
#1
Jun2905, 02:01 PM

P: 214

Here is the system of interest, S is considered at rest and S' is moving wrt to S as shown:
...a............v.............a .>><. S................................S' The magnitude of constant acceleration is given by a, and the magnitude of constant velocity is given by v. S' undergoes constant acceleration from the rest position in the frame S. At the end of the journey, the S' frame is once again at rest in the S frame. Will the time shown on the clock of S' be less than, greater than, or equal to the time shown on the clock of S when S' finishes it's journey. If it cannot be determined, then why not? EDIT: (This is a clarification of the diagram as described in a post below) S is an inertial reference frame, all accelerations/velocities of S' are measured wrt to S. S' starts with v(t=0)=0 from x=x'=0 at t=t'=0 and has a constant acceleration, a. Once S' has reached a velocity of v relative to S, it has an accleration of 0. After sometime, S' is given the acceleration, a, as indicated by the reversed arrow. It's acceleration is stopped once it is once again in the rest frame of S, hence v(t=tf)=0. 



#2
Jun2905, 02:20 PM

P: 1,294

For the record, this is the twin paradox and has been discussed to death.
The standard answer is: Because of the nonzero acceleration a, S' is motion relative to S (and not the other way around because S' cannot be considered an inertial frame). Moving clocks run slower, so the time shown on the clock in S' is less then the time shown in S. Personally I don't think this is correct, I think that the motion is symmetric in all respects and that S and S' will read the exact same time at the end of the motion. 



#3
Jun2905, 02:41 PM

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#4
Jun2905, 02:41 PM

P: 214

A simple relativity questionIf my assessment of your argument is correct, then I would have to conclude that you believe there is a preferred frame for which to make measurements for any given two frames S and S' and that this preffered frame is not neccessary either S or S' but can only be found through symmetry. I do not accept this. 



#5
Jun2905, 03:07 PM

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P: 7,433

I'm not 100% sure of the details of the problem (is S inertial? Why does it have an acceleration arrow by it in that case)  but there is a very easy answer.
Since S and S' are apparently not starting out at the same point in space, the answer depends on how you synchronize their clocks initially. 



#6
Jun2905, 03:13 PM

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P: 8,470

their clock readings at the moment they are at rest relative to each other, because of the relativity of simultaneitydifferent frames will disagree about what each clock reads at the moment they come to rest relative to each other. But if Aer is just asking about what the two clocks will read in the rest frame of S, and if my interpretation of them starting out at the same point in space is right, then S' will be behind S at the moment S' comes to rest in this frame. 



#7
Jun2905, 03:31 PM

P: 214

To be thorough, I will describe the diagram in words. S' starts with v(t=0)=0 from x=x'=0 at t=t'=0 and has a constant acceleration, a. Once S' has a velocity of v relative to S, it has an accleration of 0. After sometime, S' is given the acceleration, a, as indicated by the reversed arrow. It's acceleration is stopped once it is once again in the rest frame of S, hence v(t=tf)=0. Now since both frames are at rest wrt each other, there should not be a problem with S' sending a signal to S telling what it's time is at any time that it's relative velocity is 0 to S. Now my question is, will the time displayed by the clock of S' being less than the time displayed by the clock of S. In short I think the answer according to relativity is "yes", but I am merely asking for the official answer from anyone who has experience with these matters. 



#8
Jun2905, 04:16 PM

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#9
Jun2905, 06:16 PM

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However, if you use the U definition of simutaneity, where U is some other inertial frame moving with respect to S, the answer may not be the same. If one is to compare two clocks at different spatial positions, the method of comparison must be specified in detail, different methods will give different results. 



#10
Jun2905, 06:44 PM

P: 1,294

I don't mean to venture too far outside the mainstream, but I was trying to present an alternative answer that I prefer. 



#11
Jun2905, 10:18 PM

P: 214

Let me specify in detail how the clock comparison should be done. After S' has stopped decelerating, it sends a signal to S with the current time according to S' encoded. When S recieves this signal, it immediately returns a reply signal with the current time according to S encoded. When S' recieves this signal, it can then compute the spatial positions of S and S' based on the delay of the return signal. This recieve and reply signaling can continue on until the controlling frame (let's say S) encodes a message to stop. Would comparing the clocks in this manner work? 



#12
Jun3005, 01:43 AM

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Yes, that method of comparing clocks would work, and yield the same result as the usual method of noting the time on each clock when a flash or signal is emitted from the midpoint of S and S'.
The difficulty in synchronization arises when one introduces a moving observer, an inertial observer who has some nonzero velocity with respect to both S and S'. He has a different notion of simultaneity then the observer at S. 



#13
Jun3005, 10:31 AM

P: 214

Now here is the point I really wish to resolve. The clock of S' during it's motion as described in the diagram must be able to record a time when specific events happen such as the change from a constant acceleration, a, to no acceleration. I do not care necessarily that we are unable to compare clocks in each frame S and S' since that involves an issue of simultaneity. However, what I do wish to do is build up a "time profile" of the S' clock according to what it perceives the time to be and how this then matches up with S' agreeing with S what the clock S' should read at the end of it's journey. The first issue in doing this is to look at the acceleration, a and a. They have the same magnitude, a, and occur over the same interval (as recorded by S, I don't know what S' would say about these two intervals) but in opposite directions. Would S' percieve these two accelerations to occur over the same time interval? 



#14
Jun3005, 11:51 AM

P: 308





#15
Jun3005, 01:02 PM

P: 214





#16
Jun3005, 01:51 PM

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#17
Jun3005, 02:55 PM

P: 214

With constant acceleration, v(t) will be: v(t) = (vf  v0)*t/t1 + v0 The above formula shouldn't need any explaination. Now when we plug that in and integrate, I am claiming without showing any steps (for the sake of brevity) that the time as percieved in the accelerating frame is: t' = t1 * c/(2*(vfv0)) * [ arcsin(vf/c)  arcsin(v0/c) + (vfv0)*(c^2v0^2)^(1/2)/c^2 ] EDIT: I did in fact make a error reading the equation on my calculator, the above equation should be: [tex] t' = t1\frac{c}{2(vfv0)} \left( {sin}^{1}\left(\frac{vf}{c}\right){sin}^{1}\left(\frac{v0}{c}\right)+\frac{vf}{{c}^{2}}\sqrt{{c}^{2}{vf}^{2}}\frac{v0}{{c}^{2}}\sqrt{{c}^{2}{v0}^{2}} \right) [/tex] If you disagree with this, then I will try to prove it in detail in another post as it is possible that I made a mistake. Now from the diagram of the problem in question, let's assume numbers for the various values: c=1 v0=0 vf=.9 t1=2 These numbers yield t'=1.6801 for the 2nd acceleration case (a), the numbers follow from above as: c=1 v0=.9 vf=0 t1=2 These numbers yield t'=1.6801 I made a mistake originally, but the results should now be correct as they predict the same time interval. 



#18
Jun3005, 03:13 PM

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http://math.ucr.edu/home/baez/physic...SR/rocket.html Using this timereversal approach (which is allowed because physical laws are symmetrical under time reversal), we can say that the proper time interval representing the acceleration phase is the same as the proper time interval of the decelleration phase, as one would intiutively guess. There are some things you have to beware of. The equations given don't address the coordinate system of a person in the rocket. This turns out to be a trickier problem than it first appears. (For one thing, under some circumstances it doesn't even exist. Look up "Rindler horizon" sometime, I've talked about this a lot before in this forum, and there is some info on the web as well). Effects due to the rocket not being a single point are not modelled by the relativistic rocket equations, they require further discussion and study. As long as you are aware of this limitiation, though, the relativistic rocket equation should answer a lot of your questions. 


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