Lorentz Invariance of the Choice of $p_0 = \pm E_{p}$

Question

I'm following David Tong's notes on Quantum Field Theory and I'm struggling with proof of the Lorentz invariance of $\int d^3 p/(2E_p)$ given on page 32. The proof contains the line

"Solving for $p_0^2 = E_{\vec{p}}^2 = \vec{p}^2+m^2$, there are two branches of solutions: $p_0 = \pm E_p$. But the choice of branch is another Lorentz invariant concept."

This is something I cannot seem to grasp. Could someone give a mathematical explanation as to why this choice of branch is Lorentz invariant?

Answer 1

Consider the equation $E^2=p^2+m^2$. When $m>0$ the solution set forms two hyperboloids:

One sheet has $E\geq m>0,$ the other has $E\leq -m<0$. Note that there is a gap between the two parts.

The Lorentz transformation is continuous as long as we do not swap the time direction. Therefore it can not move a point over the gap, but only inside one of these parts. This implies that each part separately is invariant under the Lorentz transformation. Thus the choice of $E>0$ or $E<0$ is Lorentz invariant.

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My confusion comes from the first equation in this article: en.wikipedia.org/wiki/Lorentz_transformation. It appears from the first equation that you can go from positive to negative time due to the $−vx/c^2$ term in the formula. Why is this not the case?

The transformation rules for energy and momentum are similar to those for time and space: $$ E' = \gamma(E-p_x v), \quad p_x' = \gamma(p_x - Ev/c^2) $$

For $E'$ to be negative when $E$ is positive, we need to have $E-p_x v<0$ i.e. $E<p_x v.$ But this requires superluminal speeds: $$ E-p_x = \gamma mc^2 - \gamma mu_x v = \gamma m (c^2 - u_x v) > 0 $$ unless $u_x v \geq c^2.$