Question on Dirac Delta and functional derivatives

[shoehorn]

Suppose that we have a compact manifold $\mathcal{M}$ with a positive definite metric $g_{ij}$. The volume of the manifold is then

$$V = \int_\mathcal{M} d^3x \sqrt{g(x)},$$

where $x^i$ are coordinates on $\mathcal{M}$ and $\sqrt{g(x)}$ is the square root of the determinant of the metric. Suppose now that we take a functional derivative of the volume with respect to the metric, i.e., we want to calculate

$$\frac{\delta}{\delta g_{ij}(y)} V = \frac{\delta}{\delta g_{ij}(y)}\int_\mathcal{M} d^3x \sqrt{g(x)}$$

As far as I know, this calculation gives

$$\frac{\delta}{\delta g_{ij}(y)} V = \frac{1}{2}\int_\mathcal{M} d^3x \sqrt{g(x)} g^{ij}(x) \delta^{(3)}(x,y)$$

Where $\delta^{(3)}(x)$ is a three-dimensional Dirac delta function. The question I have is this: is $\delta^{(3)}(x,y)$ to be regarded as (a) a unit weight density or (b) a density of weight zero?

If the answer is (a) then the result is

$$\frac{\delta}{\delta g_{ij}(y)} V = \frac{1}{2}\sqrt{g(y)} g^{ij}(y)$$

while if the answer is (b) then the result for the functional derivative is

$$\frac{\delta}{\delta g_{ij}(y)} V = \frac{1}{2} g^{ij}(y)$$

So which one is the correct answer?

 

[jvicens]

I would first clarify the terminology being used. In mathematics, a density is a function that assigns a weight to each point in a space, while a weight density is a function that assigns a weight to each point and also varies with the coordinates. In physics, weight density is often referred to simply as density. In this context, \delta^{(3)}(x,y) should be considered a density of weight zero, as it does not vary with the coordinates.

With this clarification, the correct answer for the functional derivative would be (b), as \delta^{(3)}(x,y) is a density of weight zero and does not contribute to the weight of the manifold. Therefore, the result for the functional derivative would be

\frac{\delta}{\delta g_{ij}(y)} V
= \frac{1}{2} g^{ij}(y)

This result makes intuitive sense, as the functional derivative is a measure of how the volume changes with respect to a small change in the metric at a specific point. Since \delta^{(3)}(x,y) does not contribute to the weight of the manifold, it should not affect the calculation of the functional derivative.

 

[tacman]

The correct answer is (a), where \delta^{(3)}(x,y) is considered as a unit weight density. This is because the functional derivative with respect to a metric is defined as the variation of a functional with respect to an infinitesimal change in the metric, and the Dirac delta function represents such an infinitesimal change. Therefore, in order for the functional derivative to have the correct units, \delta^{(3)}(x,y) must be considered as a unit weight density. This is also consistent with the fact that the determinant of the metric is multiplied by \sqrt{g(x)} in the volume integral, which is a unit weight density.