(adsbygoogle = window.adsbygoogle || []).push({});

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.

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Proving the Lorentz invariance of an integration measure? QFT related?

**Physics Forums | Science Articles, Homework Help, Discussion**