Calculating Time Dilation w/ Light Clock - Implications & Mistakes?

[rgtr]

TL;DR Summary

$T_B = {T_A} {gamma}$

In this picture it shows a light clock. Let's use the moving light clock example.
Am I essentially calculating the b component of moving clock.
Assume the moving frame is the B frame.
Assume the stationary frame is the A frame
https://simple.wikipedia.org/wiki/Light_clock

Or essentially the b component of the picture below. Am I essentially calculating Light time for the vertical component?
Does this have any more profound implications?
Did I make any mistakes in my thinking ?

triangle image






In post #15 I asked the question but someone recommended I start a new post and link the the original thread. Here it is https://www.physicsforums.com/threa...ing-by-t_stationary-for-light-clocks.1015831/ .

 

[anuttarasammyak]

The length of the hypotenuse of the right triangle is the distance which the light moves from the floor to the ceiling of the car for people at rest on the Earth. Divide it by c and it's time which the clocks at rest on the Earth tick for one way, say $T_E$. Double it for go-return.

The height of those triangles, H, is the distance which the light moves from the floor to the ceiling of the car for the passengers. Divide it by c and it's time which the clocks at rest in the car tick for one way, say $T_C$. Double it for go-return.

As for one way tick counts Pythagoras theorem says
$$H^2+v^2T_E^2=c^2T_E^2$$
where v is speed of the car, and
$$H=cT_C$$
Solving the two equations
$$T_E=\frac{T_C}{\sqrt{1-\frac{v^2}{c^2}}}$$

 

[Ibix]

Am I essentially calculating Light time for the vertical component?

No. You are calculating the time for light to travel along the hypotenuse and expressing this as a ratio to the time taken for light to travel the vertical marked in your triangle diagram.

The reason for the apparent coincidence is fundamentally related to Minkowski geometry, which is the geometry of spacetime. In Euclidean geometry the distance between two points that are separated by $(\Delta x, \Delta y, \Delta z)$ is $\Delta l$, where $$\Delta l^2=\Delta x^2+\Delta y^2+\Delta z^2$$ This is true in any reference frame - if you have rotated your axes compared to mine you will have different $\Delta x, \Delta y, \Delta z$, but $\Delta l$ will be the same.

Minkowski geometry doesn't cover space, though. It covers spacetime. The equivalent to the above is that two events separated by $(\Delta t, \Delta x, \Delta y, \Delta z)$ are separated by a "distance" (formally called interval) defined by $$\Delta s^2=c^2\Delta t^2-(\Delta x^2+\Delta y^2+\Delta z^2)$$(NB: some sources define this with the opposite sign - either convention is fine and you just have to get used to checking your signs obsessively.) Again, if you are using a different frame with different coordinates you will have different $\Delta t, \Delta x, \Delta y, \Delta z$ but you will agree with me on $\Delta s^2$.

So, in the case of the light clock, one frame says that the bottom and top reflection events happened at the same place (so $\Delta x$, $\Delta y$, and $\Delta z$ are all zero) and separated by some time $\Delta t=T_A$. The other frame says that the reflection events were separated by $\Delta x'=b$, $\Delta y'=0$, $\Delta z'=0$, and time $\Delta t'=T_B$. But the two frames must agree on $\Delta s^2$, so you can calculate $\Delta s^2$ in both frames and equate them:$$c^2\Delta t^2=c^2\Delta t'^2-\Delta x'^2$$
If you rearrange that you can make it look like Pytharoras' theorem applied to one of the right triangles in your image - that's the coincidence. It follows just from the presence of the minus sign in the Minkowski equivalent of the distance formula. It's pretty specific to the case where events happen at the same location in one frame.

 

[rgtr]

I think I meant to say horizontal component. Saying vertical component was just a mistype. Whoops.

 

[anuttarasammyak]

Direction of vertical perpendicular or Horizontal transverse do not matter with the discussion when we disregard usually tiny gravity effect of GR.