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Effect of acceleration on shape and size of a light wave front 
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#1
Mar2912, 08:14 AM

P: 191

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
Mar2912, 08:38 AM

Mentor
P: 17,344

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


#3
Mar2912, 08:49 AM

P: 191

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. 


#4
Mar2912, 09:53 AM

Mentor
P: 17,344

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



#5
Mar2912, 10:06 AM

P: 1,555




#6
Mar2912, 10:51 AM

P: 191




#7
Mar2912, 10:54 AM

P: 191




#8
Mar2912, 11:04 AM

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#9
Mar2912, 11:22 AM

P: 191




#10
Mar2912, 11:29 AM

P: 1,555

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


#11
Mar2912, 12:14 PM

Sci Advisor
PF Gold
P: 5,060

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). 


#12
Mar2912, 12:39 PM

P: 191

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
Mar2912, 12:45 PM

P: 191

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
Mar2912, 01:01 PM

Sci Advisor
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P: 5,060

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 noninertial 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.] 


#15
Mar2912, 01:36 PM

P: 1,555

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


#16
Mar2912, 01:44 PM

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#17
Mar2912, 02:09 PM

P: 191

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: http://www.physicsforums.com/showthread.php?t=181421 


#18
Mar2912, 02:15 PM

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P: 5,060

[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 Ferminormal 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 FermiNormal 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 reemphasize Dalespam's point that your whole expectation of what should be true is wrong for noninertial coordinates in SR. I don't believe it makes much sense to talk about noninertial 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 noninertial 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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