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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 $p^0$.

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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 $k = (E,E,0,0)$.

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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: $$p^\mu = L(p)^\mu_{~\upsilon} k^\upsilon$$

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