I Demonstration of time dilation

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1. Nov 1, 2017

fab13

I try to find the formula of dilation of time between a reference frame (R') moving at a speed v and a fixed observer in (R). For this, I take the example that we often find as a demo : that of a train in which a light beam is emitted vertically (in the train): this vertical trajectory in the train forms, with the segment v * dt and the vertical length L that travels the light in the train, a right triangle. This is illustrated in the diagram on the left of the figure attached. This gives immediately, with Pythagore's theorem, the relation:

$$c^2 dt^2=c^2 d\tau^2 + v^2 dt^2$$

$$d\tau^2/dt^2+v^2/c^2$$ and finally : $$d\tau=dt/\gamma$$

Now, I would like to find the same formula but by placing the observer of (R) at a distance "d" from the plane (Oyz) (and with z = 0) and placed along the axis (0y) at the special place where the light reaches the bottom of the train (see the diagram on the right in the figure attached). If I take the events (in R) $t_{reception,top}$ and $t_{reception,bottom}$, I find that for the observer, the time interval between these 2 events, taking:

1) a common synchronization for (R) and (R '), that is to say at the emission of the photon from the top, $t=\tau=0$

2) $t_{reception,bottom}=\dfrac{1}{v} (\dfrac{L^2}{sin^2(\theta)}-L^2)^{1/2}+\dfrac{d}{c}$

and

$t_{reception,top}=\dfrac{1}{c }(d^2+\dfrac{L^2}{sin^2(\theta )})^{1/2}$

Finally, one gets :

$\Delta t = t_{reception,bottom}-t_{reception,top}=\dfrac{1}{v }(\dfrac{L^2}{sin^2(\theta)}-L^2)^{1/2}+\dfrac{d}{c}-\dfrac{1}{c}(d^2+\dfrac{L^2}{s in^2(\theta)})^{1/2}$

By taking $\Delta \tau = \dfrac{L}{c}$, I can't manage to find the classical relation between $\Delta \tau=\Delta t/ \gamma$

If I take d = 0 (the observer is considered to be in the plane (Oyz) in the expression of $\Delta t$ above, I find, :

$$\Delta t = t_{reception,bottom}-t_{reception,top}=0$$ which is normal because the observer receives the photon from the top at the same time as the photon of bottom is emitted and so observer receives it instantly.

In the classical demonstration with the right triangle, how can we consider that the observer instantly sees a right triangle? , that is to say we do not take into account the distance between this triangle in the plane (0yz) and the observer (which is at a certain distance on the axis (Ox) of my diagram: in my example, I take this distance equal to "d").

Any help is welcome

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2. Nov 1, 2017

FactChecker

When you talk about an observer in a reference frame, you should really think in terms of infinitely many observers along the length of the frame. They synchronize their times and record what they observed at their location. They communicate what happened after the fact. So there is usually no issue of how long it takes a single observer to see an event that happened at a distance. That would introduce complexities that are unnecessary.

3. Nov 1, 2017

fab13

So my error was to say that motionless observer in (R) can't be synchronized to the observer in the train before or at the emission of top photon ?

I don't understand very well the classical demonstration (with right triangle) where we don't take into account the distance between motionless observer and the train : by drawing this triangle, it seems this observer has an immediate way to see the emission of top photon and $\Delta t$ later (in (R) reference frame), the emission of bottom photon.

Could you tell me please where is the error in my reasoning ?

4. Nov 1, 2017

FactChecker

I'm not sure that we are talking about the same thing, but here is my two cents:
There are motionless observers all down the line. All the motionless observers have synchronized their clocks for their stationary reference frame. As the train moves, there is always a motionless observer right beside the train to record what is happening. When the experiment is done and they get together to analyse the results, they observe the time-space distortions of the train's measurements that SR specifies.

I like to think of it as a stationary "God's eye view" versus a moving "God's eye view".

5. Nov 2, 2017

Staff: Mentor

If the guy on the train notes down the time on the clock of the adjacent ground observer as he passes each of them, he determines that their clocks are running faster than his. If the observers on the ground note down the time on the clock of the guy on the train as he passes adjacent to each of them, after comparing notes, they conclude that his clock is running slower than theirs. So the observers in the two frames are in agreement. Yet, the clocks in both frames of reference are running at the same speed. This all happens because, although the clocks in each of the frames are synchronized at different locations with one another, they are not in synchronization with the clocks at different locations in the other frame.

6. Nov 2, 2017

FactChecker

I don't think this is right. Each frame is moving relatively toward the rear of the other frame and each would see the other's clock going too slow. They do not agree on who's clock is slow. Otherwise, there would be one reference frame which all agree had the slow clock (and therefore moving). That would violate the non-preference of inertial reference frame that is fundamental to SR.

7. Nov 2, 2017

Staff: Mentor

This example involves one observer in one of the frames, and the series of observers in the other frame. Certainly, as he passes adjacent to each of them, there is mutual agreement as to the times they see on one anothers clocks.

8. Nov 2, 2017

FactChecker

But they are comparing different clocks. If I was on one reference frame, I would be looking at a series of clocks toward the rear of the "moving" frame. The moving observer would be looking at a series of clocks toward what he considers to be the rear of my reference frame -- which is a set of clocks in my frame going in the other direction.
A series of clocks faster going in one direction is going slower in the other direction. Since both reference frames are going in opposite directions with respect to the other, they both see a series of clocks in the other frame that appears to be too slow.

Last edited: Nov 2, 2017
9. Nov 2, 2017

Staff: Mentor

For the guy on the train, the events occur at

x'=0, t'=0

and

x'=0, $t'=\Delta t'$

For the guys on the ground, according to the inverse Lorentz transformation, the same events occur at:

x = 0, t=0

and

$x=\gamma(0+v\Delta t')=\gamma v\Delta t'$,

$t=\gamma(\Delta t'+0)=\gamma\Delta t'$

So both the guy on the train and the observers on the ground agree that more time has passed between the two events according to the clocks he passes on the ground than according to the clock of the guy on the train.

10. Nov 2, 2017

FactChecker

Wouldn't that identify one frame as moving and the other as stationary? I thought that would violate a fundamental SR principle that there is no preferred "stationary" reference frame.

11. Nov 2, 2017

Staff: Mentor

Not really. Do you believe the LT or not? By symmetry, if the focus was on a single observer on the ground (and multiple observers on the train), the conclusion would be that the clock of the guy on the ground is running slower than the times observed on the sequence of train clocks running past him. So, from a test such as this, it is not possible to identify one frame as being stationary and the other moving.

12. Nov 2, 2017

FactChecker

I believe the LT. But every inertial reference frame can use it as though it is the stationary one. That leaves the two reference frames disagreeing about which clock is slower. The symmetry is complete.

13. Nov 2, 2017

Ibix

What @Chestermiller is doing is comparing one clock at rest in frame S to a sequence at rest in S', as you point out. If you compare a single clock at rest in S' to a sequence at rest in S you will also find that the sequence appears to tick fast - symmetric as expected. But that's a different experiment. The original experiment is indeed asymmetric between frames. But that's also to be expected because the setup is asymmetric - one clock vs many.

14. Nov 2, 2017

Bartolomeo

15. Nov 2, 2017

GrayGhost

Regarding this matter...

Consider gazillions of clocks (virtual or not) synchronized in their own frames, and let's say they are equally spaced wrt the axis-of-propagation for simplicity.

Per a stationary observer ... moving clocks tick slower, period. Observers of both POVs are correct.

Per a stationary observer ... a collection of instantaneous time readouts from momentarily co-located clocks, which includes the stationary observer's own clock and the many moving clocks passing it by over duration, will show that the time readouts of the moving clocks had steadily advanced the stationary observer's own clock.

FactChecker is saying the former.
Chestermiller (I think) is saying the later?

If so then both are correct, because the later is also true, even though moving clocks tick slower per the stationary..

Best Regards,
GrayGhost

EDIT: I now see that Ibix had posted this just prior, and so mine here is redundant to his former post.

I'd add ... this is a perfect example of how time is dilated in the moving frame relative to the stationary. For 2 non-simultaneous events that take place at the stationary clock, the moving frame records the duration between those events as stretched ... and even though both POVs hold the other frame's clocks moving and ticking slower.

Last edited: Nov 2, 2017
16. Nov 2, 2017

Bartolomeo

Two observers move relatively to each other, but there are two stationary observers. Special Relativity!
People generally like to be in the center of the universe. It took long time before they realized that Earth was rotating itself around the Sun, not vice versa. So much was resistance! How soon will they realize, that an inertial observer can also move himself? At least one of the two.
This will bring an elementary order.

17. Nov 2, 2017

FactChecker

Having a single observer in one reference frame presents a problem if he will not accept the results recorded by other observers farther back in his own reference frame and will not account for the time it takes to transmit the results (by line of sight, electronically, or stage coach) back to his location.
Ok. Maybe I missed something about some posts.

18. Nov 2, 2017

fab13

So, in the first experiment that I told in my first message, should I consider at least 2 stationnary observers, one located at the emission of top photon (but always with a distance "$d$" from the train on $(Ox)$ axis and $y=0$ on $(Oy)$ axis see my attached figure ) and the other located at the emission of bottom photon (with also at a distance "d" of $(Ox)$ axis and at a distance "$v\Delta t$" on the $(Oy)$ axis) ?

It seems that right triangle appears and can be understood with a reconstruction of the reality : it can be imagined in the case where each observer could see the emission event from the $(Oyz)$ plane and at the current $y$ position of photon on $(Oy)$, can't it ?

If yes, that's why I believed that motionless observer on the ground could had an instantaneous access of the trajectory of photon in $(R)$ reference frame.

I think I have to consider that each emission actually happens at $t_{emission}=t_{reception} - \text{distance}_{current}/c$ but the distance may be equal to $\text{distance}_{current}=(d^2+z_{current}^2)^{1/2}$ and not always necessary equal to $"d"$ (distance between observer and $(Oyz)$ plane).

Am I wrong ? Thanks

19. Nov 5, 2017

fab13

I have put the following image which is more precise :

https://imgur.com/a/35dsG

The train is moving along (Oy) axis and light starts from point A to point B. (R) is the rest reference frame and (R') is the moving reference frame.

I have understood that right rectangle has sense if I want to "rebuild" the reality, by considering the distance at which events occur and the time shift between emission of photons and reception in (R) by the 2 observers in rest reference frame.

first, could I consider only 2 observers into (R) reference drame ? It seems their position don't matter, provided that we know their mutual distance in order to be able to synchonize them before the experiment, doesn't it ?

One has also to know the distance between the 2 parallels of 2 observers line and moving train line.

I try to respect the following sequence, if there are errors, don't miss to tell it.

1) Firstly, the 2 observers synchronize their clock because they know the distance between them.

2) I follow, "Left" observer receive in (R) the photon from emission at point A in (R')
With the angle under which he receives the photon and knowing the distance orthogonal to train moving line (denoted $d$ on my figure attached), he can deduce the distance $d_{A}$ at which emission occured. Then, this emssion happened at $t=t_{A}-\dfrac{d_{A}}{c}$.

3) "Right" observer receives the photon from point B into (R). This photon has been emitted from (R') at point B. Like above, he deduces the distance $d_{B}$ at which
events occured and the time shift $t=t_{B}-\dfrac{d_{B}}{c}$

4) The 2 observers "rebuild" what has precisely happened and try to analyse the results.

For this, one can write :

$\Delta t_{reconstruction}=(t_{B}-t_{A}) - \dfrac{1}{c}(d_{B}-d_{A})$

For the moment, I take $d_{A}=0$ (it will be interesting to consider the case with $d_{A}$ not null).

that gives :

$\Delta t_{reconstruction}=\Delta t_{reception} - \dfrac{L}{c}\,\,\,\,\,\,\,$ with $\Delta t_{reception}=t_{B}-t_{A}$

$\dfrac{L}{c}=\Delta t_{reception} - \Delta t_{reconstruction}=\Delta t_{reception} - \dfrac{d}{v}$

with $d=v\Delta t_{reconstruction}$ the real distance performed by train between the interval $\Delta t_{reconstruction}$.

So Finally, I get : $L^{2}= c^{2}\Delta t_{reception} - \dfrac{2 c^{2} \Delta t_{reception}\,d}{v}+\dfrac{c^{2}d^{2}}{v^{2}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\textbf{eq(1)}$

From this part, I can't get to find the expected expression of time dilation, i.e : $L^{2}=c^{2}\Delta t_{reception}^{2}-v^{2}\Delta t_{reception}^{2}$

I suppose that $\Delta t_{reception}$ represents the $\Delta t$ into the classical and expected expression of time dilation.

Could anyone see what's wrong or help me to find, from $\textbf{eq(1)}$, the classical relation : $L^{2}=c^{2}\Delta t^{2}-v^{2}\Delta t^{2}$ with $\Delta t = \Delta t_{reception}$ ?

Any help is welcome, thanks

Last edited: Nov 6, 2017
20. Nov 6, 2017

Ibix

@fab13 - am I right in thinking that you are describing the following experiment?

There is a train moving along the y axis. There are two observers, A and B, standing by the track. As the train passes A it emits a light pulse that A receives at time $t_A$, after the pulse has travelled a distance $d_A$. As the train passes B it emits a light pulse that B receives at time $t_B$, after the pulse has travelled a distance $d_B$. You later define $d_A=0$ and seem to start using $L$ in place of $d_B$.

If the above is correct, nothing you have done will allow you to derive the invariant interval. The interval is the "distance" in spacetime between a pair of events. All frames will agree on it. You seem to be trying to construct a relationship between three events using measurements made in only one frame. You can't show invariance between frames without making measurements in two or more frames.

Also, note that light is not emitted from a frame. It's emitted at an event; a frame is just a choice of coordinates to describe that event. The same event is described in all frames.

21. Nov 6, 2017

fab13

@Ibix : The points A and B are the extremities of a vetical line into (R'), i.e into the moving reference frame. You can have an illustration on the following figure :

In my experiment, I only consider the first path, not the round trip, i.e I consider the path before the top reflection of mirror.

By taking my calculation, I am near to get the common relation of time dilation :

$L^{2}=c^{2}\Delta t^{2}-v^{2}\Delta t^{2}$

but someting is wrong; currently I get from previous reply post :

$L^{2}= c^{2}\Delta t_{reception}^{2} - \dfrac{2 c^{2} \Delta t_{reception}\,d}{v}+\dfrac{c^{2}d^{2}}{v^{2}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\textbf{eq(1)}$

I think I have to do, between my relation and the common one, the assimilation :

$\Delta t = \Delta t_{reception}$

into : $L^{2}=c^{2}\Delta t_{reception}^{2}-v^{2}\Delta t_{reception}^{2}$

But I am not sure about this assimilation : it may be actually $\Delta t =\Delta t_{reconstruction}$ ?

I hope you will better understand my issue

Last edited: Nov 6, 2017
22. Nov 7, 2017

Ibix

All you need to do is eliminate L between $\Delta t'=L/c$ (from the left hand diagram) and $L^2+v^2\Delta t^2=c^2\Delta t^2$ (from the right hand diagram), surely? Or am I misunderstanding your problem?