# A simple relativity question

1. Jun 29, 2005

### Aer

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.

Last edited: Jun 29, 2005
2. Jun 29, 2005

### Crosson

For the record, this is the twin paradox and has been discussed to death.

The standard answer is: Because of the non-zero 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. Jun 29, 2005

### JesseM

If you think the motion is symmetric, would you deny that S' feels a constant G-force due to acceleration while S experiences weightlessness?

4. Jun 29, 2005

### Aer

Let me see if I understand your argument. First of all let me state that I am unsure as to whether you are claiming you disagree with relativity or that you are claiming that relativity will predict something other than the "standard answer". Your argument seems to be that if a system of motion between S and S' can be viewed as symmetrical in some other rest frame besides that of S or S', then this "other" frame is the only frame (let's call it the preferred frame) in which calculations of time can be made and any other frame of reference will give a false result because S and S' will not show the exact same time.

If 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. Jun 29, 2005

### pervect

Staff Emeritus
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. Jun 29, 2005

### JesseM

When I read the problem, I thought S and S' started out at the same point in space (with synchronized clocks, presumably) but with a large relative velocity, and S' is experiencing constant acceleration in the opposite direction so eventually he comes to rest relative to S. If that's the case, then you can't really answer the question of the difference between
their clock readings at the moment they are at rest relative to each other, because of the relativity of simultaneity--different 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. Jun 29, 2005

### Aer

I guess my diagram is not the clearest. Frame S is an inertial frame, in fact it is the rest frame for which all other velocities/accelerations are relative to. The frame S' starts out at the same location as the frame S (the origin of the x and x' axis coincide at t=t'=0). All dashed arrows in the diagram are attributed to the motion of the frame S'.

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. Jun 29, 2005

### JesseM

In the rest frame of S, yes. In other inertial frames, maybe not. Different frames disagree about whether a given event (such as a clock striking a particular time) happens "at the same time" as another event at a different location (such as another distant clock striking another time), this is called "relativity of simultaneity".

Last edited: Jun 29, 2005
9. Jun 29, 2005

### pervect

Staff Emeritus
Using the S definition of simultaneity, in which both S and S' are at rest at the end of your acceleration period, the answer is yes, the accelerated clock will read a shorter elapsed time than the stationary clock.

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. Jun 29, 2005

### Crosson

I guess my thoughts about the issue have to do with Mach's principle. Mach's Principle basicly says that g-force is due to acceleration relative to the distant stars (a gravitational effect).

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

11. Jun 29, 2005

### Aer

Are you referring to atmospheric effects and whatnot? I do not wish to complicate the problem in such a manner. The "experiment" should be considered to be conducted in a vacuum.

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. Jun 30, 2005

### pervect

Staff Emeritus
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 non-zero velocity with respect to both S and S'. He has a different notion of simultaneity then the observer at S.

13. Jun 30, 2005

### Aer

I have not mentioned any desire to synchronize the clocks, that is make them display the same time according to each other. I assume when you refer to synchronization here, that you mean the comparison of each clock at the same instant and that both clocks must agree that this instant is in fact simultaneous.

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?

Last edited: Jun 30, 2005
14. Jun 30, 2005

### Zanket

Two links you may find helpful: This one is all about the twin paradox, which your story problem is. This one has the fairly simple equations for (noninertial) acceleration in special relativity, so you can answer questions with actual calculations. Using the equations might be easier if you change your story problem so that there is no constant velocity portion. Instead, have S’ accelerate to the halfway point and then decelerate to the finish point.

Yes, this is given by symmetry as you imply.

Last edited: Jun 30, 2005
15. Jun 30, 2005

### Aer

Thank you for the links. Regarding the equations in the 2nd link, "The acceleration of the rocket must be measured at any given instant in a non-accelerating frame of reference travelling at the same instantaneous speed as the rocket". Therefore those equations do not apply to the case presented since the constant acceleration is measured in the frame S.
Is this still true when you consider that frame S' will not see itself undergoing constant acceleration?

16. Jun 30, 2005

### JesseM

Yes, you can see this without having to think about accelerated frames, just use frame S to predict how much time will elapse on the ship's clock between the beginning and end of each acceleration period. As discussed in this section of the twin paradox page, if from the point of view an inertial frame you see an accelerating clock whose velocity as a function of time is given by v(t), then to find the time elapsed between two times t0 and t1 in that frame, just integrate $$\int_{t0}^{t1} \sqrt{1 - v(t)^2 / c^2} \, dt$$. Since the amount of time that both accelerations last is the same, and the v(t) functions will be mirror opposites, the elapsed time on the ship's clock from the beginning to the end of each acceleration will be the same.

17. Jun 30, 2005

### Aer

I assume t0 and t1 are measured in frame S. In that case, let's set t0=0 so that t1 will be the time interval as seen by frame S. In order to integrate that equation, we must first determine v(t).

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*(vf-v0)) * [ arcsin(vf/c) - arcsin(v0/c) + (vf-v0)*(c^2-v0^2)^(1/2)/c^2 ]

EDIT: I did in fact make a error reading the equation on my calculator, the above equation should be:

$$t' = t1\frac{c}{2(vf-v0)} \left( {sin}^{-1}\left(\frac{vf}{c}\right)-{sin}^{-1}\left(\frac{v0}{c}\right)+\frac{vf}{{c}^{2}}\sqrt{{c}^{2}-{vf}^{2}}-\frac{v0}{{c}^{2}}\sqrt{{c}^{2}-{v0}^{2}} \right)$$

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.

Last edited: Jun 30, 2005
18. Jun 30, 2005

### pervect

Staff Emeritus
Yes, I was using comparison and synchronization as if they were synonomous. Sorry for any confusion.

What you probably want then is the equations for the relativistic rocket

http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html

You can use the relativistic rocket equations to do this, but because they aren't written with an initial velocity, you need to be just a little bit tricky. On the outgoing trip, you use an inertial frame centered on the rocket's starting point. During the constant velocity phase, you use the Lorentz transforms. During the deceleration phase, you use an inertial frame centered on the rocket's stopping point, and reverse time.

Using this time-reversal 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.

19. Jun 30, 2005

### Aer

The rocket equation was brought forth by Zanket in which I replied: "The acceleration of the rocket must be measured at any given instant in a non-accelerating frame of reference travelling at the same instantaneous speed as the rocket". Therefore those equations do not apply to the case presented since the constant acceleration is measured in the frame S.

The text in quotes is directly from the page containing the rocket equation.

20. Jun 30, 2005