Sam5
2003-Dec-03, 04:46 PM
Glom,

And let me ask you the same question. Do you think the shift that the moving observer sees occurs “at the source” or “at the moving observer”?

Glom
2003-Dec-03, 04:51 PM
SeanF and Sam5, whoever was making these points, you're trying to interpret the Doppler shift as an event in space and time. It isn't. It is merely a product of perception due to relative motion.

Shift does not happen as a result of motion of the source. That is confusing light with mechanical waves. Mechanical waves, like sound, propagate through a medium. Light does not. If the of sound waves is moving relative to the medium, shift will occur. But in the case of light, motion is always relative. There is no ethereal frame of reference to use to judge what is and isn't moving and hence shift is simply a product of relative motion between source and observer.

Sam5
2003-Dec-03, 04:52 PM
Normandy6644 and Emspak,

I’ll ask you the same question. In the train thought experiment SeanF and I have been discussing, do you think the shift that the moving observer sees occurs “at the source” or “at the moving observer”?

Sam5
2003-Dec-03, 05:02 PM
Shift does not happen as a result of motion of the source. That is confusing light with mechanical waves. Mechanical waves, like sound, propagate through a medium. Light does not. If the of sound waves is moving relative to the medium, shift will occur. But in the case of light, motion is always relative. There is no ethereal frame of reference to use to judge what is and isn't moving and hence shift is simply a product of relative motion between source and observer.

Are you saying that light traveling in the direction of the earth but within 3-c galaxies does not travel
locally at c within those galaxies but less than c relative to the earth?

Are you saying that light does not slow down when it gets near the sun? Are you saying that since we are dealing only with “relative motion” regarding light, its speed always remains the same relative to the sun, whether it is billions of miles from it or passing right by it?

Are you saying that light passing near a “black hole” will just continue on past the black hole, traveling at c relative to it, and it won’t get “sucked into” the black hole?

Glom
2003-Dec-03, 05:08 PM
Are you saying that light traveling in the direction of the earth but within 3-c galaxies does not travel
locally at c within those galaxies but less than c relative to the earth?

First, what's a 3-c galaxy? Second, light travels at c relative to all observers in inertial frames. That's the neat thing about SR. Time, length and everything else must change so that no matter what the relative speed is, light is always perceived to travel at c. You add gravity into the picture and then you start going GR.


Are you saying that light does not slow down when it gets near the sun?

According to my tutor, the speed does change, though I thought it carried on at constant speed, just with its path bent.


Are you saying that since we are dealing only with “relative motion” regarding light, its speed always remains the same relative to the sun, whether it is billions of miles from it or passing right by it?

Are you saying that light passing near a “black hole” will just continue on past the black hole, traveling at c relative to it, and it won’t get “sucked into” the black hole?

These are great gravity wells and such beyond the realm of SR. That's what GR is for.

Sam5
2003-Dec-03, 05:24 PM
First, what's a 3-c galaxy?

I mean a “superluminal” galaxy.

SEE THIS LINK (http://nedwww.ipac.caltech.edu/level5/ESSAYS/Cohen/cohen.html)



According to my tutor, the speed does change,

Great! Then please ask your tutor to come on this thread.


These are great gravity wells and such beyond the realm of SR. That's what GR is for.

Right, and what I'm talking about in the moving train example is beyond the realm of SR too.

Glom
2003-Dec-03, 05:30 PM
Then the disagreement is that you're talking GR while SeanF is talking SR. No wonder you two can't agree.

Sam5
2003-Dec-03, 05:31 PM
Shift does not happen as a result of motion of the source.


Are you saying that Doppler was wrong when he predicted that a shift would occur in starlight if the star were moving away from the earth? Are you saying the distant high-c galaxies are “not moving” relative to the earth and their redshift is caused by some reason other than their motion?

[And I don’t care if you believe the “distance between the galaxies and us is becoming greater” because “space is expanding” or because the galaxies are “moving through space”. That doesn’t matter.]

And what about light from M-31, which exhibits a blueshift on earth. Are you prepared to say that M-31 is “not moving” in space, but our galaxy is?

Are you saying that while we see a regular shift in light from revolving binaries, and while the rate of this shift is dependent on their revolution rate, this shift is NOT due to the “motion of the source”?

kilopi
2003-Dec-03, 05:36 PM
Right now, that 98 year old theory is a law of physics. :)

It is my understanding that the “laws of physics” should match the “laws of nature”. Please correct me if I am wrong.
So far as we have been able to determine, they do. That's what experiment does for us. So you insist that the 98 year old theory violates nature?

If you're not JW, I'll eat my hat.

Sam5
2003-Dec-03, 05:37 PM
Then the disagreement is that you're talking GR while SeanF is talking SR.


The example about the “black hole”, you are free to call “GR”. But the example about the moving train and the shift occurring at the moving observer, is neither “GR” nor “SR”. It’s “Doppler”.

I have thought of starting each of my statements out with one or more of the following notations, so we can all avoid confusion:

(SR)

(GR)

(Reality)

So:

(Reality) The shift in the moving train observer example occurs “at the observer”.

Sam5
2003-Dec-03, 05:42 PM
Glom,

Some things in nature are neither GR nor SR.

Glom
2003-Dec-03, 05:55 PM
You misunderstand me. There topic of discussion was about whether the shift happens at the source or at the observer. Such a question is crap because shift is not some event in space-time. Shift is due to the motion of both the source and the observer, not one or the other, because they aren't defined individually. They are for mechanical waves where the medium of propagation serves as an ethereal frame.

You entered into a huge rant based on that one sentence when the rest of the post, I believe, demonstrated clearly I am not disputing the cause of Doppler shift. Maybe I'm not writing clearly, but I know what I meant. Doppler shift is due to the relative motion between the source and the observer.

That link about superluminal galaxies: I think I know what they're talking about. It has to do with us perceiving only one component of its motion. As I understand, we observe a galaxy at time T. We then later observe its position at T+2yrs to find that it has moved tangentially such that it appears to be 3ly from its original position. It appears as though it has travelled 3ly in 2yrs, which gives it a speed of 1.5c, violating SR. In fact, what has happened is that in that time, it has also travelled towards us a bit. Say by 4ly (to make a Pythy triangle). That means that the second pulse, received two years after the first, was released from a position 4ly nearer and hence had 4ly less far to travel. This means it was emitted 6ly after the first. The galaxy has actually travelled 5ly by Pythy (3 tangentially and 4 radially) and so that means it's actual speed relative to us is 5/6c.

This would explain why there is no mention of a modified Doppler shift eqn to accomodate true FTL and why there is no mention of Cerenkov radiation, which would occur if the galaxy were truly outrunning its own light.

SeanF
2003-Dec-03, 06:02 PM
We could look at it this way: Both observers are stationary. The lights flash. Both observers are still stationary. Thus, NO shifts occur “at the source”. Then one observer begins to move, and then he sees the first flash. And we all know that the shift occurs “at the observer”, not “at the source”.

I've got to admit, Sam5, you've gotten me thinking on this one. I've got an idea tickling at the back of my head that maybe . . . no, I don't want to say anything until I think about it some more.

Sam5
2003-Dec-03, 06:18 PM
Glom, SeanF,

One problem with SR theory is that it tends to separate different “inertial frames” into separate autonomous entities, as if they exist separately in completely separate universes that are connected only by distorted “light signals”.

So in SR theory, certain odd things happen, which don’t happen in nature, and which tend to make the statement, “1) The laws of physics are the same in all inertial frames”, appear to mean something like, “The laws of physics are the same in all inertial frames, but the frames must not interact except by means of distorted light signals, with each observer seeing distorted signals when he looks into a different frame but not when he looks into his own frame.”

However, in GR, and in reality, the different frames do interact, and in fact, we can travel from one frame to another, and we will be aware of the transition. Whereas in SR theory, we are only in one frame or the other, and there is no gradual transition between the two. We are either in one or in the other, and there is no interaction or communication between frames, except by means of the mutually distorted “light signals” that travel between the frames.

In the moving train-observer example, neither SR nor GR applies. We can have smooth physical interactions and transitions between the two frames, and between the two observers in the two frames. The same laws of nature in one frame really do apply in the other frame, and both frames are ruled by those same laws. The laws are not contradictory, meaning that whatever happens in one frame also happens in the other frame, and the two observers agree on everything, and they don’t dispute anything. When something happens in one frame, an opposite and contradictory thing doesn’t happen in the other frame. For example, the frequency shift occurs “at the observer” in both frames.

Glom
2003-Dec-03, 06:33 PM
What's wrong with that? You seem upset because you rarely get a situation that can be solved purely from SR. Well that's fine. The real world is messy and you can't just have one theory to explain a situation. We know the limitations of SR. That's why GR was created. You seem upset with SR, not because it doesn't work, but because you don't like the boundaries established for it. While you're at it, have a sulk about Newtonian mechanics; that works even less than SR. Have a sulk about Maxwell; QM did away with him.

Sam5
2003-Dec-03, 06:46 PM
That link about superluminal galaxies:

Well, I apologize if I have confused you with that link. I was talking about the most distant galaxies that appear to be moving only radially, away from us, at faster than light speed. They are also called “superluminal” galaxies, but they have no side motion that we can see. That link I gave you talks about some closer galaxies that have some side motion too, due to some things going on at the galaxies.

So, what I meant by a 3-c galaxy is one that we see that is so redshifted, we think or we suppose that it might be separating from the earth at a speed three times the speed of light.

Glom
2003-Dec-03, 06:53 PM
How can that be? The eqn for Doppler shift gives imaginary frequencies for FTL. (http://www.fourmilab.ch/cship/doppler.html)

Emspak
2003-Dec-03, 07:06 PM
Sam5, I am trying to follow what you have here, but I'll take one more stab.

Deal is, SR is a beginning, a way of approach, if you will. Many of the thought experiments in this thread and elsewhere assume that the two moving spaceships or whatever are the only objects in the universe. That's why in SR, two ships zooming away from each other have no priviledged frame -- I look at the ship zooming away and his clocks look slow to me. He looks at mine and mine look slow to him. But if we zoom back to each other at the same speed, our clocks match all of a sudden. Because you can't tell who is "really" moving.

In the real universe, of course, the situation is a bit different. One has to pick a reference frame, which can interact with others. But those interactions -- say a doppler-shifted light -- aren't happening "at" anywhere. The light shift doesn't "happen" at a certain place. It is a function of movement. It is like asking where the movement of a person on a treadmill happens. Is he on one place, moving to different places along the treadmill, or is the treadmill moving? This gets into philosphy as much as physics.

GR is designed to work in the real universe, where different frames can interact. But when they do those interactions all obey the laws of GR. Glom gave a great example about the distant galaxy that appears to move faster than light. Another might be the rotation of the Earth. In one sense, I could argue that the rest of the universe really does rotate around it every 24 hours. By both GR and SR that is a perfectly valid argument to make.

But Alpha Centauri is not emitting Cerenkov radiation, even though it appears to move faster than light. If you work backwards and figure in relativistic effects on an object 4.3 light years away appearing to circle us in 24 hours, you get an answer that demonstrates that it is not violating relativity but can still look like it is zooming around at "superliminal" speeds. (There is an old relativity text I'll look for that demonstrates this nicely, but the math is a little complicated becuase you end up having to do Lorentz tranforms in two dimensions of circular motion and that's kind of a pain).

"You pays yer money and you picks yer frame of reference. " as Joe Haldeman wrote.

Sam5, I would say that the problem you run into is that SR and GR are both pretty counterintuitive and violate everyday expectations.

You say that people on trains know if they are moving and that is correct -- they usually do. But only because they can reference something else. Similarly, you are right now moving at up to 1,000 miles an hour (depending on your latitude). But the only reference you have is the movement of the sun across the sky. If I locked you in a room in a basement there would be no experiment you could do that would demonstrate you were moving relative to anything else -- or not. The speed of light would always look the same to you. As would the acceleration due to gravity and the length of stuff in the direction of motion --assuming your room is moving, but you can't tell. That is what we have all been trying to tell you.

This is experimentally demonstrable, by the way. And you can even do it at home. Set up a simple interferometer -- basically Michelson-Morely, in your basement -- all you need is a couple of mirrors, two laser pointers, (Michelson wished he had one of those!) and a big, heavy concrete block to attach it all to. Set up the beams to bounce around mirrors and cross back. Viola! the interference pattern will demonstrate that no matter what direction the aparatus is oriented in light moves at the same speed. Your level of sophistication will be at least as good as M-M, who did this experiment in about 1890, I think. (Someone correct me on the date).

Does this help?

daver
2003-Dec-03, 07:08 PM
No. Each observer sees the other's clock slow down.


Ok, great, that?s what I?ve been saying all along that is a requirement in SR theory. Not in ?reality?, and not in GR, but in SR theory. So, now, tell me what Observer A, who is stationary relative to Clock A, will ?see? on the B clock in the first ?clock paradox? thought experiment in ?On the Electrodynamics of Moving Bodies?? Will Observer A see the B clock as running slow or running fast?

I'm lazy--i'm not going to look up your source. I'm assuming observer A watches observer B zoom by, each in an unaccelerated reference frame.. Observer A thinks Observer B's clock is running slowly. Observer B thinks Observer A's clock is running slowly.

Nobody in an unaccelerated reference frame ever thinks time in any other unaccelerated reference frame is moving quicker.

daver
2003-Dec-03, 07:56 PM
Daver, Ok, here we go. I just scanned this text from his 1907 paper:

?General remarks concerning space and time

1. We consider a number of rigid bodies in nonaccelerated motion with
velocities (i.e.. at rest relative to each other). In accordance with
principle of relativity we conclude that the laws according to which
the bodies can be grouped in space relative to each other do not change with
the change of these bodies' common state of motion. From this it follows that
laws of geometry determine the possible arrangements of rigid bodies in
accelerated motion always in the same way, independent of their common
of motion. Assertions about the shape of a body in nonaccelerated
motion therefore have a direct meaning. The shape of a body in the sense
indicated we will call its "geometric shape." The latter obviously does not
depend on the state of motion of a reference system.?

See?? There is NO change in the actual geometrical shape of a spacecraft that is moving ?relative? to the earth. The ?length contraction? in the 1905 paper is not a real thing. It is only an illusion perceived by the ?observer? who ?sees? it, and that?s because of the way he set up the theory in the first place.

Thank you for scanning it in. I don't see that it helps your position much. A person on the spacecraft doesn't see any change in the shape of his spacecraft, regardless of how fast he's going. If SR is correct, a person being passed by that spacecraft would. If you liked to play with numbers you could set up a simple algebra problem and get the length contraction and time dilation to fall out.

We have good evidence of time contraction in non-accelerated reference frames. I don't know of any evidence of length contraction. Certainly if length contraction didn't occur then SR is blown out of the water.

SeanF
2003-Dec-03, 08:14 PM
This is experimentally demonstrable, by the way. And you can even do it at home. Set up a simple interferometer -- basically Michelson-Morely, in your basement -- all you need is a couple of mirrors, two laser pointers, (Michelson wished he had one of those!) and a big, heavy concrete block to attach it all to. Set up the beams to bounce around mirrors and cross back. Viola! the interference pattern will demonstrate that no matter what direction the aparatus is oriented in light moves at the same speed. Your level of sophistication will be at least as good as M-M, who did this experiment in about 1890, I think. (Someone correct me on the date).

Does this help?

That's not going to help, Emspak. Sam5 has made comments in this thread indicating he's basically a proponent of the "local frame-dragging ether" theory where the speed of light at the Earth's surface is regulated by the Earth's atmosphere or whatever, so an M-M experiment in his basement isn't going to change his mind.

SeanF
2003-Dec-03, 08:37 PM
We could look at it this way: Both observers are stationary. The lights flash. Both observers are still stationary. Thus, NO shifts occur “at the source”. Then one observer begins to move, and then he sees the first flash. And we all know that the shift occurs “at the observer”, not “at the source”.

Sam5, I think Glom might've been on to something.


Shift does not happen as a result of motion of the source. That is confusing light with mechanical waves. Mechanical waves, like sound, propagate through a medium. Light does not. If the of sound waves is moving relative to the medium, shift will occur. But in the case of light, motion is always relative. There is no ethereal frame of reference to use to judge what is and isn't moving and hence shift is simply a product of relative motion between source and observer.

Even though the observer can consider both himself and the emitter to be at rest at the time the light is emitted, and he can consider himself to be at rest and the emitter in motion at the time the light is received at him, the fact is that the reference frame in which the light was emitted and the reference frame in which the light was received are in relative motion to each other, so there's got to be a Doppler shift.

Just like with the Twin Paradox, we can't pretend the observer stayed in the same inertial frame.

I'm sure that's probably not going to be good enough for you because you want it to be explainable with the terminology of "classical" physics (what I think you'd like to call "reality"), but I don't think it works in this case.

Which makes me think I may have been hasty in originally describing the Doppler effect as occurring "at the source" or "at the observer" - that, too, was trying to fit it in classical terminology. :-?

Eroica
2003-Dec-03, 08:40 PM
Michelson-Morely ... who did this experiment in about 1890, I think. (Someone correct me on the date).
1887 (http://scienceworld.wolfram.com/physics/Michelson-MorleyExperiment.html)

Emspak
2003-Dec-03, 09:07 PM
thanks eroica. And SeanF, you may be right about this dude. I dunno. Can't help but try tho.

Although I disagree that relativity doesn't fit with classical physics. In many ways it is a classical theory -- or "clacissist" perhaps. That is, yuo get none of the weird stuff like in quantum and the math is basic calc. (No diffy Qs with weird probability functions). Things move continuously and all that. But that's just an aesthetic judgement, I guess.

Sam5
2003-Dec-03, 10:12 PM
SeanF,

Look, if revolving binaries reveal a Doppler shift to us, the frequency of which matches their revolution periods, then I think it is safe to say that the shift takes place at the binaries. There is no reason for it to take place anywhere else in space. And if the earth sees a starlight (and distant galaxy light) blueshift and a redshift, alternating every six months, then the shift obviously takes place at or near the earth, and no place else.

What you are deathly afraid of is admitting the existence of local c-regulators that travel through space with astronomical bodies. But such things have to exist, and in fact such things are contained in the SR theory itself, and they are always “stationary” with the “frame” that Einstein calls “stationary”.

You don’t have to call them an “ether” if you don’t want to. You don’t have to call them a “propagating medium” if you don’t feel like it. Call them whatever you like, but you must admit that they exist, because there is plenty of observational evidence that they do exist. And I believe if you study the guy on the train thought experiment, you will have to come to the conclusion that the guy on the train is moving through a c-regulator that is fixed relative to the earth, and that the shifts he sees occur right at him and not any place else, and that the shifts he sees occur as a direct result of his encountering the first flash at c + v and the second one at c – v, exactly as with a sound observer moving through the air toward and away from two stationary sound emitters.

What Einstein invented in the SR theory was not so much a “relativity” theory as it was a “parallel universe” theory, and he had the two “frames” too isolated from each other. So much so that one carried its own c-regulator through space with it, and the other carried its own c-regulator through space with it, and he provided no means for the two c-regulators to “blend” somewhere in space in-between the two observers. And that is one reason why the two observers must “disagree” when their two clocks unite at the end of his clock-paradox thought experiment. They can’t agree because in the theory he provides no method for them to communicate via light signals that travel through a mutual, intermediate, and shared c-regulator. So one observer “sees” the other guy’s clock “slow down”, while the other guy “sees” the first guy’s clock “slow down”, and so they disagree about what their clocks are reading when they finally unite and both clocks are right there in their hands when the two observers are face to face. He cleared up this problem later, and in fact, he began to clear it up with his “gravitational redshift” theory of 1911.

In the train example, one c-regulator is shared by both observers. It is stationary with one, while the other is moving through it. So there is no paradox, and both observers agree about the c + v and c – v motion of the light, relative to the train observer, and both observers agree that the shift takes place at the moving observer.



Even though the observer can consider both himself and the emitter to be at rest at the time the light is emitted, and he can consider himself to be at rest and the emitter in motion at the time the light is received at him, the fact is that the reference frame in which the light was emitted and the reference frame in which the light was received are in relative motion to each other, so there's got to be a Doppler shift.


First, he can not consider himself to be “at rest” if he is moving and knows he is moving in this thought experiment. This is not a two-guys-alone-in-space experiment. And of course he will see a Doppler shift, and the question at hand is “where" and “why”? There is no reason to try to avoid considering what each and every wave or photon is doing as it travels through the air. They are all un-shifted and on their way toward the mid-point on the track. They do not shift their frequency or their wavelengths in the air. The observed frequency shift is caused by the moving observer encountering more and less waves per second because of his motion toward and away from the emitters, while their velocity relative to the embankment is regulated by the c-regulator that is stationary with the embankment.

Saying the moving guy “can consider himself to be at rest” is wrong, and he knows it’s wrong. You are trying to force a separation and an isolation of the two frames in this experiment just as they are separated and isolated in the 1905 SR theory, but you can’t do it, because here they both share one single c-regulator, and the guy on the train is moving through it.

Sam5
2003-Dec-03, 10:30 PM
Thank you for scanning it in. I don't see that it helps your position much. A person on the spacecraft doesn't see any change in the shape of his spacecraft, regardless of how fast he's going. If SR is correct, a person being passed by that spacecraft would.

True, but there is an impression given in the 1905 theory, and this impression tends to be encouraged by certain physics professors, that the two spaceships are in two different “worlds” at the same time. There is an impression given by the theory that there are two separate sets of “existences” simultaneously, one that is “contracted” and one that is not. And in fact, that is the very reason Einstein had to clear up the matter in his 1907 paper, because so many guys like me were complaining about his ambiguity in the 1905 paper.

If in 1905 he had said, “This length contraction isn’t real”, then that would have been fine. But if he had said that, then his clocks would not "really” time dilate. And that would have been even better, and we wouldn’t have wound up with the twins paradox.


We have good evidence of time contraction in non-accelerated reference frames. I don't know of any evidence of length contraction. Certainly if length contraction didn't occur then SR is blown out of the water.

Well of course there is “time contraction” and “time dilation” in real life, but this has been known for centuries, although Newton's definition of "absolute time" caused a lot of confusion over the years. But real time dilation is not related to SR theory, and some of it was known long before GR theory. As I have mentioned, the biologists have known of it a long time, and in fact the latest type of time discussed in biology papers tends to be “thermodynamic time”, which is not related to relative motion or acceleration. What a lot of physics professors need to do today (and I hope I don’t offend anyone by saying this) is read a lot of biology “thermodynamic time” papers.

Sam5
2003-Dec-03, 10:46 PM
I'm lazy--i'm not going to look up your source. I'm assuming observer A watches observer B zoom by, each in an unaccelerated reference frame.. Observer A thinks Observer B's clock is running slowly. Observer B thinks Observer A's clock is running slowly.

Nobody in an unaccelerated reference frame ever thinks time in any other unaccelerated reference frame is moving quicker.

Right, but in Einstein’s 1905 thought experiment that started all of this “twins paradox” debate, 98 years ago, he said that only one of the clocks winds up really time dilated. That is the paradox. Both “see” the other time dilate, but when they unite, only one is “really” time dilated, and that’s impossible, under the terms set forth in that theory.

It’s kind of silly for us to get all obsessed about this, but both sides of the issue get obsessed over it. I know and have discussed this topic with several physicists and professors who agree with my point of view, but, they represent only about maybe 10% to 30% of all physicists, and they have an ability to control themselves and not get into arguments about it. If this were just my idea alone, me alone, I wouldn’t say anything about it.

I’ve been wondering if this situation might have something to do with the left and right side of the brain doing part of the thinking. Let’s say, those who “understand” SR theory can switch back and forth thinking about one “frame” with the left side, and the other “frame” with the right side. But when it comes time to make that final decision about those two clocks, and which one “really” lagged behind, one side of the brain dominates the thought process. But with guys like me, we can see both sides of the issue at the same time and we can easily see the paradox and the reason for it. I don’t know if this is because we are thinking with both sides of our brains or just one side all together. I don’t even know if this is a viable idea, but something is causing both sides of this issue to think the way they do and to be fairly militant about it.

Uhh, do you know anything about how brains think and how the left side interacts with the right side?

SeanF
2003-Dec-03, 10:48 PM
SeanF,

Look, if revolving binaries reveal a Doppler shift to us, the frequency of which matches their revolution periods, then I think it is safe to say that the shift takes place at the binaries. There is no reason for it to take place anywhere else in space. And if the earth sees a starlight (and distant galaxy light) blueshift and a redshift, alternating every six months, then the shift obviously takes place at or near the earth, and no place else.

"It's safe to say" . . . famous last words, Sam5, famous last words.


What you are deathly afraid of is admitting the existence of local c-regulators that travel through space with astronomical bodies. But such things have to exist, and in fact such things are contained in the SR theory itself, and they are always “stationary” with the “frame” that Einstein calls “stationary”.

SR doesn't include any "local c-regulators," space and time themselves are "universal c-regulators."

BTW, where do the "local c-regulators" in intergalactic space come from? Or do you believe that light travels infinitely fast out there?



You don’t have to call them an “ether” if you don’t want to. You don’t have to call them a “propagating medium” if you don’t feel like it. Call them whatever you like, but you must admit that they exist, because there is plenty of observational evidence that they do exist. And I believe if you study the guy on the train thought experiment, you will have to come to the conclusion that the guy on the train is moving through a c-regulator that is fixed relative to the earth, and that the shifts he sees occur right at him and not any place else, and that the shifts he sees occur as a direct result of his encountering the first flash at c + v and the second one at c – v, exactly as with a sound observer moving through the air toward and away from two stationary sound emitters.

What Einstein invented in the SR theory was not so much a “relativity” theory as it was a “parallel universe” theory, and he had the two “frames” too isolated from each other. So much so that one carried its own c-regulator through space with it, and the other carried its own c-regulator through space with it, and he provided no means for the two c-regulators to “blend” somewhere in space in-between the two observers. And that is one reason why the two observers must “disagree” when their two clocks unite at the end of his clock-paradox thought experiment. They can’t agree because in the theory he provides no method for them to communicate via light signals that travel through a mutual, intermediate, and shared c-regulator. So one observer “sees” the other guy’s clock “slow down”, while the other guy “sees” the first guy’s clock “slow down”, and so they disagree about what their clocks are reading when they finally unite and both clocks are right there in their hands when the two observers are face to face. He cleared up this problem later, and in fact, he began to clear it up with his “gravitational redshift” theory of 1911.

Nope, nope, nope. They disagree on what their clocks are reading while they're apart and relatively moving. Whenever they are either relatively motionless or united at the same place and time they will agree on what their clocks are reading.


In the train example, one c-regulator is shared by both observers. It is stationary with one, while the other is moving through it. So there is no paradox, and both observers agree about the c + v and c – v motion of the light, relative to the train observer, and both observers agree that the shift takes place at the moving observer.

Again, space and time are c-regulators themselves, and they're "shared" by both observers in all the experiments - they just disagree on how to measure them.


Even though the observer can consider both himself and the emitter to be at rest at the time the light is emitted, and he can consider himself to be at rest and the emitter in motion at the time the light is received at him, the fact is that the reference frame in which the light was emitted and the reference frame in which the light was received are in relative motion to each other, so there's got to be a Doppler shift.

First, he can not consider himself to be “at rest” if he is moving and knows he is moving in this thought experiment.
Any inertial frame can be considered to be at rest. There is no such thing as absolutely at rest or absolutely in motion.

After all, you want to consider the embankment at rest even though it is moving through space and you know it is moving through space...

This is not a two-guys-alone-in-space experiment.
It's an example! Railcars and lightning are used because they're easy for people to visualize. SR has always been about the speed of light in empty space.

Hold on a sec . . . what do you think will happen when it is two-guys-alone-in-space? If you no longer have the Earth's "local c-regulator," what would you predict for Doppler effect?


And of course he will see a Doppler shift, and the question at hand is “where" and “why”? There is no reason to try to avoid considering what each and every wave or photon is doing as it travels through the air. They are all un-shifted and on their way toward the mid-point on the track. They do not shift their frequency or their wavelengths in the air. The observed frequency shift is caused by the moving observer encountering more and less waves per second because of his motion toward and away from the emitters, while their velocity relative to the embankment is regulated by the c-regulator that is stationary with the embankment.

Well, he's got to see it when he and the light meet each other - that's the only time he can see it. The only question is what cause he attributes it to . . .


Saying the moving guy “can consider himself to be at rest” is wrong, and he knows it’s wrong. You are trying to force a separation and an isolation of the two frames in this experiment just as they are separated and isolated in the 1905 SR theory, but you can’t do it, because here they both share one single c-regulator, and the guy on the train is moving through it.
He's only moving through space in the embankment reference frame. He's motionless in space in his own.

SeanF
2003-Dec-03, 10:53 PM
And in fact, that is the very reason Einstein had to clear up the matter in his 1907 paper, because so many guys like me were complaining about his ambiguity in the 1905 paper.

Wow, you're old! ;)



I'm lazy--i'm not going to look up your source. I'm assuming observer A watches observer B zoom by, each in an unaccelerated reference frame.. Observer A thinks Observer B's clock is running slowly. Observer B thinks Observer A's clock is running slowly.

Nobody in an unaccelerated reference frame ever thinks time in any other unaccelerated reference frame is moving quicker.

Right, but in Einstein’s 1905 thought experiment that started all of this “twins paradox” debate, 98 years ago, he said that only one of the clocks winds up really time dilated. That is the paradox. Both “see” the other time dilate, but when they unite, only one is “really” time dilated, and that’s impossible, under the terms set forth in that theory.

It's not impossible, because to get them back together one of them has to change. That breaks the reciprocity.


I don’t know if this is because we are thinking with both sides of our brains or just one side all together.
Bite your tongue, Sean, bite your tongue.

Sam5
2003-Dec-03, 10:53 PM
This is experimentally demonstrable, by the way. And you can even do it at home. Set up a simple interferometer -- basically Michelson-Morely, in your basement -- all you need is a couple of mirrors, two laser pointers, (Michelson wished he had one of those!) and a big, heavy concrete block to attach it all to. Set up the beams to bounce around mirrors and cross back. Viola! the interference pattern will demonstrate that no matter what direction the aparatus is oriented in light moves at the same speed. Your level of sophistication will be at least as good as M-M, who did this experiment in about 1890, I think. (Someone correct me on the date).


If you take a flag down into your basement, it will not wave in an 18.6 mps wind as the earth revolves around the sun at 18.6 mps.

kilopi
2003-Dec-03, 10:57 PM
Right, but in Einstein’s 1905 thought experiment that started all of this “twins paradox” debate, 98 years ago, he said that only one of the clocks winds up really time dilated. That is the paradox. Both “see” the other time dilate, but when they unite, only one is “really” time dilated, and that’s impossible, under the terms set forth in that theory.
That's not true at all. As Einstein presents it in his original paper (http://mentock.home.mindspring.com/twinrdux.htm), there is no paradox, and the conclusion is consistent with the physics we know now.


It’s kind of silly for us to get all obsessed about this, but both sides of the issue get obsessed over it. I know and have discussed this topic with several physicists and professors who agree with my point of view, but, they represent only about maybe 10% to 30% of all physicists, and they have an ability to control themselves and not get into arguments about it. If this were just my idea alone, me alone, I wouldn’t say anything about it.
Some mathematicians and physicists are capable of basic misunderstandings too. Probably the reason they control themselves is they are on shaky ground. Not every physicist is an expert at relativity theory.

But with guys like me, we can see both sides of the issue at the same time and we can easily see the paradox and the reason for it.
Most physicists understand that there is no real paradox.


I don’t know if this is because we are thinking with both sides of our brains or just one side all together. I don’t even know if this is a viable idea, but something is causing both sides of this issue to think the way they do and to be fairly militant about it.
In my experience, the ones who get the most upset about it are the ones who understand it the least. Including one individual who professed to be a math phobe--but he was still firm in his conviction that he understood the math of relativity theory well enough to stand by his conclusions.

Sam5
2003-Dec-03, 11:02 PM
It's not impossible, because to get them back together one of them has to change. That breaks the reciprocity.

You know very well that neither of them has to “change” in the 1905 paper. You are thinking about the grammar-school “twins paradox” Jack and Jill example, where Jack “blasts off” then “turns around”, but you know that situation is not in the 1905 theory.


Bite your tongue, Sean, bite your tongue.

I think what we need to do is go to a major university hospital and have brain scans done on both of us while we are debating this issue.

SeanF
2003-Dec-03, 11:05 PM
It's not impossible, because to get them back together one of them has to change. That breaks the reciprocity.

You know very well that neither of them has to “change” in the 1905 paper. You are thinking about the grammar-school “twins paradox” Jack and Jill example, where Jack “blasts off” then “turns around”, but you know that situation is not in the 1905 theory.

Yeah, it is. You're still confusing the examples with the theory, Sam5. The facts that the laws of physics hold true for all inertial frames and that the speed of light in a vacuum is a constant in all inertial frames means that the twin who goes out and then comes back will be younger, and a simple understanding of what an "inertial frame" is will tell you that he can't do that if he stays in the same inertial frame.

It may not be explicitly stated in any of Einstein's thought experiments, but it is implicitly part of the theory.



Bite your tongue, Sean, bite your tongue.

I think what we need to do is go to a major university hospital and have brain scans done on both of us while we are debating this issue.

Sounds good to me. You want to make the appointment or should I? :)

Sam5
2003-Dec-03, 11:11 PM
It's not impossible, because to get them back together one of them has to change. That breaks the reciprocity.


You also know that if Jack and Jill both “blast off” in the opposite directions, go the same distance, then “turn around”, their atomic clocks will both slow down exactly alike, as per GR, but during their unaccelerated “relative motion” SR requires that they “see” each other’s clocks “slow down” the same amount, as per SR, and when they get back to earth SR also requires that only one of their clocks “lag behind” due to SR. And in this experiment we have both of them doing exactly the same thing. And if you leave out any consideraton of the “blast off” and “turn around” accelerations, you’ve still got the same SR clock paradox.