Dirac delta expression for a sphere in Cartesian coordinates?

Question

Is this how one would express a sphere as a Dirac delta in Cartesian coordinates?

DiracDelta[Sqrt[R0^2-z^2-x^2],Sqrt[R0^2-z^2-y^2],Sqrt[R0^2-x^2-y^2]]

p 32 Barton et al gives the strong definition of the 3D (spherical coordinate) radial Dirac delta as:

$$\delta^3(\vec{r}) = \frac{\delta(r)}{4\pi r^2}\tag{1}$$

and, correspondingly in 2D (polar coordinate):

$$\delta^2(\vec{r}) = \frac{\delta(r)}{2\pi r}.\tag{2}$$

Since $4\pi r^2$ and $2\pi r$ measure the area and length of sphere and circle respectively, and the surface and line integrals of these two Dirac deltas are 1 (by definition) it seems natural to, in appropriate circumstances, use the radial Dirac delta in modeling density distributions, normalized to 1, over radial surfaces and radial lines respectively.

Answer 1

Is the thing you want something that yields the area of a spherical surface of radius r0, when integrated over all space? Then, in spherical coordinates it's r0^2 DiracDelta[r-r0], no? I believe that you may then substitute Sqrt[x^2+y^2+z^2] for r, and get what you want, formally. Unfortunately, Mathematica seems not to understand such a thing.

Answer 2

I assume you want an expression for the surface measure on a sphere having center at the origin and radius $R>0$. Let us call this measure $\sigma_R$.

Our sphere is the zero set of the function $g_R(\mathbf{x})=R^2-\lvert\mathbf x\rvert^2$. Now, according to this property of the Dirac delta distribution, we have the formula $$ \delta(g_R(\mathbf x)) = \frac{\sigma_R(\mathbf x)}{\lvert \nabla g_R(\mathbf x)\rvert} ,$$ and in our case, $\lvert \nabla g_R(\mathbf x)\rvert = 2 \lvert \mathbf x\rvert$, which reduces to $2R$ on the support of $\sigma_R$. We conclude that $$ \sigma_R (\mathbf x)= 2R\delta(R^2-\lvert \mathbf x\rvert^2).$$

In polar coordinates $(r, \theta, \phi)$, the right-hand side reads $2R\delta(R^2-r^2)$.

If we had used $f_R(\mathbf x)=R-\lvert \mathbf x\rvert$ instead of $g_R$, we would have obtained the alternative expression $$ \sigma_R(\mathbf x)=\delta(R-\lvert\mathbf x\rvert)=\delta(R-r).$$