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:(adsbygoogle = window.adsbygoogle || []).push({});

[tex]c^2 dt^2=c^2 d\tau^2 + v^2 dt^2[/tex]

which leads to :

[tex]d\tau^2/dt^2+v^2/c^2[/tex] and finally : [tex]d\tau=dt/\gamma[/tex]

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, :

[tex]\Delta t = t_{reception,bottom}-t_{reception,top}=0[/tex] 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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# I Demonstration of time dilation

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