When solving the Laplace equation on sphere coordinates you get:

$$ u(r,\theta) = \sum_{n=0}^{\infty}\left( A_n\,r^n + \frac{B_n}{r^{n+1}} \right) P_n(\cos\theta) $$

And it is clear that if you have a system with an spheric condensator and you want the potential inside the sphere you want $B_n = 0$ for all $n$ so that the potential is not infinite when $r \to 0$.

What do you do though for $r > R$? I have seen in some documents that people put $A_n = 0$ for all $n$ but my impression is that this is only necessary for $n > 1$ and so you get:

$$ u(r,\theta) = A_0\,P_0(\cos\theta) + \sum_{n=0}^{\infty}\frac{B_n}{r^{n+1}} P_n(\cos\theta) $$

So my questions are either why should $A_0 = 0$? And if that is not the case, how can one find $A_0$ and $B_0$? You can use the ortonormality of the Legendre polynomials to find all the $B_n$ for $n > 0$ but since $A_0$ and $B_0$ share the same Legendre Polynomial, I don't see how to do it.


1 Answer 1


The $n=0$ factor is just a constant, and you can always add a constant to a potential without changing the physics. So, keeping this term does not lead to anything new. One natural choice isto normalize the potential so that it is zero at infinity, which leads to $A_0=0$

  • $\begingroup$ Thanks a lot for the reply. What if there is a condition on the sphere? Like, that the potential has to be a constant on the north hemisphere and zero in the other one. Is it still possible to say that $A_0 = 0$? $\endgroup$ Commented Jun 13, 2019 at 9:21
  • $\begingroup$ In that case you need some kind of boundary conditions at $\infty$ to determine your potential completely. Otherwise, since the potential $1-1/r$ vanishes on the sphere, you can add it to any solution and get another perfectly valid solution. $\endgroup$
    – Kostya_I
    Commented Jun 13, 2019 at 9:37

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