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Consider a surface $S$ in $R^3$ given by an equation of the form $$\Phi(x,y,z)=0$$ where $\Phi$ is a sufficiently smooth function. Suppose we have a surface charge distribution on $S$ with density $\sigma(\mathbf x)$. I've seen many times the following formula $$\rho(\mathbf x)\,=\,\sigma(\mathbf x)\,|\nabla \Phi(\mathbf x)|\, \delta(\Phi(\mathbf x))\qquad \qquad (*)$$ where $\delta$ is a Dirac delta function. The formula $(*)$ presumably represents the surface charge density $\sigma(\mathbf x)$ in terms of the (singular) volume charge density $\rho(\mathbf x)$.

I have a question about the formula $(*)$. Since $\rho(\mathbf x)$ is a VOLUME charge density, the point $\mathbf x$ can be any point in $R^3$, or at least, any point in a sufficiently small neighborhood of the surface $S$. But, if you take a point very close to $S$ but NOT in $S$, then the right-hand member of equation $(*)$ is not well defined because $\sigma(\mathbf x)$ is a SURFACE charge density and hence it is only defined for points $\mathbf x \in S$. I know that $\delta(\Phi(\mathbf x))$ kills everything outside $S$ because $\delta(\Phi(\mathbf x))=0$ if $\mathbf x \notin S$, but still the right-hand member of equation $(*)$ is not well defined for points $\mathbf x \notin S$ because $\sigma(\mathbf x)$ is only defined in $S$.

On the other hand, if we restrict ourselves to points on $S$, then the formula $(*)$ works fine; in that case, however, $\rho(\mathbf x)$ would be merely a surface charge density, NOT a volume charge density. But this means that our project has failed entirely because our primary goal was precisely to construct a volume charge density from a surface charge density.

I don't want a proof of the formula $(*)$. I would simply like someone to clarify the above doubts: essentially the formula $(*)$, as it stands, seems to be meaningless.

$\mathit Note$.$\quad$By introducing a new coordinate system $(u,v,w)$ such that $\Phi=0$ becomes one of the coordinate surfaces (for this purpose I take $\Phi=w$), I can make sense of all of this. But then there is no way I can write the resulting volume charge density in the form $(*)$. I get something like $\rho \sim \sigma(u,v)\delta(w)$.

@Triatticus:

https://www.physicsforums.com/threads/writing-the-charge-density-in-the-form-of-the-dirac-delta-function.1003088/

https://www.physicsforums.com/threads/volume-charge-density-from-surface-line-charge-density.565317/ (the last answer)

Yes, $\Phi=k \,(k\in R)$ defines a family of surfaces, but $\sigma(\mathbf x)$ is only defined in $\Phi=0$.

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  • $\begingroup$ Where have you seen this identity. Just saying you've seen it gives us no context. Note that $\Phi$ can define a whole family of surfaces (take a sphere for example then $\Phi(r) = r^2 - a^2$ where $a$ is a real number.) $\endgroup$
    – Triatticus
    Commented Apr 11 at 14:01
  • $\begingroup$ I don't recognize the notation in (*). Is this some sort of inner product? $\endgroup$
    – march
    Commented Apr 11 at 15:40
  • $\begingroup$ It is the product of three terms: (1) $\sigma (\mathbf x)$, (2) the norm of the gradient $\nabla \Phi (\mathbf x)$, and (3) $\delta (\Phi (\mathbf x))$, the composition of $\Phi$ with the Dirac delta function $\delta$. $\endgroup$
    – Kubik
    Commented Apr 11 at 15:52

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The function $\sigma(\mathbf{r})$ should be taken to mean $$\begin{cases} \sigma(\mathbf{r}) & \mathbf{r} \in S \\ 0 & \mathbf{r} \notin S \end{cases}$$ as the surface charge density is zero for points not in $S$. In fact, the value of $\sigma$ outside $S$ doesn't matter because the Dirac delta function has the property $$f(\mathbf{r})\delta(\mathbf{r}) = f(\mathbf{0})\delta(\mathbf{r}).$$ So, applied here, it becomes $$f(\mathbf{r})\delta(\Phi(\mathbf{r})) = f(\mathbf{r})|_{S}\delta(\Phi(\mathbf{r})).$$ Therefore, I don't see how the values outside $S$ matter mathematically. On physical grounds, the most sensible thing is to take it as zero. Putting everything together, the function means $$\rho(\mathbf{r}) = \begin{cases} \sigma(\mathbf{r})|\nabla\Phi(\mathbf{r})|\delta(0) & \mathbf{r} \in S \\ 0 & \mathbf{r} \notin S \end{cases}.$$

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  • $\begingroup$ Yes, I had thought about this. But technically your $\sigma$ is a volume charge density because it is defined everywhere in $R^3$. And there is another issue: the singular support of your $\sigma$, considered as a distribution (generalized function) on $R^3$, is $S$ and the singular support of the distribution $\delta (\Phi)$ is also $S$. But the product of two distributions (in this case $\sigma$ and $\delta (\Phi)$) is not defined unless their singular supports are disjoint. $\endgroup$
    – Kubik
    Commented Apr 11 at 16:35
  • $\begingroup$ @Kubik "your $\sigma$ is a volume charge density because it is defined everywhere in $R^3$" This is false. $\sigma$ is just a real-valued function. Its domain says nothing about what density it represents and the implication does not follow at all. Also I don't see how your second argument works at all. $\sigma$ is a function and not a distribution and we can do this by definition of the delta function. $\endgroup$
    – Vincent Thacker
    Commented Apr 11 at 16:43
  • $\begingroup$ But then, in your approach, it is not necessary to use any δ. You can simply put: $$\rho (\mathbf x) = \sigma (\mathbf x) |\nabla \Phi (\mathbf x)| \quad for \quad \mathbf x \in S$$ $$\rho (\mathbf x) = 0 \quad for \quad \mathbf x \notin S$$ $\endgroup$
    – Kubik
    Commented Apr 11 at 16:51
  • $\begingroup$ @Kubik No. $\rho$ is infinite at the surface but zero everywhere else. That requires a delta function. You can check this by integrating over a Gaussian pillbox tangential to the surface. $\endgroup$
    – Vincent Thacker
    Commented Apr 11 at 16:55
  • $\begingroup$ Ah! Ok, Ok. I can see it now. Thanks for your reply! $\endgroup$
    – Kubik
    Commented Apr 11 at 18:16

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