# Twin Paradox: Definitive Proof That It's SR?

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• 2003-Dec-06, 12:52 PM
Sam5
Quote:

Originally Posted by Diamond
No. No. And No.

There is no such thing as "what happens in both frames at the same time" - that is covered in the chapter about the "relativity of simultaneity". Light waves do not travel as c+v or c-v but at c. There is no preferred frame of reference - each observer correctly describes the phenomena consistently with the same laws of physics.

I don’t mean they will both “see” the same events “at the same instantaneous moment”. They see the flashes at different moments, but the whole process of the flashes and the movement of the light and the observers’ individual “seeing” of the flashes at different instantaneous moments takes place in both frames “at the same time”, ie during the course of the thought experiment. When one sees a flash, something is going on in both frames at the same time, whether they both “see” the flashes at the same moment or not.

Einstein says in Chapter 9 that the moving train observer encounters the two flashes of light at c + v and c – v. That’s why that observer sees the B flash first. If the moving observer saw both flashes arrive at him at the same time, then he’d be seeing the fronts of both beams travel toward him at c. But if that happened, the embankment observer would see the A flash first at the speed of c + v.

The “relativity of simultaneity” was noticed by way back in history. Soldiers on the battlefield in Newton’s time knew that they heard cannon sound later than the actual firing of the cannon and delayed by more time the further they were away from the cannon. Renaissance scientists knew that simultaneity of “seeing” light and “hearing” sound was just relative and depended on the position and state of motion of the observer. Doppler wrote papers about this in the 1840s. In 1676 Romer knew that when he saw the moons of Jupiter, he was seeing them on a delayed basis.
• 2003-Dec-06, 01:29 PM
kilopi
Quote:

Originally Posted by Eroica
Because both twins experience exactly the same accelerations, their age differences at the end of the experiment can have nothing to do with acceleration.

I'm not sure that follows from general relativity though, does it?

Anyone remember Bell's rope?
• 2003-Dec-06, 02:02 PM
SeanF
Quote:

Originally Posted by Eroica
Quote:

Originally Posted by Eta C
The problem is that from B's point of view there has to be a time where he sees A's clock as running fast compared to his. That can't happen during the coasting periods. We've all agreed that during those periods he sees A's clock as running slow, as predicted by SR. The time when he sees A's clock running fast has to come during the non-inertial accelerations. To determine that, from B's non-inertial reference frame, requires GR.

This seems to be the nub of the problem. I for one do not agree that B sees A's clock as running slow during all the coasting periods, and fast during the accelerations. On his way out, B sees A's clock running slow and A sees B's clock running slow. Then B turns for home and immediately B sees A's clock running fast. A's clock continues to run fast from B's perspective all the way home. But A does not see B's clock running fast until some time after B's has turned for home. (Let's say B turns for home when he is 10 uncontracted light-years away: A won't see B's clock running fast until 10 years after B has turned around.)

That's not right, Eroica. If we assume normal deceleration and re-acceleration, B will see A's clock running fast during the deceleration and re-acceleration. If we assume "instantaneous" changing of direction without changing speed, B will see A's clock "jump ahead" at the switching point.

At no point, in either case, will A see B's clock running fast.

Note: The word "see" in this post does not refer to the actual detection of light waves by eyes or other detector. It refers to the perception of how and when events occurred.
• 2003-Dec-06, 02:09 PM
SeanF
Quote:

Originally Posted by Sam5
Quote:

Originally Posted by Diamond
No. No. And No.

There is no such thing as "what happens in both frames at the same time" - that is covered in the chapter about the "relativity of simultaneity". Light waves do not travel as c+v or c-v but at c. There is no preferred frame of reference - each observer correctly describes the phenomena consistently with the same laws of physics.

I don’t mean they will both “see” the same events “at the same instantaneous moment”. They see the flashes at different moments, but the whole process of the flashes and the movement of the light and the observers’ individual “seeing” of the flashes at different instantaneous moments takes place in both frames “at the same time”, ie during the course of the thought experiment. When one sees a flash, something is going on in both frames at the same time, whether they both “see” the flashes at the same moment or not.

Einstein says in Chapter 9 that the moving train observer encounters the two flashes of light at c + v and c – v. That’s why that observer sees the B flash first. If the moving observer saw both flashes arrive at him at the same time, then he’d be seeing the fronts of both beams travel toward him at c. But if that happened, the embankment observer would see the A flash first at the speed of c + v.

The “relativity of simultaneity” was noticed by way back in history. Soldiers on the battlefield in Newton’s time knew that they heard cannon sound later than the actual firing of the cannon and delayed by more time the further they were away from the cannon. Renaissance scientists knew that simultaneity of “seeing” light and “hearing” sound was just relative and depended on the position and state of motion of the observer. Doppler wrote papers about this in the 1840s. In 1676 Romer knew that when he saw the moons of Jupiter, he was seeing them on a delayed basis.

But it is accepted that in the atmosphere, the light and sound generated at the same distance simultaneously will not reach the observer at the same time.

However, under SR, it is required that two light beams generated at the same distance simultaneously will reach the observer at the same time.

If they don't, either they weren't at the same distance or they weren't simultaneous.
• 2003-Dec-06, 02:23 PM
Sam5
Quote:

Originally Posted by SeanF
But it is accepted that in the atmosphere, the light and sound generated at the same distance simultaneously will not reach the observer at the same time.

Doh.

The Chinese discovered that thousands of years ago, by observing exploding skyrockets and firecrackers at a distance.

Quote:

Originally Posted by SeanF
However, under SR, it is required that two light beams generated at the same distance simultaneously will reach the observer at the same time.

If they don't, either they weren't at the same distance or they weren't simultaneous.

That is true for observers that are stationary relative to the two emitters. But the moving train observer is not stationary relative to the two emitters. In the Chapter 9 example, the flashes were at the same distance and occurred simultaneously when the train observer was at Point M’ along the track. But since it takes time for light to travel, and since the train observer was moving during that time, he moved closer to the oncoming beam from B, whilst he was “riding on ahead” of the beam coming from A. So he saw flash B first. There is nothing mysterious about this.
• 2003-Dec-06, 04:18 PM
SeanF
Quote:

Originally Posted by Sam5
That is true for observers that are stationary relative to the two emitters. But the moving train observer is not stationary relative to the two emitters.

Doesn't matter. The speed of light is a constant regardless of the velocity of the emitters, and in SR the emitters are in motion relative to the stationary train observer.

Speaking of the constancy of c, Einstein mentions de Sitter in Chapter 7, The Apparent Incompatibility of the Law of Propagation of Light with the Principle of Relativity. I'd really like to hear how your "local c-regulator" theory gets around those binary stars.
• 2003-Dec-06, 04:45 PM
Eroica
Quote:

Originally Posted by SeanF
If we assume normal deceleration and re-acceleration, B will see A's clock running fast during the deceleration and re-acceleration. If we assume "instantaneous" changing of direction without changing speed, B will see A's clock "jump ahead" at the switching point.

At no point, in either case, will A see B's clock running fast.

I think we're arguing about two completely different meanings of the phrase running fast. I mean that if B looked back at A with his telescope, he would think that A's clock was ticking more slowly than his. But when he turned for home, A's clock would suddenly seem to be ticking faster than his.

Imagine the clocks are transmitting pulses every second. If A's clock is, say, 30 seconds behind B's, how can it jump instantaneously to being 30 seconds ahead? What happened to the 59 other pulses in between?
• 2003-Dec-06, 04:46 PM
Eroica
Quote:

Originally Posted by kilopi
I'm not sure that follows from general relativity though, does it?

I'm not sure about anything anymore! ](*,)
• 2003-Dec-06, 05:01 PM
SeanF
Quote:

Originally Posted by Eroica
Quote:

Originally Posted by SeanF
If we assume normal deceleration and re-acceleration, B will see A's clock running fast during the deceleration and re-acceleration. If we assume "instantaneous" changing of direction without changing speed, B will see A's clock "jump ahead" at the switching point.

At no point, in either case, will A see B's clock running fast.

I think we're arguing about two completely different meanings of the phrase running fast. I mean that if B looked back at A with his telescope, he would think that A's clock was ticking more slowly than his. But when he turned for home, A's clock would suddenly seem to be ticking faster than his.

Imagine the clocks are transmitting pulses every second. If A's clock is, say, 30 seconds behind B's, how can it jump instantaneously to being 30 seconds ahead? What happened to the 59 other pulses in between?

In the event of an instantaneous direction change, B would still receive all those pulses, and in the proper order. When B tries to "undo" the Doppler effect, though, figuring out where and when A was when it transmitted those pulses, things would look weird. It would almost seem (I think) kind of like A was in two different places transmitting two different pulses at the same time.
• 2003-Dec-06, 06:10 PM
Sam5
Quote:

Originally Posted by SeanF
Doesn't matter. The speed of light is a constant regardless of the velocity of the emitters, and in SR the emitters are in motion relative to the stationary train observer.

The train is not in SR, it’s on the earth. SR is a science fiction story. It should have won the Nobel Prize for Literature.

If the “the emitters are in motion relative to the stationary train observer”, then you’ve got to make a choice. If you want the B flash to travel at c relative to the train, then its speed of light has to slow down relative to the source. In Einstein’s Chapter 9 it’s traveling at c + v relative to the train observer and c relative to the source. So you want to slow it down to c at the train observer? Why would you want to do that? Einstein said, “Now in reality.....” the train observer and the light are moving together at c + v. Why would you want to change that?

Ok, then you’ve got to make it depart the moving source at less than c. It can’t both leave the source at c and encounter the train observer at c while the source is moving. If it leaves the source at c, then you’ve got to add the velocity of the source, v, to c, and that would cause the light to encounter the train observer at c + v. But if you want the light to encounter the train observer at c, then it would have to leave the moving source at c – v, relative to the source, with v being the velocity of the source. And then you would have the velocity of the light at the surface of the earth traveling less than c.

If you want to consider the distance between the two as being great, so that the Vtot light speed transforms between two different comoving spaces, then it could leave the source at c, it could gradually change its Vtot while in route by slowing down relative to the train observer, then it could reach the train observer at c. Much like what we observe in the light from the distant galaxies. But by the time it reaches the train observer, it will be traveling at c – v relative to the source.

Man, this SR stuff has really messed up your way of thinking about how waves, light, and clocks work.

Quote:

Originally Posted by SeanF
Speaking of the constancy of c, Einstein mentions de Sitter in Chapter 7, The Apparent Incompatibility of the Law of Propagation of Light with the Principle of Relativity. I'd really like to hear how your "local c-regulator" theory gets around those binary stars.

“By means of similar considerations based on observations of double stars, the Dutch astronomer De Sitter was also able to show that the velocity of propagation of light cannot depend on the velocity of motion of the body emitting the light.”

The velocity at the earth, when you view the light, when the photons finally reach us, is c relative to you and the earth. The motion of the “double stars” doesn’t change that. The Vtot change takes place in space, probably closer to the stars than to the earth, so that’s why we see red and blue shifts of revolving binaries, even though we encounter the light at c.
• 2003-Dec-06, 07:06 PM
russ_watters
I don't understand why you guys keep arguing with Sam5 about the implications of SR. His last post there clearly illustrates that he doesn't understand what SR says, nor does he understand what observations show. Until he understands that, you can't have a conversation about its implications - you can't understand the twins paradox without first understanding SR.

Sam5, I'm trying to decide if I should even attempt to explain it to you. You seem reluctant to learn. But I'll give it one little try: I'll start with the basic error in your post about C and see if we can go from there.

The speed of light is always measured by everyone everywhere - on the train, on earth, on a satellite in space - to be C. Exactly, precisely C.

Yes, that's a postulate of SR, but set relativity aside for now: this is DATA and OBSERVATIONS. It is real and its the way light works and is indisputable even without SR to explain it.
• 2003-Dec-06, 07:36 PM
SeanF
Quote:

Originally Posted by Sam5
Einstein said, “Now in reality.....” the train observer and the light are moving together at c + v. Why would you want to change that?

I've explained that "Now in reality" phrase to you twice already. Ignoring explanations and continuing to bring up the same wrong concepts after they've been explained multiple times is the kind of thing that's gotten people banned from this BB before.
Quote:

Originally Posted by Sam5
Quote:

Originally Posted by SeanF
Speaking of the constancy of c, Einstein mentions de Sitter in Chapter 7, The Apparent Incompatibility of the Law of Propagation of Light with the Principle of Relativity. I'd really like to hear how your "local c-regulator" theory gets around those binary stars.

“By means of similar considerations based on observations of double stars, the Dutch astronomer De Sitter was also able to show that the velocity of propagation of light cannot depend on the velocity of motion of the body emitting the light.”

The velocity at the earth, when you view the light, when the photons finally reach us, is c relative to you and the earth. The motion of the “double stars” doesn’t change that. The Vtot change takes place in space, probably closer to the stars than to the earth, so that’s why we see red and blue shifts of revolving binaries, even though we encounter the light at c.

Read about de Sitter and the binary stars before you attempt to dismiss it, JW. It's got nothing to do with Doppler, and it's got nothing to do with the speed of the light when it reaches us.

Start with this page (it is, after all, the first response from Google on a search for "de Sitter" and "binary stars") and get back to me on the de Sitter effect once you understand what it is.
• 2003-Dec-07, 07:33 PM
kilopi
Quote:

Originally Posted by russ_watters
I don't understand why you guys keep arguing with Sam5 about the implications of SR.

Just making sure that we understand it. Or trying to...
• 2003-Dec-08, 02:04 AM
Sam5
Quote:

Originally Posted by russ_watters
The speed of light is always measured by everyone everywhere - on the train, on earth, on a satellite in space - to be C. Exactly, precisely C.

So, uhh, this stuff about scientists measuring the speed of light passing the sun as slowing down is wrong? And so light can easily go out of "black holes" at the speed of c? Light is measured to go through glass at c and water at c?
• 2003-Dec-08, 02:43 AM
Tensor
Quote:

Originally Posted by kilopi
Just making sure that we understand it. Or trying to...

After going back and reading this thread and the relativity thread, I'm not sure I understand any of it now.... :wink:
• 2003-Dec-08, 03:13 AM
Tensor
Quote:

Originally Posted by Sam5
Quote:

Originally Posted by russ_watters
The speed of light is always measured by everyone everywhere - on the train, on earth, on a satellite in space - to be C. Exactly, precisely C.

So, uhh, this stuff about scientists measuring the speed of light passing the sun as slowing down is wrong? And so light can easily go out of "black holes" at the speed of c? Light is measured to go through glass at c and water at c?

Although you would have us believe you have a very good understanding of SR (you must if you claim it's wrong), you have either misunderstood some basic physics and GR analogies or you are being purposely obtuse with these questions. Light doesn't slow down passing the sun, its path is bent by the sun's gravity. Yes, light is moving at c trying to get out of a black hole. Depending on how you want to look at it, light's frequency is infintely redshifted or it's path is curved so much it stays within the horizon. Either way, it's not visible. And it's the time it takes photons to be absorbed and reemitted from the atoms in the water or glass that slows it down, but it's still moving at c between those interactions.
• 2003-Dec-08, 03:19 AM
Sam5
Quote:

Originally Posted by Tensor
Yes, light is moving at c trying to get out of a black hole. Depending on how you want to look at it, light's frequency is infintely redshifted or it's path is curved so much it stays within the horizon.

And so, uhh, you believe in the “tired light” theory? The light just gets tired of traveling at “c” but getting no where while trying to get out of the black hole, and it just “gives up”?
• 2003-Dec-08, 03:40 AM
Tensor
Quote:

Originally Posted by Sam5
Quote:

Originally Posted by Tensor
Yes, light is moving at c trying to get out of a black hole. Depending on how you want to look at it, light's frequency is infintely redshifted or it's path is curved so much it stays within the horizon.

And so, uhh, you believe in the “tired light” theory? The light just gets tired of traveling at “c” but getting no where while trying to get out of the black hole, and it just “gives up”?

My, my you are being obtuse. Ok, light's frequency is infintely redshifted by gravity. Now, please explain how a gravitational redshift is the same as tired light, if that's what you believe.
• 2003-Dec-08, 03:40 AM
russ_watters
Quote:

Originally Posted by kilopi
Quote:

Originally Posted by russ_watters
I don't understand why you guys keep arguing with Sam5 about the implications of SR.

Just making sure that we understand it. Or trying to...

Quote:

After going back and reading this thread and the relativity thread, I'm not sure I understand any of it now..
Ever watch a movie and think: 'jeez, I feel dumber for seeing that?' Thats what I think of this thread. I don't think you can increase your own understanding by arguing with him.
• 2003-Dec-08, 04:48 AM
Sam5
Quote:

Originally Posted by Tensor
My, my you are being obtuse. Ok, light's frequency is infintely redshifted by gravity. Now, please explain how a gravitational redshift is the same as tired light, if that's what you believe.

The "tired light" theory is silly but it was in a lot of popular physics books in the '80s and early '90s.

I’ve read two different versions of the story about the black hole and gravity. One story says a “gravitational redshift” is actually caused by a slowdown in the internal harmonic oscillation rates of atoms, which causes them to emit a lower frequency of light. So, and “infinite” redshift would seem to be caused by a non-harmonically oscillating atom that does not emit light.

The other version is that I’ve read that a light beam passing near the sun or a black hole “slows down” its speed. Something about the beam acting like a “plane wave” and being “refracted” by the “gravity well”, with the bending being caused by the closest areas of the beam to the sun slowing down the most.

In fact, both of those stories are in "On the Influence of Gravitation on the Propagaton of Light", A. Einstein, 1911, so it was Einstein who first said that light slows down when it passes near the sun. Do you think he might have been wrong about that? I think he was right.
• 2003-Dec-08, 04:55 AM
Sam5
Quote:

Originally Posted by russ_watters
Ever watch a movie and think: 'jeez, I feel dumber for seeing that?' Thats what I think of this thread. I don't think you can increase your own understanding by arguing with him.

Russ, that is a personal attack, which I think is a poor substitute for your lack of a resolution of the twins paradox.
• 2003-Dec-08, 06:03 AM
freddo
Quote:

it was Einstein who first said that light slows down when it passes near the sun.
Not my understanding of it... AFAIK it's not that the light is slowing down, but more that it has further to travel before it is observed? Something moving at a set speed and a set distance - the time it arrives can be calculated. Change the distance, your result will also change.
• 2003-Dec-08, 06:11 AM
Sam5
Quote:

Originally Posted by freddo
Not my understanding of it... AFAIK it's not that the light is slowing down, but more that it has further to travel before it is observed? Something moving at a set speed and a set distance - the time it arrives can be calculated. Change the distance, your result will also change.

I've got his 1911 paper right here. Get out your copy of it and we'll discuss it. He uses c1 and c2 for two different speeds of light.
• 2003-Dec-08, 09:10 AM
Diamond
Quote:

Originally Posted by Sam5
Quote:

Originally Posted by freddo
Not my understanding of it... AFAIK it's not that the light is slowing down, but more that it has further to travel before it is observed? Something moving at a set speed and a set distance - the time it arrives can be calculated. Change the distance, your result will also change.

I've got his 1911 paper right here. Get out your copy of it and we'll discuss it. He uses c1 and c2 for two different speeds of light.

Personally I think you're just being deliberately obtuse and argumentative for the sake of it. No-one else would spend so much time introducing debunked theories and fallacious arguments into relativity threads.

Quote:

The “relativity of simultaneity” was noticed by way back in history. Soldiers on the battlefield in Newton’s time knew that they heard cannon sound later than the actual firing of the cannon and delayed by more time the further they were away from the cannon. Renaissance scientists knew that simultaneity of “seeing” light and “hearing” sound was just relative and depended on the position and state of motion of the observer. Doppler wrote papers about this in the 1840s. In 1676 Romer knew that when he saw the moons of Jupiter, he was seeing them on a delayed basis.
...is tremendous nonsense from beginning to end. "Relativity of simultaneity" has nothing to do with two paths of information which travel at different speeds but the consequences of light being measured to be the same regardless of the inertial frame of reference used.

Personally I think you simply lurve the attention. It certainly has nothing to do with wanting to know how or why relativity is such a good scientific theory.
• 2003-Dec-08, 09:26 AM
Diamond
The Twin Paradox of Einstein is an interesting thought experiment involving two twins (who are nearly exactly the same age), one of whom sets out on a journey into space and back. Because of the time dilation effect of relativity, the twin who left experiences a slowing down of time and will actually be much younger than the twin that stayed behind. The reason that this is considered a paradox is that Special Relativity seems to imply that either one can be considered at rest, with the other moving. It does, and it doesn't.

The confusion arises not because there are two equally valid inertial rest frames, but (here's the tricky part) because there are three. A lot of explanations of the twin paradox have claimed that it is necessary to include a treatment of accelerations, or involve General Relativity. Not so.

The three inertial frames are 1) at-home twin 2) the going-away twin and 3) the coming-back twin. It doesn't make any difference that the last two are physically the same twin--they still define different inertial frames.

OK, let's see: Ann stays at home and Bob rockets away at 3/5 light speed. Time dilation is 80%. Bob lets 4 years pass. Bob returns at 3/5 light speed, again taking 4 years. Ann thinks 10 years have passed, and Ann and Bob agree that Bob is two years younger.

Important question: what is the relative speed of the two Bob frames? On first glance, it would appear that one is going 3/5c in one direction and 3/5c in the other direction, so that the difference between the two frames is 6/5c! Faster than light? No, special relativity does not add speeds this way. The actual difference is only 15/17c, fast but not faster than light. Why is this important? We'll see.

Now, since special relativity lets us use either rest frame, we assume Bob is the at-home twin. Ann speeds away at 3/5c. No problem so far. But after 4 years of waiting, Bob must change his inertial frame. If we allow Ann to return, we've only restated the problem with the names switched. In the first version, Ann stayed in an inertial frame, and she must stay in an inertial frame in this version. Bob zooms off after Ann at 15/17 light speed (now we know why it was important), and of course catches up. It takes him 4 years, and he has seen 8 years since Ann left. Ann has aged 10 years. Same result. No paradox.

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

• 2003-Dec-08, 04:08 PM
kilopi
Re: For Eroica: The Twin Paradox
Quote:

Originally Posted by Diamond

Ryback posted some of that earlier in this thread, see my comment right after.
• 2003-Dec-08, 07:01 PM
Sam5
Re: For Eroica: The Twin Paradox
Quote:

Originally Posted by Diamond
The Twin Paradox of Einstein is an interesting thought experiment involving two twins (who are nearly exactly the same age),

Actually, in the SR theory, the two clocks are both set to “00” at the start of the experiment before the relative motion begins, and they are both synchronized and synchronous, so they are exactly the "same age".

Quote:

Originally Posted by Diamond
The reason that this is considered a paradox is that Special Relativity seems to imply that either one can be considered at rest, with the other moving. It does, and it doesn't.

No, it DOES. See these statements from the actual theory:

“For this purpose we introduce a third system of co-ordinates K’, which relatively to the system k is in a state of parallel translatory motion parallel to the axis of X, such that the origin of co-ordinates of system k moves with velocity -v on the axis of X.”

And:

“Since the relations between x', y', z' and x, y, z do not contain the time t, the systems K and K’ are at rest with respect to one another, and it is clear that the transformation from K to K’ must be the identical transformation.”

So, since K’ is the same frame as K, while K sees k as “moving” to the right at +v, k sees K as “moving” to the left at the same relative speed but at -v.

One reason why the “twin paradox” is so confusing, is because most people learn about it from books or websites, but not from Einstein’s original paper. And some of the books and websites change it around, so what you read in them is quite often NOT what Einstein actually said.

I realize it is easier to just read the websites and not the original theory, but the websites can actually confuse readers more by introducing concepts that are not in the original paper.
• 2003-Dec-08, 07:31 PM
Sam5
Re: For Eroica: The Twin Paradox
Quote:

Originally Posted by Diamond
OK, let's see: Ann stays at home and Bob rockets away at 3/5 light speed. Time dilation is 80%. Bob lets 4 years pass. Bob returns at 3/5 light speed, again taking 4 years. Ann thinks 10 years have passed, and Ann and Bob agree that Bob is two years younger.

Your error here is that Ann sees 5 years pass on her clock, while she sees 4 years pass on Bob’s clock, and then you are turning Bob around when he sees 4 years pass on his clock, as seen by Bob.

But, in the SR theory, Ann sees 5 years pass on her clock, while she sees 4 years pass on Bob’s clock, while Bob sees 5 years pass on his clock while he sees 4 years pass on Ann’s clock.

So you are turning Bob around too soon. You are turning Bob around based on what Ann sees on Bob’s clock, but not based on what Bob should be seeing on his own clock at turn-around time, which should be 5 years on his own clock, not 4 years.

If you turn Bob around when he sees 4 years pass on his clock, then he will have seen only 3.2 years pass on Ann’s clock, so you wind up with the paradox. You wind up with a double Lorentz Transformation, instead of a single one. You have Ann seeing 5 years pass on her clock, but you have Bob seeing only 3.2 years pass on her clock, so you are multiplying the Lorentz Transformation that Bob sees on Ann’s clock by a factor of 2.

5 x .8 = 4, and 4 x .8 = 3.2

So you are having Bob see a double Lorentz Transformation on Ann’s clock, while you are having Ann see only a single Lorentz Transformation on Bob’s clock, and Einstein’s theory never said anything about that.

If you keep on with this, you will have Ann see 10 years on her clock when Bob returns, while she sees 8 years on Bob’s clock, but you will have Bob see only 6.4 years on Ann’s clock, and you will still have a paradox but with different “seen” numbers.

You turned Bob around too soon. He is supposed to turn around when he sees 5 years on his own clock.

If you have Bob turn around at 4 years by his own clock, he will have seen Ann age only 3.2 years, and you will have Ann aged by her clock only 4 years rather than 5.

3.2 / .8 = 4

So you’ve got Ann aging at two different rates. 5 years that she “sees” on her own clock and 3.2 years that Bob sees on her clock when he sees 4 years on his clock. Based on the Lorentz transformation equation, if Bob sees Ann age 3.2 years when he sees his own clock age 4 years, then that means Ann would see her own clock age 4 years.

So this is where your error is. She can’t see both 5 years AND 4 years on her own clock at the same time.

You’ve got Ann aging by her own clock both 5 years and 4 years at the point where Bob turns around.

You can’t use the Lorentz Transformation factor of .8 twice as “seen” by one observer going in one direction.
• 2003-Dec-08, 07:39 PM
SeanF
Bob sees his own clock tick off 4 years on the way out ( 0 to 4 ) and 4 years on the way back ( 4 to 8 ). Bob sees Ann's clock tick off 3.2 years on the way out ( 0 to 3.2 ) and 3.2 years on the trip back ( 6.8 to 10 ).

• 2003-Dec-08, 08:40 PM
Sam5
Quote:

Originally Posted by SeanF
Bob sees his own clock tick off 4 years on the way out ( 0 to 4 ) and 4 years on the way back ( 4 to 8 ). Bob sees Ann's clock tick off 3.2 years on the way out ( 0 to 3.2 ) and 3.2 years on the trip back ( 6.8 to 10 ).

Diamond said:

“Ann thinks 10 years have passed, and Ann and Bob agree that Bob is two years younger.”

And:

“It takes him 4 years, and he has seen 8 years since Ann left. Ann has aged 10 years. Same result. No paradox”

So he says the ratio is 8 to 10, but you say it is 6.8 to 10.

So the paradox not only remains, but you’ve done your math wrong. You got 6.8 when you should have gotten only 6.4.

3.2 + 3.2 = 6.4, not 6.8

I’ve had some difficulty getting this last post to show up on the board, so it might be that I will not be able to respond to any of your posts in the future, so if that turns out to be the case, I would like to say now that I’ve enjoyed our discussion.
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