Now if I have the electric potential as a function of $x$, $y$ and $z$. I use the Poisson equation to calculate the charge distribution and I obtain a charge distribution as a function of $x$, $y$ and $z$. Now, if I know that the charge lies entirely on a surface in 3d Cartesian coordinates and the potential on the surface is described by a function of its parametric variables. Can I use the Poisson equation as follows?

--- I take the laplacian of the potential function (which is a function of the parametric variables) with respect to the parametric variables and equate it to $-\rho/\epsilon$ not?

$$\frac{d^2v}{ds^2}+\frac{d^2v}{dt^2} =-\frac{\rho}{\epsilon}$$ (where $s$ and $t$ are the parametric variables.)

This will give the charge density as a function of $s$ and $t$ (the parametric variables) and is defined on the surface. Is this true? Thanks

  • $\begingroup$ It depends on precisely how your parametric variables are defined. For general curvilinear coordinates the Laplacian is a bit more complex than that, see here. $\endgroup$
    – lemon
    Jan 14, 2017 at 11:47
  • $\begingroup$ Cant we just use the regular laplacian that is defined in rectangular coordinates since we are have parametrized the surface. Hence the surface is basically derived from the flat s-t plane? $\endgroup$
    – Chandrahas
    Jan 14, 2017 at 11:57
  • $\begingroup$ Again, it depends on precisely how your coordinates are defined. If they're simply $s=x$ and $t=y$, for example, then you can... $\endgroup$
    – lemon
    Jan 14, 2017 at 12:04
  • $\begingroup$ why cant I do that if they are not x and y? can you elaborate on that? $\endgroup$
    – Chandrahas
    Jan 14, 2017 at 12:12
  • $\begingroup$ Here's a simple illustration of why. Suppose you have $s=x$ and $t=2y$, then the Laplacian (as you've written it) will not be the same as if $s=x$ and $t=y$. So your answer then becomes coordinate-dependent which it should not be! $\endgroup$
    – lemon
    Jan 14, 2017 at 17:55

1 Answer 1


In general, the Poisson equation in a Riemannian manifold $M$ is given by,

$$\nabla_i \nabla^i \phi = -\frac{\rho}{\epsilon}$$

where $\nabla_i \nabla^i$ is the Laplace-Beltrami operator, which acting on a scalar is simply,

$$\nabla_i \nabla^i \phi = \frac{1}{\sqrt{|g|}}\partial_i (\sqrt{|g|}g^{ij}\partial_j \phi)$$

for the metric $g_{ij}$ on $M$, and coordinates $\{x^i\}$ on $M$. Now, if we have a surface or sub-manifold $\Sigma$ embedded in $M$ with coordinates $\{\sigma^a\}$, then the charge density in terms of $\{\sigma^a\}$ satisfies,

$$\nabla_a \nabla^a \phi= -\frac{\rho(\sigma)}{\epsilon}$$

where $\nabla_a \nabla^a$ is now the Laplace-Beltrami operator, but for the induced metric on $\Sigma$, given by,

$$\gamma_{ab} = \frac{\partial x^i}{\partial \sigma^a}\frac{\partial x^j}{\partial \sigma^b} g_{ij}$$

which is the pullback of $g$ onto $\Sigma$ and $x^i(\sigma)$ are the embedding functions of $\Sigma$.


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