Relativistic Light Velocity

Question

I'm lay about relativity and I want to understand how does $c$ does not change between frames of reference.

Imagine a train of length $L_0$ at a relativistic speed and a light beam inside it. For an inside frame, the time taken for light to travel across the train would be $\Delta t_0 = \frac{L_0}{c}$.

Now imagine an outside still frame of reference. To him, both time and length should receive a Lorentz transformation, thus $\Delta t_R = \gamma \Delta t_0$ and $L_R = \frac{L_0}{\gamma}$. As velocity is $\frac{\Delta s}{\Delta t}$, the velocity to him would be

$$\frac{L_R}{\Delta t_R} = \frac{L_0}{\gamma^2 \Delta t_0} = \frac{c}{\gamma^2} \neq c$$

Where is the mistake?

Answer 1

You identified the mistake in your own question. You correctly stated "both time and length should receive a Lorentz transformation", but the mathematical operations that you performed were not a Lorentz transformation. Instead of doing the actual Lorentz transformation you simply multiplied or divided by the Lorentz factor. This is not a Lorentz transform.

The Lorentz transform is derived based on the postulate that the speed of light is invariant. So when used in its correct form you will always get $c$ being invariant. The Lorentz transform is given by:$$t'=\gamma \left(t-\frac{vx}{c^2}\right)$$$$x'=\gamma (x-vt)$$$$y'=y$$$$z'=z$$where$$\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$

The equation of a pulse of light going out from the origin in the primed frame is $$ c^2 t'^2= x'^2 + y'^2 + z'^2$$ This is a sphere whose radius is $ct'$, meaning that it expands at a speed of $c$. This equation is often called the light cone. By substituting the transformation equations into the light cone equation and simplifying we get $$ c^2 t^2= x^2 + y^2 + z^2$$which shows that the speed of light is the same in all reference frames.

Answer 2

The mistake is that you have assumed that in the stationary frame the light has travelled a distance equal to the contracted length of the train in that frame, which is not the case at all. The actual distance depends on the direction in which the light travels relative to the motion of the train in the stationary frame. In other words, it depends on whether the light is travelling towards or away from the front of the train. Let's assume the former case.

Suppose the rear of the train is just passing the western end of the platform when the light flash is emitted. By the time the light flash reaches the front of the train, the train has moved further down the platform, so according to the observer on the platform, the light has travelled a distance greater than the length of the train.

Answer 3

Your Lorentz transform of the time supposes $x = 0$. That means: Alice in one of the wagons (supposed the origin in the frame of the train) is comparing her clock with synchronized clocks of the outside frame of reference passing by her. She can not compare with clocks before or after her position, because $x \neq 0$ in this case.

For example, she compares her clock with two synchronized clocks passing by in the outside frame. After time interval $\Delta t_0$ she observes a time interval between that 2 outside clocks of $\Delta t_R = \gamma \Delta t_0$.

The length that she measures between these 2 clocks is $L_R = \frac{L}{\gamma}$, where $L$ is the same length measured by Bob in the outside frame. Note that $L$ is not the length of the train $L_0$, it is the length between the clocks. And:

$$\frac{L}{\Delta t_R} = \frac{L_R}{\Delta t_0}$$ showing the same relative speed in both frames.