Can Time Dilation be Determined in a Constantly Accelerating Frame?

[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:

$$x' = \gamma ( x - vt )$$
$$t' = \gamma ( t - v/{c}^{2}x )$$

$$x = \gamma ( x' - vt' )$$
$$t = \gamma ( t' - v/{c}^{2}x' )$$

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:

$$x' = \gamma (x - vt)$$
$$t' = \gamma (t - vx/c^2)$$
$$x = \gamma (x' + vt')$$
$$t = \gamma (t' + vx'/c^2)$$

...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:

$$x' = /gamma (x - vt)$$
$$t' = /gamma (t - vx/c^2)$$
$$x = /gamma (x' + vt')$$
$$t = /gamma (t' + vx'/c^2)$$

...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]

$$x' = \gamma (x - vt)$$
$$t' = \gamma (t - vx/c^2)$$
$$x = \gamma (x' + vt')$$
$$t = \gamma (t' + vx'/c^2)$$

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:

$$x' = \gamma (x - vt)$$
$$t' = \gamma (t - vx/c^2)$$

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.

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.

 

[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