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 $p^0$.


2 Answers 2


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.