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

by MikeLizzi
Tags: acceleration, effect, light, shape, size, wave
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DaleSpam
#19
Mar29-12, 02:35 PM
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Quote Quote by MikeLizzi View Post
When the twins have reunited they are in the same reference frame.
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.

Quote Quote by MikeLizzi View Post
They must now agree on the geometry of the wave front.
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.

Quote Quote by MikeLizzi View Post
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.
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.
DrGreg
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Mar29-12, 04:51 PM
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Quote Quote by MikeLizzi View Post
“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?”
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 ˝t' and in the outgoing astronaut inertial frame the sphere radius is ˝ct'. When we transform to the ingoing astronaut inertial frame, we synchronise the frame clocks with the astronaut's ˝t', 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 ˝ct'. (It will actually be c(t−˝t').) By the time the astronaut is back at Earth, a further time of ˝t' will have passed on his clock and the sphere radius will have expanded to ct.
MikeLizzi
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Mar29-12, 07:11 PM
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Quote Quote by DrGreg View Post
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 ˝t' and in the outgoing astronaut inertial frame the sphere radius is ˝ct'. When we transform to the ingoing astronaut inertial frame, we synchronise the frame clocks with the astronaut's ˝t', 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 ˝ct'. (It will actually be c(t−˝t').) By the time the astronaut is back at Earth, a further time of ˝t' will have passed on his clock and the sphere radius will have expanded to ct.
Thank you Dr Greg. I can use that analysis. I expect you will receive a critical comment from DaleSpam since in his last post he insisted that the the twins would not produce the same answer. I'll leave that fight for you.
MikeLizzi
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Mar29-12, 07:14 PM
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Quote Quote by DaleSpam View Post
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.
A series of straw man arguments an just plain wrong physics. Not worth another reply. But Dr Greg might have something to say to you.
DaleSpam
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Mar29-12, 07:20 PM
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Quote Quote by MikeLizzi View Post
A series of straw man arguments an just plain wrong physics. Not worth another reply.
If you don't want to reply that is fine, but nothing in my post was wrong. Frame variant quantitites need to be calculated in the same frame, and regardless of which frame you pick, the twins will agree on the radius if they apply the laws of physics.
PAllen
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Mar29-12, 08:25 PM
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Quote Quote by MikeLizzi View Post
A series of straw man arguments an just plain wrong physics. Not worth another reply. But Dr Greg might have something to say to you.
Actually, though you may not understand or accept it, the answers you've been given by Dalespam, Passionflower, Dr. Greg, and myself are all consistent and correct. The key point is that there is no unique answer to 'correct coordinates for an accelerating observer'. Note also, that Dr. Greg's recipe is, in fact, Fermi-Normal coordinates applied to a discontinuous trajectory, and if the turnaround is smoothed slightly, involves arbitrarily large light speed for a brief period, as I described.
pervect
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Mar29-12, 08:52 PM
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Quote Quote by MikeLizzi View Post
A series of straw man arguments an just plain wrong physics. Not worth another reply. But Dr Greg might have something to say to you.
While anyone can make a mistake, I didn't notice anything obviously wrong with Dale's physics - or with his arguments. Your comment was far from specific enough to refer to any particular piece of text in this very long thread that one could re-examine closely with a microscope.

If you're thinking that Dale's obviously wrong and your'e obviously right - I'd think again.

If you're thinking that you're getting frustrated and need to take a break from talking about this thread for a while, that's a bit more reasonable.
DrGreg
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Mar30-12, 03:14 PM
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Quote Quote by DrGreg View Post
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 ˝t' and in the outgoing astronaut inertial frame the sphere radius is ˝ct'. When we transform to the ingoing astronaut inertial frame, we synchronise the frame clocks with the astronaut's ˝t', 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 ˝ct'. (It will actually be c(t−˝t').) By the time the astronaut is back at Earth, a further time of ˝t' will have passed on his clock and the sphere radius will have expanded to ct.
Sorry, I now withdraw the above post. I made a mistake and forgot to take account of the final change-of-frame when the astronaut comes to rest on Earth. None of the values I quoted were correct.

The answer is rather more simple. The time t' was accumulated in two other inertial frames and bears no relevance to the propagation of light as measured in the Earth frame. In that frame a time of t has elapsed and the radius is ct. All observers who are at rest in that frame will agree on the radius regardless of their past history.

If you are thinking you can somehow glue together all of the inertial frames in which the astronaut has been at rest, and all the non-inertial frames when he accelerated, to produce a "composite" single coordinate system, in which the astronaut is always at rest, well, yes, it is technically possible to do this (and, as others have pointed out, there is more than one way of doing so). But this frame is not an inertial frame and so the rules of inertial frames don't apply. In particular there is no requirement for the speed of light to be constant in such a frame. It can speed up, slow down, stop*, or even go backwards* in an arbitrary coordinate system. This is because in a non-inertial coordinate system, the "time" coordinate need not tick at the same rate as a proper-time clock at rest. (This may be described as "gravitational" -- or "pseudo-gravitational" -- time dilation.)



___________
*To be mathematically precise, if the time coordinate stops or goes backwards, then it's technically incorrect to describe it as a "coordinate".
PAllen
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Apr2-12, 11:23 PM
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Quote Quote by PAllen View Post
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.]
This thread is dead, but for the sake of future searches turning it up, I have a few technical corrections to the above.

As natural as Fermi-Normal coordinates (and Rindler coordinates for uniform acceleration) are, they have a key limitation - they don't generally cover all of spacetime. In the case of Rindler, they cover just one section of spacetime. The reason for this is that they would multiply map these regions, which is not allowed for a coordinate patch. This happens because the spacelike geodesics 4-orthogonal to different points on the 'time axis' world line intersect. Wherever this occurs, you have a region that cannot be covered with these coordinates. In the case of Fermi-Normal coordinates for the twin situation with rapid or instant acceleration then coasting away, then rapid or instant turnaround, then rapid or instant deceleration to a stop, there are three patches of spacetime not covered by the coordinates (actually six for instant accelerations, so lets assume nearly instant instead). If we assume the traveler goes to the right and back, then the uncovered patches are:

1) To the left of the initial acceleration
2) To the right of the turnaround
3) To the left of the stopping acceleration

As a result, for a signal emitted by the stay at home when the traveler starts, both the left going light path and the right going light path would pass out of the coordinate patch. Thus, any concept of computing overall coordinate speed of this signal is impossible in these coordinates. It is true that using these coordinates (but not any computation of the whole path). once the travel stops, the coordinate position of the right going ray will match the stay at home twin. The left going ray will remain outside of the coordinate patch for the traveler for a long time after they have stopped, but eventually will pass to a covered region, at which point the traveler will agree with stay at home on its distance.

Radar coordinates are able to cover the whole spacetime, even for sudden turnarounds. In an earlier post, Dalespam explained you would see speed of c for the signals (part of the definition of radar coordinates - any light ray to or from the time axis world line has speed c; rays between other points, not generally); and you would determine smaller distance than stay at home. They would never come to agreement on coordinate features of these light rays, but would for later ones, after they are back together.

The upshot is that Dr. Greg's final answer (also said in several earlier posts in this thread) is the only way the traveler agrees with the stay at home: once back, they simply use stay at home measurements.
Austin0
#28
Apr3-12, 01:45 AM
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Quote Quote by PAllen View Post
T

As natural as Fermi-Normal coordinates (and Rindler coordinates for uniform acceleration) are, they have a key limitation - they don't generally cover all of spacetime. In the case of Rindler, they cover just one section of spacetime. The reason for this is that they would multiply map these regions, which is not allowed for a coordinate patch. This happens because the spacelike geodesics 4-orthogonal to different points on the 'time axis' world line intersect. Wherever this occurs, you have a region that cannot be covered with these coordinates.
Hi Am I correct in assuming that they are not allowed because they would map a reality that has no possible correspondence to the one we know.
Co-location of points with different time coordinates, , temporal disordering etc.??
Am I right that this is essentially the same as Minkowski hyper-surfaces of simultaneity wrt an accelerating frame??? [with the added complication of an internal time differential.]

If this is correct would it also be true that planes of simultaneity in Minkowski space are equally false for those regions in the case of acceleration??

Do you think that the regions of invalidity only start at the point of intersection?
Or is it more likely that the divergence from correspondence with reality simply diminishes from that point as it approaches the bounds of the physical system??
Thanks


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