Is the Delta Function a Scalar?

[kent davidge]

I read that $\delta^4(x-y)$ is invariant under Lorentz transformations. I was trying to show myself this, so I procceded as follows.

The following integrals are both equal to 1 $\int \delta^4(x-y) d^4 x$ and $\int \delta^4 (x'-y') d^4x'$ so I assume they are equal to one another, as long as we integrate them over all $\mathbb R^4$. Now $d^4 x = d^4 x'$ because the Lorentz metric has $\sqrt {| \eta |} = 1$. So I conclude that $\delta^4(x-y) = \delta^4(x'-y')$. Is this correct?

 

[Cryo]

Pretty much correct.

A more detailed explanation follows. The delta function is what Lovelock and Run called a relative tensor. In essence the delta function must transform so that $\int \delta^{(4)}\left(x\right) d^4 x$ does not depend on coordinates. However, the "volume" form $d^4 x$ does depend on coordinates. If we switch coordinates using a map $\phi: x\to\bar{x}$, then $\int_V \dots d^4 x = \int_\bar{V} \dots \frac{\partial\left(x\right)}{\partial \left(\bar{x}\right)} d^4 \bar{x}$, where $\frac{\partial\left(x\right)}{\partial \left(\bar{x}\right)}$ is the Jacobian.

It follows that:

$\frac{\partial\left(x\right)}{\partial \left(\bar{x}\right)} \delta^{(4)}\left(x(\bar{x})\right) = \delta^{(4)}\left(\bar{x}\right)$

to counteract the scaling of the volume element. With some work (Lovelock & Rund, "Tensors, Differential forms ...", Chap 4) you can show that $\frac{\partial\left(\bar{x}\right)}{\partial \left(x\right)}=\sqrt{\frac{g}{\bar{g}}}$, where $g$ is the determinant of the metric in the initial coordinate system.

Finally, you can conclude that whilst delta function is not a scalar, the following quantity:

$\frac{1}{\sqrt{g}} \delta^{(4)}\left(x\right) = \frac{1}{\sqrt{\bar{g}}} \delta^{(4)}\left(\bar{x}\right)$

is. It does not come into play for basic Lorentzian boosts, because, as you noted $\sqrt{g}=1$ at all times.