# A Lorentz invariance of the Heaviside function

1. Apr 1, 2016

### spaghetti3451

Consider the Heaviside function $\Theta(k^{0})$.

This function is Lorentz invariant if $\text{sign}\ (k^{0})$ is invariant under a Lorentz transformation.

I have been told that only orthochronous Lorentz transformations preserve $\text{sign}\ (k^{0})$ under the condition that $k$ is a time-like vector.

I would like to prove this explicitly, to convince myself of the fact. How do I start?

2. Apr 1, 2016

### vanhees71

Sure, the "direction of time" must be preserved under the Lorentz transformation for the expression $\Theta(k^0)$ to be conserved. Thus you are restricted to Lorentz-transformation matrices ${\Lambda^{\mu}}_{\nu}$ for which ${\Lambda^{0}}_{0} \geq 1$. Now for a non-like vector $k^{\mu}$, i.e., $(k^0)^2-\vec{k}^2 > 0$, i.e., (because of $k^0>0$ ) $k^0>|\vec{k}|$ you have
$$k^{\prime 0}= {\Lambda^{0}}_{\mu} k^{mu} = {\Lambda^{0}}_0 k^{0} + \vec{\Lambda} \cdot \vec{k} \geq {\Lambda^{0}}_0 k^{0} -|\vec{\Lambda}||\vec{k}|,$$
where
$$\vec{\Lambda}=({\Lambda^{0}}_j)|_{j \in \{1,2,3 \}}.$$
Further for the Lorentz transformation you have
$$\left ({\Lambda^{0}}_0 \right)^2 - \vec{\Lambda}^2=1 \; \Rightarrow {\Lambda^{0}}_{0}=+\sqrt{1+\vec{\Lambda}^2}>|\vec{\Lambda}|.$$
Thus we have
$$k^{\prime 0} \geq {\Lambda^0}_0 (k^0-|\vec{k}|) > 0.$$
In the same way you can show that if $k^0<0$ also $k^{\prime 0}<0$, and that was to be shown.