# How is $\int \frac{d^{3}\mathbf{p}}{(2\pi)^3}\frac{1}{2\sqrt{|\mathbf{p}|^2+m^2}}$ manifestly Lorentz-Invariant?

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 $$q^0$$?

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.
• I believe there is a common language error in this type of discussions. The sign of $p^0$ can change under a Lorentz transformation. The Lorentz transformations which don't do this have $\Lambda^0_0>1$. In fact, even $\text{d}^4p$ can change signs under a Lorentz transformation. Those that don't do this have $\text{det}\Lambda=1$. Regarding your answer, it is precisely the Lorentz transformations which satisfy both of these, i.e. preserve the measure, the ones that belong to the connected component of the identity in the Lorentz group. – Iván Mauricio Burbano Feb 13 at 2:47
Use the relation $$\int \frac{d^3{\bf k}}{(2\pi)^3}\frac{1}{2\sqrt{{\bf k}^2+m^2}} = \int \frac{d^4k}{(2\pi)^4}\frac{1}{k^2+m^2}$$.