# Effect of acceleration on shape and size of a light wave front

1. Mar 29, 2012

### MikeLizzi

Sometimes I get frustrated with the dialogs in this forum because there are so many misunderstandings and outright half truths being passed around (including from myself). Well, I was just over at SciForums.com warily participating in the thread “According to SR…” and it gave me a whole new perspective on frustration.

The subject of the thread at SciForums.com is quite interesting. I hope I am not violating the rules by introducing it here. Summarized, it goes something like this.

“Suppose that a flash of light is emitted from Earth at the start of the Twins problem when t=t’=0. Since the earth twin is modeled as inertial, that twin would always calculate the geometry of the light wave front to be a sphere of radius ct where t is the elapsed time on the earth twin’s clock. Now the astronaut has been calculating the location and geometry of the light wave front too (which is different during the trip). At the end of the problem, the twins are reunited and must agree on the final location and geometry of the light wave front. But the elapsed time on the astronaut clock is t’ < t. So, how can the astronaut calculate the same final radius for the light wave front?”

P.S. Over at SciForums.com, I referenced a very good and relevent thread from this forum titled “Accelerated Frames in SR” but only one of the many participants got it.

2. Mar 29, 2012

### Staff: Mentor

It is not correct that they must agree on the same final radius. They only need to agree that it is a sphere and that it propagated at c in any inertial frame. They can disagree about the radius just as they can disagree on the time.

Also, note that the second postulate refers to inertial frames. There are three relevant inertial frames in this problem, and the light cone has the appropriate geometry in each, and each matching the elapsed time in their frame but not the elapsed time in the other frames.

The traveling twin's frame is non-inertial, so it is not uniquely defined. However, if you use the definition from Dolby and Gull (http://arxiv.org/abs/gr-qc/0104077) then it will be a sphere of a radius matching the traveller's time in the non-inertial frame also. That is not necessarily the case with other defintions of the frame of a non-inertial observer.

3. Mar 29, 2012

### MikeLizzi

Here we go again.
Think about it DaleSpam. At the end of the problem the twins are coincident in the same inertial reference frame. While their clocks don't agree, they must agree on the geometry of everything in the universe from that point forward.

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4. Mar 29, 2012

### Staff: Mentor

For geometrical quantities that are frame-variant, such as the radius, that is only true if they measure those quantities with respect to the same reference frame. And if they do, then they agree, regardless of which reference frame they use, as long as it is the same.

5. Mar 29, 2012

### Passionflower

Easy, he must calculate his differential time dilation due to his acceleration and change in directions. See for instance Minguzzi's papers on differential aging.

6. Mar 29, 2012

### MikeLizzi

That's what I said in the original post. That's the problem. When the twins have reunited they are in the same reference frame. They must now agree on the geometry of the wave front. The earth twin can make the simple calculation that the radius of the wavefront is ct where t is the elapsed time of the earth twin's clock. But if the astronaut tries to do that using the time elapsed on his clock he will get a different answer. His calculation must produce the same answer.

7. Mar 29, 2012

### MikeLizzi

I'm not sure you understand the problem I posted. The problem is not how to calculate differential aging.

8. Mar 29, 2012

### Passionflower

It is because the if you know the amount of differential aging you can calculate the actual radius of the wavefront.

9. Mar 29, 2012

### MikeLizzi

The point of the post is that the twins have not aged the same and if each uses the amount they aged to calculate the radius of the wavefront, it appears to give them different results.

10. Mar 29, 2012

### Passionflower

Thee is no difference when we account for the differential aging.

So what is the issue or question here or is this perhaps some attempt to 'disprove' relativity?

11. Mar 29, 2012

### PAllen

Ah, back to this misleading concept of necessarily being 'in a frame'. I am on earth, but I can compute (and will compute for planetary problems) in the sun's frame. If I compute planetary motion in an earth centered frame, I am in for a heap of complications (forget GR, and even SR, just imagine Newtonian gravity and Galilean relativity for this analogy).

When the astronaut returns, if he wants to compute using earth's frame, he must use earth's time. If you insist that he use his own watch time, he must use a coherent coordinate chart that maps spacetime (past and ongoing), and has a coordinate time matching his watch time. There is no unique such coordinate system, but there are any number of choices (all much more complex than just choosing one inertial frame). Whichever one he picks, he will continue to label times and distances of events differently than the earthbound twin (as long as he insists on using his watch time, which is influenced by his history of deviation from inertial motion).

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12. Mar 29, 2012

### MikeLizzi

Getting close.

Of course if the astronaut uses the earth twins data he will get the earth twins answer.

The issue, which I may not have explicitly stated, is how does the astronaut use his own data to get the answer?

He can and if he does it right he will get the same answer as using the earth twins data. Knowing how to do that is the key. I referenced a post in this very forum that explains how that would be done.

13. Mar 29, 2012

### MikeLizzi

I'm not trying to disprove relativity. If anything I'm trying to prove that most of the members of this forum don't know as much as they think they know. (And I'm not averse to someone saying that statement also applys to me)

You state there is no difference when we account for differential aging. So show me how each twin would calculate the radius of the light sphere using their own data and get the same answer.

14. Mar 29, 2012

### PAllen

You didn't link to anything. I am not going to search for some thread by name, and search for which posts you are referring to.

I insist there is no unique, preferred answer to your question. Are you claiming there is only one coordinate system the astronaut is allowed to use? I have little interest in finding a particular coordinate system in which a particular light sphere has the same radius as for a particular inertial frame (note, different inertial frames may disagree on the radius of a particular light sphere; note, even meeting these constraints, it is trivially provable there are infinite possible such coordinate systems). Why not? Because I think it is very silly in SR to do anything more complex than pick some convenient inertial frame, relate measurements to it, and do all computations in that chosen frame.

[Edit: One very well known feature of non-inertial coordinates is that light speed is not constant. Thus, there is really no mystery in how a well chosen such coordinate system could get the desired agreement : proper time slower, light faster, same radius. I remain uninterested in the details of such an exercise.]

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15. Mar 29, 2012

### Passionflower

It is trivial as the radius is simply delta t times c for an inertial observer. The traveler can calculate how much his watch ran slower and when he takes that into account he gets the same answer.

I think you ask questions about something that you think is a problem but what is not a problem at all.

16. Mar 29, 2012

### PAllen

I thought Passionflower was perfectly clear about his method. Track your differential aging relative to earth twin (which can be done with only local measurements, as explained in the papers he referenced). Take your clock time, adjust for differential time, multiply by c; that is the radius you compute. You would get the same radius as earthbound twin at all points in your journey with this method.

17. Mar 29, 2012

### MikeLizzi

That's just another way of saying the astronaut should transform his observations to that of the earth twin and then make the calculations. If this were a test that would be a great way to get the right answer. But a solution like that means the poster doesn't know how to calculate the right answer using the astronauts observations.

So far no one who has responded does. No crime there, I don't know how to do it myself. But I know it can be done. User "pervect" pointed the way in his post in this thread:

18. Mar 29, 2012

### PAllen

I believe, given time, I could demonstrate many ways of doing it, but I have explained why I (and I think most anyone working in SR) would find the effort silly to expend. Doing all calculations in some convenient inertial frame is what essentially everyone does, in practice, in SR. Why do you want to make something simple seem complicated?

[Edit: Pervect is just discussing Rindler coordinates, which apply to uniform acceleration. One analog to these coordinates for a rocket path where you start at rest, accelerate away, then back, ending at rest, would be Fermi-normal coordinates. Another, as Dalespam mentioned, would be radar coordinates. Another coordinate system could be built from the Minguzzi simultaneity, as Passionfower referenced. If you used Fermi-Normal coordinates, you would find that, on return, distance to light sphere agrees with stay at home, but average light speed is greater, balancing the smaller proper time experienced. Dalespam already explained what you would get with Radar coordinates. There is no sense in which you can call one of these answers right and one wrong. I want to re-emphasize Dalespam's point that your whole expectation of what should be true is wrong for non-inertial coordinates in SR.

I don't believe it makes much sense to talk about non-inertial frames (rather than coordinates), except locally. The issue is the same as in GR. Specifically, there is a unique, simplest type of global coordinate system for each inertial frame in SR. For non-inertial frames, this is false, as it is, in general, in GR. Thus, it is clearer to treat frames as strictly local for non inertial motion, or quite generally, in GR.]

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19. Mar 29, 2012

### Staff: Mentor

I really dislike this terminology. They are always "in" every reference frame. When the twins have reunited they are at rest wrt the same reference frame.

They must ALWAYS agree on the geometry of the wave front and all other geometrical properties, both before and after the reuinion, provided that they use the same reference frame to calculate any frame variant quantities like the radius.

No, in fact, it must not produce the same answer, in that reference frame his clock has undergone significant time dilation, so the sphere must be larger than c times the time shown on his clock.

The principle of relativity states that the mathematical form of the laws of physics are the same in all inertial frames, it does not state that astronauts who make mistakes in calculations should nevertheless get the right answer. The laws of physics say that his clock ran slow while he was travelling and so his clock is not so simply related to the size of the light sphere in the earth frame.

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20. Mar 29, 2012

### DrGreg

In the standard, simple version of the twins' paradox there are 3 inertial frames (the Earth frame, the outgoing astronaut frame and the ingoing astronaut frame). Just before the astronaut reaches the turning point, the astronaut's clock is at ½t' and in the outgoing astronaut inertial frame the sphere radius is ½ct'. When we transform to the ingoing astronaut inertial frame, we synchronise the frame clocks with the astronaut's ½t', but because of the relativity of simultaneity, the original light flash did not occur at time zero in this frame but at some earlier time, so the sphere radius in this frame is more than ½ct'. (It will actually be c(t−½t').) By the time the astronaut is back at Earth, a further time of ½t' will have passed on his clock and the sphere radius will have expanded to ct.