Prove that the proper orthochronous Lorentz group is a linear group

[TaliskerBA]

Homework Statement

Prove that the proper orthochronous Lorentz group is a linear group. That is SOo(3, 1) = {a $\in$ SO(3, 1) | (ae4, e4) < 0 } where (x,y) = x$^T$$\eta$y for $\eta$ = [1 0 0 0; 0 1 0 0; 0 0 1 0; 0 0 0 -1] (sorry couldn't work out how to properly display a matrix).

Homework Equations

a is in SO(3,1) if det(a) = 1 and a$^T$$\eta$a = $\eta$

The Attempt at a Solution

To show it is a linear group I need to show that the matrices in SOo(3, 1) are invertible, and that if a matrix is in SOo(3, 1) then its inverse is in there as well, and that it is closed under matrix multiplication. I have done the latter two bits already, but the thing I am struggling with is closure by matrix multiplication.

let a$_{ij}$ be the element in the ith row and jth column of a. For a to be in SOo(3, 1) the element a$_{44}$ must be positive. Therefore if a,b $\in$ SOo(3, 1) I need to show that given a$_{44}$ and b$_{44}$ positive, then ab$_{44}$ is also positive. This is what I'm stuck on - I can't work out how to manipulate the inequality to show that it is positive. I have managed to work out that a$_{14}^{2}$ + a$_{24}^{2}$ + a$_{34}^{2}$ = a$_{44}^{2}$ - 1 from a$^T\eta$a = $\eta$. This also tells me that a$_{44}$ ≥ 1.

So, assuming my workings are correct so far, I am trying to manipulate the following:
a$_{14}^{2}$ + a$_{24}^{2}$ + a$_{34}^{2}$ = a$_{44}^{2}$ - 1
b$_{14}^{2}$ + b$_{24}^{2}$ + b$_{34}^{2}$ = b$_{44}^{2}$ - 1
a$_{44}$ ≥ 1, b$_{44}$ ≥ 1

To show that:
ab$_{44}$ = a$_{14}$b$_{41}$ + a$_{24}$b$_{42}$ + a$_{34}$b$_{43}$ > 0

Appreciate any guidance or hints in the right direction. I have been trying to use the triangle inequality so far but to no avail. Thanks.

 

[Fredrik]

(sorry couldn't work out how to properly display a matrix).

See our LaTeX FAQ.

The only real problem is to prove that $(ab)_{44}\geq 1$, right? Can you rewrite $(ab)_{44}$ in the form $a_{44}b_{44}(1+u\cdot v)$ where u and v are members of $\mathbb R^3$ and the dot is the standard dot product? Any ideas about how to argue that $1+u\cdot v$ is greater than something else?

ab$_{44}$ = a$_{14}$b$_{41}$ + a$_{24}$b$_{42}$ + a$_{34}$b$_{43}$

This right-hand side is wrong in two different ways. There's a term missing, and the three terms you've got there are all wrong.

 

[TaliskerBA]

Sorry that'd be ab$_{44}$ = a$_{41}$b$_{14}$ + a$_{42}$b$_{24}$ + a$_{43}$b$_{34}$ + a$_{44}$b$_{44}$.

So could you set $u$= (a$_{41}$/a$_{44}$,a$_{42}$/a$_{44}$, a$_{43}$/a$_{44}$)$^{T}$ and $v$= (b$_{14}$/b$_{44}$,b$_{24}$/b$_{44}$, b$_{34}$/b$_{44}$)$^{T}$?

I'm assuming we want to show the dot product is positive, but not sure of an easy way to figure it out (I'm trying to spot something more than anything). Is there are easy way to show that the direction of these vectors are within 90 degrees of each other?

 

[Fredrik]

Yes, that's how I would define u and v. It's actually sufficient to show that $1+u\cdot v\geq 0$, if you also prove that ab is a Lorentz transformation and that the 44 component of any Lorentz transformation is always either ≥1 or ≤-1.

 

[Fredrik]

Is there are easy way to show that the direction of these vectors are within 90 degrees of each other?

No. It's even possible that u=-v.

 

[Fredrik]

Did you figure out the rest?