The importance of velocity in simultaneity

[PeterDonis]

couldn't they be programmed to begin accelerating when each reads a specific time value

Yes. I was assuming that that value was $t = 0$; since both are at rest in the original inertial frame at this point, there is no ambiguity about when they start accelerating, according to either one.

wouldn't that be defined as "doing the experiments together"?

Not in the sense you mean. See below.

There is a fixed distance each must accelerate through.

In the original inertial frame, yes. Remember that distance is frame-dependent, so you have to specify which frame it is relative to.

Each could do that separately (on different days maybe) and each would presumably register the exact same time when they reached the "endpoint".

Yes, but that doesn't mean the two clocks will be in sync when they do it on the same day, starting at the same time in the original rest frame, but separated along their direction of acceleration (the $x$ direction).

how could they then get out of sync?

Because, as you have been told repeatedly, "at the same time" does not have an absolute meaning. Simultaneity is relative.

Once again, clock A starts at $x = x_A$ at time $t = 0$ in the original inertial frame (I will use $t$ to denote time according to this frame). Clock A also reads $\tau_A = 0$ (I will use $\tau$ to denote that actual times read on each clock) when it starts. Clock B starts at $x = x_B$ at time $t = 0$, and clock B reads $\tau_B = 0$ when it starts.

Now clocks A and B each accelerate for a fixed time $\Delta \tau$, i.e., clock A stops accelerating when it reads $\tau_A = \Delta \tau$, and clock B stops accelerating when it reads $\tau_B = \Delta \tau$. We assume they each had the same (constant) proper acceleration $a$ while they were accelerating. Then they will both stop at the same time $t = \Delta t$ according to the initial rest frame, and in that time, they will each have traveled the same distance $\Delta x$ according to that frame. So in the initial rest frame, clock A will stop accelerating when it reaches $x_A + \Delta x$, and clock B will stop accelerating when it reaches $x_B + \Delta x$.

Now, when you say the two clocks stop accelerating "at the same time", this is frame-dependent. In the initial rest frame, yes, they do; they both stop accelerating at $t = \Delta t$. But they are spatially separated when they stop; the events at which they stop accelerating are given by $x = x_A + \Delta x, t = \Delta t$, and $x = x_B + \Delta x, t = \Delta t$. This pair of events is simultaneous according to the initial rest frame; they both happen "at the same time" according to that frame.

But simultaneity is relative; those two events are not simultaneous in any other inertial frame. That is, only an observer (or a clock) which is at rest in the original rest frame will see those two events as simultaneous. Clock A and clock B are not at rest in that frame when they stop accelerating; they are both moving in the $x$ direction with some nonzero speed $v$. So the two events at which the two clocks stop accelerating are not simultaneous according to either clock. And at those two events, the two clocks read the same time, $\Delta \tau$. So the event at which clock A reads $\Delta \tau$ is not simultaneous, according to either clock A or clock B, with the event at which clock B reads $\Delta \tau$. This is why the clocks get out of sync; more precisely, it is what "the two clocks are out of sync" means.

 

[Micheth]

Now clocks A and B each accelerate for a fixed time Δτ\Delta \tau, i.e., clock A stops accelerating when it reads τA=Δτ\tau_A = \Delta \tau, and clock B stops accelerating when it reads τB=Δτ\tau_B = \Delta \tau. We assume they each had the same (constant) proper acceleration aa while they were accelerating. Then they will both stop at the same time t=Δtt = \Delta t according to the initial rest frame, and in that time, they will each have traveled the same distance Δx\Delta x according to that frame. So in the initial rest frame, clock A will stop accelerating when it reaches xA+Δxx_A + \Delta x, and clock B will stop accelerating when it reaches xB+Δx

This is essentially what I'm assuming, yes. In the initial rest frame (where we also we have the lightning bolts and the platform observer).
(I don't necessarily need to have them stop accelerating, just get hit by lightning bolts after equal distances have been traversed.)
So that whatever they do after that, they will continue to record the time they were hit.

The logical disconnect (for me) is that, we could have separately accelerated them on different days, the same distance, same acceleration, and doing it a hundred times, each time the experiment was conducted with either clock separately, it would show the same time elapsed, even if we did the individual experiments on different sides of the universe (the individual experiments being separated by distance), since the laws of physics apply everywhere. Each time the experiment would show the same time elapsed on the clock, right?

But somehow doing the two experiments when both clocks happen to read the same time (simultaneously in their rest frame), why only in that case would they read different times (= having gotten out of sync) at the very same points where they have traversed the very same distance as in each of the individual experiments?

 

[PeterDonis]

Each time the experiment would show the same time elapsed on the clock, right?

Yes. But that is not sufficient to show that the clocks are in sync. "In sync" means that a given reading on clock A is simultaneous, according to clock A and clock B, with the same reading on clock B.

If you do the experiment with different clocks on different days, you can't even judge whether they are "in sync", because neither clock can even compare its reading with the other's. You have to do the experiment with both clocks at once for "in sync" to even have any meaning. It makes no sense to say that your clock on Monday is "in sync" with my clock on Tuesday.

when both clocks happen to read the same time (simultaneously in their rest frame)

They don't have the same "rest frame" once they start moving.

why only in that case would they read different times (= having gotten out of sync) at the very same points

They are NOT at "the very same points". They are spatially separated. You keep on ignoring this.

 

[Micheth]

If you do the experiment with different clocks on different days, you can't even judge whether they are "in sync", because neither clock can even compare its reading with the other's. You have to do the experiment with both clocks at once for "in sync" to even have any meaning. It makes no sense to say that your clock on Monday is "in sync" with my clock on Tuesday.

Sorry if was confusing. Certainly performing on different days would not be in sync! But subjecting them to exactly defined acceleration processes through exactly defined distances, wherever they were in the universe, would cause them to register a deltaT that would always be the same, correct?
This deltaT would seem to have to be the same whether you performed the experiment separately, or when they just happened to read identical times.

They are NOT at "the very same points". They are spatially separated. You keep on ignoring this.

Perhaps the nature of the experiment got lost in the conversation. Here's how I'm proposing they are lined up:

ClockA initial position (CA)
|
(distanceD between them)
|
Clock B initial position (CB)
|
(distanceC to the platform)
|
LightningStrikeA position (LSA)
|
(distanceD)
|
LightningStrikeB position (LSB)

On one day you accelerate ClockA from CA to LSA. It registers a time elapse.
Another time you accelerate ClockB from CB to LSB. It registers exactly the same time elapse, because the distance is the same (staggered though it is).
This would have to be the case because the laws of physics don't change just because you become translated in space.
You do both a hundred times and always the same time elapse is registered.

But then you conduct both experiments not on different days but on the same day when they just happen to read the same time (synchronized in their rest frame)
Yes, they are spatially separated but they each start at the same points as before, that's what I mean by them being at the very same points.
Their times are noted when they start and when they arrive at LSA and LSB, respectively. Since they are still the same distances (as in the previous experiments) shouldn't they register the same time differences as in the individual experiments?

 

[A.T.]

Since they are still the same distances (as in the previous experiments) shouldn't they register the same time differences as in the individual experiments?

Sure, they will get stuck showing the same times, as explained here:
https://www.physicsforums.com/threa...ty-in-simultaneity.789927/page-3#post-5045831

 

[PeterDonis]

subjecting them to exactly defined acceleration processes through exactly defined distances, wherever they were in the universe, would cause them to register a deltaT that would always be the same, correct?

Yes, but that is NOT what "in sync" MEANS. Please read what I'm posting; you keep on skipping right over the key point. I'll quote it and bold the key part:

"In sync" means that a given reading on clock A is simultaneous, according to clock A and clock B, with the same reading on clock B.

In other words, to even judge whether clocks are "in sync", you need to know what events clock A and clock B see as simultaneous. None of what you have said addresses that at all. You haven't even talked about simultaneity. At most, you have implicitly used the notion of simultaneity in the original rest frame; but that is the wrong notion of simultaneity to use if you want to know whether the clocks are in sync. You have to use the clocks' own notion of simultaneity, and you have not addressed that at all.

It's useless to keep posting the same things about how your experiment is set up and what each clock reads. We don't disagree about any of that. You need to specifically address the simultaneity issue, as I have described it above.

 

[Micheth]

Yes, but that is NOT what "in sync" MEANS.
"In sync" means that a given reading on clock A is simultaneous, according to clock A and clock B, with the same reading on clock B.
In other words, to even judge whether clocks are "in sync", you need to know what events clock A and clock B see as simultaneous. None of what you have said addresses that at all. You haven't even talked about simultaneity.

As regards simultaneity, the purpose of my experiment was in fact all about simultaneity; that there stubbornly seems to be an objective simultaneity even though the rider experiences it differently.
As far as I can comprehend, the fact that the rider experiences the front lightning bolt first and the rear one later, is simply the trivial fact that the lights reach him at different times, whereas as the clock readings show, they really happened simultaneously, as can be objectively shown by the readings on the broken clocks, each of which would have traversed the same distance with identical accelerations.
(For me that's what this is all about.)
Of course I understand that the rider would not experience them in sync, my point is that, since the separate experiments always take the same amount of time, and in this experiment they are trivially simply being conducted in tandem (being started when both clocks read the same time), they "really did" get broken after an equal elapse of time, the rider simply not seeing them to happen simultaneously, but upon stopping the system, looks at the clocks' readings, knowing they performed exactly as in the previous separate experiments, and concludes (ah, well I saw them occur at different times, but the record shows it was merely an illusion since I was moving at the time).
That's my problem with the "relative simultaneity" issue.

It's not that I don't agree the moving entities experience the "out of synchness" (if the rider got shot from the front and behind he would feel them at different times, even though they were shot simultaneously (from the ground view)). It's that it seems that one of the perspectives must actually be correct given the recorded clock times and their equal acceleration histories.

I understand that SR claims differently, but isn't that merely an interpretation, between saying "it really was non-simultaneous, depending on the reference frame", as opposed to "the rider merely experienced the illusion of non-simultaneity"?

 

[PeterDonis]

the fact that the rider experiences the front lightning bolt first and the rear one later, is simply the trivial fact that the lights reach him at different times

Different times according to his clock, yes. That's what "experiences one first and the other one later" means.

I'm going to phrase things in terms of the "two clocks accelerating" version of your scenario for the rest of this post, because it's cleaner.

whereas as the clock readings show, they really happened simultaneously

No, the clock readings being the same shows that those events happened simultaneously in a particular reference frame--the frame in which the clocks were originally at rest. The clock readings being the same does not show that those events happened simultaneously in the rest frame of either of the clocks after they accelerated. (Those are two different frames, btw, because the clocks are spatially separated.) There is nothing that makes one particular frame's notion of simultaneity the "real" one.

I understand that SR claims differently, but isn't that merely an interpretation, between saying "it really was non-simultaneous, depending on the reference frame", as opposed to "the rider merely experienced the illusion of non-simultaneity"?

No, the relativity of simultaneity is not an "interpretation", it's a physical fact. To see why, consider a variation on your scenario:

Clock A and clock B are as in the previous version of the scenario: they start off both at rest and synchronized in a particular frame; they both start accelerating at $t = 0$ in that frame, at which instant both of the clocks read $\tau = 0$; each clock accelerates until it reads $\Delta \tau$, and then stops accelerating. After each clock stops accelerating, it is moving at speed $v$ in the original rest frame (the same speed for both clocks). Each clock continues to advance after it stops accelerating.

Clock C and clock D start out synchronized and at rest in a frame that is moving with speed $v$ relative to the frame in which clock A and clock B are originally at rest. Their motion is arranged so that clock C is co-located with clock A at the instant that clock A stops accelerating, and clock D is co-located with clock B at the instant that clock B stops accelerating.

Now the physical meaning of relativity of simultaneity is simple: if we suppose that clock C's "zero" of time is chosen so that it reads the same as clock A at the instant when clock C meets clock A, i.e., that clock C reads $\Delta \tau$ at that instant, then clock D will not read $\Delta \tau$ at the instant at which it meets clock B. Clock D will read an earlier time than $\Delta \tau$ (how much earlier depends on the speed $v$ and the spatial separation between the clocks). So clocks C and A will end up reading the same time from $\Delta \tau$ on (since they are now spatially co-located and at rest relative to each other), but clocks D and B will not end up reading the same time; they will remain out of sync by the same amount forever.

So now, if you insist on an absolute notion of simultaneity, you have the following paradox: clocks A and B started out synchronized, and clocks C and D started out synchronized. When clock C meets clock A, they read the same time, and continue reading the same time forever after that; when clock D meets clock B, they do not read the same time, and they stay out of sync by the same amount forever after that. So we have two clocks, C and A, both reading the same time, but they are supposedly "synchronized" with two clocks, D and B, that are reading different times. So which clock are C and A "really" synchronized with at the end--D or B?

The answer SR gives is "neither, because there is no absolute meaning to simultaneity". According to the simultaneity convention of the original rest frame of A and B, clock A (and therefore C) and clock B are synchronized; according to the simultaneity convention of the rest frame of C and D, clocks C (and therefore A) and D are synchronized. There is no contradiction because both senses of simultaneity are just conventions; but the difference in readings between clocks B and D when they are co-located is not a convention, it's a physical fact, and it directly illustrates relativity of simultaneity.

 

[Micheth]

Clock C and clock D start out synchronized and at rest in a frame that is moving with speed vv relative to the frame in which clock A and clock B are originally at rest. Their motion is arranged so that clock C is co-located with clock A at the instant that clock A stops accelerating, and clock D is co-located with clock B at the instant that clock B stops accelerating.

Now the physical meaning of relativity of simultaneity is simple: if we suppose that clock C's "zero" of time is chosen so that it reads the same as clock A at the instant when clock C meets clock A, i.e., that clock C reads Δτ\Delta \tau at that instant, then clock D will not read Δτ\Delta \tau at the instant at which it meets clock B. Clock D will read an earlier time than Δτ\Delta \tau (how much earlier depends on the speed vv and the spatial separation between the clocks).

I'm not sure i understand: Are C & D mutually accelerated in the same way and separated by the same distance as A & B?
If so then I would intuitively think that all 4 are in sync at the point where A meets C and D meets B.
But if not, that is if C & D were accelerated differently, C in such a way that it reads \Delta \tau when it meets A, but not necessarily D in the same way, then I would guess that D would not match any of the other 3 clocks anymore because of its different acceleration history and hence different time dilation with respect to the others.

 

[PeterDonis]

Are C & D mutually accelerated in the same way and separated by the same distance as A & B?

No, C and D are moving inertially the whole time. They just happen to be moving inertially in such a way that, when A and B stop accelerating, they are co-located with, and moving at the same speed as, C and D respectively.

if C & D were accelerated differently

C and D are never accelerated at all. As above, they are moving inertially the whole time. And they are synchronized in the inertial frame in which they are always at rest.

If it helps, here's how the experiment looks in the inertial frame in which C and D are always at rest: initially, A and B are moving at speed $- v$, but then A and B decelerate until they stop moving; and when they stop moving, they are co-located with C and D, respectively. In this frame, A and B do not start decelerating at the same time, or stop at the same time; B starts decelerating before A, and stops before A. Note that this does not change the time that clocks A and B display at the instants they start and stop decelerating; they both display time zero at the instant they start decelerating, and they both display time $\Delta \tau$ at the instant they stop decelerating. (This means, of course, that in this frame clocks A and B are never synchronized.)

 

[Micheth]

No, C and D are moving inertially the whole time. They just happen to be moving inertially in such a way that, when A and B stop accelerating, they are co-located with, and moving at the same speed as, C and D respectively.

If it helps, here's how the experiment looks in the inertial frame in which C and D are always at rest: initially, A and B are moving at speed $- v$, but then A and B decelerate until they stop moving; and when they stop moving, they are co-located with C and D, respectively. In this frame, A and B do not start decelerating at the same time, or stop at the same time; B starts decelerating before A, and stops before A. Note that this does not change the time that clocks A and B display at the instants they start and stop decelerating; they both display time zero at the instant they start decelerating, and they both display time $\Delta \tau$ at the instant they stop decelerating. (This means, of course, that in this frame clocks A and B are never synchronized.)

Hmm... I am having trouble visualizing this as 4 things are going on at the same time, but if i understand correctly, in C and D's frame, they are always at rest and A/B decelerate to their rest position. (correct?)
In this scenario, C and D are not moving with respect to each other? They are both initially and finally separated by the same distance separated by A and B?

 

[PeterDonis]

in C and D's frame, they are always at rest and A/B decelerate to their rest position. (correct?)

Yes.

They are both initially and finally separated by the same distance separated by A and B?

C and D are always separated by the same distance, since they are always at rest in this frame. A and B are not, because B starts decelerating before A does, so A and B end up further apart than they started in this frame. They end up as far apart as C and D (since they end up co-located with C and D), so they start out closer together than C and D are.

 

[Micheth]

Yes.
C and D are always separated by the same distance, since they are always at rest in this frame. A and B are not, because B starts decelerating before A does, so A and B end up further apart than they started in this frame. They end up as far apart as C and D (since they end up co-located with C and D), so they start out closer together than C and D are.

Well in this case, yes I totally agree and accept that A and B would be out of sync, because they moved with respect to each other.
(they end up further apart than when they started).
So either A or B experienced time dilation with respect to the other, so I agree they couldn't be in sync any longer.

But surely this is a different case from my original scenario where they accelerated in the same way and therefore never moved with respect to each other?

 

[PeterDonis]

surely this is a different case from my original scenario

No; it is exactly the same as your original scenario, just with C and D added. The motion of A and B is identical to the motion of A and B in your original scenario. My description of A and B's motion with respect to C and D's rest frame is a description of the same motion of A and B that's in your original scenario, just viewed from a different inertial frame. That's the whole point.

 

[Micheth]

No; it is exactly the same as your original scenario, just with C and D added. The motion of A and B is identical to the motion of A and B in your original scenario. My description of A and B's motion with respect to C and D's rest frame is a description of the same motion of A and B that's in your original scenario, just viewed from a different inertial frame. That's the whole point.

But, if it's the same as my original scenario, then B doesn't in fact start decelerating before A does (it just looks that way to C and D who are at rest as A&B come towards them).
I mean, if A and B each took photographic records of their clock times at the moment they began decelerating, and later gave these photos to C and D when they met, if I were C or D I would conclude, "ah, now I see that the times read the same, and A and B also report that they were originally synchronized in a rest frame, so although it looked to us like B decelerated first, the facts record that they decelerated together, so that's what happened."

 

[A.T.]

...doesn't in fact start decelerating before ... the facts record...

There is no "in fact" here, only "in frame ...". There is no "facts record" here, just "record of frame ...".

 

[PeterDonis]

f it's the same as my original scenario, then B doesn't in fact start decelerating before A does (it just looks that way to C and D who are at rest as A&B come towards them).

As A.T. said, there is no "in fact". No inertial frame is "privileged" over any other.

if A and B each took photographic records of their clock times at the moment they began decelerating, and later gave these photos to C and D when they met, if I were C or D I would conclude, "ah, now I see that the times read the same, and A and B also report that they were originally synchronized in a rest frame, so although it looked to us like B decelerated first, the facts record that they decelerated together, so that's what happened."

You're missing a key point. Before A and B decelerate, they are both at rest in one frame, and are synchronized in that frame. After they decelerate, they are both at rest in another frame (the C/D rest frame), and they are not synchronized in that frame. If, once they are at rest in the C/D frame, A and B try to perform the same procedure that they originally performed to synchronize themselves in their original rest frame, if they were still synchronized, that procedure would do nothing; it would just tell them "no adjustment needed, you're still synchronized". But in fact, they will find that they are no longer synchronized: that same procedure will now tell them that B's clock is "ahead" of A's, so one of the two will have to be adjusted to put them back in sync.

 

[Micheth]

You're missing a key point. Before A and B decelerate, they are both at rest in one frame, and are synchronized in that frame. After they decelerate, they are both at rest in another frame (the C/D rest frame), and they are not synchronized in that frame. If, once they are at rest in the C/D frame, A and B try to perform the same procedure that they originally performed to synchronize themselves in their original rest frame, if they were still synchronized, that procedure would do nothing; it would just tell them "no adjustment needed, you're still synchronized". But in fact, they will find that they are no longer synchronized: that same procedure will now tell them that B's clock is "ahead" of A's, so one of the two will have to be adjusted to put them back in sync.

That's exactly it, I don't see why A/B would not be synchronized in C/D's rest frame, once A/B reaches the point where A/B meets C/D.
I fully agree that C/D would see B's clock running ahead as A/B approach (B being in front of A), but as they got closer certainly B would appear to run less and less ahead of A until, when they reached the meeting point (side by side in the same rest frame where light from them would reach C/D essentially in tandem), light from B would no longer be reaching them earlier, and C/D would seem them as synchronized.

 

[PeterDonis]

I don't see why A/B would not be synchronized in C/D's rest frame, once A/B reaches the point where A/B meets C/D.

That's why I suggested learning the math; it makes it a lot easier to see why counterintuitive results like this are in fact true.

However, the repeated mention of relativity of simultaneity in this thread should give you a clue. Consider the relevant events:

* Event A0 is the event at which clock A starts accelerating (or decelerating, depending on which frame you are using).

* Event B0 is the event at which clock B starts accelerating (or decelerating).

* Event A1 is the event at which clock A stops accelerating (or decelerating).

* Event B1 is the event at which clock B stops decelerating.

For conciseness, we'll also label the frame in which A and B are originally at rest frame AB, and the frame in which C and D are at rest, and in which A and B are both at rest after they finish accelerating/decelerating, frame CD.

Now, by hypothesis, events A0 and B0 are simultaneous in frame AB, and events A1 and B1 are also simultaneous in frame AB. By relativity of simultaneity, that means events A0 and B0 are not simultaneous in frame CB, nor are events A1 and B1; in frame CD, event B0 happens before A0, and event B1 happens before A1. Hence, A and B are never synchronized in frame CD. This is a simple, obvious consequence of relativity of simultaneity, and it is not at all in doubt; it is an unambiguous prediction of SR.

Mathematically, you could verify the above by assigning coordinates in frame AB to all four events, and then Lorentz transforming all four sets of coordinates into frame CD, and seeing that the time coordinates of A0 and B0, and A1 and B1, are not the same in frame CD, even though they were the same in frame AB.

as they got closer certainly B would appear to run less and less ahead of A until, when they reached the meeting point (side by side in the same rest frame where light from them would reach C/D essentially in tandem), light from B would no longer be reaching them earlier, and C/D would seem them as synchronized.

Light travel time alone is not sufficient to determine clock synchronization. All it can tell you by itself is that, once clocks A and B are both at rest in frame CD, they are both running at the same rate as clocks C and D, which is true. But running at the same rate is not enough for clock synchronization; the "zero points" of both clocks also have to be the same, and they're not. They were in frame AB, but they're not in frame CD. That's what relativity of simultaneity tells you.

Once again, learning the math, or even better, learning how to draw and interpret spacetime diagrams, will make it a lot easier to see how all this works.

 

[Micheth]

Event A0 is the event at which clock A starts accelerating (or decelerating, depending on which frame you are using).
* Event B0 is the event at which clock B starts accelerating (or decelerating).
* Event A1 is the event at which clock A stops accelerating (or decelerating).
* Event B1 is the event at which clock B stops decelerating.
For conciseness, we'll also label the frame in which A and B are originally at rest frame AB, and the frame in which C and D are at rest, and in which A and B are both at rest after they finish accelerating/decelerating, frame CD.
Now, by hypothesis, events A0 and B0 are simultaneous in frame AB, and events A1 and B1 are also simultaneous in frame AB. By relativity of simultaneity, that means events A0 and B0 are not simultaneous in frame CB, nor are events A1 and B1;

(You are correct that I would have to learn to do the math to properly converse with you on this. I appreciate your efforts, though.)

What happens, though, when A and B (who have taken photographic records of their clock readings at A0 and B0, respectively, and also at events A1 and B1, respectively, which were simultaneous in their rest frame AB (they merely saw the rest of the universe accelerate toward them), and show them to C&D?

 

[PeterDonis]

at events A1 and B1, respectively, which were simultaneous in their rest frame AB

No, frame AB is not the rest frame of A and B at events A1 and B1. That's the whole point. The inertial frame AB is their rest frame at events A0 and B0, but not at events A1 and B1. At events A1 and B1, the rest frame of clocks A and B (respectively) is frame CD, not frame AB.

You could try to define a non-inertial frame N in which A and B are always at rest, even though they have nonzero proper acceleration for some period of time. (Note that there is no unique way to do this, and all of the possible ways have significant limitations.) But if you do that, you can no longer assume that A and B will stay synchronized if they both stay at rest in the frame. That assumption is only valid for inertial frames.

 

[Micheth]

No, frame AB is not the rest frame of A and B at events A1 and B1. That's the whole point. The inertial frame AB is their rest frame at events A0 and B0, but not at events A1 and B1. At events A1 and B1, the rest frame of clocks A and B (respectively) is frame CD, not frame AB.

In SR I had always thought we could consider any frame to be at rest? i.e., AB decelerating to CD which are "at rest" would be in SR equivalent to CD decelerating toward AB which are "at rest"...?

 

[A.T.]

AB decelerating to CD which are "at rest" would be in SR equivalent to CD decelerating toward AB which are "at rest"...?

Only inertial frames are equivalent.

 

[PeterDonis]

In SR I had always thought we could consider any frame to be at rest?

It depends on what properties you want a "frame" to satisfy. If you want clocks at rest in the frame to stay synchronized, then the frame must be an inertial frame, as I said before. The frame in which C and D are always at rest is an inertial frame, but the "frame" in which A and B are always at rest is not, because A and B have nonzero proper acceleration for a period of time. (I put "frame" in quotes for A and B because, as I said before, there is no unique way to define a non-inertial frame in which A and B are always at rest.)

More generally, as A.T. said, in SR, only inertial frames are all equivalent. SR certainly does not say that all frames, inertial or not, are equivalent. GR makes a more general claim of that sort, but the sense of "equivalence" being used, in terms of what properties all the "equivalent" coordinates can be assumed to satisfy, is much weaker.

 

[Nugatory]

In SR I had always thought we could consider any frame to be at rest? i.e., AB decelerating to CD which are "at rest" would be in SR equivalent to CD decelerating toward AB which are "at rest"...?

Frames aren't at rest or not, they're inertial or not. "At rest" applies to objects, and we say that an object is at rest in a frame if its spatial coordinates in that frame are constant. (Do not allow yourself to be confused by the frequently used term "rest frame of <something>" - that's just a convenient shorthand for "frame in which <something> is at rest").

The rest frame of an object experiencing no forces (let's not get into gravitational forces right now) is inertial, and all inertial frames are equivalent. However, the rest frame of an accelerating/decelerating object is not inertial.

 

[Micheth]

It depends on what properties you want a "frame" to satisfy. If you want clocks at rest in the frame to stay synchronized, then the frame must be an inertial frame, as I said before. The frame in which C and D are always at rest is an inertial frame, but the "frame" in which A and B are always at rest is not, because A and B have nonzero proper acceleration for a period of time. (I put "frame" in quotes for A and B because, as I said before, there is no unique way to define a non-inertial frame in which A and B are always at rest.)

Because A and B were accelerated, they are no longer inertial, correct? I feel I am back to square one in my logical disconnect. :-)
(I know I keep coming back to the same thing: it is not that I doubt that the math works out that way, it is the seeming logical contradiction I keep running into, i.e:
Set A to time 0, distance A0 to A1 is fixed, run experiment, at experiment end it always records a fixed time (deltaT)
Set B to time 0, distance B0 to B1 is fixed, run experiment, at experiment end it always records (the same) fixed time (deltaT)
But according to SR, if you synchronize A&B and program both to begin the experiment when they both read time 0, then when they reach A1/B1, and they are out of sync! (record different deltaT's)

Is it simply necessary to conclude, as we do in quantum weirdness, that "it doesn't necessarily make logical sense but doing the math leads us unescapably to that conclusion, so it must simply be accepted"?

 

[PeterDonis]

Because A and B were accelerated, they are no longer inertial, correct?

Yes. More precisely, they are not moving inertially while they are accelerating; before the acceleration, they are at rest in one inertial frame, but after the acceleration, they are at rest in a different inertial frame. There is no single inertial frame in which they are at rest for the entire time.

according to SR, if you synchronize A&B and program both to begin the experiment when they both read time 0, then when they reach A1/B1, and they are out of sync! (record different deltaT's)

They do not "record different deltaT's". The time elapsed on clock A between events A0 and A1 is the same as the time elapsed on clock B between events B0 and B1. The "disconnect" here is your unstated assumption that, if the time elapsed on both clocks is the same between the start and end of the acceleration, then if they were in sync before the acceleration, they must still be in sync after the acceleration. That assumption is false.

Is it simply necessary to conclude, as we do in quantum weirdness, that "it doesn't necessarily make logical sense but doing the math leads us unescapably to that conclusion, so it must simply be accepted"?

It does make logical sense, but only if you reason from correct assumptions. SR does violate assumptions that seem very natural to our pre-relativistic intuitions; the one I described above is one of them.

 

[Micheth]

They do not "record different deltaT's". The time elapsed on clock A between events A0 and A1 is the same as the time elapsed on clock B between events B0 and B1. The "disconnect" here is your unstated assumption that, if the time elapsed on both clocks is the same between the start and end of the acceleration, then if they were in sync before the acceleration, they must still be in sync after the acceleration. That assumption is false.

I could accept them reading the same times but being out of sync, if they moved with respect to each other (where there would have been time dilation between them), but in the final analysis all I see that has happened is that they had their coordinates translated by equal amounts.
We're ... not talking about time dilation (or are we?)

 

[PeterDonis]

I could accept them reading the same times but being out of sync

Um, what? "Out of sync" means what I have repeatedly said: that the two clocks have different readings at simultaneous events. Once the clocks are both at rest in frame CD (after the acceleration is complete), they have different readings at events which are simultaneous in that frame. You seem to agree that this is the case, so I really don't understand why this thread has gone on so long.

all I see that has happened is that they had their coordinates translated by equal amounts

The elapsed times on the two clocks are not coordinates; they are direct observables. The fact that the clocks read different times at events which are simultaneous in frame CD is also a direct observable. The fact that the distance each clock travels during its acceleration, in frame AB, is the same, is true, but irrelevant; that's not what determines whether the clocks are in sync.

We're ... not talking about time dilation (or are we?)

No. After the acceleration, both clocks are at rest in the same inertial frame, so neither one is time dilated relative to the other. I've already said that they both run at the same rate once they stop accelerating. But I've also explained that running at the same rate is not sufficient for them to be in sync. I get the feeling that you are not really reading my posts.

 

[Micheth]

No. After the acceleration, both clocks are at rest in the same inertial frame, so neither one is time dilated relative to the other. I've already said that they both run at the same rate once they stop accelerating. But I've also explained that running at the same rate is not sufficient for them to be in sync. I get the feeling that you are not really reading my posts.

Well, my attempts to articulate why this seems such a paradox aren't being too successful, and you're evidently finding it somewhat exasperating, so perhaps we should just let this close.
I appreciate your efforts, though.
Cheers.

 

[A.T.]

Well, my attempts to articulate why this seems such a paradox aren't being too successful

It seems like a paradox because of imprecise reasoning, which comes from imprecise language.

 

[PeterDonis]

my attempts to articulate why this seems such a paradox aren't being too successful

The word "seems" is key. I understand why it seems like a paradox to you: because you are reasoning from assumptions that, however intuitively plausible they seem, are not valid in a relativistic universe. But we do live in a relativistic universe; experiments demonstrate that. So those assumptions are simply not valid, however intuitively plausible they seem. Part of learning relativity is learning to give up intuitively plausible but invalid assumptions; we've all had to go through that at some point in the learning process.

 

[Micheth]

Once the clocks are both at rest in frame CD (after the acceleration is complete), they have different readings at events which are simultaneous in that frame. You seem to agree that this is the case, so I really don't understand why this thread has gone on so long.

(Actually I said I could agree with them being out of sync if there was time dilation (which you've pointed out there is not).

I assure you I'm reading your posts. How much I understand is a different issue, but if I went through and read your posts again anyway.

As I understand it:
AB are separated (let's say by 10 meters e.g.), and synchronized in a rest frame.
There are two points in AB's original rest frame (far, far away) that are C & D, also separated by 10 meters in that same frame.
AB then accelerate to some near-light speed toward CD.
I was originally thinking you meant that AB's clocks would simply get out of sync because they accelerated.
That was mistaken, right?
Instead, then what's really happening is that the distance between AB has increased (from CDs viewpoint) as they pass CD, and therefore they cannot respectively meet CD at the same time, correct? (because the distance AB is now different from distance CD). From AB's viewpoint, CD are no longer separated by 10 meters but by some shorter distance as they measure it.
Hence the non-simultaneity of the meeting of AC vs. BD.
If AB's clocks are fine, I'm happy with that :-)

Now, to be honest I still can't understand why distance AB would have to increase (in anybody's frame) since AB merely translated their coordinates in space (underwent identical movements in the same direction), but not to open another can of worms. :-)

 

[PeterDonis]

AB are separated (let's say by 10 meters e.g.), and synchronized in a rest frame.

Yes.

There are two points in AB's original rest frame (far, far away) that are C & D, also separated by 10 meters in that same frame.

No. C and D are two clocks that are moving, in A and B's original rest frame, with some speed $v$. C and D never accelerate; they remain in inertial motion the whole time.

AB then accelerate to some near-light speed toward CD.

No. A and B accelerate in such a way that they just match speeds with C and D, respectively, at the instants when C and D are spatially co-located with A and B, respectively. In other words, C and D are behind A and B, respectively, when A and B start accelerating; but C and D are moving faster, so they catch up to A and B, respectively (all of this is as seen by A and B), just as A and B match speeds with C and D, respectively.

I was originally thinking you meant that AB's clocks would simply get out of sync because they accelerated.
That was mistaken, right?

No. A and B's clocks do get out of sync because they accelerated. Their clocks stay in sync in their original rest frame (meaning the inertial frame in which they were originally at rest and synchronized), but A and B do not remain at rest in their original rest frame. They accelerate, hence they start moving in that frame. When they finish accelerating, they are both moving relative to that original frame, so the fact that they are synchronized in that frame is no longer relevant. See further comments below.

what's really happening is that the distance between AB has increased (from CDs viewpoint) as they pass CD

Please stop using the word "really". You keep on applying it to things that are frame-dependent. That's a very bad habit in relativity.

The correct way to state how things look in the frame in which C and D is at rest is basically what you say: A and B are moving towards C and D in this frame; then B starts decelerating; then A, a bit later, starts decelerating; then B comes to a stop right at D; then A comes to a stop right at C. Because B started decelerating first, the distance between A and B, in this frame, increases. But this is all relative to C and D's rest frame, so the word "really" is inappropriate.

(There are ways of describing the motion of A and B, and C and D, in frame-invariant form. For example, we could say that the worldlines of A and B, during the period when both are accelerating, have positive expansion; whereas the worldlines of C and D always have zero expansion. "Expansion" here is a technical term, referring to the frame-invariant mathematical description of these motions. In this sense, you could say that the distance between A and B "really" does increase as they accelerate. But then you would have to remove the phrase "from CDs viewpoint", because the expansion is frame-invariant.)

Hence the non-simultaneity of the meeting of AC vs. BD.

Non-simultaneity in frame CD. The two meetings are simultaneous in frame AB (the frame in which A and B were originally at rest). But once A and B are no longer at rest in frame AB, that frame's definition of simultaneity is no longer the right one to use to judge A and B's clock synchronization.

In other words, the reason A and B's clocks get out of sync is that they have changed inertial frames: they start out at rest in one inertial frame, and end up at rest in a different inertial frame. (They have to accelerate to do this, which is why I said above that they get out of sync because they accelerated.) That changes which definition of simultaneity is the one that determines whether their clocks are synchronized.

If all that is too abstract, consider how, physically, A and B would check that their clocks are synchronized. (This will also show how they can check the distance between them and verify that, as far as they are concerned, it has increased.) In the original rest frame, AB, before any acceleration has taken place, A and B can exchange light signals, and determine two things: (1) that the distance between them is in fact 10 meters (or whatever it turns out to be), based on the round-trip travel time of light signals between them; and (2) that their clocks are in fact synchronized, based on the fact that, when A receives a light signal from B, it shows B's clock reading exactly what A's clock read one light-travel time ago (so if they are 10 meters apart, when A receives a light signal from B, B's clock will read what A's clock read 33 nanoseconds ago, since it takes light about 33 nanoseconds to travel 10 meters). That is, the reading shown on B's clock in the light signal will be 33 nanoseconds before the reading shown on A's clock when the signal is received.

Now, once A and B have completed their acceleration, and are now co-located with C and D and moving inertially, at rest in frame CD instead of frame AB, they can repeat the above process. And when they do, they will find: (1) that the distance between them has increased, based on the round-trip travel time of light signals between them; and (2) that their clocks are no longer synchronized: when A receives a light signal from B, it shows B's clock reading something later than what A's clock read one light travel-time ago, and when B receives a light signal from A, it shows A's clock reading something earlier than what B's clock read one light travel-time ago.

For example, if A and B are now 20 meters apart (corresponding to a relative speed of about 0.87c between frames AB and CD), then when A receives a light signal from B, it will show B's clock reading, not a time 66 nanoseconds before A's (which would be one light-travel time before), but a time about 15 nanoseconds before A's, indicating that B's clock is about 51 nanoseconds ahead of A's. And when B receives a light signal from A, it will show A's clock reading, not a time 66 nanoseconds before B's, but a time about 117 nanoseconds before B's, indicating that A's clock is about 51 nanoseconds behind B's. So the two measurements are consistent (they both show the same offset between the clocks), and they clearly show the clocks not synchronized.

I still can't understand why distance AB would have to increase (in anybody's frame)

Because, as I noted above, in frame CD, B starts decelerating before A does. This has to be true because of relativity of simultaneity; A and B both start accelerating simultaneously in frame AB, therefore they cannot start accelerating (or decelerating, depending on how you view it) simultaneously in any other frame, including frame CD. Therefore, the distance between them must change in any other frame besides frame AB. (And, as the discussion above shows, they can verify that the distance has increased by exchanging light signals.)

Once again, I think part of the issue is that you are trying to reason from the wrong assumptions. Instead of starting with the known properties of relativistic spacetime (the main one here being relativity of simultaneity), and deducing what must happen in the scenario based on those properties, you are trying to start with your assumptions about how things "ought" to work, then wondering why relativity says they don't work that way.

 

[Micheth]

No. C and D are two clocks that are moving, in A and B's original rest frame, with some speed vv. C and D never accelerate; they remain in inertial motion the whole time.

Wow! That's a very long and detailed response which will take me much time to forge through since this is all very difficult to visualize (for me at least).
But first, I see we're not talking about the same thing from the start. I am talking about my original scenario as described above (where C&D are merely the points where lightning bolts will strike when A&B reach them. A&B having their clocks synchronized in that rest frame where A, B, C & D and whatever produces the lightning bolts, are all in the same frame.
(That is, where A&B are not moving with respect to C&D at the start.)
Then A&B accelerate to CD...

(I know part way through you proposed seeing A&B moving from the start and decelerating to rest at C&D, but I managed to simply become more confused, so may I suggest the original scenario.)