Why can all 4-momenta of fixed length square be constructed by applying a Lorentz transform on a “standard” 4-momentum

Question

In the subsection on 'One particle States' of Weinberg [1996], he says:

Note that the only functions of $p^\mu$ that are left invariant by all proper orthochronous Lorentz transformations $\Lambda$, are the invariant square $p^2 =\eta_{\mu\upsilon} p^\mu p^\upsilon$, and for $p^2\leq 0$ ,also the sign of $p^0$, Hence, for each value of $p^2$, and (for $p^2 \leq 0$) each sign of $p^0$, we can choose a `standard' four-momentum, say $k^\mu$, and express any $p^\mu$ of this class as
$$p^\mu = L^\mu_{~\upsilon}(p)k^\upsilon \tag{1} $$
where $L(p)$, is some standard Lorentz transformation that depends on $p^\mu$,
and also implicitly on our choice of the standard $k^\mu$.

I don't understand why equation (1) should be true for any $p^\mu$ with a given value of $p^2$ and (if $p^2\leq 0$) each sign of $p^0$.

Answer 1

Maybe it's useful to think it through in reverse? If you've got a 4-vector $p$, you can always try to put it in some standard form. For example: If $p^2 \neq 0$, you can boost to zero out the 3-momentum part, and you end up with a vector of the form $k = (m,0,0,0)$. If $\Lambda$ is the boosting Lorentz transform, then $k = \Lambda p$, so $p = \Lambda^{-1}k$. $\Lambda$ certainly depends on $p$, so it makes sense to write $p = L(p)k$. When $p^2 = 0$, you can't ever get to the particle's rest frame, but you can at least standardize to $k = (E,E,0,0)$.