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When writing integrals that look like $$ \int \frac{d^{3}\mathbf{p}}{(2\pi)^3}\frac{1}{2\sqrt{|\mathbf{p}|^2+m^2}} \ = \int \frac{d^4p}{(2\pi)^4}\ 2\pi\ \delta(p^2+m^2)\Theta(p^0) $$ it is often said that this is manifestly Lorentz invariant (where $\Theta$ is the Heaviside step function). Why is this true?

If I consider a Lorentz transformation $p \to q = \Lambda p$, then $|\det(\Lambda)| =1$ so the Jacobian is just 1, and: $$ \int \frac{d^4p}{(2\pi)^4}\ 2\pi\ \delta(p^2+m^2)\Theta(p^0) = \int \frac{d^4 q}{(2\pi)^4}\ 2\pi\ \delta(q^2+m^2)\ \Theta\big([\Lambda^{-1}]^0_{\ \nu} q^\nu \big) $$ After this transformation $p^0$ gets taken to $[\Lambda^{-1}]^0_{\ \nu} q^\nu$, and things do not look Lorentz invariant to me. If it were invariant wouldn't this get mapped to just $q^0$?

Qmechanic
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QuantumEyedea
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2 Answers2

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The measure is invariant under proper orthochronous Lorentz transformations (those continuously connected to the identity, or equivalently those that do not represent a spatial reflection or time reversal) because $\Theta(\Lambda^0_\nu q^\nu) = \Theta(q^0)$ for such transformations as they preserve the sign of the temporal component of any time-like vector.

Proof of the claim that a proper orthochronous transformation preserves the sign of the temporal component: The "mass shell hyperboloid" cut out by $p^2 = -m^2$ has two connected components, one with positive, one with negative temporal component. A transformation continuously connected to the identity cannot change the connected component of any vector without removing it from the mass shell for an intermediate transformation along the path to the identity, but the mass shell condition is invariant under all Lorentz transformations and therefore this cannot happen.

ACuriousMind
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Use the relation $$ \int \frac{d^3{\bf k}}{(2\pi)^3}\frac{1}{2\sqrt{{\bf k}^2+m^2}} = \int \frac{d^4k}{(2\pi)^4}\frac{1}{k^2+m^2} $$.

pathintegral
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