Can Time Dilation be Determined in a Constantly Accelerating Frame?

[Aer]

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' = \gamma (x - vt)
t' = \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..

Am I correct to assume that if both S and S' sperately sent out a signal at light speed each time their respective clocks ticked, that S and S' would each receive the signal from the other every 2.3 seconds? This assumes they are traveling at .9c wrt each other. In other words, if S and S' continued in this motion indefinately, would the time between received signals always be 2.3 seconds? Or is their a gradient effect due to the fact that the distance between S and S' continuously increases (I would think not).

 

[JesseM]

Am I correct to assume that if both S and S' sperately sent out a signal at light speed each time their respective clocks ticked, that S and S' would each receive the signal from the other every 2.3 seconds?

It's true that in each one's frame the other one would be sending a signal every 2.294 seconds, but you have to take into account that each signal is sent from a larger distance due to their motion (the doppler effect). During the 2.294 seconds between signals, the other ship will have moved an additional (0.9)*(2.294) = 2.065 light-seconds away, so the time between receiving signals will be 2.294+2.065=4.359 seconds. You can also see this by using the equation for the relativistic doppler effect, \nu_{observed} = \nu_{source} \sqrt{\frac{1+v/c}{1-v/c}}

(In this case, v is -0.9c, so \nu_{observed} = \nu_{source} * 0.2294, which means if \nu_{source} = 1 signal/second, \nu_{observed} = 0.2294 signals/second, which means a signal is received every 4.359 seconds.)

 

[Aer]

It's true that in each one's frame the other one would be sending a signal every 2.294 seconds, but you have to take into account that each signal is sent from a larger distance due to their motion (the doppler effect). During the 2.294 seconds between signals, the other ship will have moved an additional (0.9)*(2.294) = 2.065 light-seconds away, so the time between receiving signals will be 2.294+2.065=4.359 seconds. You can also see this by using the equation for the relativistic doppler effect, \nu_{observed} = \nu_{source} \sqrt{\frac{1+v/c}{1-v/c}}

(In this case, v is -0.9c, so \nu_{observed} = \nu_{source} * 0.2294, which means if \nu_{source} = 1 signal/second, \nu_{observed} = 0.2294 signals/second, which means a signal is received every 4.359 seconds.)

Ok, just one last point of clarification. Let's take the two examples where S' decelerates to the frame of S when S reads its own clock at 100 and the other case where S accelerates to the frame of S' when S reads its own clock at 100. If S and S' are bouncing signals back and forth each time their clock ticks 1 as previously described and they factor out the doppler effect to get the other clock's "tick interval", then they are effectively measuring time in the other frame. Note that they can only measure time in the other's frame at some point in the past as the signals necessarily take longer to reach the other as they move further apart.

Now that the setup has been defined, here is the point. When S' decelerates to S, it reads its own clock at 43.6 and can infer from the signals received by S, that it's clock is ticking in intervals of 2.3 so the S clock should read 19 (according to S') when S' begins to decelerate, during the deceleration process, S' must be able to observal all the signals coming from S so that when S' reaches the frame of S, it will agree that the clock on S should be greater than 100 (or 100 + time to decelerate). So S' sees the clock of S running incredibly fast or to put another way, it receives the signals from S at an incredibly fast rate. Is this point of view acceptable?

Likewise one can consider when S accelerates to S' to be a matter of S seeing the clock of S' running incredibly fast during the acceleration as to agree to the numbers given in a post above.

 

[JesseM]

Ok, just one last point of clarification. Let's take the two examples where S' decelerates to the frame of S when S reads its own clock at 100 and the other case where S accelerates to the frame of S' when S reads its own clock at 100. If S and S' are bouncing signals back and forth each time their clock ticks 1 as previously described and they factor out the doppler effect to get the other clock's "tick interval", then they are effectively measuring time in the other frame. Note that they can only measure time in the other's frame at some point in the past as the signals necessarily take longer to reach the other as they move further apart.

Now that the setup has been defined, here is the point. When S' decelerates to S, it reads its own clock at 43.6 and can infer from the signals received by S, that it's clock is ticking in intervals of 2.3 so the S clock should read 19 (according to S') when S' begins to decelerate, during the deceleration process, S' must be able to observal all the signals coming from S so that when S' reaches the frame of S, it will agree that the clock on S should be greater than 100 (or 100 + time to decelerate). So S' sees the clock of S running incredibly fast or to put another way, it receives the signals from S at an incredibly fast rate. Is this point of view acceptable?

S' won't actually see the signals come in incredibly fast, only in retrospect will he conclude that the event of the S-clock sending the signal of t=100 seconds happened at the same time that his own clock read 43.59 seconds. For the first 43.59 seconds, S' is receiving a signal from S every 4.359 seconds, so at t'=4.359 seconds he sees the S-clock reading 1 second, at t'=8.718 seconds he sees the S-clock reading 2 seconds, and so on, until at t'=43.59 seconds he sees the S-clock reading 10 seconds. Then he decelerates near-instantaneously, and from then on they are at rest with respect to each other so S' will see a new signal come in every 1 second...thus, when the S'-clock reads 44.59 seconds, he will see the S-clock reading 11 seconds, when the S'-clock reads 45.59 seconds he will see the S-clock reading 12 seconds, and in general when the S'-clock reads 43.59+x seconds he will see the S-clock reading 10+x seconds. So, it won't be until the S'-clock reads 133.59 seconds that he sees the S-clock reading 100 seconds. But once S' decelerates he remains at a constant distance of 90 light-seconds away from the S in their mutual rest frame, so S' will retroactively conclude that the event of the S-clock reading 100 seconds "really" happened at the same moment that his own clock read 133.59-90=43.59 seconds, in terms of the definition of simultaneity used by their current mutual rest frame.

On the other hand, from the point of view of S, light from successive ticks of the S'-clock will only come in once every 4.359 seconds, so when the S-clock reads t=x, S will see the S'-clock reading x/4.359. So when the S-clock reads 190 seconds, S will see the S'-clock reading 190/4.359=43.59 seconds, and at the moment that signal was emitted the S'-clock was 90 light-seconds away in the S frame, so S will also say in retrospect that this event "really" happened at the same time its own clock read 190-90=100 seconds.

 

[Aer]

S' won't actually see the signals come in incredibly fast, only in retrospect will he conclude that the event of the S-clock sending the signal of t=100 seconds happened at the same time that his own clock read 43.59 seconds.

How might S' come to this conclusion if his only way to measure the S clock is to measure the signals as they come in. In other words, I want to be able to determine how fast I was going (and thus how much I decelerated) based on determining what the clock of S says the time is from the constant signals that were sent out. It seems your "in retrospect" would imply I need to know how fast I decelerated from to determine what the S clock says even though we have the handy fact that S has been sending signals to S' every time the S clock ticked 1.

For the first 43.59 seconds, S' is receiving a signal from S every 4.359 seconds, so at t'=4.359 seconds he sees the S-clock reading 1 second, at t'=8.718 seconds he sees the S-clock reading 2 seconds, and so on, until at t'=43.59 seconds he sees the S-clock reading 10 seconds. Then he decelerates near-instantaneously, and from then on they are at rest with respect to each other so S' will see a new signal come in every 1 second...thus, when the S'-clock reads 44.59 seconds, he will see the S-clock reading 11 seconds, when the S'-clock reads 45.59 seconds he will see the S-clock reading 12 seconds, and in general when the S'-clock reads 43.59+x seconds he will see the S-clock reading 10+x seconds. So, it won't be until the S'-clock reads 133.59 seconds that he sees the S-clock reading 100 seconds. But once S' decelerates he remains at a constant distance of 90 light-seconds away from the S in their mutual rest frame, so S' will retroactively conclude that the event of the S-clock reading 100 seconds "really" happened at the same moment that his own clock read 133.59-90=43.59 seconds, in terms of the definition of simultaneity used by their current mutual rest frame.

On the other hand, from the point of view of S, light from successive ticks of the S'-clock will only come in once every 4.359 seconds, so when the S-clock reads t=x, S will see the S'-clock reading x/4.359. So when the S-clock reads 190 seconds, S will see the S'-clock reading 190/4.359=43.59 seconds, and at the moment that signal was emitted the S'-clock was 90 light-seconds away in the S frame, so S will also say in retrospect that this event "really" happened at the same time its own clock read 190-90=100 seconds.

I will comment on the rest of this after the issue above has been clarified.

 

[JesseM]

How might S' come to this conclusion if his only way to measure the S clock is to measure the signals as they come in.

Because he can measure the distance the clock was at the moment it emitted each signal, as measured in his own rest frame (for simplicity, assume both S and S' are moving alongside parallel rulers, one of which was at rest wrt S' before he decelerated, and one of which is at rest with respect to S, and also S' after he decelerated). And in relativity, you always assume that light travels at c in your frame. So, to find how long ago an event you are seeing now "really" happened in your frame, you just divide the distance in your frame by c.

In other words, I want to be able to determine how fast I was going (and thus how much I decelerated) based on determining what the clock of S says the time is from the constant signals that were sent out.

Well, that's a little different, I was just calculating what tick of S' was simultaneous with S ticking 100 from the perspective of the S-frame (which is also the rest frame of S' after he decelerates). For this purpose, I was assuming S' knew the distance each signal was emitted from, and obviously from this it's pretty trivial to calculate the relative speed; if you don't know the distances, I suppose you could determine the speed during any small time-interval by looking at the rate the signals were coming in during that interval, and solving for v in the relativistic doppler equation. You could then integrate the velocities during each small time-interval to find the distances, I suppose.

 

[Aer]

Because he can measure the distance the clock was at the moment it emitted each signal, as measured in his own rest frame (for simplicity, assume both S and S' are moving alongside parallel rulers, one of which was at rest wrt S' before he decelerated, and one of which is at rest with respect to S, and also S' after he decelerated). And in relativity, you always assume that light travels at c in your frame. So, to find how long ago an event you are seeing now "really" happened in your frame, you just divide the distance in your frame by c.

As you acknowledge below, this calculation is impossible since S' doesn't know how fast he was traveling.

Well, that's a little different, I was just calculating what tick of S' was simultaneous with S ticking 100 from the perspective of the S-frame (which is also the rest frame of S' after he decelerates). For this purpose, I was assuming S' knew the distance each signal was emitted from, and obviously from this it's pretty trivial to calculate the relative speed; if you don't know the distances, I suppose you could determine the speed during any small time-interval by looking at the rate the signals were coming in during that interval, and solving for v in the relativistic doppler equation. You could then integrate the velocities during each small time-interval to find the distances, I suppose.

I'm not sure I follow how one might accomplish the task as you describe it.

 

[JesseM]

As you acknowledge below, this calculation is impossible since S' doesn't know how fast he was traveling.

Not if they're moving alongside rulers as I described.

I'm not sure I follow how one might accomplish the task as you describe it.

If you know that that the other clock is emitting signals at a rate of x signals/second, and you are receiving signals at a constant rate of y signals/second, then you can solve for v in the relativistic doppler shift equation:

y = x \sqrt{\frac{1 + v/c}{1 - v/c}}
squaring both sides gives:
y^2 = x^2 \frac{1 + v/c}{1 - v/c}
now multiply both sides by (1 - v/c):
y^2 - vy^2 /c = x^2 (1 + v/c)
collect all the terms involving v on one side:
y^2 - x^2 = v (y^2 /c + x^2 /c)
which gives:
v = \frac{y^2 - x^2}{y^2 /c + x^2 /c} = c \frac{y^2 - x^2}{y^2 + x^2}

For example, if I receive a signal every 4.359 seconds, so y=0.2294 signals/second, and I know they are being sent at x=1 signal/second, then plugging in tells me the velocity is c*(0.05262 - 1)/(0.05262 + 1) = c*(-0.9474)/(1.05262) = -0.9c (and negative velocity in the doppler shift equation means the two sources are moving apart).

This will only work exactly for finite time-intervals where both S and S' are moving apart at constant velocity and thus S' is receiving signals at a constant rate, but if you assume the signals are coming in almost continuously instead of once per second then you can find small enough time-intervals where the rate the signals are coming in is very close to constant so you can get the average velocity in that time-interval pretty accurately. And once you have a pretty accurate function for the velocity as a function of time, just integrate it to find the distance as a function of time.

 

[JesseM]

Incidentally, a tricky part about using the method above to find velocity as a function of time is that when you solve for v in the doppler shift equation you're actually finding the velocity between S and S' at the moment S' received the signal, but since S' can accelerate his own clock is really showing the proper time since he departed S, while you want to find velocity as a function of time in some inertial frame so you can integrate it to get distance as a function of time in that frame.

But, if S' knows his velocity relative to S between two signals, he can figure out the time in the S-frame between his receiving those signals. If the time between signals being emitted in the S-frame is t_0, then each time S' receives a signal he is c t_0 light seconds in front of the next signal, so if his velocity is v, you can find how much time it will take the next signal to intercept him by solving for t in the equation ct = vt + c t_0, which gives t = (c t_0 ) / (c - v). It should be possible for S' to sum (or integrate, if you idealize t_0 as an infinitesimal dt) the time-intervals between the events of his receiving each signal to figure out what the current t-coordinate is in the S-frame at the moment he is receiving a given signal, and thus to determine the velocity as a function of time in the S-frame.

 

[Aer]

Incidentally, a tricky part about using the method above to find velocity as a function of time is that when you solve for v in the doppler shift equation you're actually finding the velocity between S and S' at the moment S' received the signal, but since S' can accelerate his own clock is really showing the proper time since he departed S, while you want to find velocity as a function of time in some inertial frame so you can integrate it to get distance as a function of time in that frame.

But, if S' knows his velocity relative to S between two signals, he can figure out the time in the S-frame between his receiving those signals. If the time between signals being emitted in the S-frame is t_0, then each time S' receives a signal he is c t_0 light seconds in front of the next signal, so if his velocity is v, you can find how much time it will take the next signal to intercept him by solving for t in the equation ct = vt + c t_0, which gives t = (c t_0 ) / (c - v). It should be possible for S' to sum (or integrate, if you idealize t_0 as an infinitesimal dt) the time-intervals between the events of his receiving each signal to figure out what the current t-coordinate is in the S-frame at the moment he is receiving a given signal, and thus to determine the velocity as a function of time in the S-frame.

First of all, let me state that my following comments are on special relativity in general and not your explanations. I will assume your explanations have fully been on par with special relativity unless anyone steps forth to challenge that.

I am not fully satisfied with the explanation given regarding the the accumulation of signals between S and S' when either or the other accelerate to the other frame exactly as has been previously described. To illustrate my dissatisfaction, let's analyze the following example:

Frames S1 and S2 are synonymous with frame S at the start of the experiment. Frames S1' and S2' are synonymous with frame S' at the start of the experiment. Now what I mean is S1 and S2 are stationary while S1' and S2' instantaneously accelerate to speed v wrt to frames S1 and S2. Let's assume 2 experiments are being tested independently but at the same time. That is to say that S1 and S1' compare clocks and S2 and S2' compare clocks, each set independent of the other.

The 2 experiments are started so that S1, S2, S1', S2' are at positions x1=x2=x1'=x2'=0 at t1=t2=t1'=t2'=0. S1 and S1' follow the Setup presented where S1 accelerates to the frame of S1' at the moment the clock S1 reads 100. Note that the clocks of S1 and S2 are in sync with each other up to the point before S1 accelerates, so it can also be said that S1 accelerates to the frame of S1' when the clock of S2 reads 100. Let's assume the acceleration takes a reading of .5 units of time in the S2 frame. A moment after S1 reaches the frame of S1', they both decelerate to the frame of S2 and S1' brings S2' along for the deceleration. They clocks of S1' and S2' should be the same while the clocks of S1 and S2 should be different.

However, if my understanding is correct of the results predicted by special relativity for the two experiments, then the clocks of S1' and S2' should not be the same. I will be thoroughly impressed if you can show what the results would be for each clock in this experiment in a non-contradictory way. If you cannot, then I believe you must acknowledge what I have said prior:

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]

Frames S1 and S2 are synonymous with frame S at the start of the experiment. Frames S1' and S2' are synonymous with frame S' at the start of the experiment. Now what I mean is S1 and S2 are stationary while S1' and S2' instantaneously accelerate to speed v wrt to frames S1 and S2.

How can frames instantaneously accelerate? An inertial frame is just a coordinate system for assigning coordinates to events, and it's assumed that the origin of this coordinate system never accelerates, otherwise it'd be non-inertial. Also, why is it necessary to define two identical frames S1 and S2, and likewise the identical frames S1' and S2'? Let's just say that frame S1 is the one where S is initially at rest and S' is moving at velocity v, while frame S1' is the one where S' is initially at rest and S is moving with velocity -v. Now, the symmetry of the situation is illustrated by the following:

1. If in frame S1, S' decelerates to zero velocity at the moment his clock reads 43.59, this will happen at coordinate time t=100, and the S-clock will read 100 at that time.

2. If in frame S1', S decelerates to zero velocity at the moment his clock reads 43.59, this will happen at coordinate time t'=100, and the S'-clock will read 100 at that time.

3. If in frame S1, S accelerates to velocity v at the moment his clock reads 43.59, this will happen at coordinate time t=43.59, and the S'-clock will read 19 at that time.

4. If in frame S1', S' accelerates to velocity -v at the moment his clock reads 43.59, this will happen at coordinate time t'=43.59, and the S-clock will read 19 at that time.

It's all completely symmetrical. If there's one of these statements you think is wrong, I can go into it in more detail. Also, if you still think you see some sort of contradiction, could you describe it in terms of the terminology I'm using, where the two ships are S and S' while the two frames are S1 and S1'?

 

[Aer]

How can frames instantaneously accelerate?

It appears we are considering two different situations on a technicality. Let's define S1 and S2 as mother ships that exist at the position x1=x2=0 at the time t1=t2=0 in the frame S. S1' and S2' are scout ships for S1 and S2 respectively that exist at the position x1'=x2'=0 at the time t1'=t2'=0 in the frame S'. For the time being, the mother ships exist at x=0 for all t>0 in the S frame (They both can be at x=0 by having separate offsets in the y dimension).

Now your concern is a valid one in which how can we consider a frame to instaneously accelerate? I am not claiming that S' is the same frame before acceleration as it is after acceleration, it is just the instanteous frame according to S1' and S2' at any instant in time. Now if you have a problem with this, then I will refer you to the rocket equation presented earlier in this discussion. It uses the same concept.

Now that the initial setup has been fully defined, the rest of the problem as previously presented should be clear. However, I will reiterate it as follows:

The clocks of S1 and S2 are ticking in sync with each other as they share a common rest frame S. When the clocks of S1 and S2 read 100 as defined by the frame S, the S1 ship instanaeously accelerates to the frame of ships S1' and S2' which at this point is S' defined as a frame moving with velocity v=.9c wrt frame S. Now we know that any acceleration is not instanaeous in the literal sense. There is a finite interval over which it occurs and for our purposes here, we'll say that this finite interval is .5 unit of time as measured in the S frame. As soon as S1 reaches the frame of S', it, along with S1' and S2' decelerate to the frame of S. This deceleration also takes an interval of .5 unit of time as measured in the S frame. Therefore it should be clear that the clock of S2 reads 101 when all ships are back in the frame of S. My question to you is, what does the clock of S1 and S1' say right after S1 enters the S' frame and right before S1 starts its deceleration back to the frame S. And what is the time on the S2 and S2' clock right after the S2' ship is decelerated back to the S frame.

Please note the points in time of interest and do not confuse the issue with other points in time. The points in time have been defined in which frame they are considered, if you contest this issue, please explain why.

An inertial frame is just a coordinate system for assigning coordinates to events, and it's assumed that the origin of this coordinate system never accelerates, otherwise it'd be non-inertial. Also, why is it necessary to define two identical frames S1 and S2, and likewise the identical frames S1' and S2'? Let's just say that frame S1 is the one where S is initially at rest and S' is moving at velocity v, while frame S1' is the one where S' is initially at rest and S is moving with velocity -v. Now, the symmetry of the situation is illustrated by the following:

1. If in frame S1, S' decelerates to zero velocity at the moment his clock reads 43.59, this will happen at coordinate time t=100, and the S-clock will read 100 at that time.

2. If in frame S1', S decelerates to zero velocity at the moment his clock reads 43.59, this will happen at coordinate time t'=100, and the S'-clock will read 100 at that time.

3. If in frame S1, S accelerates to velocity v at the moment his clock reads 43.59, this will happen at coordinate time t=43.59, and the S'-clock will read 19 at that time.

4. If in frame S1', S' accelerates to velocity -v at the moment his clock reads 43.59, this will happen at coordinate time t'=43.59, and the S-clock will read 19 at that time.

It's all completely symmetrical. If there's one of these statements you think is wrong, I can go into it in more detail. Also, if you still think you see some sort of contradiction, could you describe it in terms of the terminology I'm using, where the two ships are S and S' while the two frames are S1 and S1'?

I redefined the terminology as I meant it, sorry about any confusion.

 

[JesseM]

It appears we are considering two different situations on a technicality. Let's define S1 and S2 as mother ships that exist at the position x1=x2=0 at the time t1=t2=0 in the frame S. S1' and S2' are scout ships for S1 and S2 respectively that exist at the position x1'=x2'=0 at the time t1'=t2'=0 in the frame S'.

We're assuming that the frame S' has a velocity of v relative to frame S, correct? So are you assuming the mother ships start out at rest in frame S, and the scout ships start out at rest in frame S', so the scout ships start out moving at velocity v relative to the mother ships?

Now your concern is a valid one in which how can we consider a frame to instaneously accelerate? I am not claiming that S' is the same frame before acceleration as it is after acceleration, it is just the instanteous frame according to S1' and S2' at any instant in time.

But in that case S1' and S2' will have different instantaneous inertial rest frames before and after the acceleration, you can't use the name S' to refer to both unless you want S' to be the label for a non-inertial frame. Isn't it easier just to analyze this problem using two different inertial frames, S and S'?

The clocks of S1 and S2 are ticking in sync with each other as they share a common rest frame S. When the clocks of S1 and S2 read 100 as defined by the frame S, the S1 ship instanaeously accelerates to the frame of ships S1' and S2' which at this point is S' defined as a frame moving with velocity v=.9c wrt frame S. Now we know that any acceleration is not instanaeous in the literal sense.

It's not, but you can consider the limit as the acceleration time goes to zero. If you want the acceleration time nonzero, then to solve the problem exactly you'll have to give a function for the velocity as a function of time (as seen by some inertial frame) during the acceleration period, and we'll have to do an integral of \sqrt{1 - v^2/c^2} to find the clock-reading at the end of the acceleration. This seems needlessly complicated, why can't we just assume we're looking at the limit as the acceleration goes to zero?

As soon as S1 reaches the frame of S', it, along with S1' and S2' decelerate to the frame of S.

If S1 just accelerates to v briefly and then immediately decelerates back to 0, what's the point? Why not just assume S2 accelerates quasi-instantaneously from 0 to v while S1 does not accelerate at all? What specific contradiction do you see when a finite period of acceleration is involved that wouldn't be present if we assumed quasi-instantaneous acceleration?

This deceleration also takes an interval of .5 unit of time as measured in the S frame. Therefore it should be clear that the clock of S2 reads 101 when all ships are back in the frame of S. My question to you is, what does the clock of S1 and S1' say right after S1 enters the S' frame and right before S1 starts its deceleration back to the frame S. And what is the time on the S2 and S2' clock right after the S2' ship is decelerated back to the S frame.

If you think the acceleration is essential to formulating the problem you're having, then I need more details on how the acceleration works--are you assuming that both the acceleration and deceleration take 0.5 seconds of proper time rather than time as measured in some inertial frame? Also, are you assuming the acceleration is constant over that that time as seen by the ships doing the accelerating (ie they feel a constant G-force)?

 

[Aer]

We're assuming that the frame S' has a velocity of v relative to frame S, correct? So are you assuming the mother ships start out at rest in frame S, and the scout ships start out at rest in frame S', so the scout ships start out moving at velocity v relative to the mother ships?

Since the S1' and S2' frames necessarily start out in the same frame as S1 and S2, they do not reach the S' frame until their acceleration stops. Yes it is true we can ignore this part of the problem, but then the entire problem is not entirely defined, it is just as easy to leave it in and have no confusion as to the origins of S1' and S2'.

But in that case S1' and S2' will have different instantaneous inertial rest frames before and after the acceleration, you can't use the name S' to refer to both unless you want S' to be the label for a non-inertial frame. Isn't it easier just to analyze this problem using two different inertial frames, S and S'?

S2' has the same motion as S1' in every aspect, so it is not true to say that they have different instaneous frames at any instaneous point in time as defined by either S1' or S2' since they will agree that all moments in time as measured by the other are simultaneous with their own clock. The entire point of having the two ships is that only S1 and S1' can communicate and likewise only S2 and S2' can communicate (that is, read each others clock by sending signals).

It's not, but you can consider the limit as the acceleration time goes to zero. If you want the acceleration time nonzero, then to solve the problem exactly you'll have to give a function for the velocity as a function of time (as seen by some inertial frame) during the acceleration period, and we'll have to do an integral of \sqrt{1 - v^2/c^2} to find the clock-reading at the end of the acceleration.

As has been shown, the time measured by the accelerating ship over the acceleration period is negligible when you consider the acceleration to be instaneous. We know that this time is not 0, but ignoring it does not give our approximation much of an error. The issue is not the acceleration.

This seems needlessly complicated, why can't we just assume we're looking at the limit as the acceleration goes to zero?

Agreed.

If S1 just accelerates to v briefly and then immediately decelerates back to 0, what's the point?

Agreed, it is a stupid mission by the fact that it is stupidly simply. But the point is not about the usefulness of the mission as it was purposefully made this simple to keep the calculations to a minimum.

Why not just assume S2 accelerates quasi-instantaneously from 0 to v while S1 does not accelerate at all?

What is the purpose of this change to the problem as stated? Was there an error in the problem statement that makes the calcuations impossible? I don't think so or otherwise I am missing a very important concept of special relativity that you haven't pointed out (please do if it exists).

What specific contradiction do you see when a finite period of acceleration is involved that wouldn't be present if we assumed quasi-instantaneous acceleration?

None, the contradiction does not lie within the acceleration - you've missed the point entirely. I've stated specificly if you do the calculations as I presented, S1' and S2' cannot have the same time at the end of the experiment. This is a claim I am making and asking you to disprove, please note it is not a simple matter of just saying "well of course they will have the same time, you are wrong...", you must either show that the problem statement is in error (it very well could be, though I don't believe it is) or that the calculations I asked for will not show a contradiction as I plainly stated they will. You must show all calculations asked for to prove this.

If you think the acceleration is essential to formulating the problem you're having, then I need more details on how the acceleration works--are you assuming that both the acceleration and deceleration take 0.5 seconds of proper time rather than time as measured in some inertial frame? Also, are you assuming the acceleration is constant over that that time as seen by the ships doing the accelerating (ie they feel a constant G-force)?

The acceleration is not essential as I've stated several times. I even told you the acceleration times were all measured as .5 units of time in the S frame, that is .5 units of proper time in the S frame (which can be considered instaneous as the coasting period was measured as 100 units of proper time in the S frame).

 

[JesseM]

Since the S1' and S2' frames necessarily start out in the same frame as S1 and S2, they do not reach the S' frame until their acceleration stops.

Your language is confusing--can you distinguish between frames and physical objects like ships? I thought S1' and S2' were the labels for the two scout ships, not "frames". So I think what you mean here is that the S1' and S2' ships start out in the same frame as S1 and S2, namely the frame S.

Yes it is true we can ignore this part of the problem, but then the entire problem is not entirely defined, it is just as easy to leave it in and have no confusion as to the origins of S1' and S2'.

Why does it matter that the scout ships started out at rest relative to the mother ships? It's easier if you assume the two scout ships just whizzed by the mother ships at constant velocity, and at the moment they passed each other they synchronized their clocks to all read 0.

But in that case S1' and S2' will have different instantaneous inertial rest frames before and after the acceleration, you can't use the name S' to refer to both unless you want S' to be the label for a non-inertial frame. Isn't it easier just to analyze this problem using two different inertial frames, S and S'?

S2' has the same motion as S1' in every aspect, so it is not true to say that they have different instaneous frames at any instaneous point in time as defined by either S1' or S2' since they will agree that all moments in time as measured by the other are simultaneous with their own clock.

That's not what I meant--I didn't say they'd have different instantaneous frames from each other, just that each one would have different instantaneous inertial rest frames at different moments during the acceleration, since each one's instantaneous velocity is changing. And again, it's a lot easier to analyze the problem in SR if you look at everything from the point of view of two inertial frames S and S', including the acceleration.

This seems needlessly complicated, why can't we just assume we're looking at the limit as the acceleration goes to zero?

Agreed.

OK, so can I treat the acceleration as instantaneous and thus skip the whole business of S1 accelerating and then immediately decelerating back to zero in S? In other words, can I rewrite the problem as follows?

"The clocks of S1 and S2 are ticking in sync with each other as they share a common rest frame S. When the clocks of S1 and S2 read 100 as defined by the frame S, S1' and S2' decelerate to the frame of S. My question to you is, what does the clock of S1 and S1' say right after S1' and S2' decelerate to the frame of S? And what is the time on the S2 and S2' clock right after the S2' ship is decelerated back to the S frame."

What is the purpose of this change to the problem as stated? Was there an error in the problem statement that makes the calcuations impossible? I don't think so or otherwise I am missing a very important concept of special relativity that you haven't pointed out (please do if it exists).

The purpose is just that instantaneous accelerations are a lot easier to deal with mathematically than accelerations over a finite period, if you insist on making the acceleration finite we have to do an integral to see how much time elapses on the clock of S1 during the acceleration (assuming it lasts for 0.5 seconds of coordinate tiem in frame S).

I've stated specificly if you do the calculations as I presented, S1' and S2' cannot have the same time at the end of the experiment.

Why? Didn't they both decelerate at exactly the same position and time? In that case they will definitely have the same time at the end of the experiment. This calculation is simple if you assume instantaneous acceleration, if not it will be a bit complicated, but either way I'm willing to do it if you tell me which one you want.

The acceleration is not essential as I've stated several times. I even told you the acceleration times were all measured as .5 units of time in the S frame, that is .5 units of proper time in the S frame (which can be considered instaneous as the coasting period was measured as 100 units of proper time in the S frame).

Your language is confusing again--"proper time" only refers to time as measured by a clock along some worldline, if S1 accelerates then his proper time will not match up with time as measured in the S frame. If you want to refer to time in the S frame, that's coordinate time, not proper time. But again, if the acceleration is not essential, then why can't we drop this business with the acceleration lasting for a finite time period of 0.5 seconds?

 

[Aer]

Your language is confusing--can you distinguish between frames and physical objects like ships? I thought S1' and S2' were the labels for the two scout ships, not "frames". So I think what you mean here is that the S1' and S2' ships start out in the same frame as S1 and S2, namely the frame S.

Yes, I should have said "Since the S1' and S2' ships necessarily start out in the same frame..."

Why does it matter that the scout ships started out at rest relative to the mother ships? It's easier if you assume the two scout ships just whizzed by the mother ships at constant velocity, and at the moment they passed each other they synchronized their clocks to all read 0.

Yes, the calculation is the same either way since we are ignoring the time it takes to accelerate. If you want, the problem can be changed - but this change is only a change in the problem statement, not a change in the end result or the calculations required for the end result.

That's not what I meant--I didn't say they'd have different instantaneous frames from each other, just that each one would have different instantaneous inertial rest frames at different moments during the acceleration, since each one's instantaneous velocity is changing. And again, it's a lot easier to analyze the problem in SR if you look at everything from the point of view of two inertial frames S and S', including the acceleration.

Ok.

OK, so can I treat the acceleration as instantaneous and thus skip the whole business of S1 accelerating and then immediately decelerating back to zero in S? In other words, can I rewrite the problem as follows?

"The clocks of S1 and S2 are ticking in sync with each other as they share a common rest frame S. When the clocks of S1 and S2 read 100 as defined by the frame S, S1' and S2' decelerate to the frame of S. My question to you is, what does the clock of S1 and S1' say right after S1' and S2' decelerate to the frame of S? And what is the time on the S2 and S2' clock right after the S2' ship is decelerated back to the S frame."

Regarding specificly "can I treat the acceleration as instantaneous and thus skip the whole business of S1 accelerating and then immediately decelerating back to zero in S". No, this part of the problem can not be left out. Again, if you want the problem could be changed so that S1 does not immediately accelerate back to the frame of S, that is, it coasts in the S' frame for a finite amount of period before decelerating back. But I do not see the point of this requirement, and unless you can prove that it is in fact required, then I do not see any reason to change the problem. The point is that the clocks of S1 and S1' should read ~100 and ~230 right before the deceleration back to frame S. That was the entire point of which I wanted to make and you want to leave out.

The purpose is just that instantaneous accelerations are a lot easier to deal with mathematically than accelerations over a finite period, if you insist on making the acceleration finite we have to do an integral to see how much time elapses on the clock of S1 during the acceleration (assuming it lasts for 0.5 seconds of coordinate tiem in frame S). Why? Didn't they both decelerate at exactly the same position and time? In that case they will definitely have the same time at the end of the experiment. This calculation is simple if you assume instantaneous acceleration, if not it will be a bit complicated, but either way I'm willing to do it if you tell me which one you want.

Are you considering the problem I stated or the problem you redefined which I've pointed out is not the same problem I stated.

Your language is confusing again--"proper time" only refers to time as measured by a clock along some worldline, if S1 accelerates then his proper time will not match up with time as measured in the S frame. If you want to refer to time in the S frame, that's coordinate time, not proper time. But again, if the acceleration is not essential, then why can't we drop this business with the acceleration lasting for a finite time period of 0.5 seconds?

Again, the proper times I noted were all measured by S2 which is always in the S frame, S2 never moves. I think your argument about S1 accelerating is irrelevent.

 

[JesseM]

Regarding specificly "can I treat the acceleration as instantaneous and thus skip the whole business of S1 accelerating and then immediately decelerating back to zero in S". No, this part of the problem can not be left out. Again, if you want the problem could be changed so that S1 does not immediately accelerate back to the frame of S, that is, it coasts in the S' frame for a finite amount of period before decelerating back. But I do not see the point of this requirement, and unless you can prove that it is in fact required, then I do not see any reason to change the problem. The point is that the clocks of S1 and S1' should read ~100 and ~230 right before the deceleration back to frame S. That was the entire point of which I wanted to make and you want to leave out.

OK, how about this--at time t=100 in the S-frame, S1 instantaneously accelerates to velocity v so it's at rest in the S' frame, and coasts there for 1 second in the S-frame (0.229 seconds in the S' frame), then instantaneously decelerates back to velocity 0 at t=101.

If that's acceptable, then when do you want S1' and S2' to decelerate to velocity 0 in the S frame, at t=100 or t=101?

Again, the proper times I noted were all measured by S2 which is always in the S frame, S2 never moves. I think your argument about S1 accelerating is irrelevent.

I wasn't making an argument, just saying that the problem would involve less messy calculations if we assumed instantaneous acceleration. See what you think of the modification I just suggested. Also, "proper time" can really only be measured along an observer's own worldline, if S2 is measuring the times of distant events not on its own worldline, then this should still be called coordinate time (in frame S) rather than proper time.

 

[pervect]

I'm afraid I can't keep up with this thread, there are just too many long detailed posts for me to keep trying to follow it.

I do think that Aer is off on the wrong track and making the problem unnecessarily complicated, but I'm not quite sure how to communicate this and get things on the right track - especially with the large number of posts flying by, I feel like I'm about a week behind the discussion. Anyway, good luck to all participants.

 

[Aer]

OK, how about this--at time t=100 in the S-frame, S1 instantaneously accelerates to velocity v so it's at rest in the S' frame, and coasts there for 1 second in the S-frame (0.229 seconds in the S' frame), then instantaneously decelerates back to velocity 0 at t=101.

This is an acceptable change to the problem but I don't see why it is necessary.

If that's acceptable, then when do you want S1' and S2' to decelerate to velocity 0 in the S frame, at t=100 or t=101?

t=101 as for the case you outlined above. The point has always been that S1, S1', and S2' decelerate from the S' frame to the S frame simultaneously as seen by the S frame, that is the S2 mother ship. All I've asked for are the time for S1' before it instantaneously decelerates and the time for S2' after it instantaneously decelerates. Do not forget the details of the problem when you make these calculations.

 

[Aer]

This is an irresponsible post on your part.

I'm afraid I can't keep up with this thread, there are just too many long detailed posts for me to keep trying to follow it.

I do think that Aer is off on the wrong track and making the problem unnecessarily complicated, but I'm not quite sure how to communicate this and get things on the right track - especially with the large number of posts flying by, I feel like I'm about a week behind the discussion. Anyway, good luck to all participants.

No one has presented a valid contradiction to the problem statement as I presented it. The problem statement has only been changed in terminology, not it any substantial way that changes the calculations required to answer the problem statement. Only if you can show the problem statement to be in error in this fashion, then you may post that I am off on the wrong track.

Besides, the discussion JesseM and I have been having is irrelevant to the discussion between you and I which ended with this:

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]

This is an acceptable change to the problem but I don't see why it is necessary.

It isn't "necessary", but as I said a few times, it makes the math simpler. If you wanted a period of finite acceleration, you'd have to first find the velocity as a function of time in some inertial frame during the acceleration, then integrate \int^{t_1}_{t_0} \sqrt{1 - v(t)^2/c^2} \, dt in that frame to find out how much time will elapse on the ship's clocks during the acceleration. If you specify constant acceleration from the point of view of the ship, then the rate of acceleration will not be constant from the point of view of an inertial reference frame, so v(t) won't just be a*t.

t=101 as for the case you outlined above. The point has always been that S1, S1', and S2' decelerate from the S' frame to the S frame simultaneously as seen by the S frame, that is the S2 mother ship. All I've asked for are the time for S1' before it instantaneously decelerates and the time for S2' after it instantaneously decelerates. Do not forget the details of the problem when you make these calculations.

Sounds good. Assuming all three decelerate at the same moment in frame S, the clock of S1 will read 100 + (0.4359)(1) = 100.4359 seconds, while S1' and S2' will read (101)*(0.4359) = 44.025 seconds. In frame S', the event of S1 decelerating (I'll keep calling it 'decelerating' even though in this frame the velocity increases) happens at a different time than the event of S1' and S2' decelerating; S1 decelerates at time t'=229.852 in this frame (found by doing a Lorentz transform on the coordinates x=0.9, t=101 in the S-frame), while S1' and S2' decelerate 185.827 seconds earlier at t'=44.025 (found by doing a Lorentz transform on the coordinates x=90.9, t=101 in the S frame). Since they run at 0.4359 the normal rate after decelerating in this frame, they will read 44.025 + (0.4359)*(185.827) = 125.025 at the moment S1 decelerates, in the S' frame.

On the other hand, you might want to specify that all three decelerate at the same moment in frame S' instead. In this case, it's still true that the clock of S1 will read 100.4359 seconds, but now S1' and S2' will read 229.852 seconds at the moment they decelerate (again, found by doing a Lorentz transform on x=0.9, t=101 in the S frame). Now in the S frame these events no longer happen at the same moment--instead, at the moment S1 decelerates, S1' and S2' have not yet decelerated, and they only read (0.4359)*(101) = 44.025 seconds at that moment.

 

[Aer]

Assuming all three decelerate at the same moment in frame S, the clock of S1 will read 100 + (0.4359)(1) = 100.4359 seconds, while S1' and S2' will read (101)*(0.4359) = 44.025 seconds. In frame S', the event of S1 decelerating (I'll keep calling it 'decelerating' even though in this frame the velocity increases) happens at a different time than the event of S1' and S2' decelerating; S1 decelerates at time t'=229.852 in this frame (found by doing a Lorentz transform on the coordinates x=0.9, t=101 in the S-frame), while S1' and S2' decelerate 185.827 seconds earlier at t'=44.025 (found by doing a Lorentz transform on the coordinates x=90.9, t=101 in the S frame). Since they run at 0.4359 the normal rate after decelerating in this frame, they will read 44.025 + (0.4359)*(185.827) = 125.025 at the moment S1 decelerates, in the S' frame.

So, S1' and S2' decelerate to the S frame before S1 enters the S' frame according to an observer stationary in the S' frame. Meanwhile, while S1 is in the S frame before accelerating to the S' frame, an observer in the S frame will say that S1' and S2' are still in the S' frame. So no matter what, over this infintesimally small instant of acceleration, S1 managed to escape being in the same frame as S1' and S2' even though S1' and S2' didn't decelerate to the S' frame at the same time according to either an observer in the frame S or S' as S1 accelerated to the S' frame. I am not saying this is a contradiction per say, I am only asking if this is an accurate assessment of the situation or if it needs to be revised.

On the other hand, you might want to specify that all three decelerate at the same moment in frame S' instead. In this case, it's still true that the clock of S1 will read 100.4359 seconds, but now S1' and S2' will read 229.852 seconds at the moment they decelerate (again, found by doing a Lorentz transform on x=0.9, t=101 in the S frame). Now in the S frame these events no longer happen at the same moment--instead, at the moment S1 decelerates, S1' and S2' have not yet decelerated, and they only read (0.4359)*(101) = 44.025 seconds at that moment.

Let's not consider two different scenarios.

 

[JesseM]

So, S1' and S2' decelerate to the S frame before S1 enters the S' frame according to an observer stationary in the S' frame.

Yes.

Meanwhile, while S1 is in the S frame before accelerating to the S' frame, an observer in the S frame will say that S1' and S2' are still in the S' frame. So no matter what, over this infintesimally small instant of acceleration, S1 managed to escape being in the same frame as S1' and S2' even though S1' and S2' didn't decelerate to the S' frame at the same time according to either an observer in the frame S or S' as S1 accelerated to the S' frame.

It's a little ambiguous to talk about what happens over an infinitesimally small period of time--the velocity is really undefined at the exact moment of an instantaneous acceleration, and so the instantaneous inertial rest frame at that moment is also undefined. If you made the acceleration non-instantaneous, then S1 would have a different instantaneous rest frame at each moment during the acceleration, and each of these instantaneous rest frames would have a velocity midway between S and S'. Note that this would be just as true in Newtonian mechanics as it is in relativity.

 

[pervect]

This is an irresponsible post on your part.

Sorry you feel that way. I'm about 5-10 posts behind, so perhaps you and Jesse have already addressed the things that were bothering me. I just don't know, because I don't have time to follow this thread in the amount of detail that it would take me to figure out whether or not you two had already covered my misgivings or not.

I thought I'd just let everyone involved know that I had to "abandon ship" due to time constraints, rather than just walk away from the thread without telling anyone.

 

[Aer]

Yes. It's a little ambiguous to talk about what happens over an infinitesimally small period of time--the velocity is really undefined at the exact moment of an instantaneous acceleration, and so the instantaneous inertial rest frame at that moment is also undefined. If you made the acceleration non-instantaneous, then S1 would have a different instantaneous rest frame at each moment during the acceleration, and each of these instantaneous rest frames would have a velocity midway between S and S'. Note that this would be just as true in Newtonian mechanics as it is in relativity.

Very well, my purpose to this thread was to gather a better understanding of all the implications of special relativity as I've merely only had an introductory lecture on it many years ago. I still have concerns about the notions of space contraction as described by the lorentz transformations as I alluded to in another thread as well as the issue of simultaneity as illustrated in this thread. These concerns originated from an idea (I would hesitate to refer to it as any type of theory at this point) on the fundamental nature of matter which would be a better topic for the Strings, Branes, & LQG forum. This idea would explain a lot of the experiments on special relativity but I noticed right away it would not allow for some of the ideas purported by special relativity. Anyway, thank you for your time.

Sorry you feel that way. I'm about 5-10 posts behind, so perhaps you and Jesse have already addressed the things that were bothering me. I just don't know, because I don't have time to follow this thread in the amount of detail that it would take me to figure out whether or not you two had already covered my misgivings or not.

I thought I'd just let everyone involved know that I had to "abandon ship" due to time constraints, rather than just walk away from the thread without telling anyone.

I can understand time restraining anyone from delving deeply into a certain topic, there is no need for you to apologize for this.