View Full Version : dilation for fictional gravity

grav

2009-Jun-21, 01:46 AM

Let's say Alice and Bob are stationary to each other in the same place, then Bob begins to constantly accelerate away from Alice. When Alice's clock reads that some time has passed on her clock, she would say that the same amount of time has passed for Bob, but time dilated. In terms of just SR, that would be t' = Int dt sqrt[1 - (v/c)^2], time dilated only according to the instantaneous relative speed, where v/c = (a t / c) / sqrt[1 + (a t / c)^2], so

t' = Int dt sqrt[1 - (a t / c)^2 / (1 + (a t / c)^2)]

= Int dt sqrt[((1 + (a t / c)^2) - (a t / c)^2) / (1 + (a t / c)^2)]

= Int dt / sqrt[1 + (a t / c)^2]

= (c / a) * asinh-1(a t / c)

= (c / a) * ln[(a t / c) + sqrt(1 + (a t / c)^2)]

Does all of that look correct? Here is the problem. It accounts for time dilation in SR, but does not seem to account for gravitational time dilation with the fictional gravity that acts upon the accelerating observer in order to accelerate. So shouldn't there be another term in the integration for that as well?

grav

2009-Jun-21, 05:20 PM

Any comments or opinions on this? I need it to work something out.

publius

2009-Jun-21, 07:05 PM

Grav,

That looks correct at first blush -- you're using hyperbolic motion, constant proper acceleration.

Grav, the fictional gravity doesn't appear in the frame of inertial observer. That only comes in the frame of the accelerating observer. Alice sees nothing but Minkowski and simple SR. Bob, in a Rindler frame, is the one who sees fictitious gravity.

Carefully consider what you're calculating. Alice is calculating what Bob's clock reads according to her own sense of "now", simultaneity. Bob's simultaneity is something else (indeed, his lines of simultaneity are curved, not staight, from Alice's POV). Bob would calculate what Alice is doing according to his own, curved, sense of simultaneity.

Between two inertial observers in flat space-time, there is sort of a familiar reciprocity between the two. That reciprocity is broken in non-inertial frames.

It's much simpler to do things from inertial frames. :)

-Richard

grav

2009-Jun-21, 07:35 PM

Grav, the fictional gravity doesn't appear in the frame of inertial observer. That only comes in the frame of the accelerating observer. Alice sees nothing but Minkowski and simple SR. Bob, in a Rindler frame, is the one who sees fictitious gravity.

Carefully consider what you're calculating. Alice is calculating what Bob's clock reads according to her own sense of "now", simultaneity. Bob's simultaneity is something else (indeed, his lines of simultaneity are curved, not staight, from Alice's POV). Bob would calculate what Alice is doing according to his own, curved, sense of simultaneity.

Between two inertial observers in flat space-time, there is sort of a familiar reciprocity between the two. That reciprocity is broken in non-inertial frames.Thanks, Richard. :) Bob is the one experiencing the proper acceleration from fictitious gravity, but Alice also observes a time dilation for Bob due to that, right? I mean, if Bob were stationary in a gravitational field and Alice were observing from far away, she would say his clock is dilated due to gravitational time dilation. Referring to the equivalence principle, that would also apply to fictitious gravity, wouldn't it? And if we were trying to find the time dilation due to the proper acceleration and that due to the relative motion, shouldn't both be included somehow?

grav

2009-Jun-21, 10:08 PM

I've been trying to find another way to get the time dilation using simultaneity shifts, but now I've got another problem. Let's say Bob is in the center of a ship, the back of which is stationary to Alice and coincides with her. At T=0, two things happen. Bob begins to accelerate away from Alice in the ship and at the same time two beams of light are flashed from either end of the ship. Now, at T=0, the acceleration has just begun, so everything is still simultaneous in the original frame and the light is travelling an equal distance across the ship in opposite directions toward Bob, according to both Alice and Bob. After acceleration has begun, however, from Bob's point of view, the light should still be travelling at c over equal distances at all times relative to him and coincide in the center of the ship with him. From Alice's perspective, however, after acceleration has begun, the two beams of light will coincide in the center of where the ship began in the original frame, but now that the ship has begun accelerating forward over the time it took for the light beams to coincide, it will not be observed by her to coincide in the center of the ship with Bob, but more toward the back of the ship. All observers must agree on the events that coincide in the same place. So how do we fix this situation?

grav

2009-Jun-21, 11:16 PM

Okay, I have figured out that dilemma from the last post. Apparently, light doesn't always travel at c according to an accelerating observer, but the accelerating observer conceives the light also accelerating toward them from in front and away in back, due to the simultaneity shifts while accelerating. Well, there goes my equivalence principle in regards to light always travelling at c to any inertial observer, so that it should while accelerating between frames as well, because it doesn't take into account the simultaneity shift that is only observed while accelerating, so goodbye to that. Although now it seems so simple I should have realized it before.

So here is an example. An observer is 10 light-seconds away from an emitter that is stationary to the observer. At T=0, the emitter begins emitting pulses once per second. When the observer receives the first pulse that reads T=0, he instantly accelerates toward the emitter to .6 c . The pulse that read T=5 from the emitter was halfway across the original distance when the observer accelerated. After accelerating, the emitter is pushed tl = L Z d v / [c^2 (1 - (v/c)^2)] = (.8) (.8) (10) (.6) / (1 - (.6)^2) = 6 seconds future forward due to the simultaneity shift. The observer still observes the pulse that reads T=0, but now conceptualizes the clock of the emitter to read T=10 + 6 = 16. So the T=5 pulse that was halfway to the observer originally, at a distance of 5 light-seconds, is now 5 / 16 of the new distance to the emitter of d' = L d = (.8) (10) = 8 light-seconds, so the T=5 pulse is now (5/16) (8) = 2.5 light-seconds away, at half the original distance it was. That gives half the original wavelength so twice the original frequency, which agrees with the relativistic Doppler shift of sqrt[(1+v/c) / (1-v/c)] = 2.

If the observer instantly accelerated away from the emitter, the observer would conceptualize the emitter's clock pushed tl = 6 light-seconds past-back, so the observer says it is currently emitting at T = 10 - 6 = 4 while the observer stil observes the pulse from T = 0, over a distance of 8 light-seconds, so the wavelength becomes twice as great and the frequency halves, agreeing with the relativistic Doppler shift of sqrt[(1-v/c) / (1+v/c)] = .5 . The pulse T=5 hasn't been emitted yet according to the new frame, but if the emitter were further away, it would now be at a distance of 10 light-seconds.

Okay, so even light itself is simultaneity shifted. Now I guess I need to work out an apparent acceleration of light observed according to an accelerating observer.

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