Lorentz transformations of energy and relativistic 3-momentum

Question

My question is how do energy and the relativistic 3-momentum change under a Lorentz transformation? I know that the 4-momentum transforms as $P' = \Lambda P$ where $\Lambda$ is a $4\times 4$ Lorentz transformation matrix, and I thought that using this should tell me how the energy and momentum transform as the 4-momentum encapsulates both of these quantities, but I am unsure on how to proceed with this as there are many $4\times 4$ Lorentz transformation matrices, and doing the algebra with the 'boost along $x$-axis matrix' didn't seem to get me anywhere.

Answer 1

In order to match the dimensions of momentum, $p^0$ has to be defined as $p^0 = E/c$ where $E$ is the relativistic energy (=rest + kinetic). Taking a boost in the $x$ direction with speed $v$, let $\beta = v/c$ and $\gamma = (1-\beta^2)^{-1/2}$. Then \begin{align} \frac{E'}{c} &= \gamma (\frac{E}{c} - \beta p^1)\\ p^1{'} &= \gamma(p^1-\beta\frac{E}{c})\\ p^2{'} &= p^2\\ p^3{'} &= p^3 \end{align}

The Lorentz group has 4 components, with the identity component $SO^+(1,3)$ called the proper orthochronous Lorentz group. It is generated by 6 one-parameter subgroups, each generated by an infinitesimal generator of the Lie algebra. Three of these are Euclidean rotations in the $xy, yz, zx$ planes. The other three are boosts in the hyperbolic planes $tx, ty, tz$. (notice that $\gamma = \cosh \varphi$, $\beta = \tanh \varphi$ and $\gamma \beta = \sinh \varphi$ for some $\varphi$ called the rapidity.)

A convenient alternative view is given as follows. Let $A\in SL_2(\mathbb{C})$ the group of $2$ by $2$ matrices with determinant $1$. Now represent a vector $(ct, x, y, z)$ in Minkowski space as $$X = \begin{pmatrix} ct + z & x - iy\\ x + iy & ct - z \end{pmatrix}$$

(this is the space of all Hermitian $2$ by $2$ complex matrices.) Notice that $\det(X)$ is the Lorentz metric. Now consider the action of $A$ on $X$ given by $$X \mapsto A X A^*$$ Since $\det(A) = 1$ we have $\det(AXA^*) = \det(X)$ so this preserves the Lorentz metric. The image of this map from $SL_2(\Bbb{C})$ to the Lorentz group is $SO^+(1,3)$, and the map is a double cover (note that $-A$ acts the same as $A$.) This map is the universal cover of $SO^+(1,3)$ which has fundamental group $\Bbb{Z}/2\Bbb{Z}$ while $SL_2(\Bbb{C})$ is simply connected.

If you want $A$ to act on a momentum 4-vector then take $$P = \begin{pmatrix} E/c + p^3 & p^1 - ip^2\\ p^1 + ip^2 & E/c - p^3 \end{pmatrix}$$ and the action $P\mapsto A P A^*$.

Answer 2

The equations that I used when I learned the theory of relativity more than 30 years ago were: $$ x = \gamma(x'+t'v), \quad y=y', \quad z=z', \quad t=\gamma(t'+x'v/c^2) $$ for the coordinates. Here the primed system is moving with speed $v$ in the positive $x$ direction. As usual, $\gamma=1/\sqrt{1-v^2/c^2}.$

For energy and momentum we then have $$ p_x = \gamma(p_x'+E'v/c^2), \quad p_y=p_y', \quad p_z=p_z',\quad E=\gamma(E'+p_x'v). $$