# Lorentz invariance of $E_p \delta ({\bf p}- {\bf q})$

LayMuon
Can anybody help me with the proof that $E_p \delta ({\bf p}- {\bf q})$ is a Lorentz invariant object?

I did a boost along z axes and used the formula $$\delta (f(x)) = \frac{\delta(x-x_0)}{|f'(x_0)|}$$ and the factor in front of the delta function indeed is invariant but within the function I have something like this:

$$E_p \delta (p_z -(v(E_q-E_p)+q_z))$$

but not $E_p \delta (p_z- q_z)$

Thanks.

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LayMuon

Nobody? :(

What makes you think it's Lorentz invariant?

LayMuon
Because in QFT we specifically choose a normalization in such a way as to get scalar product invariant.

Ok, now I understand! To distinguish between 3-dimensional and 4-dimensional quantities, let me call the energy-momentum 4-vector p, with components (E, k) and invariant norm p2 = m2.

To begin with, the 4-dimensional delta function δ4(p - p0) is clearly Lorentz invariant. Write this as δ(E - E0) δ3(k - k0). And now change variables on the first factor. In place of the variable E, I want to use the variable p2. Well, one can show that δ(p2 - m2) = (1/2E0) δ(E - E0). [There's a second term with δ(E + E0) if you want to include the negative energies.]

This gives us δ4(p - p0) = 2E0 δ(p2 - m2) δ3(k - k0), which is Lorentz invariant.

LayMuon
Thank you, I got it. I made a mistake in calculation.

That's good. By the way, the expression as you originally gave it is not Lorentz invariant. Only if you include the other delta function, δ(p2 - m2). And the variable you choose to replace E makes all the difference too. If you use some other f(E), you'll get something else instead of 2E0 as the factor in front.

LayMuon
But $\delta (p^2-m^2)$ is Lorentz invariant.