Temporal distortion of a moving train

• davidong3000
In summary: If the clocks were synchronized after the train accelerated will the front still be farther ahead in time than the rear section relative to the stationary frame? or the otherway round?In summary, the discussion revolves around the concept of time dilation and the distortion of time on a moving train at near light speeds. The question is posed about whether the front or rear carriage will be further ahead in time from a stationary frame of reference, assuming that the clocks were synchronized when the train was stationary. The conclusion is that the clocks will no longer be synchronized after the train accelerates, and the front and back sections will have different perceptions of time.
davidong3000
Hi i have a question about how time is distorted along the length of a train moving at near light speeds.

Will the front carriage be farther foward in time than the rear carriage from stationary frame of reference? or the otherway round?
Assume that the front and rear sections had clocks that were synchronized when the train was stationary and the train took time to accelerate to near light speeds.

If the clocks were synchronized after the train accelerated will the front still be farther ahead in time than the rear section relative to the stationary frame? or the otherway round?

Dave

davidong3000 said:
Hi i have a question about how time is distorted along the length of a train moving at near light speeds.

Will the front carriage be farther foward in time than the rear carriage from stationary frame of reference? or the otherway round?
Assume that the front and rear sections had clocks that were synchronized when the train was stationary and the train took time to accelerate to near light speeds.
Dave

Good questions.

Foreward clocks always lag aftward clocks, so at any instant in the stationary embankment frame, the moving engine clock will lag the moving caboose clock by gamma(-vx/c^2). This is the temporal offset portion of the time transform ... T=gamma(t-vx/c^2). The -vx/c^2 relates to Minkowski's frame rotation. Gamma is related to the required Fitzgerald contraction.

davidong3000 said:
If the clocks were synchronized after the train accelerated will the front still be farther ahead in time than the rear section relative to the stationary frame? or the otherway round?
Dave

It doesn't matter when the train's clocks are synchronized. Clocks in sync in the train frame before acceleration, remain in sync per the train passengers even during train accelrational periods. So as long as the clocks are synchronized in the train at some point before the embankment observer inspects them on driveby, the moving engine clock will lag the moving caboose clock by gamma(-vx/c^2) units of time.

pess

Geoffrey A Landis disagrees

Geofrrey A Landis a NASA scientist disagreed tho, do a google search for him if u want his contact details and find out who he is but here is what he wrote to me:

Here's a quick thought experiment.

Suppose your "train" consists of two super-incredible rockets, one
sitting next to kilometer marker zero, one sitting next to kilometer
synchronized, and there's no ambiguity about what we mean by time t=0.

So, at time t=0, the front rocket (piloted by Francis) and the back
rocket (piloted by Buck) both turn on their rockets. They accelerate
at 30 quadrillion gravities for one nanosecond. This gets them up to
eighty some percent of the speed of light.

Now, suppose their clocks were still synchronized in the accelerated
frame. In that case, since they were always synchronized in their
own frame, and they both were accelerating the same, then in that
frame they would still be exactly one kilometer apart.

Francis takes a photo out her window, and it shows the 1 kilometer
marker (since, despite their acceleration, she moved only a foot or
so in the nanosecond of acceleration). Buck takes a photo out his
window, and it shows the zero kilometer marker. They compare notes,
and Francis radios to Buck "Lorentz contraction is false! The two
kilometer markers are moving past our one-kilometer long train at
relativistic speed, but they are exactly one kilometer apart in our

Meanwhile, somebody on the ground standing at kilometer marker zero
takes a photo, and somebody else standing at kilometer marker one
takes a photo, both at one nanosecond after ignition. They then say
"Lorentz contraction is false! The ships are one kilometer apart in
their own frame, and the photos prove that they're one kilometer
apart in our frame, too! They're not Lorentz contracted at all!"

Paradox? No. The assumption that the clocks in the front ship and
the back ship are still synchronized is false. And the assumption
that the two ships are rigorously the same distance apart (in their
own frame) is false, since the clocks aren't synchronized.

Clocks synchronized before acceleration are not synchronized after
acceleration (assuming that they are at different positions x, and
the acceleration is in the x direction-- y and z are unaffected).

--
Dr. Geoffrey A. Landis
Ronald E. McNair-NASA Visiting Professor of Astronautics at MIT
http://mit.edu/aeroastro/www/people/landis/landis.html

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Hi,

Right. Once they start moving, their clocks lose simultaneity, and
when the clocks begin to differ, the front and back no longer agree
on things like acceleration.

The ship in back sees the clock on the ship in front as running
fast. The ship in front sees the clock on the ship in back as
running slow. The clocks at the front and back are no longer
synchronized in anybody's reference frame.

The classic way to analyze this problem is assuming that each ship
emits a pulse at regular intervals, and look at when the pulses
arrive at a central (fixed) station. Any special relativity book
with a discussion of the twin paradox should do this analysis.

Oddly, last week a friend forwarded me this, which is a different
discussion of the same thing. I'm not sure I really like his
analysis, but he gets to the same result.

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

--
Dr. Geoffrey A. Landis
Ronald E. McNair-NASA Visiting Professor of Astronautics at MIT
http://mit.edu/aeroastro/www/people/landis/landis.html

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davidong3000 said:
Hi i have a question about how time is distorted along the length of a train moving at near light speeds.

Will the front carriage be farther foward in time than the rear carriage from stationary frame of reference? or the otherway round?
Assume that the front and rear sections had clocks that were synchronized when the train was stationary and the train took time to accelerate to near light speeds.
First point: I hope you realize that "farther forward in time" depends on who is doing the observing.

Second point: The only way you can maintain synchronization of the front and rear clocks with respect to the station frame as the train accelerates is by accelerating the front and rear of the train uniformly with respect to the station. That will introduce forces that will rip the train apart. (On the other hand, if the train accelerates uniformly with respect to its own instantaneous frames, then the clocks will maintain synchronization--according to train observers.)

If the clocks were synchronized after the train accelerated will the front still be farther ahead in time than the rear section relative to the stationary frame? or the otherway round?
As pess5 explained, if the clocks are synchronized in the train frame they will be unsynchronized in the station frame, the clock in front will lag the clock in the rear--according to station observers.

Doc Al said:
As pess5 explained, if the clocks are synchronized in the train frame they will be unsynchronized in the station frame, the clock in front will lag the clock in the rear--according to station observers.

when u say the front clock will lag the rear, do u mean the front clock will read behind the rear clock?

Doc Al said:
First point: I hope you realize that "farther forward in time" depends on who is doing the observing.

(On the other hand, if the train accelerates uniformly with respect to its own instantaneous frames, then the clocks will maintain synchronization--according to train observers.)

.

Why - if the train is accelerated, the effect is the same as a G field, the clock at the lower potential will be viewed as being in a lower gravitational potential - and therefore from the perspective of the front clock, will be seen to run slower; the forward clock will be seen to run faster from the standpoint of the caboose clock. Or did I misinterpret the problem?

davidong3000 said:
when u say the front clock will lag the rear, do u mean the front clock will read behind the rear clock?
That's right. In the train frame, the clocks are synchronized and both strike 3:00pm at the same time. But in the station frame, when the rear clock strikes 3:00pm the front clock reads 2:55pm (for example).

Doc Al said:
Second point: The only way you can maintain synchronization of the front and rear clocks with respect to the station frame as the train accelerates is by accelerating the front and rear of the train uniformly with respect to the station. That will introduce forces that will rip the train apart.

Hi Doc Al,

I agree in that the train would be ripped apart.

However, it seems to me that clocks synchronized in the train frame can never remain in sync per the station observer, given they are in motion. Are you saying that the clocks remain in sync, but that they slow down identically as the train accelrates?

pess

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Doc Al said:
(On the other hand, if the train accelerates uniformly with respect to its own instantaneous frames, then the clocks will maintain synchronization--according to train observers.)

If the train accelerates such that its length, measured by a train observer, remains constant, then the clocks will not be synchronised, neither to train observers nor station observers.

See posts #3 and #4 in this thread.

pess5 said:
I agree in that the train would be ripped apart.

However, it seems to me that clocks synchronized in the train frame can never remain in sync per the station observer, given they are in motion. Are you saying that the clocks remain in sync, but that they slow down identically as the train accelrates?
That's what makes sense to me. Do you disagree? Since--according to the track observers--the clocks undergo the same acceleration for the same time, on what basis would they have different readings? (Again: according to the track frame.)

Let me know if I'm missing something.

Doc Al said:
(On the other hand, if the train accelerates uniformly with respect to its own instantaneous frames, then the clocks will maintain synchronization--according to train observers.)

DrGreg said:
If the train accelerates such that its length, measured by a train observer, remains constant, then the clocks will not be synchronised, neither to train observers nor station observers.

See posts #3 and #4 in this thread.
I don't see how you get that from posts #3 and #4. In those posts, the two rockets are given equal "bursts" with respect to the track. That is not an example of the acceleration being uniform with respect to the "train". Despite the claim that "To start with, nobody's moving, so everybody's clock is synchronized, and there's no ambiguity about what we mean by time t=0.", that's only true for the first instant. As soon as they start moving, the rockets no longer agree that they are undergoing equal accelerations. (Instead of a single nanosecond burst, it may be helpful to think of it as a series of 1000 picosecond bursts.)

By the way, I highly recommend Bell's paper "How to teach special relativity", in which he describes--and properly analyzes--the infamous Bell spaceship paradox.

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Doc Al said:
That's what makes sense to me. Do you disagree? Since--according to the track observers--the clocks undergo the same acceleration for the same time, on what basis would they have different readings? (Again: according to the track frame.)

Let me know if I'm missing something.

Doc Al,

Interesting, because I had never given it much thought before about a scenario such as this one.

I just checked it by inspecting Minkowski worldlines wrt another scenario. The clocks must drop out of sync per the station observers even if the train stretched in such a way to always maintain the same length per the station observer. In fact, the worldlines suggest that the de-synchronization becomes worse, and not better ... that is, they are further out of sync than had the train not stretched at all. The rotation of the frames doesn't change in your scenario, since the train is v per the station whether the train maintain's its original length or not as it accelerates. Although the rotation of frames doesn't change, the clock separations do change since the train lengthens. Hence, the clocks must become more desynchronized than had the train not stretched.

And as DrGreg mentioned, it seems the train clocks must also de-synchronize since they are moving wrt each other if the train stretches, per the train passgengers.

Does this sound reasonable to you?

pess

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pess5 said:
Interesting, because I had never given it much thought before about a scenario such as this one.

I just checked it by inspecting Minkowski worldlines wrt another scenario. The clocks must drop out of sync per the station observers even if the train stretched in such a way to always maintain the same length per the station observer. In fact, the worldlines suggest that the de-synchronization becomes worse, and not better ... that is, they are further out of sync than had the train not stretched at all. The rotation of the frames doesn't change in your scenario, since the train is v per the station whether the train maintain's its original length or not as it accelerates. Although the rotation of frames doesn't change, the clock separations do change since the train lengthens. Hence, the clocks must become more desynchronized than had the train not stretched.
Why would the clock separations change with respect to the station? At all times both clocks move with the same speed--according to the station observers--so they remain at the same separation. (Of course, in their own frame the clock separation does increase and the clocks lose synchronization.)

And as DrGreg mentioned, it seems the train clocks must also de-synchronize since they are moving wrt each other if the train stretches, per the train passgengers.
See my response to DrGreg above. Certainly the train clocks, if accelerated uniformly with respect to the station, will lose synchronization in their own frame.
Does this sound reasonable to you?
Not yet!

During acceleration,

will the train's rear clock appear to tick faster than the train's front clock from stationary frame of reference ?

im asking because i don't see how else the train's rear clock would read farther into the future than the train front clock. Either the rear clock would appear to speed up or the front clock appear to slow down or both.

2nd question: if yes to above, will the train's rear clock tick faster than stationary clock from stationary frame of reference during acceleration?

regarding the latest debate , are u guys taking into account that the train is made up of atoms that are electrostatically attracted to each other and prevent a trully uniform acceleration like that of 2 rockets 1 in front of each other that have no electrostatic forces between them to interfere with their simultaneous accelerations? I am assuming absence of any electrostatic forces between the 2 rockets means the space between rockets do not get lorenz contracted from station frame, but each individual rocket body is lorenz contracted?

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Doc Al said:
Why would the clock separations change with respect to the station? At all times both clocks move with the same speed--according to the station observers--so they remain at the same separation. (Of course, in their own frame the clock separation does increase and the clocks lose synchronization.)

You're right, the clocks do not change in separation per the station. A poor choice of wording on my behalf. Let me word it differently here then ...

The rotation of frames is the same no matter if the train accelerates normally whereby the train contracts in length by 1/gamma, versus the train accelerating in a way to maintain the train length per the station. The degree of rotation between frames is a function of v alone, and drives the de-synchronization of the moving clocks. So given the same frame rotation either way, a larger x means greater desynchronization ...

Since v is the same per either accelerational method, then the desynchronization must become more pronounced if the train maintains its length per the station during train acceleration. This because the separation of moving clocks (x) is greater than it would be have been had the train accelerated normally with clock separation becoming contracted by 1/gamma...

t'=gamma(t-vx/c^2)​

Since v & c are the same no matter if the moving train maintains its length as it accelerates, as opposed to the train accelerating normally whereby the train contracts, then the only variable is x...

Since x is larger per the station if the moving train maintains its length as it accelerates (as opposed to the train accelerating normally whereby the moving train contracts), then the desynchronization must be larger than had the station seen the moving train contract in length.

pess

davidong3000 said:
regarding the latest debate , are u guys taking into account that the train is made up of atoms that are electrostatically attracted to each other and prevent a trully uniform acceleration like that of 2 rockets 1 in front of each other that have no electrostatic forces between them to interfere with their simultaneous accelerations?

davidong3000,

The scernario is impossible of course. I was assuming a train made of stretchable rubber, for example. Actually, I am envisioning a continuous string of clocks which would (say) be sitting in train windows facing outward so the station observer could see them ... then just assume the train vanished but clocks remain. So a string of self powered moving clocks that could emulate where they would be in a rubber train had it stretched.

pess

how can the train accelerate and at the same time remain uncontracted? surely the atoms would each lorenz contract if an external force were to manually stretch the train during acceleration? if this happens wouldn't the train "try" to contract it self electrostatically between atoms? (from stationary frame)

im assuming that during normal acceleration, rear atoms are electrostatically attracted to atoms infront of it that are farther in the future and thus farther infront (had more accelerating time) causing rear atoms to speed up electrostatically to catch up while foward atoms slow down a little as the forces of atoms behind them pulling them back are farther behind both in time and space.

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ok above msg posted or edited out of sync of other above ones lol but correct any mistakes in it please if any.

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repeating my question from prev page...

During acceleration,

will the train's rear clock appear to tick faster than the train's front clock from stationary frame of reference ?

I am asking because i don't see how else the train's rear clock would read farther into the future than the train front clock. Either the rear clock would appear to speed up or the front clock appear to slow down or both.

2nd question: if yes to above, will the train's rear clock tick faster than stationary clock from stationary frame of reference during acceleration?

pess5 said:
You're right, the clocks do not change in separation per the station. A poor choice of wording on my behalf. Let me word it differently here then ...

The rotation of frames is the same no matter if the train accelerates normally whereby the train contracts in length by 1/gamma, versus the train accelerating in a way to maintain the train length per the station. The degree of rotation between frames is a function of v alone, and drives the de-synchronization of the moving clocks. So given the same frame rotation either way, a larger x means greater desynchronization ...

Since v is the same per either accelerational method, then the desynchronization must become more pronounced if the train maintains its length per the station during train acceleration. This because the separation of moving clocks (x) is greater than it would be have been had the train accelerated normally with clock separation becoming contracted by 1/gamma...

t'=gamma(t-vx/c^2)​

Since v & c are the same no matter if the moving train maintains its length as it accelerates, as opposed to the train accelerating normally whereby the train contracts, then the only variable is x...

Since x is larger per the station if the moving train maintains its length as it accelerates (as opposed to the train accelerating normally whereby the moving train contracts), then the desynchronization must be larger than had the station seen the moving train contract in length.
We seem to be talking past each other. Allow me to take another crack at it.

To me, it's clear that two clocks that undergo the same acceleration for the same time period (according to the station frame) must be affected in the same way according to the station frame. If they are synchronized according to the station frame prior to the acceleration, they will remain synchronized according to the station frame once they've reached their final speeds. Symmetry seems to demand this. If you dispute this, please tell me what the difference is between the physical procedures applied to each clock.

It's certainly true that the clocks will not be synchronized in the train frame, once they have reached their final speeds. And it's also true that the degree to which the train frame and station frame disagree about the synchronization of clocks does not depend on how they were accelerated; it only depends on the speed of the clocks and the proper distance between them as follows:

If two clocks are synchronized in the train frame and are separated by a distance D in the train frame, then according to the station frame the clock in front lags behind the clock in the rear by an amount T = Dv/c^2.​

What this means for this example is that after the (now stretched) train reaches its final speed, train observers will find that their clocks have lost synchronization by the amount given above.

And it's also true, as I think you point out, that since the proper distance between clocks ends up greater when the train is uniformly accelerated (with respect to the station), the amount of desynchronization between frames is also greater. But that doesn't change the fact that according to the station frame the clocks remain synchronized.

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Doc Al said:
If they are synchronized according to the station frame prior to the acceleration, they will remain synchronized according to the station frame once they've reached their final speeds. Symmetry seems to demand this. If you dispute this, please tell me what the difference is between the physical procedures applied to each clock.

I believe the electrostatic nature of atoms along the train causes the train to behave like a string of magnets attracted to each other. This forces the rear most atoms of the train to accelerate even before the whistle is blown to signal the train to start moving (it anticipates the future) because they are electrostatically attracted to atoms infront of them that are farther in the future that have already started accelerating. This i believe plays a part in lorenz contraction too.

2 rockets simultaneously accelerating on the other hand are not electrostatically attracted to each other, they can be treated a separate units. The front rocket in the future does not attract/drag the rear rocket in the past foward in space.

So the train although setup to accelerate uniformly is unable to do so. The rear of train moves first (before whistle is blown) and i suspect the clock there (rear) appears to tick faster than the front clock in train (from station ref during acceleration) and also ticks faster than station clock from station ref during acceleration.

The train's front section in the future will cause the rear section in the past to move before any force is applied to the rear section from rear's period in time.

please jump in and correct me if I am wrong about any of this.

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Doc Al said:
We seem to be talking past each other. Allow me to take another crack at it.

To me, it's clear that two clocks that undergo the same acceleration for the same time period (according to the station frame) must be affected in the same way according to the station frame. If they are synchronized according to the station frame prior to the acceleration, they will remain synchronized according to the station frame once they've reached their final speeds. Symmetry seems to demand this. If you dispute this, please tell me what the difference is between the physical procedures applied to each clock.

Doc Al,

Yes, let us assume a velocity of 0.866c, so gamma=2, and an engine clock separated from the caboose clock by a proper length "x", say 5.196 lh (light hrs). In motion at 0.866c, this proper length x must be recorded as a moving 50% contracted length of x/2.

scenario 1 ... train accelerates normally to speed v, and contracts with acceleration per the station by 1/gamma. So the fwd & aft clocks are separated by x/gamma=x/2=5.195lh/2=2.598lh. Then the LT is ...

t'=gamma(t-vx/c^2) = gamma(t-v(x/2)/c^2)​

After acceleration, v is contant at 0.866c. Far as the temporal offset component of the equation goes [ie gamma(-vx/c^2)], the only variable in the equation is "x". So gamma=2 here since v is 0.866c after acceleration is completed. So the clock in the engine & caboose must be separated by "x/gamma", a moving length 50% contracted wrt the proper separation, so 5.196lh/2=2.598lh. The offset is then ... offset=2(-0.866c(5.196/2)/1^2) = 4.5hr. So the fwd clock lags the aft clock by 4.5hr at any station instant.​

scenario 2 ... train accelerates to speed v while always maintaining its original stationary length per the station. Then the LT is ...

t'=gamma(t-vx/c^2)​

After acceleration, v is contant at 0.866c, same as in scenario 1. Far as the temporal offset component of the equation goes [ie gamma(-vx/c^2)], the only variable in the equation is "x". And gamma=2 here as well, as in scenario 1 since v is the same after acceleration. So the clock in the engine & caboose must be separated by a moving "x", numerically identical to the proper length of 5.196lh. The offset is then ... offset=2(-0.866c(5.196)/1^2) = 9.0hr. So the fwd clock lags the aft clock by 9hr at any station instant.​

Look right?

pess

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pess5 said:
Yes, let us assume a velocity of 0.866c, so gamma=2, and an engine clock separated from the caboose clock by a proper length "x", say 5.196 lh (light hrs). In motion at 0.866c, this proper length x must be recorded as a moving 50% contracted length of x/2.

scenario 1 ... train accelerates normally to speed v, and contracts with acceleration per the station by 1/gamma. So the fwd & aft clocks are separated by x/gamma=x/2=5.195lh/2=2.598lh. Then the LT is ...

t'=gamma(t-vx/c^2) = gamma(t-v(x/2)/c^2)​

After acceleration, v is contant at 0.866c. Far as the temporal offset component of the equation goes [ie gamma(-vx/c^2)], the only variable in the equation is "x". The offset is then ... offset=2(-0.866c(5.196/2)/1^2) = 4.5hr. So the fwd clock lags the aft clock by 4.5hr at any station instant.​

scenario 2 ... train accelerates to speed v while always maintaining its original stationary length per the station. Then the LT is ...

t'=gamma(t-vx/c^2)​

After acceleration, v is contant at 0.866c, same as in scenario 1. Far as the temporal offset component of the equation goes [ie gamma(-vx/c^2)], the only variable in the equation is "x". And gamma=2 here as well, as in scenario 1 since v is the same after acceleration. So the clock in the engine & caboose must be separated by a moving "x", numerically identical to the proper length of 5.196lh. The offset is then ... offset=2(-0.866c(5.196)/1^2) = 9.0hr. So the fwd clock lags the aft clock by 9hr at any station instant.​

Look right?
The issue is not the amount of offset (our calculations agree), but who sees the offset. In general, the offset can be interpreted as I described earlier:

(a) If two train clocks are synchronized in the train frame,

(b) then they will be seen by the station frame as being offset by the amount you have calculated.​

In scenario 1, the train clocks are synchronized in the train frame, so the offset is seen by the station frame.

In scenario 2, however, the train clocks are not synchronized in the train frame, but remain synchronized in the station frame since both clocks were identically accelerated according to the station frame. (Again, if you disagree with this, please tell me what was done to one clock that wasn't also done to the other clock.) In this scenario, it's the train observers who note that the the train clocks have become out of synch by the offset amount.

Doc Al said:
The issue is not the amount of offset (our calculations agree), but who sees the offset. In general, the offset can be interpreted as I described earlier:

(a) If two train clocks are synchronized in the train frame,

(b) then they will be seen by the station frame as being offset by the amount you have calculated.​

In scenario 1, the train clocks are synchronized in the train frame, so the offset is seen by the station frame.

In scenario 2, however, the train clocks are not synchronized in the train frame, but remain synchronized in the station frame since both clocks were identically accelerated according to the station frame. (Again, if you disagree with this, please tell me what was done to one clock that wasn't also done to the other clock.) In this scenario, it's the train observers who note that the the train clocks have become out of synch by the offset amount.

Interesting. I do see your point. I'll think on it as to whether the clocks should appear in sync versus become desynchronized in some different fashion, and get back.

pess

Doc Al said:
(Again, if you disagree with this, please tell me what was done to one clock that wasn't also done to the other clock.) In this scenario, it's the train observers who note that the the train clocks have become out of synch by the offset amount.

Doc Al,

I studied it by Minkowski's worldlines. Indeed, the clocks remain in sync as you stated. As I originally had asked about, the clocks must slow down during acceleration identically, thus keeping the clocks in sync per the station. The clocks drop out of sync per the train passengers. Once inertial, the clocks remain in sync at some constant slower rate per the station and wrt the station's clock rate. So although the clocks appear in sync per the station, no 2 clocks actually reside in the same instant "of train time" after acceleration commences.

How's that sound?

thanx,
pess

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pess5 said:
I studied it by Minkowski's worldlines. Indeed, the clocks remain in sync as you stated. As I originally had asked about, the clocks must slow down during acceleration identically, thus keeping the clocks in sync per the station. The clocks drop out of sync per the train passengers. Once inertial, the clocks remain in sync at some constant slower rate per the station and wrt the station's clock rate. So although the clocks appear in sync per the station, no 2 clocks actually reside in the same instant "of train time" after acceleration commences.

How's that sound?
Perfect! (I knew we'd get there eventually. )

Doc Al said:
Perfect! (I knew we'd get there eventually. )

Great minds think alike, even when the one is not as quick as the other.

pess

1. How does the speed of a train affect temporal distortion?

The speed of a train has a direct impact on temporal distortion. As the train moves faster, the time measured on the train will be slower compared to an observer on the ground. This is due to the effects of time dilation, a concept in Einstein's theory of relativity.

2. Can temporal distortion occur on any moving train?

Yes, temporal distortion can occur on any moving train as long as it is moving at a significant speed. However, the effects may not be noticeable unless the train is travelling at a very high speed, close to the speed of light.

3. How can temporal distortion be measured on a moving train?

Temporal distortion can be measured by using precise clocks on the train and comparing them to clocks on the ground. The difference in time between the two clocks will indicate the amount of distortion occurring.

4. What other factors can affect temporal distortion on a moving train?

Aside from speed, the gravitational pull of the Earth can also affect temporal distortion. This is known as gravitational time dilation and can cause a difference in time between the front and back of the train due to the varying strength of gravity.

5. Are there any practical applications of understanding temporal distortion on a moving train?

Yes, understanding temporal distortion can have practical applications in fields such as GPS technology and satellite communication. These systems take into account the effects of time dilation to ensure accurate measurements and data transmission.

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