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A question about the equivalence principle.

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jartsa
#19
May25-12, 10:54 PM
P: 476
Let's say we have a non-length-contracting accelerating rocket.

If a clock is thown from the rear of the rocket to the front of the rocket, it will be time dilated during the flight an amount that makes the reading of the clock to be the same as if the clock would have been in a gravity well.

If this same clock is thrown back to the rear from the front, it will be time dilated less than everything else in the rocket during the flight, and this effect is larger than the effect of the first throw, and the reading of the clock will be the same as if the clock had visited upper regions of a gravity well.
Austin0
#20
May26-12, 12:18 AM
P: 1,162
Quote Quote by jartsa View Post
Let's say we have a non-length-contracting accelerating rocket.

If a clock is thown from the rear of the rocket to the front of the rocket, it will be time dilated during the flight an amount that makes the reading of the clock to be the same as if the clock would have been in a gravity well.

If this same clock is thrown back to the rear from the front, it will be time dilated less than everything else in the rocket during the flight, and this effect is larger than the effect of the first throw, and the reading of the clock will be the same as if the clock had visited upper regions of a gravity well.
Hi Take a look at yuiop's diagrams a couple of posts back. Compare the relative path lengths between forward and backward travel and see if you still think the front to back travel would have greater effect on elapsed time than the travel from back to front.
jartsa
#21
May26-12, 12:57 AM
P: 476
Quote Quote by Austin0 View Post
Hi Take a look at yuiop's diagrams a couple of posts back. Compare the relative path lengths between forward and backward travel and see if you still think the front to back travel would have greater effect on elapsed time than the travel from back to front.

We did have an accelerating and a non-length-contracting rocket, so in this rocket distances stay the same, while everything is slowing down, because of time dilation ... so the travel bacwards takes a little bit more time, or a lot more time if there is a large rocket velocity difference.

Interestingly, if the clock is thrown at relativistic speed, this doesn't seem to work.

Well I'll check the diagrams now.
stevendaryl
#22
May26-12, 11:07 AM
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Quote Quote by Austin0 View Post
Hi WHy do you thnk there would be a difference between a single rocket and two rockets if the distance remained constant in the reference frame in both cases ?
In the two rocket case, the distance between the two rockets remains constant (in the launch frame), while in the one rocket case, the distance from the bottom of the rocket to the top decreases with time (in the launch frame). But in the latter case, the rate of decrease starts off zero, so it makes no difference to the Doppler shift initially.
Austin0
#23
May26-12, 06:55 PM
P: 1,162
Quote Quote by stevendaryl View Post
In the two rocket case, the distance between the two rockets remains constant (in the launch frame), while in the one rocket case, the distance from the bottom of the rocket to the top decreases with time (in the launch frame). But in the latter case, the rate of decrease starts off zero, so it makes no difference to the Doppler shift initially.
Gotcha. I thought we were talking about the no contraction case for both instances as is sometimes done in the Bell or Born scenarios.
Austin0
#24
May26-12, 08:34 PM
P: 1,162
Quote Quote by DaleSpam View Post
Yes. In fact, the fact that the clock "lower" in the accelerating system slows down can be derived purely using SR without any reference to the equivalence principle or the Einstein field equations.

The elegant way to do it is to determine the transform between the inertial and accelerating systems and then evaluate the spacetime interval for a stationary object.

The brute force way to do it is to calculate the Doppler shift between the upper and lower clocks.

Hi I encountered this question in an old thread and I later tried a ballpark calculation like this:
c=300,000km/s
A rocket R with a rest length of 1 km in the launch frame . With a constant coordinate acceleration of 1000g= 10km/s2 in LaunchF

Accelerating between 0.6 c and 0.7 c the displacement arising from contraction is 0.09km
in 3,000s Average v=10-10 c
with an Additive avg. v =1.7316 e-10 within the ship.

The gamma comes up NaN for the relative velocity in the ship frame (HypPhy calculater). back then I got a gamma of 1 +( 2.9484 x 10 -21 ) but I can no longer remember how I arrived at it.

It is 1.315903389919538 for the average gamma for the front of the ship and LF

It is 1.3159033900676484 for the average gamma for the back of the ship and LF

With a difference of 1.48110401x 10-10 between the two in the LF

I tried to get a handle on the Rindler math to work out a comparison (actually you tried to help me with that) but never actually got a figure.

Looking at the Pound-Rebka results for 22.5 m and 1 g ,,giving a factor on the orderof 5

x 10-15 it is hard to tell if 1,000m and 1,000g with a resulting difference

factor of 1.48110401 x 10-10is equivalent or not? Or even close.

Any hints???

Is this aything along the lines of what you were talking about above??

Where did the original Rindler calculations come from, what was the basis?

Do you know of any treatment done through Doppler that has been worked out for review??

The brute force approach as you called it. It certainly seems messy in principle and hard to interpret but I would like to know the method of approach..


Thanks
stevendaryl
#25
May26-12, 09:25 PM
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P: 2,150
Okay, I did some calculations and I think I have a good grasp now on the clock discrepancies in the accelerated rocket. There are two effects going on:
  1. Because of relativity of simultaneity, if two clocks are synchronized in the "launch" frame, then the forward clock will be farther ahead, as measured in the instantaneous rest frame.
  2. The front clock is traveling slightly slower than the rear clock (as measured in any inertial frame), and so experiences slightly less time dilation.

Here's what's interesting: Early on, effect number 1 is the most important, and effect number 2 is negligible. Much later, when the rocket is traveling very close to the speed of light, effect number 2 is most important, and effect number 1 is negligible.
Austin0
#26
May26-12, 10:13 PM
P: 1,162
Quote Quote by stevendaryl View Post
Okay, I did some calculations and I think I have a good grasp now on the clock discrepancies in the accelerated rocket. There are two effects going on:
  1. Because of relativity of simultaneity, if two clocks are synchronized in the "launch" frame, then the forward clock will be farther ahead, as measured in the instantaneous rest frame.
  2. The front clock is traveling slightly slower than the rear clock (as measured in any inertial frame), and so experiences slightly less time dilation.

Here's what's interesting: Early on, effect number 1 is the most important, and effect number 2 is negligible. Much later, when the rocket is traveling very close to the speed of light, effect number 2 is most important, and effect number 1 is negligible.
If by 1) you mean relative to a MCIRF I think your right but that doesn't really address the question of the mechanism of why the clocks would have different rates, which relative velocity does. It also brings up the complication that the system in a sense doesn't have a single momentarily comoving frame
Just looking at synchronization alone lets assume the there is no differential dilation.
The ship starts out with a certain synch. If there is no resynchronization it will maintain that relationship throughout the course , right?
So initially there will be small discrepancy between it and succeeding MCIRF's But as velocity increases that discrepancy will also increase because the relationship of the ship to the MCIRF is the same as the relationship of that MCIRF to the initial frame. So adding in the time dilation acts to increase that discrepancy , the front clock keeps running even further ahead of the rear and the front of the MCIRF
So to figure out dilation you would have to eliminate the ever increasing effect of desynchronization if you are going to use a MCIRF as a reference. Or so it seems to me.
stevendaryl
#27
May26-12, 11:41 PM
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Quote Quote by Austin0 View Post
If by 1) you mean relative to a MCIRF I think your right but that doesn't really address the question of the mechanism of why the clocks would have different rates, which relative velocity does.
Well, consider the events
e_1: the rear clock shows time 12:00
e_2: the front clock shows time 12:00

If there is no length contraction, then e_1 and e_2 are simultaneous in the launch frame. But in the frame in which the rocket is momentarily at rest, e_2 happens before e_1. So in that frame, the front clock runs ahead of the rear clock.

If there is length contraction, then it means that even in the launch frame, event e_2 happens before e_1.
stevendaryl
#28
May26-12, 11:57 PM
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Quote Quote by Austin0 View Post
The ship starts out with a certain synch. If there is no resynchronization it will maintain that relationship throughout the course , right?
So initially there will be small discrepancy between it and succeeding MCIRF's But as velocity increases that discrepancy will also increase because the relationship of the ship to the MCIRF is the same as the relationship of that MCIRF to the initial frame. So adding in the time dilation acts to increase that discrepancy , the front clock keeps running even further ahead of the rear and the front of the MCIRF
So to figure out dilation you would have to eliminate the ever increasing effect of desynchronization if you are going to use a MCIRF as a reference. Or so it seems to me.
I'm not exactly sure what you are saying. But if I did the calculation correctly, the exact result for the time τ on a clock aboard a Born-rigid rocket of constant proper acceleration is this:

τ = R/c arcsinh(ct/R)

where t is the time in the launch frame, and where R is a distance computed by:

R = c2/a + x

where x = the height of the clock above the rear of the rocket, and a = the acceleration felt by the rear of the rocket.
Austin0
#29
May27-12, 12:13 AM
P: 1,162
Quote Quote by stevendaryl View Post
Well, consider the events
e_1: the rear clock shows time 12:00
e_2: the front clock shows time 12:00

If there is no length contraction, then e_1 and e_2 are simultaneous in the launch frame. But in the frame in which the rocket is momentarily at rest, e_2 happens before e_1. So in that frame, the front clock runs ahead of the rear clock.

If there is length contraction, then it means that even in the launch frame, event e_2 happens before e_1.
I thought we were no longer talking about the case of non contraction.
And I agreed with your thought regarding the desynchronization relative to the momentarily comoving frame. I disagreed with your assumption that that effect would dimish over time.
IMO it would increase throughout the complete course of acceleration. Why do you think it would diminish?
Austin0
#30
May27-12, 12:19 AM
P: 1,162
Quote Quote by stevendaryl View Post
I'm not exactly sure what you are saying. But if I did the calculation correctly, the exact result for the time τ on a clock aboard a Born-rigid rocket of constant proper acceleration is this:

τ = R/c arcsinh(ct/R)

where t is the time in the launch frame, and where R is a distance computed by:

R = c2/a + x

where x = the height of the clock above the rear of the rocket, and a = the acceleration felt by the rear of the rocket.
I wish I could handle the math. I have essentially completely forgotten basic trig and never knew the hyperbolic functions.
So did you derive this from your calculations involving the momentarily comoving inertial frames? Aren't these the Rindler coordinate functions?
jartsa
#31
May27-12, 01:09 AM
P: 476
Quote Quote by Austin0 View Post
Hi Take a look at yuiop's diagrams a couple of posts back. Compare the relative path lengths between forward and backward travel and see if you still think the front to back travel would have greater effect on elapsed time than the travel from back to front.

I have no idea what the diagrams mean.

I think we can say this:

In all cases when the clock did make extra ticks, because of the visit it made upstairs, in these cases the bacward travel had greater effect on elapsed time.
jartsa
#32
May27-12, 01:34 AM
P: 476
Quote Quote by stevendaryl View Post
Okay, I did some calculations and I think I have a good grasp now on the clock discrepancies in the accelerated rocket. There are two effects going on:
  1. Because of relativity of simultaneity, if two clocks are synchronized in the "launch" frame, then the forward clock will be farther ahead, as measured in the instantaneous rest frame.
  2. The front clock is traveling slightly slower than the rear clock (as measured in any inertial frame), and so experiences slightly less time dilation.

Here's what's interesting: Early on, effect number 1 is the most important, and effect number 2 is negligible. Much later, when the rocket is traveling very close to the speed of light, effect number 2 is most important, and effect number 1 is negligible.


OK. But effect 1 is not a time dilation effect. Clocks aren't in anyway effected by effect 1.

In our accelerating rocket case clocks are effected only by velocity time dilation.


I have tried to suggest a very simpe thing: After the rocket has accelerated and stopped accelerating, somebody carries a clock from the rear of the rocket to the front of the rocket, where he compares the clock to a clock that was at the front of the rocket all the time, and he observes that the clocks have ticked at different rates, which he may explain by gravitational time dilation, while an outside observer must explain it by velocity time dilation, which was caused by two things: 1: length contraction 2: carrying
stevendaryl
#33
May27-12, 07:29 AM
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P: 2,150
Quote Quote by Austin0 View Post
I thought we were no longer talking about the case of non contraction.
Well, I was trying to see the contributions of length contraction and relativity of simultaneity to the discrepancy between the front and rear clocks in the comoving frame. Obviously, if there is no length contraction, then it is all due to relativity of simultaneity.

And I agreed with your thought regarding the desynchronization relative to the momentarily comoving frame. I disagreed with your assumption that that effect would dimish over time.
IMO it would increase throughout the complete course of acceleration. Why do you think it would diminish?
Okay, let's pick a time t after the two clocks have been accelerating. Let the event e1 be an event taking place at the rear clock at time t, and let e2 be an event taking place at the front clock at time t. Let the coordinates of e1 be (x1,t) and let the coordinates of e2 be (x2,t), as measured in the launch frame. Let the corresponding coordinates in the momentary inertial rest frame be (x1',t1') and (x2',t2'). Letting δt' be the difference between t1' and t2', and letting δx be the difference between x1 and x2, the Lorentz transforms tell us that:

δt' = γ (δt - v/c2 δx)

We chose the two events so that they are simultaneous in the launch frame, so δt = 0. So we have:

δt' = -γ v/c2 δx

But δx is the distance between the front and the rear, as measured in the launch frame. So by length contraction, that is L/γ, where L is the length of the rocket in its comoving frame. So we have:

δt' = -γ v/c2 L/γ
= - v/c2 L

As time goes on, v→c, so this expression approaches

δt' = -L/c

So the desynchronization effect doesn't keep growing, it approaches a fixed constant (which happens to be the length of time required for light to travel from the rear to the front, in the comoving frame; hmm, not sure what the significance of that is).

In contrast, the discrepancy due to length contraction keeps getting bigger and bigger.
DaleSpam
#34
May27-12, 07:34 AM
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P: 17,343
Quote Quote by Austin0 View Post
Do you know of any treatment done through Doppler that has been worked out for review??

The brute force approach as you called it. It certainly seems messy in principle and hard to interpret but I would like to know the method of approach.
I posted it a few years back in this post, but the LaTeX has changed and so now it is all math processing errors:
http://www.physicsforums.com/showthr...=236880&page=4

I will try to re-work it and re-post it here.
stevendaryl
#35
May27-12, 07:56 AM
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P: 2,150
Quote Quote by Austin0 View Post
I wish I could handle the math. I have essentially completely forgotten basic trig and never knew the hyperbolic functions.
So did you derive this from your calculations involving the momentarily comoving inertial frames? Aren't these the Rindler coordinate functions?
Yeah, it's the Rindler coordinates. The Rindler coordinates X and T relate to the coordinates x and t of the launch frame via:

X = √(x2 - (ct)2)

T = c/a arctanh(ct/x)
= c/a arcsinh(ct/X)

where a = the acceleration felt by the rear of the rocket, and where x is not measured from the rear of the rocket, but is instead measured so that the rear of rocket is at x=c2/a; that just makes the formulas come out nicer).

T is not the same as the time shown on clocks aboard the accelerated rocket; that time, τ, is related to T through

τ = T aX/c2
= X/c arcsinh(ct/X)
stevendaryl
#36
May27-12, 09:45 AM
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P: 2,150
Quote Quote by DaleSpam View Post
I posted it a few years back in this post, but the LaTeX has changed and so now it is all math processing errors:
http://www.physicsforums.com/showthr...=236880&page=4

I will try to re-work it and re-post it here.
It seems like the Doppler derivation is the worst of all possible worlds, as far as difficulty; you have to take into account everything: the changing distances between the two clocks, the time dilations experienced by the clocks, and the transit time for light.

Of course, right after launch, you can compute the Doppler shift approximately by assuming that the clocks are still moving nonrelativistically.


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