# The set of Lorentz boosts and space rotations form a group

1. Dec 2, 2011

### Demon117

Ok. I understand that the set of Lorentz boosts and space rotations is equivalent to the set of Lorentz transformations. I understand that they form a group, but what I cannot seem to grasp is this. What the explicit form of such 4x4 matrices? One needs to know this in order to show that the properties of a group hold. The way I thought they were represented is as follows:

$L_{x}[\beta]=\left(\begin{array}{cccc} \gamma & -\beta \gamma & 0 & 0 \\-\beta \gamma & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right)$

The Lorentz boosts in the y and z directions would have similar elements in different entries of the matrix. Is this all I have to work with to show that the Lorentz transformations form a group?

2. Dec 3, 2011

Staff Emeritus
You don't absolutely need the matrix form to prove things are a group. If you still want to do it this way, you need to find the matrix for an arbitrary boost, not just the boost along one axis, which is a mess - why you usually don't see it.

3. Dec 3, 2011

### DrGreg

If you really want to see the general form, you can find it on Wikipedia at Lorentz transformation#Boost in any direction. See also the end of the next section "Composition of two boosts" which gives the same result using 3-vectors.

(Note that this is a boost in any arbitrary direction, but doesn't include any rotation of the spatial axes.)

4. Dec 5, 2011

### Demon117

So this general form represents a boost in any direction. If we multiply that by some arbitrary rotation along the spatial axes will we in turn find the boost in some specified direction? I feel like I am confusing myself.

5. Dec 5, 2011

### Fredrik

Staff Emeritus
"boost in any arbitrary direction" means "boost in some specified direction". The specified direction is the direction of the velocity (3-)vector $\vec v$.