# 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 $$p^0$$.

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

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