# Normalization of a delta function in curved spacetime

1. Aug 23, 2008

### jdstokes

Which of the following are true in curved spacetime?

$\int d^4 x \delta^4(x - x_0) = 1$ (1)

$\int d^4 x \sqrt{-g} \delta^4(x - x_0) = 1$ (2)

I think the first one is incorrect in curved spacetime, or in general when the metric is non-constant. I would argue this by saying that the delta function does not transform, whereas the fourth-order differential transforms in the opposite way to $\sqrt{-g}$, so the whole thing transforms as a scalar as it must.

I've also heard that $\delta^4$ is not a scalar, which suggests that (1) is the correct statement. However, this seems strange to me as I would think that (1) will fail to hold in curvilinear coordinates e.g.

2. May 14, 2011

### vaibhavtewari

actually the right difinition is
$$\int_M F(x^{\mu})[\frac{\delta^{(4)}(x^{\sigma}-y^{\sigma})}{\sqrt{-g}}]\sqrt{-g}d^4x=F(y^{\sigma})$$

3. May 15, 2011

### Bill_K

The delta function is a scalar density. You don't need curved space or four dimensions to see this, it follows from the identity δ(f(x)) = (1/|f '(x0)|) δ(x-x0) where f(x0) = 0. For example δ(3x) = (1/3) δ(x). Or take plane polar coordinates: δ(x) ≡ δ(x) δ(y) = (1/r) δ(r) δ(Θ).

Last edited: May 15, 2011
4. May 15, 2011

### haushofer

I'd say that

$\int d^4 x \delta^4(x - x_0) = 1$

is the usual definition. The RHS is trivially a scalar. The measure on the LHS is a density. So the delta distribution is also a density, as was mentioned by others here.

You can "tensorize" the delta distribution by defining

$$\delta^4(x - x_0) \rightarrow \frac{\delta^4(x - x_0) }{\sqrt{g}}$$