# Orbits of the Lorentz group

I want to determine the orbits of the proper orthochronous Lorentz group $SO^{+}(1,3)$.

If I start with a time-like four-momentum $p = (m, 0, 0, 0)$

with positive time-component $p^{0} = m > 0$,

the orbit of $SO^{+}(1,3)$ in $p$ is given by:

$$\mathcal{O}(p) \equiv \lbrace \Lambda p \mid \Lambda \in SO^{+}(1,3) \rbrace$$

Now the point is: how do you show that

$$\mathcal{O}(p) = \lbrace q \mid q^{2} = m^{2}, q^{0} > 0 \rbrace$$ ?

Essently, the question is: why does a Lorentz transformation $\Lambda \in SO^{+}(1,3)$ exist
such that two four-vectors $p$ and $q$ with $p^{2} = q^{2} = m^{2}$ and $p^{0}, q^{0} > 0$ are related via $q = \Lambda p$ ?

In fact, it is possible to answer my question(s) by brute-force calculations. But I am searching for an "elegant way", e.g. with the help of group theory.

I already searched in the literature, but in most cases it seems to be "trivial" for the authors
and I see now explicit proof.

Does anyone know of anything like that?

## Answers and Replies

Isn't that the very definition of the group SO+(1,3)?

WannabeNewton
I agree with the above. It's trivial according to those authors because it's the definition.

Isn't that the very definition of the group SO+(1,3)?

No, it is not the definition of $\mathrm{SO}^{+}(1,3)$. It is:

$$\mathrm{SO}^{+}(1,3) = \lbrace \Lambda \in \mathrm{GL}(4, \mathbb{R}) \mid \Lambda^{t} \eta \Lambda = \eta, \mathrm{det} \, \Lambda = 1, \Lambda^{0} \, _{0} \geq 1 \rbrace$$
The elements of this group will not change $p^{2}$ and the sign of $p^{0}$ by definition.

So we can say that the set
$$\lbrace q \in \mathbb{M} \mid q^{2} = m^{2}, q^{0} > 0 \rbrace$$ remains invariant under Lorentz transformations.

But this does not necessarily mean that for any two fourvectors $p, q$ with $p^{2} = q^{2} = m^{2}$ and $p^{0}, q^{0}$ there exists a transformation $\Lambda \in \mathrm{SO}^{+}(1,3)$ with $q = \Lambda p$. In other words this would mean that $\mathrm{SO}^{+}(1,3)$ acts transitivly on $\lbrace q \in \mathbb{M} \mid q^{2} = m^{2}, q^{0} > 0 \rbrace$. And you need to show this in order to proof that
$$\mathcal{O}_{p} = \lbrace q \in \mathbb{M} \mid q^{2} = m^{2}, q^{0} > 0 \rbrace$$

Actually I also don't see how to do this without using brute force. I think it is not that trivial.