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grav
2007-Sep-09, 12:48 AM
Um, I been thinking quite a bit about this and it seems to me that in order for all observers to see light travelling at the same speed, then the rate of time for what we observe of another observer with a relative speed must be faster, not slower. Now, if length contraction shortens the other in the line of travel by gamma and time dilation also makes their clocks tick slower by gamma, then it would seem to make sense at first that light travelling over a lesser distance in a lesser time would make all observers see light travelling at 'c', right? But if time dilation is meant to counteract the length contraction for this purpose, where the other's rulers shrink in proportion to the length contraction, then they should actually measure a greater distance by laying the shrunken rulers end to end along that distance, shouldn't they?

Consider this. We send a signal to another observer with some relative speed. Let's say that they are travelling away from us at such a speed that the contraction shrinks them to half length. Then by their own shrunken rulers, they should measure the distance between us and them as twice as great, shouldn't they? Therefore, they would measure the speed of light as twice as great since it is travelling over twice the measured distance in the same time. So the only way to counteract that is with time dilation, whereas, their clocks are ticking twice as fast, not half as great. So in the same time that we see the light travelling to them at 'c' over some distance from our point of view (since we cannot go by theirs because to them nothing has changed), they will see the light travelling twice the distance over twice the time, since twice the time has passed for them, in order to also measure the speed of light at 'c'.

Even if we consider the alternate scenario where a pulse is sent between two observers upon opposite ends of a moving rod while we also observe, we would still see the pulse travelling at 'c' from one to the other, but over the shorter distance across the rod, while the travellers think the length of the rod has remained the same, so has travelled over twice the distance for them in the same time. The only way to counteract this, then, so that all observers see the light travelling at 'c' is for them to also measure the transit over twice the time, so that their clocks must be ticking twice as fast, not half as great.

This also makes sense if we consider gravitational time dilation. The stress upon a stationary object in a gravitational field will physically shrink the object somewhat. So the light will travel from one end to the other in a lesser time, and processes occur quicker, clocks tick faster, etc., in inverse proportion to the contraction of the object along the line of contraction, although I'm not sure how that would translate for the other axes.

So what am I missing here? What would the mainstream say about this?

cjl
2007-Sep-09, 01:49 AM
The problem is that to them, the ship isn't what contracts, the distances along the axis of the ship's travel contracts. So, they see earth as closer than it really is, not farther. At least that's how I understand it...

EvilEye
2007-Sep-09, 01:55 AM
If they were traveling at the speed of light in opposite directions, they would always see themsleves as the same distance apart......until they slowed.

Communication during lightspeed would be impossible because time had stopped while at lightspeed.

Only observers outside the ships would notice the difference.

Ken G
2007-Sep-09, 02:14 AM
Um, I been thinking quite a bit about this and it seems to me that in order for all observers to see light travelling at the same speed, then the rate of time for what we observe of another observer with a relative speed must be faster, not slower. Now, if length contraction shortens the other in the line of travel by gamma and time dilation also makes their clocks tick slower by gamma, then it would seem to make sense at first that light travelling over a lesser distance in a lesser time would make all observers see light travelling at 'c', right? But if time dilation is meant to counteract the length contraction for this purpose, where the other's rulers shrink in proportion to the length contraction, then they should actually measure a greater distance by laying the shrunken rulers end to end along that distance, shouldn't they?This is a very common pitfall to fall into. Think of it this way. Let's say you both have a 1 light-second-long ruler, and two clocks, one at each end of your ruler, and you are measuring the speed of light, each by sending a light signal along your own ruler and testing how long it takes. So apparently you both need to get 1 second as the answer. If you think the other guy is moving away from you in the direction of his ruler, then his ruler is short, so you think the light is really not travelling a full light second when it goes from one end of his ruler to the other. But you are forgetting that the end of the ruler is moving away from you, so moves while the light is in transit. You have to include that. But that's not all-- you're also forgetting the role of the simultaneity shift at the ends of his moving ruler from your reference frame! People are always forgetting the simultaneity shift. To you, his two clocks are not synchronized. So we have a host of effects that all have to be included: the ruler is short, the end of it moves during propagation of the light, and his clocks are not synchronized from your point of view. Put all that together, and you find he still doesn't get 1 second for his answer unless his clocks are ticking slowly.

grav
2007-Sep-09, 04:05 AM
This is a very common pitfall to fall into. Think of it this way. Let's say you both have a 1 light-second-long ruler, and two clocks, one at each end of your ruler, and you are measuring the speed of light, each by sending a light signal along your own ruler and testing how long it takes. So apparently you both need to get 1 second as the answer. If you think the other guy is moving away from you in the direction of his ruler, then his ruler is short, so you think the light is really not travelling a full light second when it goes from one end of his ruler to the other. But you are forgetting that the end of the ruler is moving away from you, so moves while the light is in transit. You have to include that. But that's not all-- you're also forgetting the role of the simultaneity shift at the ends of his moving ruler from your reference frame! People are always forgetting the simultaneity shift. To you, his two clocks are not synchronized. So we have a host of effects that all have to be included: the ruler is short, the end of it moves during propagation of the light, and his clocks are not synchronized from your point of view. Put all that together, and you find he still doesn't get 1 second for his answer unless his clocks are ticking slowly.Could you give an example?

Ken G
2007-Sep-09, 04:24 AM
The best way to do it is actually to get rid of length contraction and simultaneity issues by simply aligning the ruler perpendicular to the motion. Then all you need is the Pythagorean theorem to derive time dilation, because as you watch the other guy do his ruler experiment, you think the light is travelling farther (the hypoteneuse instead of the leg of the triangle) than he does. Thus to get the same speed as him, his clock must be slower. Once you have derived it that way, you can rotate the ruler along the direction of motion, and since that will not affect his clock rate but it does change the length of the ruler, you can then derive what desynchronization is consistent with the slowed time you just calculated. If you think about it, there are actually three unknowns here, the length contraction, the time dilation, and the desynchronization. There are only two independent alignments of the ruler for which you can apply the fixed c constraint, so you need a third constraint to solve for all three of those unknowns consistently. That third constraint is the other postulate of relativity-- the symmetry of the observers. That means you have to get the same answer if you conceptualize the receiver as moving as if you conceptualize the source as moving.

grav
2007-Sep-13, 02:13 AM
Thanks Ken. Sorry I didn't get back to you sooner but I wanted to run through a few calculations first. As far as aligning a ruler perpendicular to the motion as you mentioned, it seems an observer would actually see the light travelling a lesser distance, not farther, but with the same result, the other's time would run at a slower pace. It would be like watching him in slow motion, where his clock ticks slower. I didn't want to use a ruler for this example, though, because what the observer sees would also then might depend upon the angle to each point along it as far as the time dilation and contraction are concerned, especially since gamma itself can be regarded in terms of angles as 1/sqrt(1-(v/c)^2) = cos(sin-1(v/c)). Not that it means anything in particular, but I didn't want to take any chances.

So instead I used a black hole, where a photon might continually orbit it at some distance in a perfect circle. So in respect to the black hole, the light is moving at 'c' in orbit around it. But to an observer moving away from the black hole at a constant speed 'v', perpendicular to the path of the photon at all times, if all observers must measure the speed of light as 'c', then the speed that observer would view the photon travelling around the black hole to be would be c', where c = sqrt(c'^2 + v^2), since the square of the speeds for the vectors for the photon as it moves away from the observer at v while orbitting the black hole at c' must always add up to the square of the speed of light. This gives the speed of the photon in respect to the orbit only as c' = c * sqrt(1-(v/c)^2), so that the rate of time that passes for the black hole would appear to the observer to be lesser by sqrt(1-(v/c)^2)=1/gamma, exactly as relativity stipulates, as long as there is no contraction in any direction perpendicular to the direction of motion between the black hole and observer, and this is what is stated with the Lorentz contraction, so let's take a look at that.

With the Lorentz contraction, objects contract in the line of motion by 1/gamma, but not perpendicular to that motion. This is based upon the MM experiment, so we can apply it to that. We have a light beam split in perpendicular directions, travelling a distance of 'd' in each direction and brought back together again. No interference is detected with this, so we figure that the light has travelled in each direction and back in the same time. But if all observers in all reference frames observe the speed of light at 'c', then we have a problem for observers with some relative speed to the apparatus classically. (I know you know all of this, but I'm just trying to run through all of the bases). Lorentz, however, found a solution, that the length contracts in the direction of motion to an observer. So let's work through this. As you said, the time of the travel of light would also depend upon the motion of the point of reception. For an observer stationary with the apparatus, the light travels along all paths back and forth again in the time t = d/c each, but when the light travels forward at 'c' to the observer, it does so in the time,

c*t1 = x*d + v*t1, where x is the contraction in the line of motion, and v*t1 is the distance the receiving point has moved forward in the same time. For the light travelling back again, it is

c*t2 = x*d - v*t2, since in this case the receiving point is moving toward the direction of the light and lessens the distance it has to travel. Perpendicularly, it is

(c*t3)^2 = d^2 + (v*t3)^2 forward and back, where the length does not contract by 'x' in this case since the lioght is travelling perpendicularly to the line of motion. The resulting times are

t1 = xd/(c-v)
t2 = xd/(c+v)
t3 = d/sqrt(c^2-v^2)

Now, since the times forward and back in each direction must be the same, we have t1 + t2 = 2*t3, so

xd/(c-v) + xd/(c+v) = 2d/sqrt(c^2-v^2)

x*[1/(c-v) + 1/(c+v)] = 2/sqrt(c^2-v^2)

2c*x/[(c-v)(c+v)] = 2/sqrt(c^2-v^2)

c*x/(c^2-v^2) = 1/sqrt(c^2-v^2)

x = sqrt(c^2-v^2)/c

x = sqrt(1-(v/c)^2) = 1/gamma

This demonstrates the contraction as 1/gamma, also exactly according to relativity. And placing this back into the formulas for the time gives us

t = t1 + t2 = 2*t3

t = xd/(c-v) + xd/(c+v)

t = sqrt(1-(v/c)^2)*d*2c/(c^2-v^2)

t = (d/c)/sqrt(1-(v/c)^2)

t = (d/c) * gamma

And the time for observers stationary with the apparatus measure the passage of time as just d/c, so the observer sees a greater time passing according to their own clock and therefore less for the other, exactly as relativity says. So that's three for three for relativity, right? This is how it is all said to be. So all's good so far, though, right?

Okay, now here's my dilemma. Everything works out fine when we consider the times back and forth together. But let's take each of these times, back or forth, not both, using the same formulas as above, apply relativity, and compare them to those in the stationary frame individually. Now we have

t1 = xd/(c-v)
= sqrt(1-(v/c)^2)*d/(c-v)
= sqrt[(1+v/c)/(1-v/c)]*(d/c)

t2 = xd/(c+v)
= sqrt(1-(v/c)^2)*d/(c+v)
= sqrt[(1-v/c)/(1+v/c)]*(d/c)

t3 = d/sqrt(c^2-v^2)
= 1/(1-(v/c)^2)*(d/c)
= gamma*(d/c)

Now, according to this, then, the times we observe passing for the apparatus and stationary observers would be proportional to their Relativistic Doppler shift, not gamma, except in the perpendicular direction only. This is partly the same dilemma I had in the Lorentz contraction thread towards the end, where from one perspective, the observed time is directly related to gamma, but from another, it should be directly related to relativistic Doppler. Adding them together for the total time forward and back in the direction of motion all falls in line with relativity, but not individually, as the ratio of the rate of times for one time which pass for the apparatus and observer would be different from the other, depending upon which way we consider the light to be travelling, one faster and one slower by the same amount, which should not be the case.

I am currently attempting to work through this, using three unknowns; length contraction, time dilation, and the speed of light from a moving source. To find these I am applying various scenarios such as using relativistic Doppler for its simplicity, assuming it is correct, which one link says has been recently verified very accurately, although the measurements were very difficult to make the way they had to do it, while another says that an error in relativistic Doppler explains the Pioneer anomaly. Another might be assuming a lesser time for what each observer sees of the other might be completely light travel time, where light travels at a speed other than 'c' from a moving source according to observers. Others might be reference to an ether or an expansion of space and/or of objects while the light is in transit, etc. Results are so far inconclusive, however. There are many such possible scenarios to run through, at least twenty or thirty that I can think of offhand, with perhaps a couple hundred variations between them. I will have to find ways to narrow them down.

Ken G
2007-Sep-13, 04:33 AM
Now, according to this, then, the times we observe passing for the apparatus and stationary observers would be proportional to their Relativistic Doppler shift, not gamma, except in the perpendicular direction only.

That sounds fine to me, I don't see why you consider this a "dilemma". This is just what you should get, because the calculation you are doing is exactly the same as what you'd get if you consider a receiver moving through a light wave of wavelength d in the source frame. The gamma factor comes from time dilation instead of length contraction, but the expressions are exactly the same, and you of course expect to get the relativistic Doppler shift. Everything here is just what it should be.

grav
2007-Sep-13, 12:24 PM
That sounds fine to me, I don't see why you consider this a "dilemma". This is just what you should get, because the calculation you are doing is exactly the same as what you'd get if you consider a receiver moving through a light wave of wavelength d in the source frame. The gamma factor comes from time dilation instead of length contraction, but the expressions are exactly the same, and you of course expect to get the relativistic Doppler shift. Everything here is just what it should be.I was afraid you might think that. That's why I ran through the whole thing. It would seem it might be that way at first glance, but look a little deeper. The ratio of the rate of times that pass for the moving observer and for the apparatus and observers stationary with it are just the ratio of the rate of times that pass for the photon on its two way trips back and forth for the moving and stationary observers, which gives 1/gamma for the rate of time of the stationary observers according to the moving one. Those last times, however, are only the one way times of the photon, which give different ratio of rates of times forward and back, sqrt[(1-v/c)/(1+v/c)] forward and sqrt[(1+v/c)/(1-v/c)] back, the same as with relativistic Doppler, but these are the actual ratio of the rates of times observed upon each of the observers' clocks, not the ratio of the frequency of pulses, which would be found by comparing the rate of times between successive pulses and then mutiplying that by the ratio of the rate of times as 1/gamma as each observer views the other's clock as the pulses are emitted. The ratio of the rate of times itself is found in the same way as that for the ratio of the rate of times for the entire trip that originally gave us 1/gamma, but which now gives a different ratio of rate of times between the clocks stationary with the apparatus and the moving observer, depending upon whether the photon is travelling forward or back, which should not be. When the photon is travelling forward, the moving observer sees a time of sqrt[(1+v/c)/(1-v/c)](d/c) passing upon his clock while the stationary observers see just d/c passing upon theirs, so the moving observer would say that the rate of time is passing (d/c) / [sqrt[(1+v/c)/(1-v/c)](d/c)] = sqrt[(1-v/c)/(1+v/c)] more slowly for the stationary observers, and the inverse of this when observing the photon travelling back.

Now, you also mentioned simultaneity issues here, which also led to relativistic Doppler in the other thread, but it seemed then also that the formula for the simultaneity should also give the ratio of the rate of times as sqrt[(1-v/c)/(1+v/c)] and sqrt[(1+v/c)/(1-v/c)], but this was only after the ratio of the rate of times was already multiplied in as 1/gamma, so it could not be both. That formula gives different solutions depending also upon the distance between observers, but not upon light travel times, since we get the same times observed between observers which are both stationary, which incorporates but does not include a lag of times in the result. The simplest way to possibly avoid such a simultaneity issue might be to just run the photon out some distance and then bring it back to the same original point, as is done with the MM experiment, when considering the entire two way paths of the light forward and back. It's possible the observed time for the one way path of the photon might be affected by simultaneity issues in some way, although I don't see how yet specifically, but I'll keep working on it.

Ken G
2007-Sep-13, 03:05 PM
It would seem it might be that way at first glance, but look a little deeper. The ratio of the rate of times that pass for the moving observer and for the apparatus and observers stationary with it are just the ratio of the rate of times that pass for the photon on its two way trips back and forth for the moving and stationary observers, which gives 1/gamma for the rate of time of the stationary observers according to the moving one. Those last times, however, are only the one way times of the photon, which give different ratio of rates of times forward and back, sqrt[(1-v/c)/(1+v/c)] forward and sqrt[(1+v/c)/(1-v/c)] back, the same as with relativistic Doppler, but these are the actual ratio of the rates of times observed upon each of the observers' clocks, not the ratio of the frequency of pulses, which would be found by comparing the rate of times between successive pulses and then mutiplying that by the ratio of the rate of times as 1/gamma as each observer views the other's clock as the pulses are emitted.I know all that, and there is still no problem at all.
The ratio of the rate of times itself is found in the same way as that for the ratio of the rate of times for the entire trip that originally gave us 1/gamma, but which now gives a different ratio of rate of times between the clocks stationary with the apparatus and the moving observer, depending upon whether the photon is travelling forward or back, which should not be. Why on Earth not? Are you suggesting a new "relativity of time ratios"? No such principle. Once again you are imagining that time dilation is the only thing that happens to time. Do I have to mention one more time that there is also a shift in synchronicity when you change frames? That does violence with the ratios you expect to be preserved. They are not. Please, never again apply time dilation without the simultaneity shift, that single oversight creates the vast majority of all relativity "paradoxes" I've ever seen.

Now, you also mentioned simultaneity issues here, which also led to relativistic Doppler in the other thread, but it seemed then also that the formula for the simultaneity should also give the ratio of the rate of times as sqrt[(1-v/c)/(1+v/c)] and sqrt[(1+v/c)/(1-v/c)], but this was only after the ratio of the rate of times was already multiplied in as 1/gamma, so it could not be both.Think again, it's all the simultaneity shift.
That formula gives different solutions depending also upon the distance between observers, but not upon light travel times, since we get the same times observed between observers which are both stationary, which incorporates but does not include a lag of times in the result.The difference in distance comes out in the wash because it appears in both simultaneity shifts, and all that ends up mattering is d, or if you like, the light travel time. Just put it all in, and you will get it all out. Stop trying to tell it what it "should" do, you'll be forever getting yourself in a tizzy-- just follow the Lorentz transformation, that's what it's there for.

grav
2007-Sep-13, 06:00 PM
Although it may have sounded otherwise, I was actually sort of agreeing with you with that last paragraph, that I may still be overlooking something with the simultaneity shift, especially since I'm thinking about it as a direct result of time dilation to begin with, and I want to make sure. So I guess what I should ask is if the ratio of the rate of times observed appears greater when the light travels in one direction and smaller in the other, then precisely how would one apply simultaneity effects in order to correct for this?

Ken G
2007-Sep-13, 08:09 PM
The simultaneity shifts are not applied to "correct for" this, there is no need to correct for it-- the simultaneity shifts are what causes the discrepancy, but there's nothing wrong with
that discrepancy. Think of it this way. Let's say a brother, A, gives his sister, B, a treasured family heirloom, and years later B returns it to A. A and B are like the mirrors of the MM experiment. Let's further say that this transaction is witnessed by two other people, X and Y, who are like your reference frames. Now, person X tends to put great personal value in memorabilia, so thinks the value of what was exchanged was spectacular, while person Y looks only at what price it could get at an auction, so values it much less. The analog of this value is like the elapsed time between mirrors. So we have a ratio there in the value as perceived in the reference frames of X and Y.

Now the question is, does that value ratio necessarily have to be the same in both directions that the transaction occurred? The answer is no, not if there was a value shift that Y saw and X didn't. For example, perhaps X sees the value as unchanged by the transaction because it still represents a treasured memory, while Y thinks that an heirloom in the hands of someone (B) that is not the original owner (A) will reduce its value at auction. So that shift is like a simultaneity shift that one person sees and the other doesn't-- and it will change the ratio of the value as seen from reference frame X versus Y when the heirloom is going to B then when it is returned to A. So here we have an everyday analogy that can achieve the same effect-- ratios of things are not necessarily time reversible if there is a differential shift that occurs at one end of the process.

grav
2007-Sep-22, 11:24 AM
Sorry I took so long to respond again. I've been going over this some, so let's see what I can come up with, and how it sounds to you.

Okay, so let's see. The simplest solution to this would be that light travels with the additional speed of the source. Then, although observers stationary with the apparatus see light travelling at c in all directions since the source is moving with them, another observer that sees the apparatus moving at some relative speed away from them would see light travelling at c+v from the source to the mirror further away while the mirror is also moving away at the same time that the light is catching up, and c-v back again, where the original mirror of separation of the beams would be moving toward the light according to the other observer, so that

(c+v)*t = d + v*t
c*t = d

(c-v)*t' = d - v*t'
c*t' = d

t=t'

and the same thing goes for the perpendicular beam. But Einstein used the postulate, from Maxwell's equations, that all observers see the speed of light at c, and relativity matches observations so far. As well as this, light travelling with the additional speed of the source would make a particle phenomenom.

So let's say we have three observers, two of which are stationary to each other and a third with a relative speed to them, so that the first two are both positioned along the line of travel. The first, closest to the third, sends a pulse to the second. The second also sends a pulse to the first. To them, each receives the other's pulse in the same time, and records it as such according to their own clocks. Furthermore then, each should see the other receiving each pulse when the other's clock reads the same thing their own clock would read for the distance travelled, minus the time of flight of light for this observation, of course. The third observer, however, would see the pulses still travelling at c in either direction, first to second and second to first, but while the pulse from the first travels toward the second, the second observer is moving away, so the light takes longer to reach the second than it does the from the second to the first, while the first would be moving toward the pulse. The third observer must read the same times upon each of their clocks when the pulses are actually received, however, since at that instant the pulses would be synchronous with each of the two moving observers. That means that the third observer must see the second observer's time lag behind the first's, since the light takes longer to reach the second observer from the first than it does to the first from the second according to the third observer. Since the first and second observers are moving together, there is no time dilation between them, though, just the time lag, which then becomes the simultaneity issue, as far as I can tell.

So how would would this simultaneity issue come about? Let's see. If we begin with two observers which are stationary to each other, then we can synchronize their clocks. Now let's move them apart some distance from each other. As we do so, their will be a relative speed between them. So each sees the other's clock moving more slowly, with more than just the time of flight lag due to separation, but in a relativistic way. This will build a time lag between them that each observes of the other while they are being separated. So when we bring them to rest in respect to each other again after some separation, this time lag will remain, as well as the time of flight lag, even though each of their clocks are once again ticking at the same rate.

To a third observer, as the other two separate, the second observer (the one furthest away) would appear to be moving away from the third at a greater rate than the first to the third, and so the time dilation between the second and third would be greater. This would mean a greater time lag would build up during the time of separation between the second and third than between the first and third during the time of separation. So when the third observer sees the pulse travelling from the first to the second, it would appear to take longer, and quicker from the second to the first, but the lag on the second observer's clock would compensate for that, so that all observers see the light being received between the first and second in the same times according to their own clocks when the light is received.

Does any of this sound reasonable so far, as far as the simultaneity shift is concerned? I hope so. It's the only thing I've been able to come up with so far. But I still have a few concerns. For instance, the simultaneity issue for how a third observer would perceive two others that are moving together should only depend upon the relative speed to the third and the separation of the first two, so the same physics can be applied to any such scenario, regardless of how they got there. But the time lag that would occur between the first two while separating would depend upon the instantaneous relative speeds between them at each instant as they separate. So it would depend upon how they separate. Also, one or the other would experience an additional time lag depending upon who is actually accelerating in order to separate in the first place. So this is different than that which is just typical of the relative speeds between them while separating, where each would view the other with a lesser time equally, according to SR. But I guess this is where GR steps in. Of course, each of the first two would also see this additional time dilation as well as the third. We can eliminate this, however, by considering each of the first two to be accelerating away from each other in the same way and then decelerating to come back to rest. The third observer will still see an additional time dilation, but it will be the same with each of the first two.

Okay. So the first two accelerate at the same rate away from each other. This negates any additional time dilation between them due to the acceleration, right? Then they should only each observe a time dilation for the other according to their instantaneous speeds. But the instaneous speeds would also depend upon the acceleration in the first place. The third must see the same time lag according to the separation between the first and second and their relative speed to the third. Also, if the first and second experience a time lag between them, while viewing the speed of light travelling at c from one to the other, then each will see the light reaching the other in a lesser time, so they would measure a lesser distance. Now, I could probably work all of this out, as to find what the perceived acceleration would have to be relative to each other as compared to the acceleration in the local frame, in order to produce the same time lag for the same distance of separation according to the third in each case, but I still see another potential problem. As the first two separate, the third observer will have a lesser relative speed to the first than to the second. So the second, the one furthest away, will always experience a greater time lag than the first according to what the third observes. But if the first two were travelling toward the third, the light would take longer to travel from the second to the first than the other way around, which means the first should experience the greatest time dilation according to the third. So what am I still missing here? What is the correct way to perform this so that it works out for all observers in either direction?

grav
2007-Sep-22, 11:52 AM
As the first two separate, the third observer will have a lesser relative speed to the first than to the second. So the second, the one furthest away, will always experience a greater time lag than the first according to what the third observes. But if the first two were travelling toward the third, the light would take longer to travel from the second to the first than the other way around, which means the first should experience the greatest time dilation according to the third. So what am I still missing here? What is the correct way to perform this so that it works out for all observers in either direction?
Oh, wait. As the first two are moving toward the third, the first observer will obtain a greater relative speed toward the first and the second observer a lesser one upon separating. So the first observer would time dilate at a greater rate relative to the third than the second relative to the third, and so a greater lag would be produced for the first than the second in this case, so that their times can still be read in the same way due to the time lag, regardless of their direction of travel. Cool. Does that all sound about right, then?

Ken G
2007-Sep-22, 07:13 PM
The third observer must read the same times upon each of their clocks when the pulses are actually received, however, since at that instant the pulses would be synchronous with each of the two moving observers. That means that the third observer must see the second observer's time lag behind the first's, since the light takes longer to reach the second observer from the first than it does to the first from the second according to the third observer.Right, that is a good way to see why motion has to induce a synchronicity shift. You, the third observer, conclude the two clocks, though they read the same time when the respective pulses arrive, must not be synchronized with each other, even though they are both dilated by the same speed. Velocity, in the presence of separation, causes a synchronicity shift-- the forgotten term in the Lorentz transformation.


So how would would this simultaneity issue come about? Let's see. If we begin with two observers which are stationary to each other, then we can synchronize their clocks. Now let's move them apart some distance from each other. As we do so, their will be a relative speed between them. So each sees the other's clock moving more slowly, with more than just the time of flight lag due to separation, but in a relativistic way. Actually, you can move them apart slowly enough that there's no problem with synchronicity shifts. The synchronicity shift comes not just from the separation, but from the separation and the fast velocity-- it appears during the acceleration, not the separation. If you imagine the acceleration is first and the separation comes later, then the shift does come during the separation, but it's much harder because you have to consider the additional acceleration and subsequent deceleration you need to prepare that state.

Basically, there are very different ways to think of how you have set up this situation, but they all give the same answer in the end. For example, you can imagine that you separate and synchronize 1 and 2, then accelerate 3, or you can imagine that you separate 1 and 2, then accelerate them, then re-synchronize them in their own frame. Constrast those-- in the first case, the shift (from 3's point of view) comes in when you accelerate 3-- it thinks it is held fixed in a "gravitational field" that both accelerates and differentially shifts the separated clocks 1 and 2 the way gravity always shifts things at different "heights" in the well. The upshot is that the trailing clock ends up ahead of the leading clock from 3's point of view. Or, in the second case, the shift comes in when they re-synchronize themselves. Observer 3 says, why are you changing your times, you were synchronized before! But 1 and 2 say "you were in free fall so you didn't notice that big gravitational field we felt as we accelerated. That field ruined our synchronization and we have to re-establish it." So in the second case, the shift is manual, and in the first case, it happens automatically. The situations are distinguishable by all parties, yet the shift at the end is the same-- 3 thinks the trailing clock is ahead.

Or you could first accelerate either 3, or 1 and 2, and then separate 1 and 2. Either way there's no relative shift during that main acceleration, but there now has to be additional acceleration coming once they are already moving at high speed. So that's the case where you will have to track the effects of the time dilation, but you'll also have to worry about the tricky acceleration/deceleration to get the separation, and if you do it gradually, you'll have to do it at large distance from 3, which will mitigate your attempt to reduce its importance. There's no way around a tough calculation-- don't do it that way!


This will build a time lag between them that each observes of the other while they are being separated. So when we bring them to rest in respect to each other again after some separation, this time lag will remain, as well as the time of flight lag, even though each of their clocks are once again ticking at the same rate.
This is the situation to avoid, it's too complicated to analyze easily. If it was 3 that was originally accelerated, youcan do the above gradually enough that 1 and 2 still think they are synchronized with each other, but you still need to see how 3 disagrees with that, and that's the hard part.

So when the third observer sees the pulse travelling from the first to the second, it would appear to take longer, and quicker from the second to the first, but the lag on the second observer's clock would compensate for that, so that all observers see the light being received between the first and second in the same times according to their own clocks when the light is received. That's the ultimate result, yes.


For instance, the simultaneity issue for how a third observer would perceive two others that are moving together should only depend upon the relative speed to the third and the separation of the first two, so the same physics can be applied to any such scenario, regardless of how they got there.Correct, but it can be easy to leave out something important.

But the time lag that would occur between the first two while separating would depend upon the instantaneous relative speeds between them at each instant as they separate. So it would depend upon how they separate. Also, one or the other would experience an additional time lag depending upon who is actually accelerating in order to separate in the first place.True-- that all has to "come out in the wash" for a self-consistent theory. That's why you use mathematics like the Lorentz transformation, it's good at coming out consistent.


Okay. So the first two accelerate at the same rate away from each other. This negates any additional time dilation between them due to the acceleration, right? In some reference frames, yes, but not in others. Each frame sees very different physics happening, but the same answer at the end.


But if the first two were travelling toward the third, the light would take longer to travel from the second to the first than the other way around, which means the first should experience the greatest time dilation according to the third. So what am I still missing here? The leading clock does experience more time dilation, as it has to be going faster to end up "leading", and that also explains why it reads an earlier time, according to 3.

Here's another way to infer that the leading clock lags the trailing clock, for someone observing a pair of rapidly approaching clocks that are synchronized in their own frame. Let's say the two clocks have a 1 foot ruler that spans between them, and they are moving at very high speed indeed. Observer 3 says that the ruler that spans them is very shrunken, perhaps is only an inch long let's say. Now the leading clock sees observer 3 arrive, and then the trailing clock sees observer 3 arrive a light-foot in time later. But observer 3 says, nope, the trailing clock got to me only a light-inch in time later, the only reason it read a light-foot later is that it was synchronized to be already ahead of the leading clock by 11 light-inches.

Ken G
2007-Sep-22, 07:14 PM
Cool. Does that all sound about right, then?

Yes, that all sounds right, I think you have a consistent picture.

grav
2007-Sep-23, 01:52 PM
Thanks, Ken. I appreciate the help. I'm going to try to work through some simple scenarios to see what sort of mathematical relationships might come out of it. I will still need the results scanned over when I'm done.

Ken G
2007-Sep-23, 03:14 PM
I think you're on the right track now. Of course, there are always surprises in relativity, I can't trust my own conclusions until I've thought them through pretty carefully or gotten feedback from others. Then I go to my colleagues and pose the questions to them, and I find the same problem is widespread!

John Mendenhall
2007-Sep-26, 06:58 PM
I think you're on the right track now. Of course, there are always surprises in relativity, I can't trust my own conclusions until I've thought them through pretty carefully or gotten feedback from others. Then I go to my colleagues and pose the questions to them, and I find the same problem is widespread!

That's a good description of how to approach SR and GR problems. There is an entire thread on BAUT about a GR idea that was passed around at CERN for review. The CERN physicists all said "No, it doesn't work that way." Yet a lot of our better mainstream posters went for it lock, stock, and barrel, despite the violation of a basic postulate. I will leave the thread unnamed for the time being, since I intend to reopen it at some point after accumulating enough ammo to sink a fleet. It is wise to take anything you read about GR and SR with a grain of salt; even the best folks stumble on it. Regardless, the fundamental ideas of SR and GR are sound and correct, and are verified continually by their application in our technology.

publius
2007-Sep-26, 08:24 PM
There is an entire thread on BAUT about a GR idea that was passed around at CERN for review. The CERN physicists all said "No, it doesn't work that way." Yet a lot of our better mainstream posters went for it lock, stock, and barrel, despite the violation of a basic postulate. I will leave the thread unnamed for the time being ...

You'd better name it right now, or I'll go crazy wondering what it is. :)


-Richard

grant hutchison
2007-Sep-26, 09:21 PM
You'd better name it right now, or I'll go crazy wondering what it is. :)I believe it's probably the old rockets-and-rope chestnut, most recently rehearsed in detail here (http://www.bautforum.com/questions-answers/62037-lorentz-contraction.html). But it has had at least one previous incarnation, courtesy of a previous incarnation of hhEb09'1, here (http://www.bautforum.com/against-mainstream/13-rockets-rope.html).
(John Mendenhall believes the rope doesn't break.)

Grant Hutchison

publius
2007-Sep-26, 09:42 PM
Ah, so. That crossed my mind, but John said "GR" so I was wondering if this was something else involving some gravitational thing, since most would consider the rope and rocket to be SR.

-Richard

grant hutchison
2007-Sep-26, 09:53 PM
Ah, so. That crossed my mind, but John said "GR" so I was wondering if this was something else involving some gravitational thing, since most would consider the rope and rocket to be SR.I think you (and IIRC, Ken to a lesser extent) took a largely GR stance on it. (Something which would have gratified the heart of Sam5, of course, were he with us still.)

Grant Hutchison

publius
2007-Sep-26, 10:28 PM
I think you (and IIRC, Ken to a lesser extent) took a largely GR stance on it. (Something which would have gratified the heart of Sam5, of course, were he with us still.)

Grant Hutchison


:lol: That is worthy of some clarification (or rambling, however you look at it). The current view of the high priests is that GR = Riemann curvature, which is invariant and thus has "more meaning". If space-time is flat, one is doing SR, regardless of the coordinates.

The other view is that SR's domain is the *Minkowski metric*, where the familiar, more simple rules apply. If one is doing and explaining things from some non-inertial frame (in flat space-time), one is doing "more than SR".

Me, I think I've come to the position that we should forget about SR vs. GR, and just call it 'R', Relativity, which we'll roughly define as "the machinery for doing calculations about stuff in space-time". Then we'll define "gravity" as the curvature that happens to space-time due to mass-energy as described by the EFE.

-Rchard

publius
2007-Sep-27, 02:05 AM
You know, I think the best way to put the rope and rocket thing is the following, which uses Euclidean geometry.

It should be no more surprising that an accelerating rocket must experience gradients in proper acceleration (the force felt at any point by an observer there) across its length than the fact that different points on a spinning flywheel must experience different acceleration in order for it to remain rigid. The centripetal acceleration is w^2*r directed radially, increasing with radius.

If we were to specify that two points at different radii moving in concentric circles must feel the same force, then they could not remain stationary with respect to each.

In fact, mathematically, these two cases are very similiar. Constant proper acceleration is hyperolic, and hyperbolas in Minkowski are like circles in Euclid. Same constant "distance" (norm/interval) from a given point.

So, I don't see any basic postulates being violated at all. It is what our good friend PM (Principia Mathematica :lol: ) tells us follows from our basic postulates.

-Richard

grav
2007-Sep-28, 12:50 AM
Well, here are some of my preliminary results. I have started with the single simple postulate that two observers will observe the same thing of each other. Basically, this just means that each will see the same thing regardless of which way the light travels. Potentially, this could even mean that light can travel at different speeds in different frames of reference or with different relative speeds between observers, just as long as both observers experience the same physics between them. This postulate is the same thing Relativity says, but on a more basic level, although without a medium to move relative to, the physical constants must also remain the same in every frame of reference, because there is no relative motion to anything concrete which might change them due to that relative motion, and so the speed of light would remain the same to all observers as well.

I have already run into a snag with my first scenario using Relativity, however. I synchronized three observers which are stationary to each other in the same position, then accelerated one to some relative speed to the others for some time and then accelerated another to the same relative speed. They are accelerated in such a way that any time dilation due to this acceleration is neglible, either by accelerating them for a very short duration of time to the stationary observer or by running the first accelerated observer out for a very long duration of time, over a large distance, whereas the lag in times due to the relative speeds over this distance would far outway any due to acceleration. Since any distance whatsoever can be aquired this way, any time dilation due to acceleration can be ignored. Also, each observer carries their own fuel supply by which to accelerate, so does not affect the other observers in any way while doing so, whereby conservation of energy and momentum doesn't enter into this either.

So let's say the first observer travels out a distance of 'd' at a relative speed of 'v'. To the stationary observer (the third observer that isn't accelerated), the first will appear to be time dilated by t(1/y), where t is the time according to the stationary observer's clock and 1/y is the time dilation experienced by the first according to the SO (stationary observer). But the SO will also time dilate to the same degree according to the first, as well as with the second observer, which hasn't accelerated yet. The first and second observers, then, will now observe some time lag upon each others' clocks. Then the second observer is made to aquire the same relative as the first. At this point, the first and second will have the same time lag relative to each other, each lesser to the other, but their clocks will now tick again at the same rate, so no further time lag is aquired.

The problem here is that if each has now obtained a time lag in respect to the other, and we now only consider their relative distance and motion, and therefore can forget about the SO, then if we were to send a pulse of light from one to the other, then according to the observer that sent the pulse, if light always travels at a constant c, then the other's clock should read a lesser time when the pulse is received over some distance than when the other were to send a pulse to the observer, since each observer's clock reads a greater time than the other from their own perspective, and so the speed of light could not be measured the same in either direction since the observer would measure different times for when the pulse was emitted and received over the same distance either way according to what they observe on the other's clock when the pulse is simultaneous to the other's position, seen by the other by simply raising their hand when the pulse is received or emitted by them, or some other similar method. So different times in either direction for the same distance will give different values for the speed of light.

I tried various ways to get around this, but nothing has worked out so far. Finally, I gave in and tried time dilations which vary, not with the relative speed between observers, but with respect to a stationary frame. This has worked out for the most part. It basically just means that instead of each observer observing a lesser time lag for the other, one's time is actually less and the other greater. In this last scenario, for instance, the first observer's time would dilate at t(1/y) according to the stationary clock, where 't' is the time of travel, whereas the second observer which is initially stationary during this time would experience the full time of t. After the second observer is accelerated to the same relative speed to the SO, the time dilation will remain the same thereafter, and so their clocks will tick at the same rate, but a time lag of t(1-1/y) will persist, the first observer's clock showing this much less time than the second. To the SO, t=d/v, so we get tlag = (d/v)(1-1/y). The speed of light in the stationary frame is always c, so according to the SO, for the first observer sending a pulse to the second, we get

c*t1 = d + v*t1
t1 = d/(c-v)

While for the second to the first, we get

c*t2 = d - v*t2
t2 = d/(c+v)

The time the SO sees pass on their clocks between emission and reception, then, is just t1*(1/y) - tl and t2*(1/y) + tl. These two times must be the same for the moving observers in order to observe the same physics either way, namely that each will observe the same speed of light over the same distance, so that they must also observe the same times. So we have

t1*(1/y) - tl = t2*(1/y) + tl
[d/(c-v)](1/y) - tl = [d/(c+v) + tl
2*tl = [d/(c-v)]*(1/y) - [d/(c+v)]*(1/y)
2*tl = [2*d*v/(c^2-v^2)]*(1/y)
tl = (1/y)*d*v/(c^2-v^2)

and since we found before that tl = (d/v)(1-1/y), then

(d/v)(1-1/y) = (1/y)*d*v/(c^2-v^2)
1-1/y = (1/y)*v^2/(c^2-v^2)
1 = (1/y)*[1 + v^2/(c^2-v^2)]
1 = (1/y)*[c^2/(c^2-v^2)]
1/y = (c^2-v^2)/c^2
1/y = 1 - (v/c)^2

Now, this is the square of what relativity says, but I haven't been able to get even as far as this without considering a stationary frame. Each time I work through such a scenario, however, I attempt to consider what it would mean in terms of the relative speed between observers only, in order to keep Relativity on even grounds, but it has led to nothing but dead ends so far when I try to apply it that way.

For this last scenario, when we place the time dilation we found into the formula for the time lag we find

tl = (d/v)(1-1/y)
tl = (d/v)[1-(1-(v/c)^2)]
tl = (d/v)(v/c)^2
tl = d*v/c^2

Then using that for the times observed, which are t1*(1/y)-tl and t2*(1/y)+tl in each direction, which must be equal, we get

T = t1*(1/y) - tl
T = [d/(c-v)][1 - (v/c)^2] - dv/c^2
T = [d/(c-v)][(c^2-v^2)/c^2] - dv/c^2
T = d(c+v)/c^2 - dv/c^2
T = d/c

This means that each observer measures a time of T for the time of the pulse to reach the other or sent from the other according to the times read upon their clocks at each end of the trip, over the same distance 'd', giving a speed of the pulse of d/T = c. So not only is the speed of the pulse the same both ways, it is always measured at the same speed to all observers, at least those which are stationary to each other. Other scenarios have led to the same thing, so apparently observers that measure the speed of light as c in all frames isn't that difficult to come by. Furthermore, even though each observer experiences a time lag in respect to the other, they wouldn't actually observe this. Since they each measure the same speed of light over the same distance in either direction, they would assume that their clocks are also running at the same rate when stationary to each other. But other scenarios have led to different results for the time lag, however, some similar, some not, so it is still lacking some fundamental underlying principle that I am now searching for.

For instance, if all three observers were initially stationary to each other and synchronized, and then each of the first two observers are accelerated to different speeds simultaneously and the first sends pulses to the second at regular intervals according to their own clock, we would get

c*t = d - v2*t
t = d/(c+v2)

c*t' = d + v1*(ti/(1/y1)) - v2*(ti/(1/y1)) - v2*t',
t' = [d + (v1-v2)(ti/(1/y1))]/(c+v2)

where for the succeeding pulse, sent at an interval of 'ti' according to the first observer, so ti/(1/y1) to the SO, while the second closes in the original distance between them over the interval of time and while the light is in transit while the first also moves away from the second during the interval of time, increasing the distance accordingly, the SO sees the time interval of the reception of the pulses, then, as

ti' = (t' + ti/(1/y1)) - t
ti' = (v1-v2)(ti/(1/y1))/(c+v2) + ti/(1/y1)
ti' = (v1-v2)(ti/(1/y1))/(c+v2) + (c+v2)(ti/(1/y1))/(c+v2)
ti' = (c+v1)(ti/(1/y1))/(c+v2)

The emitted frequency is just 1/ti, so the ratio of observed to emitted frequency is then

fo/fe = (c+v2)(1/y1)/(c+v1)

To the second observer, however, the one actually receiving the pulses, which experiences a time dilation of 1/y2, this would further become

fo/fe = [(c+v2)(1/y1)]/[(c+v1)(1/y2)]

Running through this again in the opposite direction, where the pulse is sent from the second to the first, we find

fo/fe = [(c-v1)(1/y2)]/[(c-v2)(1/y1)]

So since this must be the same in either direction, so that the physics remains the same, we get

[(c+v2)(1/y1)]/[(c+v1)(1/y2)] = [(c-v1)(1/y2)]/[(c-v2)(1/y1)]
(c^2 - v2^2)(1/y1)^2 = (c^2 - v1^2)(1/y2)^2
(1/y1)^2/(1/y2)^2 = (c^2 - v1^2)/(c^2 - v2^2)

which will work out for 1/y1 = sqrt(1-(v1/c)^2) and 1/y2 = sqrt(1-(v2/c)^2) and all observers measure the speed of light as 'c', but it gives a Doppler shift between them of sqrt[(1-v1/c)/(1+v1/c)] * sqrt[(1+v2/c)/(1-v/c)] instead of according to their relative speed.

As I said, other various scenarios give varying results for this so far also. Some give 1-v/c for the time dilation, for instance. So I will keep working at trying to find and apply some underlying principle that produces the same results for every possible scenario. If anyone has any ideas about this, please let me know. Whether that requires a stationary frame for this to happen, or is aquired in a purely relativistic manner, it matters not. When the results are the same in every case, I will know I have it.

Ken G
2007-Sep-28, 02:07 PM
The other view is that SR's domain is the *Minkowski metric*, where the familiar, more simple rules apply. If one is doing and explaining things from some non-inertial frame (in flat space-time), one is doing "more than SR".
Yes, I think the problem with the terminology is there are really 3 flavors of relativity, not 2, so there is endless confusion about which side we should include the "middle one" in!


Me, I think I've come to the position that we should forget about SR vs. GR, and just call it 'R', Relativity, which we'll roughly define as "the machinery for doing calculations about stuff in space-time". Then we'll define "gravity" as the curvature that happens to space-time due to mass-energy as described by the EFE.That is certainly in the spirit of unifying R rather than further fragmenting into R#1, R#2, and R#3. But there is the problem that we tend to only teach R#1 to undergraduates, and R#3 to graduate students, and R#2 everyone is kind of on their own to piece together! (which they do either by transforming to a global inertial frame, doing the physics there, and transforming back, which makes it seem like SR, or including the two transformations automatically in the metric, which makes it seem like GR).

Ken G
2007-Sep-28, 02:11 PM
It should be no more surprising that an accelerating rocket must experience gradients in proper acceleration (the force felt at any point by an observer there) across its length than the fact that different points on a spinning flywheel must experience different acceleration in order for it to remain rigid. I agree, this is a good analogy, though I confess I hate how rotating systems have that big headache about being the individual points being more than just orbiting, but also themselves rotating!

So, I don't see any basic postulates being violated at all. It is what our good friend PM (Principia Mathematica :lol: ) tells us follows from our basic postulates.

I agree, and am waiting for John Mendenhall to come through on his promise to sink that battleship. My prediction: the "CERN" answer will only be correct if you make an assumption that is not natural to make in the rope problem as we analyzed it.

Jerry
2007-Sep-28, 05:39 PM
Regardless, the fundamental ideas of SR and GR are sound and correct, and are verified continually by their application in our technology.
Not quite.

The equations that fall out of relativistic theory have proven to be excellent predictqrs of past and future events...and we await further confirmation in the detection of gravity waves, not to mention a more comprehensive, unifying theory.

Ken G
2007-Sep-29, 04:33 AM
I tried various ways to get around this, but nothing has worked out so far. Finally, I gave in and tried time dilations which vary, not with the relative speed between observers, but with respect to a stationary frame.

Yes, I strongly suggest that if there is no real gravity, you find a global inertial frame to do all the calculations in. Then you can just use SR, and transform back later, and you'll always know what you are doing.

publius
2007-Sep-29, 05:20 AM
Yes, I think the problem with the terminology is there are really 3 flavors of relativity, not 2, so there is endless confusion about which side we should include the "middle one" in!

..............

That is certainly in the spirit of unifying R rather than further fragmenting into R#1, R#2, and R#3. But there is the problem that we tend to only teach R#1 to undergraduates, and R#3 to graduate students, and R#2 everyone is kind of on their own to piece together! (which they do either by transforming to a global inertial frame, doing the physics there, and transforming back, which makes it seem like SR, or including the two transformations automatically in the metric, which makes it seem like GR).

That's very true. The difference between R#1 and R#2 is exactly the same thing as the difference between Cartesian coordinates and "curvy" coordinates, such as polar in the plane. It's just that time gets thrown in the act in a non-Euclidean (but still "flat" :) ) way and makes things a lot more complicated. But it is the exact same thing. For example, write F = ma in spherical coordinates, and you get a lot of extra stuff (and that extra stuff if what Christoffel symbols are, although they are more general that what you get playing with curved coordinates in Euclid). The machinery to deal with extra stuff gets tedious enough mathematically even for Euclid.

Anyway, no one but a masochist would do things in polar coordinates if they weren't actually needed (such as equations being easier to solve there). And that's pretty much why #2 isn't explicity taught as something of itself, I think. The additional complexity is just too much to bother with at that stage -- things are hard enough without it. :)


Going to #3, is then taking that flat plane and curving it, which can be visualized by imagining the plane as the proverbial rubber sheet embedded in a 3D space, and then curving it around via that 3rd dimension. When you're sitting on that curved sheet, you've got to deal with the "extra stuff" for certain then. If you don't know how, you're up the creek without a paddle. Cartesian simplicity has abandoned us.

And so to go to #3 you have to deal with it all, and that distinction between #2 and #3 gets a bit blurred, at least until you figure it out.


-Richard

Ken G
2007-Sep-30, 03:16 PM
Anyway, no one but a masochist would do things in polar coordinates if they weren't actually needed (such as equations being easier to solve there). And that's pretty much why #2 isn't explicity taught as something of itself, I think. The additional complexity is just too much to bother with at that stage -- things are hard enough without it.


Yes, I think we may do a disservice there. We always leap to a coordinate system first, rather than teach the physics first and pick coordinates last. It's a little harder, but it would really pay dividends when we get to GR. Given that Newtonian physics is now taught in elementary college classes, if we are to imagine GR one day being in that spot, then it will come after we invert the order of learning physics/coordinates. I'm not necessarily convinced that students find coordinates easier to understand than their underlying abstract generators-- a lot of people certainly never get to the point of being able to manipulate coordinates in any meaningful way.

Now, if only John Mendenhall will come through on his promise-- I'm chomping at the bit!

grav
2007-Oct-03, 01:52 AM
Well, I still seem to be having some tremendous problems with this for some reason. I'm coming up with a solution, but it is just the same one as before, that light travels with the additional speed of the source. I've applied it in every way I can think of and I keep coming up with the same thing, so any more help is appreciated.

Let's take two scenarios. Scenario A has three observers originally moving together, stationary to each other, and synchronized. Then the first observer accelerates quickly to some relative speed to the other two until it reaches some distance from the others and then the second observer accelerates in the same way to the same relative speed to the third, which remains stationary. In this scenario, the first observer will have a time lag to the other two as it travels the distance due to the time dilation with the relative speed during its travel, and that time lag will remain after the second observer has accelerated to the same speed.

Scenario B has the first and second observers accelerate in the same way as in scenario A, but in opposite directions until they gain the same distance between each other, and then the second observer quickly accelerates in the other direction until both observers are travelling in the same direction at the same speed.

Now, with Relativity, any time lag produced will affect how we measure the speed of light between them, so such a time lag should be the same for the same relative distance between them and the same relative speed. But with scenario A, we cannot have the same time lag between one observer travelling at some relative speed for some time according to the stationary observer and two travelling away from each other at twice the relative speed for half the time according to the same observer. This is not how Relativity is set up, especially since speeds do not add together between the first and second observers as they would for the third. But just to be sure, let's say that the first, in scenario A, accelerates to a relative speed of v and travels to a distance of d before the second accelerates to the same speed. The time lag in this case would be tl = t(1-1/y), where t is the time that passes in the stationary frame and 1-1/y is the difference in times that pass between observers. t=d/v according to any observers, so we get tl = (d/v)(1-1/y). In the scenario B, the first observer accelerates to v, and so does the second, but in the opposite direction, until they reach the same distance from each other. According to Relativity, this would give a relative speed of w = (u+v)/(1+u*v/c^2) between them, so since u=v, then w = 2v/[1+(v/c)^2)]. For the same distance travelled according to either, this gives a time lag of tl = (d/w)(1-1/y'). The time lag should be the same in every scenario that results in the same distance and relative speed between observers, regardless of the method of separation, in order for things to be truly relative, at least as far as the inertial speeds are concerned, so we have (d/v)(1-1/y) = (d/w)(1-1/y'), each observer seeing this distance between them as the same in either scenario. From here, we find

(1/v)(1-1/y) = (1/w)(1-1/y')
1/v - 1/w = 1/vy - 1/wy'
c/v - c/w = c/vy - c/wy'

1/y = sqrt[1-(v/c)^2)]
1/y' = sqrt[1-(w/c)^2)]

c/v - [1+(v/c)^2]*c/2v = c*sqrt[1-(v/c)^2]/v - c*[1+(v/c)^2]*sqrt[1-4v^2/(1+(v/c)^2)c^2]/2v

I have tried this for some values of v/c and it doesn't work out, as it should for all values. Also, the acceleration while changing frames wouldn't affect this since the distance can always be made greater for a greater time lag due to a greater separation while any time lag during the same acceleration remains the same.

I have found that one solution to this would be that with a reference to an absolute frame where the time dilation is greater in one direction and smaller in the other, but I don't see how that could work out for all directions of space, especially perpendicularly then, where there would be no time dilation at all in that case, so we might as well figure that there is no time dilation in any direction to begin with. Another is that objects somehow retain a memory of the amount of time lag they have undergone relative to the absolute frame, and will regain that upon acceleration, the entire amount upon returning to the stationary frame, but reading a greater time if continuing to accelerate. It would have the additional effect of regaining the same time as the stationary frame, regardless of speed, when coming back to the same point from which it originally departed from the third observer with which it was originally stationary. The problem with these, though, is that I would would no idea what could produce such effects in order for it to work out this way.

Now, going back to the MM experiment, what started all of this, we must have the same times for light to travel some distance forward as back, and according to Relativity, without an absolute frame, all observers should see light travelling at c, not relative to an absolute frame but to the observer. But if there is a time lag between two observers, then they would measure the speed of light as d/(T+tl) in one direction between them, and d/(T-tl) in the other. Even with a Lorentz contraction, this still becomes dL/(T+tl) and dL/(T-tl), which cannot be equal for the speed of light to be measured the same unless tl=0. Simulataneity, however, also says that the time lag between two observers in the same frame is zero, regardless of the distance that separates them. To the third observer, then, who always sees the speed of light as c in their own frame as well, they would measure the time between light sent from the second observer to the first as c*t1 = d + v*t1, so t1 = d/(c-v), and from the first to the second as c*t2 = d - v*t2, so t2 = d/(c+v). Without a time lag between them, if these two times are the same to the moving observers, then so should they be for the stationary one, regardless of the time dilation involved. So we get

d/(c-v) = d/(c+v)
1/(c-v) = 1/(c+v)

which obviously doesn't work out. Since there is only one variable here, v, but that is already set as the given, then we must try c as a variable also. That gives us

1/(c1-v) = 1/(c2+v)

Now, we can immediately see that c1=c+v and c2=c-v would work out, whereas light would be emitted with the additional speed of the source, but let's be sure. So far we really just have c1 = c2 + 2v. Applying this back into scenario A, for light travelling in either direction between the first and second observers, we get

tl = (d/v)(1-1/y)

c1*t1 = d + v*t1
t1 = d/(c1-v) - tl

c2*t2 = d - v*t2
t2 = d/(c2+v) + tl

d/(c1-v) - tl = d/(c2+v) + tl

c1 = c2 + 2v

d/(c1-v) - (d/v)(1-1/y) = d/(c2+v) + (d/v)(1-1/y)
1/(c1-v) - (1/v)(1-1/y) = 1/(c2+v) + (1/v)(1-1/y)
1/(c1-v) - 1/(c2+v) = (2/v)(1-1/y)
1/(c1-v) - 1/(c2+v) = (2/v)(1-1/y)

c1 = c2 + 2v

(2/v)(1-1/y) = 0
1-1/y = 0
1/y = 1

So according to this, there is no time dilation, no time lag, and light travels with the additional speed of the source. So what am I still overlooking? It doesn't seem that this should work as well as Relativity does in explaining things such as precession and gravitational lensing and such, being so very different. I suppose I will have to go ahead and try running some similations for it, though, just in case. But regardless if light works this way or not, it wouldn't seem that gravity should, since bodies don't so much emit gravity anyway, as far as I know, but pretty much simply manipulate what already exists within that space. Does anybody have any other ideas about all of this, or another fundamental principle that might be applied?

Urbane Guerrilla
2007-Oct-03, 08:28 AM
A point of verbal usage unrelated to the math (which I doubt I can follow anyway): dilation is the popular usage, but shouldn't the word be dilatation -- a slowing down?

hhEb09'1
2007-Oct-03, 03:47 PM
A point of verbal usage unrelated to the math (which I doubt I can follow anyway): dilation is the popular usage, but shouldn't the word be dilatation -- a slowing down?In American English (http://dictionary.reference.com/browse/dilatation), dilatation doesn't mean slowing down, it means ... dilation. :)

grant hutchison
2007-Oct-03, 04:53 PM
In American English (http://dictionary.reference.com/browse/dilatation), dilatation doesn't mean slowing down, it means ... dilation. :)Likewise in British English, according to the Oxford English Dictionary. The entry for dilatation makes no mention of the sense "slowing down", only of various forms of stretching and expansion.
(In fact, "Delay, procrastination, postponement" is given as an obsolete meaning of dilation, from which we get dilatory, "tending to cause delay".)
So I think both time dilation and time dilatation refer to the "stretching" of time.

Grant Hutchison

grav
2007-Oct-11, 10:38 PM
Well, I have found so far that observers cannot observe the same physics with or without an absolute frame unless light is emitted with the additional speed of the source. The only other possibility following this premise was that time was somehow wound up like a spring so that the time lag between two observers is conserved upon accelerating to a speed different from an absolute frame and then released upon decelerating back to it, to regain the same time as that of the absolute frame, and then continues to wind the other way if the observer continues to accelerate in the other direction. But if an observer regains his stationary status in the absolute frame for a time and then accelerates again, he could either wind up or back down again equally, unless he winds up in one direction and down in the other, but then space wouldn't be symmetrical.

So that didn't really make much sense, but I still had to explore the possibility. Besides that, though, I found a set of scenarios where this can't work anyway. Three observers are stationary. One observer accelerates to some speed in one direction and continues at that speed for a time, and then another observer accelerates to the same speed in the opposite direction. When they reach some distance from each other, they each emit light to the other and the times measured for the light to travel this distance will depend upon the time lag between them. Compare this to that of the first and second observers accelerating to the same speed in opposite directions simultaneously. Or to the first travelling out for a shorter or longer time before the second. When they reach the same distance from each other, the time lag will vary in each particular case, so they would measure the speed of light differently in each case as well, even between two sets of observers in the same frames as the corresponding observers in each set, simply because the time lags can be arbitrarily different in each case. The acceleration and time getting "wound up" do not matter because the acceleration is the same in every case, and therefore any time lag produced by it, and since neither observer doubles back to unwind it as well.

So what are we left with? Under this premise, light must travel with the additional speed of the observer. That's it. No time lag, no time dilation, no contraction, and no gamma. I haven't begun exploring this in detail yet, but I will soon. I will apply it to precession and gravitational lensing. The only thing here is that this is found for the way light travels in order for all observers to observe the same physics, and the speed of light always as c, in accordance with the MM experiment, but it says nothing of gravity itself, which may operate under somewhat different principles, and most all of the verification of Relativity relates to gravity. Anyway, if I manage to figure anything out along these lines, I will post it in ATM.

In the meantime, for the rest of this thread, I still want to find something that agrees better with the way Relativity is laid out. So I want to see if there is anything about the premise for these scenarios that might vary. Any assumptions about the way physics operates or any other underlying principle that has not been taken into account. I cannot change the original premise that all observers observe the same physics, because then we wouldn't have Relativity at all, nothing based solely upon the relative distance and speed, but instead we could get any arbitrary result depending upon how one got there, and the physics for this would have to be determined through experiment in each case. Surely the universe is kinder to mathematicians than that, even amateur ones. :)

Since that is the only real premise here, maybe something about the physics itself. Each set has been based upon the assumption that the observers accelerate to some speed and then continue in a straight line inertially. Since the acceleration doesn't matter and the inertial part is the only other physics that applies, then I am looking at that. First, we could say that the observers do not follow a straight line. But then there is nothing to determine what other path they would follow, in what direction, at least not without becoming assymetrical. And if the observers are to observe the same physics, then the situations should be the same and symmetrical in this regard. So that one's out.

The only other thing I can see, then, is that the speed is not inertial. The observers must either accelerate or decelerate as they travel through space in such a way as to gain the same time lag in each case. If they decelerate, then this might be caused by some density they must travel through. But this density must be the same mechanism that produces the speed of light also. If not, then the density and resulting deceleration could have any arbitrary value whatsoever, therefore even zero, which would bring us back to our original scenarios. So if the density increased, and therefore the deceleration was made greater, then the speed of light would have to change with it, in order for all things to remain relative between observers.

I have tried this for the original set of scenarios, however, and it's looking like the observers must actually accelerate to remain relative with a time lag. This could mean that the space between them is increasing while they travel, but again, it would have to directly relate to the speed of light as well, so may depend upon some expanding space or negative pressure, which might still contain a density of some sort. If the space is expanding so that the density decreases, whereas the observers would separate more than just inertially due to the expansion while decelerating due to the density, but decelerating less and less as the density decreases also, well, then that would make for some interesting calculations. :wall::eek:

grav
2007-Oct-16, 06:42 PM
Well, relating the time dilation of the observers to an absolute frame, with some sort of acceleration or expanding distance, was easy. If we take that last set of scenarios, then the second one says that the two observers accelerate and then travel inertially in the same way in opposite directions, so their motions relative to the absolute frame are symmetrical. Any additional acceleration or expansion of the distance that is experienced by one, relative to the absolute frame, will also be experienced by the other in the same way as well, then. So their motions are symmetrical no matter what, and so will be their time dilations relative to the absolute frame also. Therefore, there can be no time lag between them regardless.

Relating this to the first of the scenarios, where one accelerates and travels out some distance before the other accelerates in the opposite direction, then, for the physics to be the same for the same relative distance and relative speed as with the second scenario, there can be no time lag here either. The only way to do that with the first observer travelling out with some time dilation relative to the absolute frame before the second does, so that the first does not experience any time lag relative to the second, is if the times of all observers remains the same, so that there is no time dilation to begin with. So we are back to the c+/-v relative to the source thing with that as well, in regards to an absolute frame.

I still have to try to relate it between the observers themselves, however, instead of an absolute frame, as with Relativity, where the time dilation takes place between the observers. It does not work with inertial speeds, but although I haven't found a way that it can work with an acceleration or expanding distance yet, I have not shown that it cannot yet either. I'll keep working on it.