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1. Feb 4, 2016

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?

2. Feb 4, 2016

CrazyNinja

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

3. Feb 4, 2016

idea2000

Hi,

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

Last edited: Feb 4, 2016
4. Feb 4, 2016

5. Feb 4, 2016

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?

6. Feb 4, 2016

jartsa

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)

Last edited: Feb 4, 2016
7. Feb 4, 2016

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.)

8. Feb 4, 2016

jartsa

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.

9. Feb 4, 2016

stevendaryl

Staff Emeritus
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.

10. Feb 4, 2016

jartsa

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

11. Feb 4, 2016

Janus

Staff Emeritus
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.

12. Feb 4, 2016

Staff: Mentor

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:

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.

13. Feb 4, 2016

pervect

Staff Emeritus
[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.

Last edited by a moderator: Feb 4, 2016
14. Feb 6, 2016

m4r35n357

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.

15. Mar 21, 2016

idea2000

Does the twin paradox still work if clocks behind you are travelling slower? When the travelling twin accelerates away, shouldn't the clock of the person on earth run slower? The travelling 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?

16. Mar 21, 2016

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 travelled 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.

17. Mar 21, 2016

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...

18. Mar 21, 2016

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.

19. Mar 21, 2016

jartsa

Distance is the thing you are ignoring.

clock's time dilation = acceleration * distance to clock

(This seems to be explained in post #1)

Last edited: Mar 21, 2016
20. Mar 21, 2016

m4r35n357

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.