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If in an inertial frame $S$, $p^0$ is positive, then it is claimed that under a proper orthochronous Lorentz transformation ${\rm SO^+(3,1)}$ (i.e., those Lorentz transformations which are continuously connected to identity) $p^0$ does not change sign.The assertion that $p^0$ does not change sign is crucial to prove that the integral $\int \frac{d^3p}{(2\pi)^3} \frac{1}{E_p}$ is Lorentz invariant.

However, the way in which $p^0$ changes from frame S to frame $S^\prime$ is given by $$p^{0\prime}=\Lambda^{0}{}_{0}p^0+\Lambda^{0}{}_{1}p^1+\Lambda^{0}{}_{2}p^2+\Lambda^{0}{}_{3}p^3$$ with the constraints $\Lambda^{0}{}_{0}\geq 1$ and $\sum\limits_{i=1}^{3}(\Lambda^{0}{}_{i})^2\geq 0$. Though $p^0>0$, there is no such constraint on $p^1,p^2$ and $p^3$ which can be negative.

Therefore, I don't understand why cannot $p^{0\prime}$ change sign? . I'm looking for a simple proof that $p^0$ cannot change sign under a proper orthochronous Lorentz transformation. The existing answer by Qmechanic is too complicated to be tractable.

SRS
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2 Answers2

16

Sometimes, diagrams are more useful than equations!

Note that $p^2 = - (p^0)^2 + |\vec{p}|^2$ remains unchanged under any Lorentz transformation. Suppose that $p^2 < 0$ (say $p^2 = -1$). A plot of this hypersurface is shown below

enter image description here

Lorentz transformations that are connected to the identity are continuous transformations take a single point on this line to a different point on this line. Clearly, if I start off on the branch of the line with $p^0 > 0$, then no continuous transformation will ever take me to the $p^0 < 0$ branch. Thus, if $p^2 < 0$, Lorentz transformations cannot change the sign of $p^0$. Note that this is not true for the discrete Lorentz transformations such as $T$ which, of course, does change the sign of $p^0$.

On the other hand, if $p^2 > 0$ (say $p^2 = 1$), then the hypersurface looks like

enter image description here

It is now possible to have a continuous Lorentz transformation that changes the sign of $p^0$.

Prahar
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One needs to include the additional condition that the 4-vector $p^{\mu}$ should be a non-zero non-spacelike vector. Then it is true that the sign of the zero-component $p^0$ is preserved by an orthochronous Lorentz transformation. See e.g. the first part of my Phys.SE answer here.

Qmechanic
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