# Theory of relativity and Lorentz transformations

Question is at the end

I am reading the Einstein's book on relativity, but my math courses are a bit far away in time (lol) and something bugs me. I am reading the book in french so if any of the expressions I use are wrong, please let me know in the comments and I will edit asap.

# Lorentz transformation

Given two frames of reference $K$ and $K'$, moving at a relative speed $v$. In this example, $K$ is the ground and $K'$ is a train moving straight.

We trigger an event after $t$ seconds from $K$ point of view, $t'$ being the time elapsed before the event from $K'$ point of view.

We also have $x, y, z$ and $x', y', z'$ representing the distance separating the origins of $K$ and $K'$ from the event on each axis.

If we know when and where the event happened from $K$, we can determine when and where it happened from $K'$ with the following equations (Lorentz transformation).

$$x' = \frac{x - vt}{\sqrt {1-\frac{v^2}{c^2}}}$$ $$y' = y$$ $$z' = z$$ $$t' = \frac{t - \frac{v}{c^2}x}{\sqrt {1-\frac{v^2}{c^2}}}$$

# Calculating space difference

We have a straight stick, which, in the frame of reference $K'$, is 1 meter long. We have $x_0' = 0$ as the start of the stick and $x_1' = 1$ as the end of it. What is the size of the stick viewed from $K$ ?

From the first equation, we can say that: $$x_0 = 0\cdot\sqrt {1-\frac{v^2}{c^2}}$$ $$x_1 = 1\cdot\sqrt {1-\frac{v^2}{c^2}}$$ So the stick is $\sqrt {1-\frac{v^2}{c^2}}$ meters long for $K$.

Now what I understand is that to find this last equation, we took the 1st Lorentz transformation and did the math for $t = 0$, as $t$ is not taken in account here. Is it the case? If it is, then I have a problem with the next equation.

# Calculating time difference

Now let's consider a clock placed on $K'$ staying at $x' = 0$.

$t_0'$ is the moment at which the seconds display $0$, and $t_1'$ the moment the seconds display $1$.

$t_0'$ and $t_1'$ are separated by 1 second when observed from $K'$. How much time will it be from $K$?

HERE IS MY QUESTION

If I apply the same reasoning as before, which is taking the 4th Lorentz transformation and setting $x = 0$, I find $t_0 = 0$ and $t_1 = 1\cdot \sqrt {1-\frac{v^2}{c^2}}$ seconds.

Now I know this is wrong, because in the book Einsteins says: $$t_1 = \frac{1}{\sqrt {1-\frac{v^2}{c^2}}}$$ How does he come to this conclusion? Where did I go wrong? I tried using the same 4th transformation, but changing it by using $x = vt$ instead of $x = 0$, but I still don't come to Einstein's conclusion.

This has been bugging me for some time, any help will be appreciated :)