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To prove $\frac{\mathrm{d}^3\vec{p}}{E}$ is Lorentz invariant is to prove $$\frac{\mathrm{d}^3\vec{p}}{E} = \frac{\mathrm{d}^3\vec{p}'}{E'} \qquad(\mathrm{d}^3\vec{p} := \mathrm{d}p_x \mathrm{d}p_y \mathrm{d}p_z). \tag{1}$$

Suppose reference system $\Sigma'$ is moving at speed $v=\beta$ ($c=1$) along $z$ axis of $\Sigma$. We have $$E' = \gamma(E - \beta p_z), \qquad p_{x,y}' =p_{x,y}, \qquad p_z' = \gamma(p_z-\beta E), $$

then $$\det\frac{\mathrm{d}\vec{p}'}{\mathrm{d}\vec{p}} = \gamma \Big( 1 - \frac{\mathrm{d}E}{\mathrm{d}p_z} \Big).$$

If $$ \frac{\mathrm{d}E}{\mathrm{d}p_z} = \frac{\partial E}{\partial p_z} = \frac{p_z}{E}, \tag{2} $$ we can prove (1). However, I don't understand the first equality in (2).

Qmechanic
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Doris
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1 Answers1

1

Start from $d^4p = dp_{\mu}\,dp^{\mu}$ which is manifestly Lorentz invariant. In order to obtain the actual measure to integrate against you have to pair this up with the mass-shell condition for the particle to have positive energy and lie on the mass-shell hyperboloid $$ dp_{\mu}\,dp^{\mu}\,\theta(k^0)\,\delta(k^2-m^2). $$ Use now $k^2={k^0}^2 - k_ik^i$ and the Jordan lemma to integrate on the contour where $k^0$ is positive. You can then integrate out the $k^0$ component and re-write the denominator in the residue in terms thereof, which leads to the equation you want to prove, plus $2\pi$ according to how you the define the residues.

A very simple walkthrough of the above is in this book: http://www.amazon.co.uk/Quantum-Field-Theory-Lewis-Ryder/dp/0521478146 (or any other introductory book in QFT).

gented
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