Question 2.4: The Lorentz-Invariant Phase-Space Factor Show that the phase......
The Lorentz-Invariant Phase-Space Factor
Show that the phase-space factor
dLips(s;\,p_3,\,p_4)
=(2\pi)^{4}\,\delta^{4}(p_{3}+p_{4}- p_{1}-p_{2})\frac{1}{(2\pi)^{3}}\frac{\mathrm{d}^{3}p_{3}}{2E_{3}}\frac{1}{(2\pi)^{3}}\frac{\mathrm{d}^{3}p_{4}}{2E_{4}} (1)
is Lorentz invariant.
The phase-space factor is apparently invariant under spacial rotations. We must therefore investigate its behavior under proper Lorentz transformations, which are induced by the matrix Λ(v_{\mathrm{b}}) with a boost velocity v_{\mathrm{b}}. Owing to rotational invariance one can, without loss of generality, put v_{\mathrm{b}} parallel to the z axis. This considerably simplifies the dependence of the particle coordinates in the system at rest p_{\mathrm{n}} (n = 1, 2, 3, 4) on the new coordinates p_{n}^{\prime} in the moving reference system. In general this dependence is
p_{n}^{\prime}= Λ(v_{\mathrm{b}})p_{n}\ . (2)
with the inversion
p_{n}=Λ^{-1}(v_{\mathrm{b}})p_{n}^{\prime}= Λ(-v_{\mathrm{b}})p_{n}^{\prime}\ . (3)
For the differentials \mathrm{d}p_{n\prime}^{x},\,\mathrm{d}p_{n\prime}^{y},\,\mathrm{d}p_{n\prime}^{z}, we obtain \left(\gamma=1/{\sqrt{1-v_{\mathrm{b}}^{2}}}\right)\!\!:
\mathrm{d}p_{n}^{x}=\mathrm{d}p_{n}^{x\prime}\ ,\\ \mathrm{d}p_{n}^{y}=\mathrm{d}p_{n}^{y\prime}\ ,\\ \mathrm{d}p_{n}^{z}=\gamma\left(\mathrm{d}p_{n}^{z\prime}+\vert v_{b}\vert\mathrm{d}E_{n}^{\prime}\right)\\ =\gamma {\mathrm{d}}p_{n}^{z^{\prime}}\left(1+\vert v_{\mathrm{b}}\vert\frac{p_{n}^{z^{\prime}}}{E_{n}^{\prime}}\right)=\mathrm{d}p_{n}^{z\prime}{\frac{E_{n}}{E_{n}^{\prime}}}. (4)
Here we have employed the relations
E_{n}\equiv p_{n}^{0}=\left(p_{n}^{2}+m_{n}^{2}\right)^{\frac{1}{2}}=\gamma(E_{n}^{\prime}+\vert v_{b}\vert p_{n}^{z^{\prime}})~, (5)
which verifies the last step in equation (4), and
\frac{\mathrm{d}E_{n}^{\prime}}{\mathrm{d}p_{n}^{z\prime}}=\frac{\mathrm{d}}{\mathrm{d}p_{n}^{z\prime}}\sqrt{m^{2}+\left(p_{n}^{x\prime}\right)^{2}+\left(p_{n}^{y\prime}\right)^{2}+\left(p_{n}^{z\prime}\right)^{2}}=\frac{p_{n}^{z\prime}}{E_{n}^{\prime}}~. (6)
For the volume element \mathrm{d}^{3}p_{n}=\mathrm{d}p_{n}^{x}\mathrm{d}p_{n}^{y}\mathrm{d}p_{n}^{z} we therefore have
\frac{\mathrm{d}^{3}p_{n}}{E_{n}}=\frac{\mathrm{d}^{3}p_{n}^{\prime}}{E_{n}^{\prime}}~. (7)
It remains to prove that the four-delta function is a Lorentz scalar. By definition we have
\int\mathrm{d}^{4}p\delta^{4}(p)=1 (8)
in any reference frame. Owing to the properties of the matrix Λ(v) the volume element is a Lorentz scalar too:
\mathrm{d}^{4}p=\left|{\frac{\partial p^{\mu}}{\partial p^{\prime\nu}}}\right|\mathrm{d}^{4}p^{\prime}=|\mathrm{det}(Λ(-v_{\mathrm{b}}))|\mathrm{d}^{4}p^{\prime}=\mathrm{d}^{4}p^{\prime}\ . (9)