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So, first off, I'm thinking Lorentz invariant quantities are the same in any inertial frames S and S' regardless of their relative velocity.

I'm thinking I need to show that

[itex]\frac{d^3k}{(2\pi)^32E(\vec{k})} = \frac{d^3k'}{(2\pi)^32E'(\vec{k'})}[/itex] where the primed & unprimed quantities denote different frames.

We also have [itex] E(\vec{k}) = \sqrt{\vec{k}^2 + m^2}[/itex] in the denominator. This isn't Lorentz invariant - the mass term is, but the 3 - momentum [itex]\vec{p} = \hbar \vec{k}[/itex] is not, therefore E is not.

[itex]E \neq E'[/itex] where [itex] E'(\vec{k'}) = \sqrt{\vec{k}'^2 + m^2}[/itex].

This is making me think that neither E nor d

^{3}k are Lorentz invariant, so that when they are used in this fraction, the Lorentz invariance from each one somehow cancels out overall.

Then we come to the hint. I 'm really not sure what to make of this, I mean, I can write [itex]\delta(x^2 - x_0^2) = \frac{1}{2|x|}(\delta(x-x_0) + \delta(x+x_0))[/itex] with k's and m's: [itex]\delta(k^2 - m^2) = \frac{1}{2|k|}(\delta(k-m) + \delta(k+m))[/itex]

to get an apparently Lorentz invariant expression: [itex] \frac{d^4k}{(2\pi)^3}\delta(k^2 - m^2)\theta(k_0) = \frac{d^4k}{(2\pi)^3}\frac{1}{2|k|}(\delta(k-m) + \delta(k+m))\theta(k_0) [/itex]

but I'm not sure where this gets me, other than guessing that

[itex] \delta(k^2 - m^2)\theta(k_0) [/itex] is also Lorentz invariant seeing as I am told d

^{4}k is.

I have no idea how to proceed with this. I don't know what to make of the delta functions either.