A special point in special relativity

[n_ktt]

I've been wondering abut the following situation:

Let us imagine that there is a long train that moves with speed of v=sqrt(3/4)*c
For v=sqrt(3/4)*c
gamma=2

The clocks are synchronized when the railway engine meets the observer in reference frame.


When a one unit of time in reference frame has passed (t=1), then the observer
looks on the car that is at position:
x = sqrt(1/3)

Then the oberver is supprised, because the clock in the car shows the same time as its clock:

t' = gamma(t-v/c2 * x)
gamma=2
t=1
x=sqrt(1/3)

t'=2(1-sqrt(3/4)*sqrt(1/3) = 1

Can anybody explain what that it means?

 

[ghwellsjr]

I've been wondering abut the following situation:

Let us imagine that there is a long train that moves with speed of v=sqrt(3/4)*c
For v=sqrt(3/4)*c
gamma=2

The clocks are synchronized when the railway engine meets the observer in reference frame.When a one unit of time in reference frame has passed (t=1), then the observer
looks on the car that is at position:
x = sqrt(1/3)

Then the oberver is supprised, because the clock in the car shows the same time as its clock:

t' = gamma(t-v/c2 * x)
gamma=2
t=1
x=sqrt(1/3)

t'=2(1-sqrt(3/4)*sqrt(1/3) = 1

Can anybody explain what that it means?

It means you transformed an event that is not on the engine but on a car behind the engine. If you also calculate x' you will see that it is equal to -sqrt(1/3) but the engine at t=1 is located at x=sqrt(3/4) and the clock in the engine is at 1/2. This shows the relativity of simultaneity. Events that are simultaneous in the train frame are not necessarily simultaneous in the ground frame and vice versa.

 

[Chestermiller]

In addition to what George said, when you say that "The clocks are synchronized when the railway engine meets the observer in reference frame," this does not mean that the clocks on the ground will be synchronized with the clocks on the train. Even if the clocks on the ground could be instantaneously synchronized with one another when the engine meets the observer, and the clocks on the train could be instantaneously synchronized with one another when the engine meets the observer, this does not mean that the clocks on the train will be synchronized with the clocks on the ground.

 

[n_ktt]

Thank for your answers. You are absolutely right about it. Let me focus on it for a while.

The LT equations in standard form are:

(1) x' = gamma*(x-v*t)
(2) t' = gamma*(t-v/c2 * x)

Supposing that I'm the observer, I've noticed that at every time moment I can find a
car in which clock shows the same time as mine. The point can be named xt
The point xt can be easily found from the second LT equation:

t' = gamma*(t-v/c2 * x) => xt = gamma/(gamma+1) *v*t

The quantity eta can be defined as:

eta = gamma/(gamma+1)

Then xt point can be calculated from:

xt = eta*v*t


eta = 0.5 for v=0
eta goes to 1 when v goes to c

It means that at every time moment I can find a car with the same time as mine.
It is not the same car at different time moments.

 

[n_ktt]

Recently I managed to learn the $\tex$ typescript. So I’ve rewritten my last post with $\tex$ to make it more clear. I hope that a reader won’t be confused.

The LT equations in standard form are:

(1) $x' = \gamma\left(x - v t\right)$
(2) $t' = \gamma\left(t-\frac{v}{c^2} x\right)$

Supposing that I'm the observer, I've noticed that at every time moment I can find a
car in which clock shows the same time as mine. The point can be named $x^t$
The point $x^t$ can be easily found from the second LT equation:

$t' = \gamma \left( t - \frac{v}{c^2} x \right) \quad \rightarrow \quad x^t = \frac{\gamma}{\gamma+1} v t$

The quantity $\eta$ can be defined as:

$\eta = \frac{\gamma}{\gamma+1}$

Then $x^t$ point can be calculated from:

$x^t = \eta v t$

$\eta = 0.5$ for $v = 0$
$\eta \rightarrow 0$ when $v \rightarrow c$

It means that at every time moment I can find a car with the same time as mine.
It is not the same car at different time moments.

 

[ghwellsjr]

I'm not sure what is special about this point. But one thing that concerns me is that you seem to think that an observer can actually see that the time on a remote clock is what you calculated, correct?

For example, in your first post you said: the observer looks on a particular car and he is surprised, because the clock in the car shows the same time as his own clock. In fact, because it takes time for the image of the clock to get to him, his clock will be much later when he actually sees the time that matched his clock according to your calculation.

So maybe if you were the observer, you might be surprised to see that the remote clock does not match your clock like you expected.

Does this make sense to you?

 

[n_ktt]

Thank ghwellsjr for your answer and question.
Yes, you are right, there is a problem with observation in LT. I've already discussed it with Chestermiller already a little bit.

If an event has happened then the information about the event has to reach the observer.
And a time is needed for a signal to reach the observer. Usually the signal of specific event moves with light speed.

In my analysis I prefer the following kind thinking:
I’m the observer. I know the Lorentz Transformation and I know that LT is correct.
I’m standing by the rails and observe the moving train (engine and cars).
When the engine reaches my position my clock and engine’s clock are set to 0.
When $t$ time has passed then I look at the LT equations and on the basis of that I know the distance from me to the car with the same time $t$.

 

[ghwellsjr]

I’m the observer. I know the Lorentz Transformation and I know that LT is correct.
I’m standing by the rails and observe the moving train (engine and cars).
When the engine reaches my position my clock and engine’s clock are set to 0.
When $t$ time has passed then I look at the LT equations and on the basis of that I know the distance from me to the car with the same time $t$.

But do you understand that the distance and the times are arbitrary and due to the definition that you applied to them and that when you use the LT equations to calculate that same distance and those same times in another Inertial Reference Frame moving with respect to you, a different distance and times will be determined that are just as correct as the ones you established?

 

[Dale]

It means that at every time moment I can find a car with the same time as mine.
It is not the same car at different time moments.

Yes, this is correct. It is a consequence of time dilation together with the relativity of simultaneity.

 

[n_ktt]

To answer ghwellsjr’s question I propose to look at the situation from other point.
We can always find similar point from the engine point of view: a point where $t=t'$


$t = \gamma \left( t’ + \frac{v}{c^2} x’ \right) \quad \rightarrow \quad {x^t}’ = -\frac{\gamma}{\gamma+1} v t$

As it was before from mine point of view the kind of point is:

$x^t = \frac{\gamma}{\gamma+1} v t$

It means that distance from me to $x^t$ is the same as from engine to ${x^t}’$. The two points have the same time. If I’m not wrong I would say that they are the same points in space.

Moreover, let’s look at the LT equation:

$x' = \gamma\left(x - v t\right)$

It means that $x’$ is a function of $x$ and $t$:

$x'(x,t)= \gamma\left(x - v t\right)$

What happens when we evaluate it for $x=x^t$:

$x'(x=x^t,t)= \gamma\left(\frac{\gamma}{\gamma+1} v t - v t\right) = \gamma\left(\frac{\gamma}{\gamma+1}-1\right) v t = -\frac{\gamma}{\gamma+1} v t = {x^t}’$

It means that

$x’(x^t,t)={x^t}’$

So finaly it can be said that $x^t$ is really a special point.

 

[Chestermiller]

I guess I agree with ghwellsjr. I fail to see why this is such a special point, particularly since it hinges on the mathematically arbitrary choice of coordinate and time settings that are used to define a particular event at which the engine passes a specific location on the ground (in this case your x = 0). You could equally well have chosen these settings as x = 5, ct = 3, x' =-15, ct' = 37 (rather than x =0, ct =0, x' =0, ct' =0). This would just have added a few constant terms to the equations for the LT. But the mathematical "cuteness" of your results would be totally destroyed by this choice.

 

[ghwellsjr]

It means that distance from me to $x^t$ is the same as from engine to ${x^t}’$. The two points have the same time. If I’m not wrong I would say that they are the same points in space.

It is meaningless to compare the Coordinate Times from two different frames or the Coordinate Locations from two different frames. They are arbitrarily assigned. They are just numbers.

What you are doing makes as much sense if you found some relationship between the coordinates between orthogonal coordinates and circular coordinates or spherical coordinate.

Or how about this: did you know that the temperature of -40 degrees is a special temperature? Yes, it's very special because it has the same value on both the Celsius scale and the Fahrenheit scale. Really, it's nothing more than a curiosity as a result of the arbitrary decisions to define the two scales in the way they were. Any two scales defining the same parameter will have a point where the values are the same. That's all you are doing.

Furthermore, once you have a single scale to define your measurements of a physical parameter, you don't need any other scale but if there is one and you want to convert the values you get from one scale to another scale it is just a mathematical exercise and does not reflect anything physically significant.

In the same way, once you define a reference frame, you don't need any other reference frame, it won't add any more content or information or understanding than what you have from the first reference frame. Mathematically transforming to a different frame does not reflect anything physically significant.

 

[DrGreg]

If I've understood this correctly, the only significance of what you call "$x^t$" is that if there were another observer passing through all of the events $(t,x^t)$ (for all $t$), then, relative to that observer, the ground and the train would have equal but opposite velocities.

 

[Dale]

So finaly it can be said that $x^t$ is really a special point.

It is the spatial origin of a third frame, half way between the other two frames. It is special in the same way that the bisector of an angle is special.

 

[n_ktt]

Thank for your answers and your understanding of special meaning (or not) of $x^t$ point. Probably you are right that such a point has no special meaning, but it is specific at least, I suppose.
All I try to point out are the facts that:
- At every time moment $t$ I can find a point $x^t=\frac{\gamma}{\gamma+1} v t$ where clocks in my frame and in moving frame show the same time
- The same point can be set up for engine ${x^t}’=-\frac{\gamma}{\gamma+1} v t$
- In means that $x^t$ is a kind of ‘mid-point’ (as some of you noticed) between $x$ and $x’$ frames
- In my opinion $x^t$ is not a third frame but just a point in $x$ frame that can be find at arbitrary time moment ($x^t$ is not connected with additional observer)
I agree with ghwellsjr and DeleSpam that it can’t be set physical significance to the point $x^t$ because two frame has something in common at the point.

I am curious all the time, if $x^t$ has other meaning anyway?

 

[Dale]

- In my opinion $x^t$ is not a third frame but just a point in $x$ frame that can be find at arbitrary time moment ($x^t$ is not connected with additional observer)

It is not a third frame, it is the spatial origin of a third frame. I.e. The line in spacetime formed by setting x=y=z=0 and letting t vary. That line bisects the angle between the spatial origins of the first two frames.

EDIT: maybe I should call it the time axis rather than the spatial origin. Apologies for any confusion.

 

[Chestermiller]

Thank for your answers and your understanding of special meaning (or not) of $x^t$ point. Probably you are right that such a point has no special meaning, but it is specific at least, I suppose.
All I try to point out are the facts that:
- At every time moment $t$ I can find a point $x^t=\frac{\gamma}{\gamma+1} v t$ where clocks in my frame and in moving frame show the same time
- The same point can be set up for engine ${x^t}’=-\frac{\gamma}{\gamma+1} v t$
- In means that $x^t$ is a kind of ‘mid-point’ (as some of you noticed) between $x$ and $x’$ frames
- In my opinion $x^t$ is not a third frame but just a point in $x$ frame that can be find at arbitrary time moment ($x^t$ is not connected with additional observer)
I agree with ghwellsjr and DeleSpam that it can’t be set physical significance to the point $x^t$ because two frame has something in common at the point.

I am curious all the time, if $x^t$ has other meaning anyway?

Suppose that, when the engine meets the observer, the observer's location and time are x₀ and t₀, rather than 0 and 0, and the engine's location and time are x_0' and t_0', rather than 0 and 0. And, of course, all the clocks on the train are synchronized with one another and all the clocks on the ground are synchronized with one another. I would like you to write down the LT equations for this situation. After writing these, I would like you to determine whether there is still a "special point" according to your rationale.

 

[n_ktt]

To answer Chestermiller I would put it In such a way:
There is a train (engine with cars) that moves at speed $v$. Let us assume that when the engine is at distance $x_0$ from observer at ground the clocks at ground and in train are synchronized in reference to time $t_0$. In such a situation the LT has the form:

1) $x’=\gamma\left( (x-x_0) – v (t-t_0) \right)$
2) $t’-t_0 = \gamma \left( (t-t_0) - \frac{v}{c^2} (x-x_0) \right)$

The ‘mid-point’ $x^t$ is in the point where $t=t’$, so from the second LT equation we get:

$t - t_0 = \gamma \left( (t-t_0) - \frac{v}{c^2} (x^t-x_0) \right)$

So finally:

$x^t = \frac{\gamma}{\gamma+1} v \left(t-t_0\right) + x_0$

 

[Chestermiller]

To answer Chestermiller I would put it In such a way:
There is a train (engine with cars) that moves at speed $v$. Let us assume that when the engine is at distance $x_0$ from observer at ground the clocks at ground and in train are synchronized in reference to time $t_0$. In such a situation the LT has the form:

1) $x’=\gamma\left( (x-x_0) – v (t-t_0) \right)$
2) $t’-t_0 = \gamma \left( (t-t_0) - \frac{v}{c^2} (x-x_0) \right)$

The ‘mid-point’ $x^t$ is in the point where $t=t’$, so from the second LT equation we get:

$t - t_0 = \gamma \left( (t-t_0) - \frac{v}{c^2} (x^t-x_0) \right)$

So finally:

$x^t = \frac{\gamma}{\gamma+1} v \left(t-t_0\right) + x_0$

No, no, no. For the problem I posted, the equations should read:

1) $x’-x_{0'}=\gamma\left( (x-x_0) – v (t-t_0) \right)$
2) $t’-t_{0'} = \gamma \left( (t-t_0) - \frac{v}{c^2} (x-x_0) \right)$

Note the primes on the initial coordinates on the left hand side. What do you get now for your "special point", and, in this case, is it really that special?

 

[n_ktt]

Chestermiller,
That is all the same.
$x_0$ – it is a distance from observer on the ground to the frame on the ground
$x_0’$ – similary, but in train, there is observer on train and its frame as at distance $x_0’$ from him

$t_0$ – it is a reference time for the observer on the ground
$t_0’$ – it is a reference time for the observer in the train

$t_0$ and $t_0’$ do not need to be the same. In LT a value connected with time passing is taken, any way, not just clock time showing (it can be understand that stoppers are used).
You can always define the following quantities:

$\Delta x = x-x_0 \quad , \quad \Delta x’ = x’-x_0’$
$\Delta t = t-t_0 \quad , \quad \Delta t’ = t’-t_0’$

and put it to LT. Then LT has the following form that sometimes is used:

1) $\Delta x’=\gamma\left(\Delta x - v \Delta t\right)$
2) $\Delta t’=\gamma\left(\Delta t - \frac{v}{c^2} \Delta x \right)$

Then the ‘special’ point is taken from relation:

$\Delta t=\gamma\left(\Delta t - \frac{v}{c^2} \Delta x^t \right)$

Then

$\Delta x^t= \frac{\gamma}{\gamma+1} v \Delta t$
$x^t= \frac{\gamma}{\gamma+1} v \Delta t + x_0$

Because it may make some mess in analysis, usually without loose of generality, deltas are skipped (it means the initial values are set to zero: $x_0=0$, $x_0’=0$, $t_0=0$, $t_0’=0$)
And that is all

 

[Chestermiller]

Yes. This is all done correctly. And, in terms of the Δ's, your relationships are properly recovered. But, I guess beauty is in the eye of the beholder. You look at the results and say that they represent something special. I look at the results and say "So what?"

 

[n_ktt]

Thanks Chestermiller. You are right that there is a point and ‘what of that’.
But a certain interesting relation connected with ‘special’ point $x^t$ Has came to my mind.

The standard LT equation is:

1) $x’ = \gamma \left(x – v t \right)$

I can always rewrite to the form:

2) $x (x’,t) = \frac{x’}{\gamma} + v t$

It can be interpreted in the following way: I’m staying on the ground and observe the moving train. I have a clock. At time moment $t$ I know the distance of the engine from me:

$x(x’=0,t) = v t$

Similarly, the position of the end of the train (supposing the train has $L'$ length) is:

$x(x’=-L’,t) = \frac{-L’}{\gamma} + v t$

And so on.
So for me the train has the length:

$L = x(x’=0,t) – x(x’=-L’,0) = \frac{L’}{\gamma}$ (the standard Lorentz contraction)

What's more, I can always define the one as:

$1 = \frac{\gamma+1}{\gamma} \frac{\gamma}{\gamma+1}$

Then I can multiply the second part of equation 2) by the one:

$x = \frac{x’}{\gamma} + \frac{\gamma+1}{\gamma} \frac{\gamma}{\gamma+1} v t$

Because $x^t$ point is defined as

$x^t = \frac{\gamma}{\gamma+1}$

I have:

$x = \frac{x’}{\gamma} + \frac{\gamma+1}{\gamma} x^t$

So finally I get:

$x = x^t + \frac{x^t + x’}{\gamma}$

So, using the $x^t$ point I’ve got LT equation in quite other form.

 

[n_ktt]

On the context of the special point $x^t$, in which $t=t’$, a question has risen in my mind.

Let us suppose that at time moment $t$ the train stops.
When the train stops it is said (in special relativity – twin paradox) that engine’s clock shows the time $t’=\frac{t}{\gamma}$.
When the train stops it means that the time in all cars is the same, especially in the car, in which the time was the same as on mine clock $t=t’$ just before stopping.

The question is:
Just before stopping the clocks in all cars and in engine show different time moments.
So why after stopping the time $t’$ in the whole train is the same as in engine?
Why the time is not the same as was, for example, in $n$-th car just before stopping?
Why the time in the train is not the same as was on the car in which $t=t’$?

 

[Dale]

Let us suppose that at time moment $t$ the train stops.
When the train stops it is said (in special relativity – twin paradox) that engine’s clock shows the time $t’=\frac{t}{\gamma}$.
When the train stops it means that the time in all cars is the same

No. If the whole train stops simultaneiously at time t in one frame then each part of the train stops at a different time in other frames.

 

[ghwellsjr]

No. If the whole train stops simultaneiously at time t in one frame then each part of the train stops at a different time in other frames.

And the train will be ripped into parts, correct?

 

[Dale]

And the train will be ripped into parts, correct?

Or crushed, depending on the details. Unless, of course, it is sufficiently elastic.

 

[ghwellsjr]

Or crushed, depending on the details. Unless, of course, it is sufficiently elastic.

The detail being which frame we say the train stopped simultaneously in, correct? And if we assume n_ktt meant the train's rest frame, then we have that detail, correct?

 

[Dale]

Correct. But off the top of my head I don't know which way it winds up going, I would have to work it out. Either way, it would be messy for anyone on the train!

 

[ghwellsjr]

Well, off the top of my head, there is no difference between acceleration and deceleration and applying brakes on every car of the train is the same as rockets being fired on every car and the latter would be like Bell's Paradox which results in the string breaking so I would think all the couplers would break at a minimum.

 

[n_ktt]

I seen your point, ghwellsjr.
But still the question can't come out of my mind about the time in all cars after their stop.
Even if the couplers are broken, the cars eventually stops.
The engine driver, after stopping, goes along all the cars and looks at the clocks in the cars.
All clocks should show the same time, or I’m wrong.
So the question is what is the time shown by all the clocks in the cars?

 

[Dale]

All clocks should show the same time, or I’m wrong.

You are wrong. See post 24.

The key is understanding the relativity of simultaneity.

 

[Chestermiller]

When the train is at rest with respect to the ground (prior to its initial acceleration), its clocks can be synchronized with one another and with the clocks on the ground. But, during the acceleration of the train, the clocks on the train will fall out of synchronization with one another (and with the clocks on the ground). This is analogous to the gravitational time dilation effect.

After the acceleration is complete, and the train is now traveling at constant speed, its clocks can be resynchronized with one another. At this point, the locations on the train and the clock times on the train can be related to those on the ground by the Lorentz transformation. But, when the train next decelerates down to zero speed relative to the ground, its clocks will again fall out of synchronization, so that when it stops, they will again have to be resynchronized with one another and, if desired, with those on the ground. If, after the initial acceleration, the clocks on the train are not resynchronized with one another, then my guess is that, after the final deceleration down to zero speed, the clocks on the train will again be in synchronization with one another and with those on the ground. However, I am not sure of this, since I have not analyzed the problem in detail. Maybe another responder with more experience than I could address this.

Chet