Lorentz invariance of volume element from the four-volume element: why on-shell?

Question

In Srednicki's QFT book, in chapter 3 (eqn. 3.16 onwards) he talks about the lorentz invariance of the volume element. For this he writes $d^3k/f(k)$ should be invariant under lorentz transformations. He then tells us that an integration measure that's manifestly invariant under such a transformation is $d^4k \delta (k^2+m^2)\theta(k^0)$.

I know that $d^4k$ is manifestly invariance under lorentz transformations since $det \Lambda = 1 or -1$. Why do we multiply the on-shell condition and the theta function to it? And then why do we integrate over $k^0$ to finally obtain $f(k)\propto \omega$?

Why can't we simply write $d^3k = d^4k/E_k$ transforming from the spatial element to a space-time element?

I'm looking for rationale (physical and mathematical) for going through the particular hoops that Srednicki does.

Answer 1

Writing something like $d^3k = d^4k/E_k$ simply would not make sense, since the left-hand side is a volume element in three dimensions whereas the right-hand side is a volume element in 4 dimensions. In order to obtain the three-dimensional volume element, then, you need to do something else, and that's precisely integrating over $k^0$. The theta function ensures that we are only interested in accounting for Lorentz transformations that preserve the orientation of the timelike basis vector, which amounts to preserving the sign of the time component. The mass-shell condition is just what it is: it ensures that the particle satisfies the relativistic dispersion relation. I don't know the book you are following, but this should be pretty straightforward.