# Orthochronous Lorentz Transformations

1. Feb 21, 2010

### vertices

How can I convince myself of the following statement:

If x2<0, there exists L orthochronous Lorentz tranformation such that:

Lx = -x

My concern is this:

If for example, we take xµ=(1,0,0,0), then Lx in component form is:

Λµβxβµ0x0
=(Λ00, Λ10, Λ20, Λ30).

By definition, if it is an orthochronous LT, we must have Λ00≥1 so it is clear that the statement above is false for this example (as the RHS in the case of our example vector, can't be positive).

Can anyone think of an example xµ that would actually satisfy the condition Lx = -x?

2. Feb 21, 2010

### Fredrik

Staff Emeritus
You're asking the first question in a way that strongly suggests that the statement is supposed to be true for any x with x2<0. (And the fact that you tried x=(1,0,0,0) must mean that you're using a -++++ metric). But Λµβxβ=(Λ00x0,0,0,0) proves that the statement is false.

Are you really looking for a specific choice of Λ and x such that Λx=-x? It looks impossible to me, but I haven't proved it. I suggest that you try playing around with the explict formula for Λ in 1+1 dimensions:

$$\Lambda=\gamma\begin{pmatrix}1 & -v\\ -v & 1\end{pmatrix}$$

$$\gamma=\frac{1}{\sqrt{1-v^2}}$$

(This is in units such that c=1).

Hm...or maybe not. Maybe it won't work in 1+1 dimensions but will work in 2+1 and 3+1 because then you will have the opportunity to flip the sign of two spatial components with a rotation.