# Lorentz Transformations in Special Relativity

### The Lorentz Transformations in Special Relativity

Defining
$\beta \equiv \frac{v}{c}$
($v$ is the relative, uniform velocity between the two reference frames),
and $\gamma$ as the Lorentz factor:
$\gamma \equiv \frac{1}{\sqrt{1- (v/c)^{2}}} = \frac{1}{\sqrt{1-\beta^{2}}}.$
The Lorentz transformations are:

$t’ \ = \ \gamma \ [t – (xv/c^{2})]$
$x’ \ = \ \gamma \ [x – vt]$
$y’ \ = \ y$
$z’ \ = \ z$

The Lorentz transformations are symmetric in the sense that each frame of reference perceives the other to behave in the same way. But how do you remember which way around these apply since they are not symmetric in the coordinates? Well, light must travel in a straight line in the frame in which it was emitted so a light ray in the “other frame” will have to travel a longer distance in your frame. However, the speed of light has to be the same, so times in your frame must be stretched out compared to times in the other frame. In other words, every tick of a clock in the other frame will appear to take longer in your frame. So distance and time in the other frame appear to be smaller than in your frame. Unfortunately, time dilation and the Lorentz-Fitzgerald contraction have been conventionally defined in different frames. But just remember that in order to keep the speed of light the same, and to keep the relative velocity between frames the same, both distance and time must change in the same sense. So lengths and time intervals in the other frame are $t’/\gamma$ and $x’/\gamma$, but time dilation is defined via.

$\Delta t’ = \gamma \Delta t$

and the Lorentz-Fitzgerald contraction via.

$\Delta x = \Delta x’ /\gamma$

There is some wierdness in the length contraction because there is a subtle difference between appearance (via. rays of light) and actual measurement. The equation refers to appearance. If the ends of a rod in the other frame emitted light and “burned” a trace in your frame, you would measure the length of the rod to be the same in both frames.

Energy ($E$) and momentum ($p$) in your frame are

$E’ = \gamma E$
$p’ = \gamma p$

The energy-momentum equation relating to the invariant $mc^{2}$:
$E^{2} – p^{2}c^{2} = m^{2}c^{4}.$