# 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 :)

## 1 Answer

Firstly, you are correct in thinking of taking t=0, but not because it doesn't come into play. (There are other methods of deriving length contraction as well) If we have the Lorenz transformations we can then look at measuring the stick and using those transformations. In measuring the stick we have to measure where one end is and then where the other end is and take the difference as the length of the stick. We say that the measurements of the ends are done simultaneously (change in time is 0) and that the first end is at x=0 (second end coordinate is the length). Taking the transformations of the coordinates leads to the length contraction formula. Now, for the clock on the train if you take the change in x to be 0 then the train isn't moving and you find that the time for each observer is the same. If the train is moving, then you have a change in x and we need to consider further. A good example is Einstein's light clock. Here we have a clock made of two mirrors that light bounces back and forth between on the train where we take the time the light travels between the mirrors to be a unit of time. The observer on the train sees the light bounce back and forth like you might imagine, but the observer on the ground sees the light travel in the horizontal direction as well due to the trains movement. This makes the path the light travels longer as seen by the ground observer. Considering the lengths of the paths of the light as seen be the ground observer and the change in time you can come to the time dilation formula. Using the 4th transformation to come to the same conclusion as your book works the same way. However, the transformation you wrote is for measuring in the ground frame and transforming into the moving frame. Going the opposite direction the transformation is changed slightly by moving the primes to the other side and changing the sign from minus to plus. Using that transformation will give you the same answer as the book.

• Thanks for your answer! You helped me understand the 1st part very clearly. However I still have trouble with the second one. I understand your logic and how you get to your solution, but I can't seem to connect it with the 4th Lorentz transformation. Would it be possible to come to the same conclusion using only the equations above, without imagining a new experiment? – Down Apr 12 '18 at 13:22
• I edited in some clarification. @Down – greenlf89 Apr 12 '18 at 18:28