How Do Lorentz Transformations Relate Time-like Four-Momenta in SO^{+}(1,3)?

[parton]

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?

 

[dauto]

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.

 

[SophusLie]

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.