Lorentz invariance of [itex] E_p \delta ({\bf p}- {\bf q}) [/itex]

[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.

 

[LayMuon]

Nobody? :(

 

[Bill_K]

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.

 

[Bill_K]

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.

 

[Bill_K]

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.

 

[Bill_K]

Consider the domain where these functions are nonzero. δ⁴(p - p₀) is nonzero at just a single point - of course that's a Lorentz invariant domain. And δ(p² - m²) is nonzero on a hyperboloid - that's Lorentz invariant too.

But δ³(k - k₀) is nonzero on an entire line - and that's not Lorentz invariant! It singles out a rest frame. And putting E₀ in front of it doesn't change the fact.

It's only when you multiply the two of them together, δ(p² - m²) and δ³(k - k₀) -- then the product is nonzero on the intersection of the line and the hyperboloid, namely a single point again -- that you get back to a Lorentz invariant domain.