Understanding the GR Twin Paradox: Explained in a Concise Video

[idea2000]

Hi, I just saw a video about the twin paradox, explained by using GR. I was wondering whether I understood the video correctly.

The video states that when the rocket twin is first accelerating away from earth, his clock and his Earth twin's clock are different, but, roughly the same because of how close together they are. But, on the way back, because there is so much distance between the two twins, the rocket twin would see the Earth twin's clock flying forward in time much faster than his. So, my question is, in GR, does how fast the stationary observer's clock appear to be moving depend not only on how fast the rocket observer is accelerating but also on how close he is to the stationary observer? So, the farther away the stationary observer is, the faster his clock appears to move?

 

[CrazyNinja]

I do not remember there being such a case. The clocks only depend on how fast the corresponding containers are moving.

 

[idea2000]

Hi,

Here is a link to the video that I watched, just for reference.

 

[Samy_A]

A very clear and detailed explanation of what each of the twins sees in the twin paradox was posted by @jtbell : https://www.physicsforums.com/threads/time-difference-cameras.69214/#post-510214

 

[idea2000]

Many thanks for your reply. =) I have seen the doppler effect explanation for the twin paradox before, but then I ran into this video that used GR to explain what was going on and now I feel confused. My overall question doesn't really have to do wth the twin paradox, but, more so about GR. In GR, does how fast the stationary observer's clock appear to be moving depend not only on how fast the rocket observer is accelerating but also on how close he is to the stationary observer? In other words, does distance matter?

 

[jartsa]

in GR, does how fast the stationary observer's clock appear to be moving depend not only on how fast the rocket observer is accelerating but also on how close he is to the stationary observer? So, the farther away the stationary observer is, the faster his clock appears to move?

That's how it is in GR. I mean if the word "appear" is removed, then the above is correct.

When Bob's acceleration is 1 g, then according to Bob objects ahead of him accelerate at 1g * distance to object * some constant.

But It does not appear so, because Bob is seeing old light that was emitted when he was not accelerating. (appear=what eyes see)

 

[idea2000]

So, just to clarify, if i am in a rocket frame, and there are no planets or large masses around me, and I accelerate towards two observers in the same "stationary" frame, one of them being 10 light years away, and the other one being 20 light years away, would I see their two clocks moving at different rates, simply because they are different distances from me? And, would I see the clock of the "stationary" observer that is 20 light years away going much faster than the clock of the "stationary" observer who is 10 light years away? (From the "doppler shift" explanation of the twin paradox, I was under the impression that only acceleration mattered, and not distance. But, from the GR explanation I just watched, distance does seem to matter. That's why I'm confused.)

 

[jartsa]

So, just to clarify, if i am in a rocket frame, and there are no planets or large masses around me, and I accelerate towards two observers in the same "stationary" frame, one of them being 10 light years away, and the other one being 20 light years away, would I see their two clocks moving at different rates, simply because they are different distances from me? And, would I see the clock of the "stationary" observer that is 20 light years away going much faster than the clock of the "stationary" observer who is 10 light years away?

No. Because you are seeing what the clocks were doing 10 and 20 years ago. And if you happened to be accelerating 10 and 20 years ago, you still do not see any ridiculously fast running clocks 10 or 20 years after, because of some reason.

to see = to see light with eyes

The distant clocks run ridiculously fast when you accelerate. You never see them running ridiculously fast.

 

[stevendaryl]

Near a planet, if you choose a coordinate system centered on the planet (the Schwarzschild coordinates), then time dilation can be approximately described by:

$\delta \tau = (1 - \frac{GM}{c^2 r} - \frac{1}{2} \frac{v^2}{c^2}) \delta t$

where $\delta \tau$ is the elapsed time for a clock traveling at speed $v$ at a location $r$ away from the center of the planet, and $\delta t$ is the elapsed time for a clock stationary with respect to the planet, far away from the planet (where the gravity due to the planet is negligible).

So there are (from the point of view of this coordinate system) two effects on the clock's rates:

1. The faster the clock moves through space, the slower the clock runs.
2. The farther away from the planet, the faster the clock runs.

So for a clock in orbit around the planet, both effects come into play.

 

[jartsa]

When Bob's acceleration is 1 g, then according to Bob objects ahead of him accelerate at 1g * distance to object * some constant.

Correction: When Bob's acceleration is 1 g, then according to Bob objects ahead of him accelerate this much:

1 g * 1 g * distance to object * some constant

In other words: Bob's acceleration * potential difference * some constant

 

[Janus]

Hi, I just saw a video about the twin paradox, explained by using GR. I was wondering whether I understood the video correctly.

The video states that when the rocket twin is first accelerating away from earth, his clock and his Earth twin's clock are different, but, roughly the same because of how close together they are. But, on the way back, because there is so much distance between the two twins, the rocket twin would see the Earth twin's clock flying forward in time much faster than his. So, my question is, in GR, does how fast the stationary observer's clock appear to be moving depend not only on how fast the rocket observer is accelerating but also on how close he is to the stationary observer? So, the farther away the stationary observer is, the faster his clock appears to move?

First off, as long as you aren't dealing with gravitational fields, you don't have to invoke GR; sometimes you will hear that you need to use GR to deal with acceleration, but this is not the case, SR is capable of dealing with acceleration, you just have to be careful when doing so.
Second, you have to clarify what you mean by "appears". Do you mean "actually see with your eyes" or do you mean the time that clock reads at this moment compared to your own clock(what you would see after factoring out light propagation delays) If you are not clear by what you mean, you are going to get conflicting answers, because some people will think you mean one and some the other.

If you mean the second case, then yes, the distance between your clock (as well as its direction from you) does affect the rate that that clock appears to run compared to your own. The greater the separation along the line of acceleration, the greater this effect (of course, since the "stationary" clock also is going to have a velocity with respect to you, you will also have to factor in the time dilation due to that also. This not only applies to clocks that you accelerating with respect to, but also clocks that are accelerating along with you. A clock in the nose of your rocket will run fast compared to a clock in the tail as measured by anyone in your rocket even though they are not moving with respect to each other.

 

[PeterDonis]

(From the "doppler shift" explanation of the twin paradox, I was under the impression that only acceleration mattered, and not distance. But, from the GR explanation I just watched, distance does seem to matter. That's why I'm confused.)

One important thing to keep in mind is that in order to even define the "clock rate" of an object that is spatially separated from you, you need to choose a simultaneity convention--that is, you need to choose a convention for which events on the other object's worldline, and which events on your worldline, happen "at the same time". This is not something physics can tell you; it is something you have to arbitrarily choose. When people talk about distant clocks "running slower" or "running faster" than yours, they are always implicitly choosing some simultaneity convention.

In the case of an inertial observer in flat spacetime, i.e., someone who is always in free fall, never accelerating, and where there is no gravity, then there is a "natural" choice of simultaneity convention, the one used by standard inertial frames in SR. But for an accelerating observer, there is no unique "natural" choice of simultaneity convention; there are different possible ones, and all of them have significant limitations. The viewpoint which says that, for an accelerating observer, distance matters--i.e., that the clock rate you assign to a distant object, relative to yours, depends on how far away from you it is--implicitly adopts a particular simultaneity convention for accelerating observers, which works ok for this particular scenario but does not work well for others (for example, it breaks down for objects that are too far "below" the accelerating observer, as opposed to "above").

There is another viewpoint which avoids all this trouble, btw: spacetime geometry. This viewpoint is described for the standard twin paradox in this page from the Usenet Physics FAQ article on the twin paradox:

http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_spacetime.html

The advantage of the spacetime geometry viewpoint is that it is completely general; it works for any scenario whatsoever. But it also requires you to accept that questions that seem natural to you to ask, like "how fast is that other clock far away running, relative to mine?", simply do not have unique, well-defined answers. Some people find that hard to accept.

 

[pervect]

[Moderator's note: a portion of this post has been removed.]

I'm going to assume that the whole concept of "see" and more importantly "clock rates" is understood differently by different posters. This observation is probably true, but not as helpful as I would like :(. Unfortunately, I'm not sure how to make more helpful observations at this point. I would think it would be best to get the original poster to explain his interpretations of what "see" and "clock rate" means to him in an operational fashion, then an unambiguous answer can perhaps be given.

 

[m4r35n357]

Hi,

Here is a link to the video that I watched, just for reference.

That video is an impressive and fairly exhaustive attempt at explaining the "Twin Paradox" in terms of GR for beginners. It is also an excellent demonstration that it is far easier to analyze the TP using SR! Notice the lengthy explanations, particularly the concentration on the minutiae of the turnaround and launch procedures, together with the absence of any real numbers to back up the analysis.

Contrast this with the simple evaluation of three spacetime intervals in SR with constant velocity motion . . .

More pedantically, the argument also makes use of the "moving clocks run slower" argument, which IMO is a pedagogical (I enjoy using that word!) cul-de-sac when one moves on to what moving clocks really look like.

 

[idea2000]

Does the twin paradox still work if clocks behind you are traveling slower? When the traveling twin accelerates away, shouldn't the clock of the person on Earth run slower? The traveling twin should only see clocks ahead of him as going faster. When he decelerates and turns around, only the clocks he is facing should run faster, but, the clocks that are behind him should run slower. When he decelerates towards earth, his Earth twin's clock should again run slower.

So where is his net gain or loss in time?

 

[Ibix]

You are mixing time dilation with the Doppler shift. If you correct for the naive Doppler effect (i.e. subtract out the component due to velocity in your rest frame) you will find that all clocks that are moving relative to you are ticking slowly.

I'm afraid people are sometime a bit casual about the use of the word "see" in this context. You will, indeed see things as you describe. However, you will deduce that moving clocks run slowly, and this is also sometimes described (slightly generously) as what you will see.

The age difference comes from the fact that the two twins have traveled different paths through spacetime, and those paths have different intervals. Interval is a concept in Minkowski geometry equivalent to length in Euclidean geometry. It turns out, in this case, to be equal to the time experienced by each twin. In short, the different ages is no more surprising than two cars built in the same place at the same time can have different odometer readings when they meet up again.

 

[idea2000]

All moving clocks run slower, as long as there is no acceleration. But, when there is acceleration, clocks in the direction of travel should move faster, while clocks behind the traveler should run slower. So why should one twin experience more or less time than the other? I've seen the "doppler effect" explanation of the twin paradox that uses only sr, but I'm pretty sure I'm missing something here...

What am I not seeing? Thanks in advance for any help you can provide...

 

[Ibix]

When you are accelerating there is no universally acceptable way to define what "clocks running faster" or "clocks running slower" means. The reason for the different ages is the different path lengths through spacetime. Therereally isn't a simpler explanation than that.

 

[jartsa]

All moving clocks run slower, as long as there is no acceleration. But, when there is acceleration, clocks in the direction of travel should move faster, while clocks behind the traveler should run slower. So why should one twin experience more or less time than the other?

Distance is the thing you are ignoring.

clock's time dilation = acceleration * distance to clock

(This seems to be explained in post #1)

 

[m4r35n357]

All moving clocks run slower, as long as there is no acceleration. But, when there is acceleration, clocks in the direction of travel should move faster, while clocks behind the traveler should run slower. So why should one twin experience more or less time than the other? I've seen the "doppler effect" explanation of the twin paradox that uses only sr, but I'm pretty sure I'm missing something here...

What am I not seeing? Thanks in advance for any help you can provide...

What you are not seeing is that the twins meet up again. That point is crucial. They both therefore travel between the same two events (starting and meeting up again) via different routes through spacetime (one does nothing whilst the other goes away and comes back). That is the cause of the discrepancy in as few words as I can manage.

 

[pixel]

Hi, I just saw a video about the twin paradox, explained by using GR. I was wondering whether I understood the video correctly...So, the farther away the stationary observer is, the faster his clock appears to move?

I just came across that same video very recently. There are two things going on here. First, by the equivalence principle, when the twin on the rocket accelerates, for him it is indistinguishable from being in a gravitational field. Second, in a gravitational field, a clock that is higher up will run fast compared to one that is lower.

 

[Jilang]

Pixel, have you heard of the triplets paradox? You don't need acceleration to explain the effect.

 

[pixel]

Jilang: No I haven't; will have to look it up.

Also, I remember Mermin's book on special relativity has a good discussion of the twin paradox without using GR.

 

[jartsa]

Pixel, have you heard of the triplets paradox? You don't need acceleration to explain the effect.

Blah. Different scenarios, different effects, different explanations.

 

[jartsa]

Blah. Different scenarios, different effects, different explanations.

If anybody can explain how the effect is explained in triplet paradox, I'm willing to listen.

 

[Nugatory]

If anybody can explain how the effect is explained in triplet paradox, I'm willing to listen.

The Doppler shift explanation and the spacetime interval explanation both work just fine for the triplet paradox. If this is not clear to you, start a separate thread - but try working it out yourself first, using either of those techniques.

 

[Janus]

Pixel, have you heard of the triplets paradox? You don't need acceleration to explain the effect.

What version of the Triplets paradox are you referring to? The most common version I've seen involves a set of triplets A, B and C. A stays home while B and C fly off in opposite directions and then return home. Since A and B have to accelerate to return back to A, acceleration is indeed involved. Or do you have another scenario that you are thinking about?

 

[Jilang]

One twin stays at home, the second twin heads off and the third twin is heading home. The second and third twin synchronise watches as they pass each other, some distance away from home.

 

[Janus]

One twin stays at home, the second twin heads off and the third twin is heading home. The second and third twin synchronise watches as they pass each other, some distance away from home.

It thought this might be what you were referring to.
First off, to eliminate acceleration completely, you can't have your second "Twin" head off from "home" since he would have to accelerate in order to leave "home" .
So instead we will have his ship flying by the first twin and they synchronize watches as they do so.

So we have A, B and C. B is flying past A towards C and C is flying towards A from some distance. A and B sync watches as they pass and we will assume that according to A, C's clock reads the same time when A and B pass each other. B and C have the same speed with respect to A, according to A, A and B's clocks run at the same rate and they should read the same time when they meet. According to B however C's clock runs slower than his, and according to C, B's clock runs slower than his. This seems to lead to the paradox that when they meet, B should read his clock as being ahead of C's and C should read his clock as being ahead of B's.
This is not the case however, The apparent paradox arises from assuming that at the when B passes A, all three agree that C's clock reads the same as A and B. In fact, this is only true according to A, B and C conclude otherwise. B concludes that C's clock reads some time later than his, and C concludes that B's clock reads the same amount ahead of his own(Relativity of Simultaneity). The end result being that even though each determines that the other clock is running slower as they fly towards each other, they end up reading the same when they meet, and this reading will be the same as they read according to A.

 

[Jilang]

Janus, that is a good explanation. I am thinking that it must be the version where two of the triplets head out and return, but one goes further before turning round. Both the traveling triplets undergo the same acceleration, but their ages will be different when they meet up.