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A question about the equivalence principle. 
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#19
May2512, 10:54 PM

P: 476

Let's say we have a nonlengthcontracting 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. 


#20
May2612, 12:18 AM

P: 1,162




#21
May2612, 12:57 AM

P: 476

We did have an accelerating and a nonlengthcontracting 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. 


#22
May2612, 11:07 AM

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#23
May2612, 06:55 PM

P: 1,162




#24
May2612, 08:34 PM

P: 1,162

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/s^{2} 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 PoundRebka 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^{10}is 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 


#25
May2612, 09:25 PM

Sci Advisor
P: 2,118

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:
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. 


#26
May2612, 10:13 PM

P: 1,162

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. 


#27
May2612, 11:41 PM

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P: 2,118

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. 


#28
May2612, 11:57 PM

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P: 2,118

τ = R/c arcsinh(ct/R) where t is the time in the launch frame, and where R is a distance computed by: R = c^{2}/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. 


#29
May2712, 12:13 AM

P: 1,162

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? 


#30
May2712, 12:19 AM

P: 1,162

So did you derive this from your calculations involving the momentarily comoving inertial frames? Aren't these the Rindler coordinate functions? 


#31
May2712, 01:09 AM

P: 476

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. 


#32
May2712, 01:34 AM

P: 476

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 


#33
May2712, 07:29 AM

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P: 2,118

δt' = γ (δt  v/c^{2} δx) We chose the two events so that they are simultaneous in the launch frame, so δt = 0. So we have: δt' = γ v/c^{2} 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/c^{2} L/γ =  v/c^{2} 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. 


#34
May2712, 07:34 AM

Mentor
P: 17,317

http://www.physicsforums.com/showthr...=236880&page=4 I will try to rework it and repost it here. 


#35
May2712, 07:56 AM

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P: 2,118

X = √(x^{2}  (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=c^{2}/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/c^{2} = X/c arcsinh(ct/X) 


#36
May2712, 09:45 AM

Sci Advisor
P: 2,118

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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