Lorentz transformation (I)

Problem: Let $p^{\mu}\equiv[E, p^1, p^2, p^3]^T$ be the four-momentum of a particle with the rest mass $M$, assuming $c=1$. In the rest frame, such four-momentum becomes to $k^{\mu}\equiv[M, 0,  0,  0]^T$.  The problem is to find a Lorentz transformation (matrix) $L^{\mu}_{\,\,\nu}(p)$ such that $p^{\mu} =L^{\mu}_{\,\,\nu}(p)\,k^{\nu}$.

(This problem is about the proof of Eq. (2.5.24) in Weinberg's QFT book, Volume I)

Solution:

Recall that the vector form of Lorentz boost of four-momentum in two inertial frames with relative velocity $\mathbf{v}$: \begin{equation}\begin{split} E'&=&\gamma\left(E - v\,\mathbf{n}\cdot\mathbf{p}\right)\,,\\ \mathbf{p}'&=&\mathbf{p} + (\gamma-1)\left(\mathbf{n}\cdot\mathbf{p}\right)\mathbf{n}-\gamma\,E\, v\,\mathbf{n}\,,\end{split}\tag{1}\end{equation} where $v$ is the magnitude of $\mathbf{v}$ and $\mathbf{n}\equiv \mathbf{v}/v$ is the unit vector of  $\mathbf{v}$. $\gamma\equiv \frac{1}{\sqrt{1-v^2}}$ is the Lorentz factor.

In terms of ${p'}^{\mu} \equiv [E',\, \mathbf{p}']^T$ and $p^{\mu} \equiv [E,\, \mathbf{p}]^T$, the above Lorentz boost (1) can be written as ${p'}^{\mu} = L^{\mu}_{\,\,\nu}\left(\mathbf{v}\right)p^{\nu}$, where \begin{equation}\begin{split} L^{0}_{\,\,0}\left(\mathbf{v}\right)&=&\gamma\,,\\ L^{0}_{\,\,i}\left(\mathbf{v}\right)&=&-\gamma\,v\,n_i\,,\\ L^{i}_{\,\,0}\left(\mathbf{v}\right)&=&-\gamma\,v\,n^i\,,\\ L^{i}_{\,\,j}\left(\mathbf{v}\right)&=&\delta^i_{\,\,j}+(\gamma-1) n^in_j\,.\end{split}\tag{2}\end{equation}

Now let's apply the general formula (1) to our specific problem, we have relations \begin{eqnarray} E &=& \gamma\,M\,,\tag{3.1}\\ \mathbf{p}&=&-\gamma\,M\, v\,\mathbf{n}\,.\tag{3.2}\end{eqnarray}

Note that $E=\sqrt{\mathbf{p}^2+M^2}$, from Eq. (3.1) we can solve that \begin{equation}\gamma=\frac{\sqrt{\mathbf{p}^2+M^2}}{M}\,.\tag{4}\end{equation}

We then submit \begin{equation}v =\frac{1}{\gamma} \sqrt{\gamma^2-1} = \frac{1}{\gamma} \frac{|\mathbf{p}|}{M}\tag{5}\end{equation} into Eq. (3.2) and obtain the relation \begin{equation}\mathbf{n}=-\frac{\mathbf{p}}{|\mathbf{p}|}\equiv -\hat{\mathbf{p}}\,, \tag{6}\end{equation} where $\hat{\mathbf{p}}$ is the unit vector of $\mathbf{p}$.

Finally, submitting Eq. (5)-(6) into Eq. (2) gives final answer to our problem: \begin{equation}\begin{split} L^{0}_{\,\,0}\left(\mathbf{p}\right)&=&\gamma\,,\\ L^{0}_{\,\,i}\left(\mathbf{p}\right)&=&\hat{p}_i\sqrt{\gamma^2-1}\,,\\ L^{i}_{\,\,0}\left(\mathbf{p}\right)&=&\hat{p}^i\sqrt{\gamma^2-1}\,,\\ L^{i}_{\,\,j}\left(\mathbf{p}\right)&=&\delta^i_{\,\,j}+(\gamma-1) \hat{p}^i\hat{p}_j\,,\end{split}\end{equation} with $\gamma$ in Eq. (4).

Note:

In special relativity, a particle with rest mass $M$ and velocity $\mathbf{u}$ with respect to the ground has the momentum $\mathbf{p}=\gamma M \mathbf{u}$. The velocity $\mathbf{v}$ appeared in above Eq. (3) , (5) and (6) is just the relative velocity of the ground with respect to the particle: $\mathbf{v}=-\mathbf{u}=-\mathbf{p}/(\gamma M)$.