Can Time Dilation be Determined in a Constantly Accelerating Frame?

[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.

 

[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.

 

[JesseM]

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.

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

 

[Aer]

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.

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 necessary either S or S' but can only be found through symmetry.

I do not accept this.

 

[pervect]

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.

 

[JesseM]

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.

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.

 

[Aer]

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.

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.

 

[JesseM]

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 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".

 

[pervect]

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.

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.

 

[Crosson]

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

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.

 

[Aer]

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.

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 receives this signal, it immediately returns a reply signal with the current time according to S encoded. When S' receives this signal, it can then compute the spatial positions of S and S' based on the delay of the return signal. This receive 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?

 

[pervect]

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.

 

[Aer]

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.

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?

 

[Zanket]

INow here is the point I really wish to resolve.

Two links you may find helpful: This one is all about the twin paradox, which your story problem is. http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html 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.

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?

Yes, this is given by symmetry as you imply.

 

[Aer]

Two links you may find helpful: This one is all about the twin paradox, which your story problem is. http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html has the fairly simple equations for (noninertial) acceleration in special relativity, so you can answer questions with actual calculations.

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 traveling 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.

Yes, this is given by symmetry as you imply.

Is this still true when you consider that frame S' will not see itself undergoing constant acceleration?

 

[JesseM]

Is this still true when you consider that frame S' will not see itself undergoing constant acceleration?

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.

 

[Aer]

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.

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 explanation. Now when we plug that in and integrate, I am claiming without showing any steps (for the sake of brevity) that the time as perceived 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&#039; = 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.

 

[pervect]

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.

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

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.

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

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

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?

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 modeled 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.

 

[Aer]

As long as you are aware of this limitiation, though, the relativistic rocket equation should answer a lot of your questions.

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 traveling 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.

 

[Ich]

Your equation doesn´t work.

t' = t1 * c/(2*(vf-v0)) * [ arcsin(vf/c) - arcsin(v0/c) + (vf-v0)*(vf^2-v0^2)^(1/2)/c^2 ]

(vf^2-v0^2)^(1/2) will be imaginary.

 

[pervect]

I didn't realize you wanted a constant acceleration in frame S, I thought you wanted a constant "proper" acceleration.

A constant accelration in frame S will be impossible after a certain amount of time, because you can't acclerate beyond the speed of light.

You should able to use the invariance of the Lorentz interval to calculate the amount of proper time that elapses on the rocket.

With 't' being the coorinate time of the inertial observer, and \tau being the proper time of the accelerated observer, we can write

d\tau = \sqrt{1-(at/c)^2}dt

Plugging this into my favorite symbolic integrator, we get

<br /> \tau = K+\frac{1}{2}\,t\sqrt {1-{\frac {{a}^{2}{t}^{2}}{{c}^{2}}}}+\frac{1}{2}\,\arctan \left( <br /> \sqrt {{\frac {{a}^{2}}{{c}^{2}}}}t{\frac {1}{\sqrt {1-{\frac {{a}^{2}<br /> {t}^{2}}{{c}^{2}}}}}} \right) \frac{c}{a}<br /> <br /> <br />

where K is any constant, K=0 if t=0 when tau=0.

a*t/c must be less than 1, of course. I'm not sure what the "felt" or "proper" acceleration would be, but I'm not sure if anyone is interested either.

 

[Aer]

Your equation doesn´t work.

(vf^2-v0^2)^(1/2) will be imaginary.

You are right, that wouldn't work and is copy error on my part, it should be (c^2-v0^2)^(1/2)

perhaps I should have used the copy and paste feature instead of retyping it out, from my MATLAB code, I have:
t=t1*c/(2*(vf-v0))*(asin(vf/c)-asin(v0/c)+(vf-v0)/c^2*(c^2-v0^2)^.5)

 

[Aer]

<br /> \tau = K+\frac{1}{2}\,t\sqrt {1-{\frac {{a}^{2}{t}^{2}}{{c}^{2}}}}+\frac{1}{2}\,\arctan \left( <br /> \sqrt {{\frac {{a}^{2}}{{c}^{2}}}}t{\frac {1}{\sqrt {1-{\frac {{a}^{2}<br /> {t}^{2}}{{c}^{2}}}}}} \right) \frac{c}{a}<br /> <br /> <br />

where K is any constant, K=0 if t=0 when tau=0.

From your equation of v(t)=at, v0 is necessarily equal to 0. I did this same integration in post #17 but using v(t) = (vf - v0)*t/t1 + v0 and assuming t0=0 so t1 is the time interval in the frame S.

By the way, what symbolic integrator are you using? I had to use a TI-89 which gave it in a funky way which I simplified to what is shown.

 

[Zanket]

Is this still true when you consider that frame S' will not see itself undergoing constant acceleration?

No, sorry, I didn't notice that in your scenario.

 

[pervect]

From your equation of v(t)=at, v0 is necessarily equal to 0. I did this same integration in post #17 but using v(t) = (vf - v0)*t/t1 + v0 and assuming t0=0 so t1 is the time interval in the frame S.

I probably ought to go back and read this thread more carefully, I guess, I've been a bit preocuppied with some other threads (which have since wound down). I see that JesseM has also already derived a very similar formula, so this is about the third time the same (or a very similar) formula has been posted to the thread.

Note you can still apply this simpler formula simply by choosing the point at which the object stops as your reference point, and by inverting time, as I mentioned in a previous post.

I'm not quite sure why we are interested in a constant value of coordinate acceleration, rather than the constant physical (proper) acceleration as given by the rocket equation. A constant value of coordinate acceleration is definitely non-physical, it cannot be maintained indefinitely even in theory.

By the way, what symbolic integrator are you using? I had to use a TI-89 which gave it in a funky way which I simplified to what is shown.

I used an old version of Maple. It's output was a bit funky too, it divided the result by sqrt(a^2/c^2) rather than mutiplying it by c/a, so I massaged it just a little bit. I also expressed 1/2 via a call to \frac{}{} rather than the straight division the direct latex output function gave me.

 

[Aer]

If the acceleration portion of S' is well defined as given above, then the constant velocity portion can be analyzed. First of all I want to state that I am only concerned with the proper time as seen by S'. It is my understanding that the times computed above, t', is the proper time that S' will see between the intervals of acceleration.

Now, as someone had pointed out earlier, the constant velocity portion should be given by the lorentz transformation. However, if we ignore the acceleration parts and only look at S' in it's constant velocity portion, then S' can be thought of as an inertial frame during this portion. And the time it records from a=0 to right before a=(0-.9)/2 = -.45 should be given in it's proper time. Also, during this time, it would perceive the S frame clock as slower than it's own.

Obviously this appears to be a point of confusion for me, because people always seem to claim that during the acceleration to the rest frame of S, the clock of S' will match what an observer in the frame of S says it should be. But it seems that this depends on how long S' is moving at constant velocity and the time interval for the acceleration portion was in no way shown to be dependent on this.

 

[Aer]

I'm not quite sure why we are interested in a constant value of coordinate acceleration, rather than the constant physical (proper) acceleration as given by the rocket equation. A constant value of coordinate acceleration is definitely non-physical, it cannot be maintained indefinitely even in theory.

It was how I originally defined the problem. I do see your point that it eventually becomes physically impossible.

 

[pervect]

Obviously this appears to be a point of confusion for me, because people always seem to claim that during the acceleration to the rest frame of S, the clock of S' will match what an observer in the frame of S says it should be. But it seems that this depends on how long S' is moving at constant velocity and the time interval for the acceleration portion was in no way shown to be dependent on this.

Remember that there is a constant of integration, K, in the formula that relates tau to t. (Here t is the coordinate time of the observer in S, and tau is the proper time of a clock on the accelerating rocket).

The formula we all derived earlier relates dt to dtau. The integrated formula always incorporates a constant of integration that must be applied to match the boundary conditions. If you note, I specified that t=0 when tau=0 to derive the boundary conditons that I used.

It sounds to me like you are confused because you are not including the proper value of K in your formula when there is "coasting". You have to adjust K, the constant of integration, so that the boundary condition (t,tau) satisfies the formula at the instant that the deacceleration begins. When you use the proper value of K, the formula will give the correct answers as long as the object follows the specified curve.

To use the simple approach, you just redefine t=0 and tau=0 to be the time at which the rocket stops. You can then use the principle of time reversal and the simple formula that I and JesseM derived to find -tau as a function of -t.

 

[Aer]

Remember that there is a constant of integration, K, in the formula that relates tau to t. (Here t is the coordinate time of the observer in S, and tau is the proper time of a clock on the accelerating rocket).

Ok, very well, so my equation should be t' = t*(...) + K
If I want t'=0 at t=0, then isn't K necessarily 0? In other words, are you contesting the results that I've come up with in post 17

To use the simple approach, you just redefine t=0 and tau=0 to be the time at which the rocket stops. You can then use the principle of time reversal and the simple formula that I and JesseM derived to find -tau as a function of -t.

What formula did JesseM derive?

 

[Aer]

Ok, very well, so my equation should be t' = t*(...) + K
If I want t'=0 at t=0, then isn't K necessarily 0? In other words, are you contesting the results that I've come up with in post 17

OK, I get K=0 for the first case and K=1.6801 for the second case. Here is the MATLAB code (equations for K and t' are given as k and t respectively):
t1=2;
c=1;

v0=0;
vf=.9;
k=-1/c*((c^2-v0^2)^.5*v0+asin(v0/c)*c^2)*t1/(2*(vf-v0))
t=t1*c/(2*(vf-v0))*(asin(vf/c)-asin(v0/c)+vf/c^2*(c^2-vf^2)^.5-v0/c^2*(c^2-v0^2)^.5)+k

v0=.9;
vf=0;
k=-1/c*((c^2-v0^2)^.5*v0+asin(v0/c)*c^2)*t1/(2*(vf-v0))
t=t1*c/(2*(vf-v0))*(asin(vf/c)-asin(v0/c)+vf/c^2*(c^2-vf^2)^.5-v0/c^2*(c^2-v0^2)^.5)+k

And here are the results:
k =
0
t =
1.6801
k =
1.6801
t =
3.3602

EDIT: Something seems to be wrong here, the second t is twice what it should be.

 

[Aer]

Remember that there is a constant of integration, K, in the formula that relates tau to t. (Here t is the coordinate time of the observer in S, and tau is the proper time of a clock on the accelerating rocket).

Here are the results I get with your equation

MATLAB Code:

c=1;

tau=0;
t=0;
a=.45;
K = tau - 1/2*t*(1-(a*t/c)^2)^.5 +1/2*atan((a^2/c^2)^.5*t/(1-(a*t/c)^2)^.5)*c/a
t=2;
tau = K + 1/2*t*(1-(a*t/c)^2)^.5 +1/2*atan((a^2/c^2)^.5*t/(1-(a*t/c)^2)^.5)*c/a

tau=0;
t=0;
a=.45;
K = tau - 1/2*t*(1-(a*t/c)^2)^.5 +1/2*atan((a^2/c^2)^.5*t/(1-(a*t/c)^2)^.5)*c/a
t=-2;
tau = K + 1/2*t*(1-(a*t/c)^2)^.5 +1/2*atan((a^2/c^2)^.5*t/(1-(a*t/c)^2)^.5)*c/a


Results:

K =

0


tau =

1.6801


K =

0


tau =

-1.6801

 

[JesseM]

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 explanation. Now when we plug that in and integrate, I am claiming without showing any steps (for the sake of brevity) that the time as perceived in the accelerating frame is

You're doing a lot of unecessary work if you actually integrate that--it's enough to notice that the period of acceleration is the same in both cases, and the magnitude of the acceleration is also the same, so it's obvious that if the first acceleration is from t0 to t1 and the second is from t1 to t2, then the velocity at t0+t during the first period will be the same as the velocity at t2-t; so, obviously the integral will be the same in both cases.

But if you want to do the integral, then it might be easier in this form:

for the outbound period from t0 to t1, v(t) = a*(t - t0) + v0
for the inbound period from t1 to t2, v(t) = -a*(t2 - t) - v0

So, the first integral is:

\int^{t_1}_{t_0} \sqrt{1 - (a*(t - t_0 ) + v_0 )^2 / c^2 } \, dt

and the second is:

\int^{t_2}_{t_1} \sqrt{1 - (-a*(t_2 - t ) - v_0 )^2 / c^2 } \, dt
= \int^{t_2}_{t_1} \sqrt{1 - ((a*(t_2 - t ) + v_0)*(-1))^2 / c^2 } \, dt
= \int^{t_2}_{t_1} \sqrt{1 - (a*(t_2 - t ) + v_0)^2 / c^2 } \, dt
= \int^{t_1}_{t_0} \sqrt{1 - (a*(t_1 - t ) + v_0)^2 / c^2 } \, dt
= \int^{t_0}_{t_1} \sqrt{1 - (a*(t - t_0 ) + v_0)^2 / c^2 } \, dt


So you can see why these integrals will come out the same.

 

[pervect]

The advantage of my approach is precisely that K does equal zero, because t=0 when tau=0.

The end result is that you have a peak time dilation factor of 2.29 (with v=.9c). The "average" time dilation factor for the acceleration period is much more modest, somewhere around 1.2 if your numbers are correct (2/1.68). This does not seem unreasonable, you spend most of your time at low velocities.

If you coast, though, the full time dilation factor of 2.2 will apply during the coasting period.

 

[Aer]

The end result is that you have a peak time dilation factor of 2.29 (with v=.9c). The "average" time dilation factor for the acceleration period is much more modest, somewhere around 1.2 if your numbers are correct (2/1.68). This does not seem unreasonable, you spend most of your time at low velocities.

Very well, are these proper times in their respective frames? That is while S sees S' accelerating, 2 units of time elapsed and while S' knows it is accelerating, it records 1.68 units of time to have elapsed.

If you coast, though, the full time dilation factor of 2.2 will apply during the coasting period.

If we consider two inertial frames moving at .9c wrt each other, then we clearly cannot distinguish between which is at rest - or rather we are free to choose either of them to be at rest.

So the proper time of each frame is measured according to each frame assuming that itself is at rest and the other frame is moving. Therefore each frame will conclude that less time has passed for the other frame.

If this is true, then how can we measure the proper time of S', while it is not accelerating, in the frame of S.

First of all let me state at this point that the explanation that has transpired (unanimously from all those who have responded) for how S' will measure proper time does not agree with the explanation given by a physics professor (as memory serves, since this was over 4 years ago). But I do not wish to get into that now. I'll assume your explanation is correct.

Now let me reiterate my point of contention. You do not explicitly state that you are measuring the proper time of S' when you say

If you coast, though, the full time dilation factor of 2.2 will apply during the coasting period.

but I assume you are responding to my question about what will S' measure the proper time to be between the intervals of acceleration, that is the interval under which it has constant velocity. I explicitly stated the following so that there would be no confusion:

If the acceleration portion of S' is well defined as given above, then the constant velocity portion can be analyzed. First of all I want to state that I am only concerned with the proper time as seen by S'.

Your answer seems contradictory to the rule of Special Relativity that I cited above: "So the proper time of each frame is measured according to each frame assuming that itself is at rest and the other frame is moving. Therefore each frame will conclude that less time has passed for the other frame." The time each frame concludes passes for the other frame is not the proper time of that frame since proper time by definition is measured in a frame at rest for the said frame at rest.

I guess my question for you is this, during the time of constant velocity and during this time only, are you saying that S' is not an inertial frame?

 

[JesseM]

Suppose S' ends up coasting at v=0.9c in the S-frame. You can certainly analyze the entire problem from the point of view of an inertial observer who moves at 0.9c relative to S throughout the entire experiment; during the coasting period, this observer will say that the clock of S' ticks at the same rate as his own clock while the clock of S is slowed down, but when the entire trip is considered, he'll get the same answer to the question about whose clock is behind when S and S' reuinite.

 

[Aer]

Suppose S' ends up coasting at v=0.9c in the S-frame. You can certainly analyze the entire problem from the point of view of an inertial observer who moves at 0.9c relative to S throughout the entire experiment; during the coasting period, this observer will say that the clock of S' ticks at the same rate as his own clock while the clock of S is slowed down, but when the entire trip is considered, he'll get the same answer to the question about whose clock is behind when S and S' reuinite.

This conclusion is implied from the very statement of the problem in the first post of this thread. However this conclusion does not deal in anyway with the question at hand. And that is, what will S' record as the proper time in it's own frame (redundant, I know) at the event it observes when the acceleration changes from a=.45 to a=0 and what will S' record as the proper time in it's own frame (redundant, I know) at the event it observes when the acceleration changes from a=0 to a=-.45?

Please let me know if there is any ambiguity in this question in anyway.

 

[JesseM]

This conclusion is implied from the very statement of the problem in the first post of this thread. However this conclusion does not deal in anyway with the question at hand. And that is, what will S' record as the proper time in it's own frame (redundant, I know) at the event it observes when the acceleration changes from a=.45 to a=0 and what will S' record as the proper time in it's own frame (redundant, I know) at the event it observes when the acceleration changes from a=0 to a=-.45?

Please let me know if there is any ambiguity in this question in anyway.

Yeah, I don't quite understand what you're asking. How do you measure the proper time "at" an individual event? Are you asking about how much time S' will record between the events of stopping its acceleration in one direction and starting it in the other (in other words, how long the coasting period lasts in its own frame?) If so, how long does the coasting period last in the S-frame? However long it lasts in the S-frame, it will last a shorter amount of time in the S'-frame, with the time shrunk down by a factor of \sqrt{1 - v^2/c^2}, where v is the velocity of S' relative to S.

 

[pervect]

Very well, are these proper times in their respective frames? That is while S sees S' accelerating, 2 units of time elapsed and while S' knows it is accelerating, it records 1.68 units of time to have elapsed.
If we consider two inertial frames moving at .9c wrt each other, then we clearly cannot distinguish between which is at rest - or rather we are free to choose either of them to be at rest.

So the proper time of each frame is measured according to each frame assuming that itself is at rest and the other frame is moving. Therefore each frame will conclude that less time has passed for the other frame.

If this is true, then how can we measure the proper time of S', while it is not accelerating, in the frame of S.

There are several ways of doing this, the Lorentz transform is the easiest. I thought you already knew how to do this?

<snip>

Now let me reiterate my point of contention. You do not explicitly state that you are measuring the proper time of S' when you say but I assume you are responding to my question about what will S' measure the proper time to be between the intervals of acceleration, that is the interval under which it has constant velocity. I explicitly stated the following so that there would be no confusion:

Your answer seems contradictory to the rule of Special Relativity that I cited above: "So the proper time of each frame is measured according to each frame assuming that itself is at rest and the other frame is moving. Therefore each frame will conclude that less time has passed for the other frame." The time each frame concludes passes for the other frame is not the proper time of that frame since proper time by definition is measured in a frame at rest for the said frame at rest.

I guess my question for you is this, during the time of constant velocity and during this time only, are you saying that S' is not an inertial frame?

I read this passage several times, and I still don't understand what you are after.

Let's look at the specific quesion you ask at the end first:

During the time of constant velocity, S' is not accelrating. This makes it essentially an inertial frame. If we imagine that S' is equipped with a radar, for instance, it can measure distances with its radar as long as the radar signal is emitted and returns during the interval during which S' is not accelerating. This won't be sufficient to allow S' to assign a (time,distance) coordinate pair to every point in space. We can improve S' ability in this respect by adding an observer who never accelerates that is co-located with S' during the interval when S' is not accelerating.

Now for some other points.

Your "rule of relativity" seems to be wrong as stated as well, since it states that we cannot compute something that we just computed! It is certainly possible for an inexperienced erson to incorrectly compute the proper time, (or for an experienced person to make a computational blunder), but we have just demonstrated the methods by which proper time can be computed. The proper time is, as you state, the time elapsed on a clock in its own frame, but there is absolutely no law that says that this quantity cannot be computed by someone else who is not in that frame. However, computations that do not correctly utilize relativity will not be correct.

Are you perhaps asking at this late date how an inertial observer sets up a coordiante system? Since you've been getting the right answers, I've been assuming that you knew how to do this. You don't seem to be grasping the idea that different inertial observers set up coordinate systems with different notions of simultaneity, (in spite of this being mentioned several times), so perhaps that is indeed the source of your confusion.

 

[Aer]

Yeah, I don't quite understand what you're asking. How do you measure the proper time "at" an individual event? Are you asking about how much time S' will record between the events of stopping its acceleration in one direction and starting it in the other (in other words, how long the coasting period lasts in its own frame?)

The proper time at the start of the journey was defined as t=0 to S and t'=0 to S'. Therefore, recording the proper time at these events implies recording the time interval between them.

If so, how long does the coasting period last in the S-frame? However long it lasts in the S-frame, it will last a shorter amount of time in the S'-frame, with the time shrunk down by a factor of \sqrt{1 - v^2/c^2}, where v is the velocity of S' relative to S.

Very well, if I define the coasting time to be 100 units of time in the S frame, then the S' frame will be 100(1-.9^2)^.5 = 43.6. Now that I know the proper time in the S' frame, how do I find the proper time in the S frame? (I know, this question is stupid, we found the S' proper time from the S proper time - now why are we finding the S proper time from the S' proper time)

But it is not stupid because you will have to acknowledge the contradiction. If neither frame knows which frame is going to accelerate so that their relative velocity is 0, then how can you make this a priori statement that the proper time of S' is dependent on the proper time of S at the time of acceleration. I could easily redefine the problem to have S accelerate to S' after S records a proper time of 100. What would you say the proper time of S' is for this case?

 

[JesseM]

The proper time at the start of the journey was defined as t=0 to S and t'=0 to S'. Therefore, recording the proper time at these events implies recording the time interval between them.

As I understand it, the term "proper time" only refers to time-intervals between pairs of events on a given worldline, it doesn't mean the value of the time-coordinate in the object's rest frame.

Very well, if I define the coasting time to be 100 units of time in the S frame, then the S' frame will be 100(1-.9^2)^.5 = 43.6. Now that I know the proper time in the S' frame, how do I find the proper time in the S frame?

The proper time between what two events on the S worldline? Remember, the question of what event on the S worldline happened at the "same time" as the event of S' ending his coasting will depend on which frame's definition of simultaneity you use.

But it is not stupid because you will have to acknowledge the contradiction. If neither frame knows which frame is going to accelerate so that their relative velocity is 0, then how can you make this a priori statement that the proper time of S' is dependent on the proper time of S at the time of acceleration. I could easily redefine the problem to have S accelerate to S' after S records a proper time of 100. What would you say the proper time of S' is for this case?

In this case, it would depend on what event on the S' worldline you want to use to find the proper time between it and the event of S' beginning to coast. If you use an event that happened at the "same time" that S began to accelerate in the original S-frame, you'll get one answer; if you use an event that happened at the "same time" that S began to accelerate in the S'-frame, you'll get a different answer.

 

[Aer]

Your "rule of relativity" seems to be wrong as stated as well, since it states that we cannot compute something that we just computed! It is certainly possible for an inexperienced erson to incorrectly compute the proper time, (or for an experienced person to make a computational blunder), but we have just demonstrated the methods by which proper time can be computed. The proper time is, as you state, the time elapsed on a clock in its own frame, but there is absolutely no law that says that this quantity cannot be computed by someone else who is not in that frame. However, computations that do not correctly utilize relativity will not be correct.

Do you disagree with these mathematical statements:

x' = γ ( x - vt )
t' = γ ( t - v/c^2x )

x = γ ( x' + vt' )
t = γ ( t' + v/c^2x' )

If so, can you show me a derivation of the lorentz tranformation that doesn't require this? Please not that most derivations will not explicitly show this, but nevertheless, I have not found a derivation that does not make the same assumptions that require this to be true, but then proceed to do it in a similar but not explicitly same way. This includes Einstein's derivation. I have found derivations that don't require this, but all of those have contained errors that I can prove if necessary. I did not intend this discussion to go into the issue of whether reciprocity is required by special relativity. I thought this was a generally accepted fact. Maybe you can elaborate on how the above equations are not in fact the reciprocity that special relativity requires or that these equations do not imply what "my rule" says they imply.



This link mentions the concept of reciprocity.


Here is Einstein's derivation

 

[JesseM]

Do you disagree with these mathematical statements:

x&#039; = \gamma ( x - vt )
t&#039; = \gamma ( t - v/{c}^{2}x )

x = \gamma ( x&#039; - vt&#039; )
t = \gamma ( t&#039; - v/{c}^{2}x&#039; )

If so, can you show me a derivation of the lorentz tranformation that doesn't require this?

You've got the equations wrong, they should be:

x&#039; = \gamma (x - vt)
t&#039; = \gamma (t - vx/c^2)
x = \gamma (x&#039; + vt&#039;)
t = \gamma (t&#039; + vx&#039;/c^2)

...where v is the velocity of the origin of the x',t' system as measured in the x,t system (it's assumed that the two origins are moving along each other's x-axes, and that x=0,t=0 matches up with x'=0,t'=0).

Anyway, these equations aren't required to derive the lorentz transformation, they are the equations for the lorentz transformation between the two coordinate systems. To derive them, you just have to assume that the laws of physics should look the same in both coordinate systems, and that the speed of light is always c in both coordinate systems.

 

[Aer]

You've got the equations wrong, they should be:

x&#039; = /gamma (x - vt)
t&#039; = /gamma (t - vx/c^2)
x = /gamma (x&#039; + vt&#039;)
t = /gamma (t&#039; + vx&#039;/c^2)

...where v is the velocity of the origin of the x',t' system as measured in the x,t system (it's assumed that the two origins are moving along each other's x-axes, and that x=0,t=0 matches up with x'=0,t'=0).

Yes, I realized that, it was an unintentional error

 

[Aer]

x&#039; = \gamma (x - vt)
t&#039; = \gamma (t - vx/c^2)
x = \gamma (x&#039; + vt&#039;)
t = \gamma (t&#039; + vx&#039;/c^2)

Anyway, these equations aren't required to derive the lorentz transformation, they are the equations for the lorentz transformation between the two coordinate systems. To derive them, you just have to assume that the laws of physics should look the same in both coordinate systems, and that the speed of light is always c in both coordinate systems.

I have not found a derivation that does not make the same assumptions that require this to be true, but then proceed to do it in a similar but not explicitly same way.

Sorry, I am being incoherent at the present. "but then proceed to do it in a similar but not explicitly same way" is meant that, the procedure used in the derivation is exactly the same as if those equations above were used instead for that particular part of the derivation.

In other words, we have the following from Einstein's derivation with interpretations included added by myself:

x-ct=0 [1]
x'-ct'=0 [2]
(x'-ct')=l(x-ct) [3]

Equation 3 is "fulfilled in general" which means that x and x' are not the same in equations 1 & 2 as they are in equation 3. Equations 1 & 2 desiginate the coordinates of a photon fired from x=x'=0 at t=t'=0 whereas equation 3's x & x' need not be the coordinate of the said photon (Otherwise equation 3 would simplify from 1 & 2 as 0=l*0 and be useless). Instead, equation 3 is the relative position of the photon to an object at an arbitrary postion x or x'.

assumption:
(1) x=x'=0 at t=t'=0

x'+ct'=m(x+ct) [4]

Equation 4 is an expression that accounts for the relative positions x and x' from a photon fired in the opposite direction, that is along the negative x-axis. It is the use of this equation that causes reciprocity for these two condtions: (1) frame S' moving with speed v relative to S and (2) frame S moving with speed -v relative to S'. This equation is invalid unless the following assumption is made.

assumption:
(2) Light travels at a constant speed in all directions for any inertial frame.

Equations 3 & 4 can be added and subtracted to get equation set 5:

x'=ax-bct
ct'=act-bx
where: [5]
a=(l+m)/2
b=(l-m)/2

 

[Aer]

Sorry, I didn't see this post.

As I understand it, the term "proper time" only refers to time-intervals between pairs of events on a given worldline, it doesn't mean the value of the time-coordinate in the object's rest frame.

Ok, I will refer to them as proper time intervals from now on to avoid confusion.

The proper time between what two events on the S worldline? Remember, the question of what event on the S worldline happened at the "same time" as the event of S' ending his coasting will depend on which frame's definition of simultaneity you use.

The proper time interval between when S' had no acceleration and when S' was programmed to decelerate. However S' did not decelerate because it's decelerator malfunctioned. S "instantaneously" accelerated at the moment S' was suppose to decelerate (remember, you said this should be given by the S world line). S had the intention of sabotaging the experiment... However, we are left with the result that S accelerated to the S' inertial frame. I don't think my question has been answered, tell me if there is something fundamentally wrong with the scenerio I described.

 

[JesseM]

The proper time interval between when S' had no acceleration and when S' was programmed to decelerate.

Proper time interval along whose worldline, S or S'?

However S' did not decelerate because it's decelerator malfunctioned. S "instantaneously" accelerated at the moment S' was suppose to decelerate (remember, you said this should be given by the S world line).

If it's along the S worldline, then what is the initial point that you're using in the interval? Is it the tick of the S-clock that is simultaneous with the event of S' beginning to coast in the S-frame, or in the S'-frame?

For any definite pair of events on a given worldline of an observer who is moving inertially between those events, the proper time between the events will be larger than the time between the events as measured in every other frame. But I don't understand how you think this leads to a contradiction, exactly.

 

[Aer]

Proper time interval along whose worldline, S or S'? If it's along the S worldline, then what is the initial point that you're using in the interval? Is it the tick of the S-clock that is simultaneous with the event of S' beginning to coast in the S-frame, or in the S'-frame?

For any definite pair of events on a given worldline of an observer who is moving inertially between those events, the proper time between the events will be larger than the time between the events as measured in every other frame. But I don't understand how you think this leads to a contradiction, exactly.

Let's only look at the constant velocity part which was said to be in the frame of S over an interval of 100 units in time. This means the acceleration time is almost negligible. Therefore when S' decelerates, it's time says 43.6 if we neglect the time it took to decelerate. Now if we assume S' does not decelerate at this instant but that S does accelerate at this instant, what effects regarding world-lines, simultaneity, and anything else do we have to consider to get the time on the clock of S as soon as it reaches the inertial frame of S'. Is this an ill-posed problem? Remember that we defined S' to accelerate at a certain time according to the frame of S, now we are considering S to be accelerating instead according to the when S preprogrammed S' to decelerate.

 

[JesseM]

Let's only look at the constant velocity part which was said to be in the frame of S over an interval of 100 units in time. This means the acceleration time is almost negligible. Therefore when S' decelerates, it's time says 43.6 if we neglect the time it took to decelerate. Now if we assume S' does not decelerate at this instant but that S does accelerate at this instant

Does S accelerate "at this instant" as defined by its previous rest frame, or as defined by the rest frame of S'? If it accelerates at the same moment that S' reads 43.6 as defined by the original S-frame, then the time on its own clock when it accelerates will be 100. On the other hand, if it accelerates at the same moment that S' reads 43.6 as defined by the S' rest frame, then the time on its own clock when it accelerates will be 0.19 (I'm assuming for the sake of the argument that S' was at the same position as S when S' began to coast--if not the numbers would be different).

 

[Aer]

Does S accelerate "at this instant" as defined by its previous rest frame, or as defined by the rest frame of S'? If it accelerates at the same moment that S' reads 43.6 as defined by the original S-frame, then the time on its own clock when it accelerates will be 100.

What is the difference between these two statements?

S has a proper time of 100 and S' has a proper time of 43.6.

S clock says 100 and S' clock says 43.6.

x=x'=0 at t=t'=0 is implied in the second statement. If there is no difference, then what is the difference between S accelerating at t=100 or S' decelerating at t'=43.6

On the other hand, if it accelerates at the same moment that S' reads 43.6 as defined by the S' rest frame, then the time on its own clock when it accelerates will be 0.19 (I'm assuming for the sake of the argument that S' was at the same position as S when S' began to coast--if not the numbers would be different).

Can you elaborate on where 0.19 came from?

 

[JesseM]

Does S accelerate "at this instant" as defined by its previous rest frame, or as defined by the rest frame of S'? If it accelerates at the same moment that S' reads 43.6 as defined by the original S-frame, then the time on its own clock when it accelerates will be 100.

What is the difference between these two statements?

The two frames have different definitions of simultaneity, so they disagree about what the reading on S was "at the same time" that S' read 43.6.

S has a proper time of 100 and S' has a proper time of 43.6.

In the frame of S', the events of S ticking 100 and S' ticking 43.6 are not simultaneous.

x=x'=0 at t=t'=0 is implied in the second statement.

OK, then you can just use the Lorentz transform to see what I'm talking about. In the S-frame, the coordinates of S' at the time it reads 43.6 are x=90, t=100, and the coordinates of S at the time its clock reads 100 are x=0, t=100. Since they have the same time-coordinate, these events are simultaneous in this frame. But now transform these two events into the S'-frame, using the Lorentz transform:

x&#039; = \gamma (x - vt)
t&#039; = \gamma (t - vx/c^2)

Let's assume we're in a unit system where c=1, so v=0.9. Also, \gamma is equal to 1/0.4359, or 2.294. So, the coordinate x=90,t=100 becomes x'=0,t'=43.6, while the coordinate x=0,t=100 becomes x'=-206.5,t'=229.4. So you can see that in the S' frame, the event of the S-clock ticking 100 happens at t'=229.4, long after the event of its own clock ticking 43.6 which of course happened at t'=43.6 in its frame.

Can you elaborate on where 0.19 came from?

Should be 19, sorry--to get that, I just multiplied 43.6 by 0.436, since in the S'-frame the S-clock is ticking at 0.436 the normal rate. But you can see this works using the Lorentz transform again. If the event of the S-clock ticking 19 happens at coordinates x=0,t=19 in the S-frame, then in the S'-frame this becomes x'=-39.2,t'=43.6, so in the S' frame this happened at the same time coordinate as the S'-clock ticking 43.6.