Why can all 4-momenta of fixed length square be constructed by applying a Lorentz transform on a "standard" 4-momentum? 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 $p^0$.
 A: 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 $k = (E,E,0,0)$.
A: We note that for each class, if the assertion in the book is true for a particular choice of $k^\mu$, then the assertion is true for arbitrary choice of standard momentum. This can be seen from the following:
Let $p'^\mu$ is standard momentum for which the assertion is true. Then for any other member of the class
$$ p^\mu= L'(p)^\mu_{~\upsilon} p'^\upsilon$$
now, if we take any other standard momentum $k^\mu$, we have the following equation from the assertion of the book which is true for $p$ we have assumed:
$$k=\Lambda p$$
combining the last two equations, we have that
$$ p^\mu= (L'(p)\Lambda^{-1})^\mu_{~\upsilon} k^\upsilon$$
and we know that $(L'(p)\Lambda^{-1})$ is a Lorentz Transformation, proving the assertion for arbitrary $k$
We ,therefore, will prove the assertion only for particular choice of the standard momentum.
Now, in each class $L(p)$ can be written in the form with suitable choice of $k^\mu$ (Assume that the $p$ is arbitrary element in the class):
$$ L(p) =R_z(\phi)R_y(\theta)L_z(\zeta) \tag{1} \label{1}$$
where  $R_z(\phi),R_y(\theta),L_z(\zeta)$ are respectively a rotation in the $z$ direction, rotation in the $y$ direction, boost in the $z$ direction with $\phi, \theta$ being respectively azimuthal angle, polar angle of the vector $(p^1,p^2,p^3)$, and the value of $\zeta$ is collected in the table below.
(I am using the same metric as Weinberg [1996], for the length of the vector $\mathbf{P}$ the notation $\mathbf{P}^2$ and M is a positive number with the properties mentioned in the table)
In the table below, I am writing down the choice of $k^\mu$ and $\zeta$ mentioned above for each  class:




The Class
Standard $k^\mu$
Note




$ \mathbf{P}^2 < 0$  and $P^0 <0 $
$(-M,0,0,0)$
here $M = \sqrt{-\mathbf{P}^2}$ and $\sinh \zeta=\frac{\sqrt{(p^1)^2+(p^2)^2+(p^3)^2}}{-M}$


$\mathbf{P}^2 < 0$  and $P^0 >0$
$(M,0,0,0)$
here  $M = \sqrt{-\mathbf{P}^2}$ and $\sinh \zeta =\frac{\sqrt{(p^1)^2+(p^2)^2+(p^3)^2}}{M}$


$\mathbf{P}^2 =0$  and $P^0 >0 $
$(\omega,0,0,\omega)$
$\omega $ any positive number and $p^0= \omega e^\zeta$


$\mathbf{P}^2 =0$  and $P^0 <0 $
$(-\omega,0,0,\omega)$
$\omega $ any positive number and $p^0= -\omega e^{-\zeta}$


$\mathbf{P}^2 > 0$
$(0,0,0,M)$
here   $M = \sqrt{\mathbf{P}^2}$ and $\sinh \zeta=\frac{p^0}{N} $




using the above values and equation \eqref{1}, it is trivial to prove that for any choice of $p$ in each class:
$$p^\mu = L(p)^\mu_{~\upsilon} k^\upsilon$$
