# Electric potential due to a charged, deformed sphere

Lord Rayleigh, in an 1882 paper (Philosophical Magazine XIV, pp 184-186), writes that the potential due to a sphere of radius $$a_0$$ and a charge $$Q$$ on it is $$Q/a_0$$. He then states that if the sphere is slightly deformed so that the polar equation of its surface is expressed as a Laplace series $$r = a(1 + F_1 + F_2 + \cdots)$$ then $$$$V = \frac{Q}{a_0}\left(1 - \sum(n-1)\iint \frac{F_n^2}{4\pi}d\sigma\right),$$$$ where $$d\sigma$$ is the area element on the deformed sphere.

The only information I could get about a Laplace series is that its terms are written in terms of spherical harmonics. If this is correct then perhaps $$$$F_n = \sum_{m=-n}^n Y_n^m(\theta, \phi).$$$$ But with this assumption, I am unable to get the expression for $$V$$. Can anyone offer help in getting it? A screenshot of the paper is shown below.

• The expressions for the Laplace series are a bir more complicated: mathworld.wolfram.com/LaplaceSeries.html – Lucas Baldo Sep 6 '19 at 16:53
• Looks like in Lord Rayleigh's times the definition of Laplace series was a bit different. The $F_n$ in this paper are not spherical harmonics. You may want to look at Issac Todhunter's 'An Elementary Treatise on Laplace's functions, Lame's functions and Bessel's functions' for more details. – Amey Joshi Sep 6 '19 at 16:58
• I did see the mathworld link. That's the only information I could get on the internet. However, it didn't take me too far. – Amey Joshi Sep 6 '19 at 16:59

The polar equation of a slightly deformed sphere is $$r = r(\theta, \phi)$$ so that the position of a point on its surface is given by $$$$\tag{1}\label{e1} \vec{r} = r(\theta, \phi)\hat{e}_r.$$$$ We follow the physicists' convention of $$\theta$$ being the co-latitude and $$\phi$$ being the longitude of a point. A small area element on the surface of such a sphere can be described by a parallelogram formed of vectors $$d\vec{r}_1$$ and $$d\vec{r}_2$$ where $$\begin{eqnarray} d\vec{r}_1 &=& \frac{\partial\vec{r}_1}{\partial\theta}d\theta \tag{2} \\ d\vec{r}_2 &=& \frac{\partial\vec{r}_2}{\partial\phi}d\phi \tag{3}, \end{eqnarray}$$ the area being $$d\vec{A} = d\vec{r}_1 \times d\vec{r}_2$$. Now, $$\begin{eqnarray} \frac{\partial\vec{r}_1}{\partial\theta} &=& \frac{\partial r}{\partial\theta}\hat{e}_r + r\frac{\partial\hat{e}_r}{\partial\theta} = \frac{\partial r}{\partial\theta}\hat{e}_r + r\hat{e}_\theta \tag{4} \\ \frac{\partial\vec{r}_2}{\partial\phi} &=& \frac{\partial r}{\partial\phi}\hat{e}_r + r\frac{\partial\hat{e}_r}{\partial\phi} = \frac{\partial r}{\partial\phi}\hat{e}_r + r\sin\theta\hat{e}_\phi \tag{5} \end{eqnarray}$$ so that $$$$\tag{6}\label{e6} d\vec{A} = \left(r\sin\theta\frac{\partial r}{\partial\theta}\hat{e}_\theta - r\frac{\partial r}{\partial\phi}\hat{e}_\phi - r^2\sin\theta\hat{e}_r\right)d\theta d\phi.$$$$ The surface area of the deformed sphere is $$$$\tag{7}\label{e7} A = \int_0^\pi\int_0^{2\pi}\left(1 + \frac{1}{r^2}\left(\frac{\partial r}{\partial\theta}\right)^2 + \frac{1}{r^2\sin^2\theta}\left(\frac{\partial r}{\partial\phi}\right)^2\right)^{1/2}r^2\sin\theta d\theta d\phi.$$$$ Simplifying the integrand using the binomial theorem and retaining only the lowest ordered derivatives, $$$$\tag{8}\label{e8} A = \int_0^\pi\int_0^{2\pi}\left(1 + \frac{1}{2r^2}\left(\frac{\partial r}{\partial\theta}\right)^2 + \frac{1}{2r^2\sin^2\theta}\left(\frac{\partial r}{\partial\phi}\right)^2\right)r^2\sin\theta d\theta d\phi.$$$$ We now express $$r$$ as a Laplace series $$$$\tag{9}\label{e9} r(\theta, \phi) = a_0 + \sum_{n \ge 1}a_n Y_n(\theta, \phi).$$$$ where $$Y_n$$ is the Laplace function. It is a solution of the partial differential equation $$$$\tag{10}\label{e10} \frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\left\{\sin\theta\frac{\partial Y_n}{\partial\theta}\right\} + \frac{1}{\sin^2\theta}\frac{\partial^2Y_n}{\partial\phi^2} + n(n+1)Y_n = 0.$$$$ We further assume that $$a_n \ll a_0$$ in equation \eqref{e9}. From equation \eqref{e9} $$\frac{\partial r}{\partial\theta} = \sum_{n \ge 1}a_n\frac{\partial Y_n}{\partial\theta}$$ so that $$$$\tag{11}\label{e11} \left(\frac{\partial r}{\partial\theta}\right)^2 = \sum_{m,n \ge 1}a_m a_n \frac{\partial Y_m}{\partial\theta}\frac{\partial Y_n}{\partial\theta}.$$$$ Similarly $$$$\tag{12}\label{e12} \left(\frac{\partial r}{\partial\phi}\right)^2 = \sum_{m,n \ge 1}a_m a_n \frac{\partial Y_m}{\partial\phi}\frac{\partial Y_n}{\partial\phi}.$$$$ We now express the surface area of the deformed sphere as $$$$\tag{13}\label{e13} A = A_1 + A_2 + A_3,$$$$ where $$\begin{eqnarray} A_1 &=& \iint_S r^2\sin\theta d\theta d\phi \\ A_2 &=& \frac{1}{2}\iint_S \left(\frac{\partial r}{\partial\theta}\right)^2 \sin\theta d\theta d\phi \\ A_3 &=& \frac{1}{2}\iint_S \frac{1}{\sin\theta}\left(\frac{\partial r}{\partial\phi}\right)^2 d\theta d\phi, \end{eqnarray}$$ where $$S$$ denotes the surface of the deformed sphere. From equation \eqref{e9} $$r^2 = a_0^2 + 2a_0\sum_{n \ge 1}a_nY_n + \sum_{m, n \ge 1} a_m a_n Y_m Y_n$$ so that $$A_1 = 4\pi a_0^2 + 2a_0\sum_{n \ge 1}a_n \iint_S Y_n\sin\theta d\theta d\phi + \sum_{m, n \ge 1}a_m a_n \iint_S Y_m Y_n \sin\theta d\theta d\phi.$$ From the last equation in Article 170 of Todhunter's book "An Elementary Treatise on Laplace's Functions, Lamé's Functions, and Bessel's Functions", $$$$\tag{17}\label{e17} \int_0^{2\pi}Y_n d\phi = 2\pi P_n(\cos\theta)P_n\cos(\theta^\prime)$$$$ so that $$\iint_S Y_n\sin\theta d\theta d\phi = 2\pi P_n(\cos\theta^\prime)\int_0^\pi P_n(\cos\theta)\sin\theta d\theta$$ Using the usual substitution $$x = \cos\theta$$, we get $$$$\tag{18}\label{e18} \iint_S Y_n\sin\theta d\theta d\phi = 2\pi P_n(\cos\theta^\prime)\int_{-1}^1P_n(x)dx = 0,$$$$ where the last equality is a consequence of the property of Legendre functions. The last equation of article 187 and the first equation of article 196 of Todhunter's book can be combined to get $$$$\tag{19}\label{e19} \iint_S Y_n Y_m \sin\theta d\theta d\phi = \frac{4\pi}{2n+1}\delta_{mn}.$$$$ Using equations \eqref{e18} and \eqref{e19} in the expression to get $$A_1$$ we get $$$$\tag{20}\label{e20} A_1 = 4\pi a_0^2 + \sum_{n \ge 1}a_n^2 \iint_S Y_n^2\sin\theta d\theta d\phi.$$$$ We now evaluate the sum $$$$\tag{22}\label{e21} A_2 + A_3 = \frac{1}{2}\sum_{m, n \ge 1} a_ma_n \iint_S\left(\sin\theta \frac{\partial Y_m}{\partial\theta}\frac{\partial Y_n}{\partial\theta} + \frac{1}{\sin\theta}\frac{\partial Y_m}{\partial\phi}\frac{\partial Y_m}{\partial\phi}\right) d\theta d\phi.$$$$ The integral in this equation can be readily evaluated using the partial differential equation \eqref{e10}. We multiply it by $$Y_m$$ and integrate the result over $$S$$. Thus, we have $$$$\tag{22}\label{e22} \iint_S Y_m\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial Y_n}{\partial\theta}\right)d\theta d\phi + \iint_S\frac{Y_m}{\sin\theta}\frac{\partial^2 Y_n}{\partial\phi^2}d\theta d\phi = -n(n+1)\iint_S Y_m Y_n \sin\theta d\theta d\phi.$$$$ The first integral in the above equation can be expressed as $$\int_0^{2\pi}\left(Y_m \sin\theta\frac{\partial Y_n}{d\theta}\Big|_0^{\pi} - \int_0^\pi \sin\theta \frac{\partial Y_n}{\partial\theta}\frac{\partial Y_m}{\partial\theta}d\theta\right)d\phi = -\iint_S \frac{\partial Y_m}{\partial\theta}\frac{\partial Y_n}{\partial\theta}\sin\theta d\theta d\phi.$$ Similarly, the second integral can be simplified to $$\int_0^\pi\frac{1}{\sin\theta}\left(Y_m\frac{\partial Y_n}{\partial\phi}\Big|_0^{2\pi} - \int_0^{2\pi}\frac{\partial Y_n}{\partial\phi}\frac{\partial Y_m}{\partial\phi}d\phi\right)d\theta = -\iint_S\frac{1}{\sin\theta}\frac{\partial Y_m}{\partial\phi}\frac{\partial Y_n}{\partial\phi}d\theta d\phi.A$$ Equation \eqref{e22} thus becomes $$$$\tag{23}\label{e23} \iint_S\left(\sin\theta \frac{\partial Y_m}{\partial\theta}\frac{\partial Y_n}{\partial\theta} + \frac{1}{\sin\theta}\frac{\partial Y_m}{\partial\phi}\frac{\partial Y_n}{\partial\phi}\right) d\theta d\phi = n(n+1)\iint_S Y_m Y_n \sin\theta d\theta d\phi.$$$$ From equations \eqref{e19}, \eqref{e20} and \eqref{e23}, we get $$$$\tag{24}\label{e24} A_2 + A_3 = \frac{1}{2}\sum_{n \ge 1}a_n^2 n(n+1)\iint_S Y_n^2 \sin\theta d\theta d\phi.$$$$ From equations \eqref{e13}, \eqref{e20} and \eqref{e24}, $$$$\tag{25}\label{e25} A = 4\pi a_0^2\left(1 + \sum_{n \ge 1} \frac{a_n^2}{a_0^2} \left(1 + \frac{n(n+1)}{2}\right) \iint_S \frac{Y_n^2}{4\pi} \sin\theta d\theta d\phi\right).$$$$

The volume element of a deformed sphere is $$d\tau = d\vec{r}_3 \cdot (d\vec{r}_1 \times d\vec{r}_3)$$, where $$d\vec{r}_3 = \partial\vec{r}/\partial r = \hat{e}_r dr$$. From equation \eqref{e6}, it follows that $$$$\tag{26}\label{e26} d\tau = r^2\sin\theta drd\theta d\phi.$$$$ The volume of the deformed sphere is $$$$\tag{27}\label{e27} \tau = \iint_S \frac{r^3}{3}\sin\theta d\theta d\phi.$$$$ From equation \eqref{e9} $$r^3 = a_0^3 + 3a_0^2\sum_{n \ge 1} a_n Y_n + 3 a_0 \sum_{m, n \ge 1}a_m a_n Y_m Y_n + O(a_n^3).$$ We ignore terms of $$O(a_n^3)$$ so that $$r^3 = a_0^3 + 3a_0^2\sum_{n \ge 1} a_n Y_n + 3 a_0 \sum_{m, n \ge 1}a_m a_n Y_m Y_n.$$ Substituting this in \eqref{e27} and using equations \eqref{e18} and \eqref{e19} we get $$\tau = \frac{4\pi a_0^3}{3} + a_0 \sum_{n \ge 1} a_n^2 \iint_S Y_n^2 \sin\theta d\theta d\phi.$$ or $$$$\tag{28}\label{e28} \tau = \frac{4\pi a_0^3}{3}\left(1 + 3 \sum_{n \ge 1} \frac{a_n^2}{a_0^2}\iint_S \frac{Y_n^2}{4\pi}\sin\theta d\theta d\phi\right).$$$$

We now find the potential at the origin due to a charge $$Q$$ on the surface of a slightly deformed spherical drop of fluid. Let the charge be spread uniformly with a density $$\sigma$$. An element of area $$dA$$ will have a charge $$\sigma dA$$ and will produce a potential $$dV$$ at the origin given by $$dV = \sigma\frac{dA}{r}$$ From equation \eqref{e6}, $$dV = \frac{\sigma}{r}\left(1 + \frac{1}{2r^2}\left(\frac{\partial r}{\partial\theta}\right)^2 + \frac{1}{2r^2\sin^2\theta}\left(\frac{\partial r}{\partial\phi}\right)^2\right)r^2\sin\theta d\theta d\phi$$ or $$dV = \sigma\left(1 + \frac{1}{2r^2}\left(\frac{\partial r}{\partial\theta}\right)^2 + \frac{1}{2r^2\sin^2\theta}\left(\frac{\partial r}{\partial\phi}\right)^2\right)r\sin\theta d\theta d\phi$$ We now express the potential as a sum $$$$\tag{29}\label{e29} V = V_1 + V_2 + V_3,$$$$ where $$\begin{eqnarray} V_1 &=& \sigma\iint_S r\sin\theta d\theta d\phi \\ V_2 &=& \frac{\sigma}{2}\iint_S \frac{1}{r}\left(\frac{\partial r}{\partial\theta}\right)^2 \sin\theta d\theta d\phi \\ V_3 &=& \frac{\sigma}{2}\iint_S \frac{1}{r\sin\theta}\left(\frac{\partial r}{\partial\phi}\right)^2 d\theta d\phi. \end{eqnarray}$$ From equations \eqref{e9} and \eqref{e18} it is clear that $$$$\tag{33}\label{e33} V_1 = 4\pi\sigma a_0.$$$$ Before considering the other two terms, we note that $$\frac{1}{r}\left(\frac{\partial r}{\partial\theta}\right)^2 = \frac{1}{a_0}\left(1 - \sum_{n \ge 1}\frac{a_n}{a_0}Y_n\right)\sum_{l,m \ge 1}a_la_m\frac{\partial Y_l}{\partial\theta}\frac{\partial Y_m}{\partial\theta}.$$ If we ignore the terms in third order of $$a_n$$, $$$$\tag{34}\label{e34} \frac{1}{r}\left(\frac{\partial r}{\partial\theta}\right)^2 = \frac{1}{a_0} \sum_{l,m \ge 1}a_la_m\frac{\partial Y_l}{\partial\theta}\frac{\partial Y_m}{\partial\theta}.$$$$ Similarly, $$$$\tag{35}\label{e35} \frac{1}{r}\left(\frac{\partial r}{\partial\phi}\right)^2 = \frac{1}{a_0\sin\theta} \sum_{l,m \ge 1}a_la_m\frac{\partial Y_l}{\partial\phi}\frac{\partial Y_m}{\partial\phi}.$$$$ Using a procedure similar to the one used to get equation \eqref{e24}, we get $$$$\tag{36}\label{e36} V_2 + V_3 = \frac{\sigma}{2a_0}\sum_{n \ge 1}a_n^2 n(n+1)\iint_S Y_n^2 \sin\theta d\theta d\phi.$$$$ Therefore, $$$$\tag{37}\label{e37} V = 4\pi\sigma a_0\left(1 + \frac{1}{2}\sum_{n \ge 1}\frac{a_n^2}{a_0^2}n(n+1)\iint_S\frac{Y_n^2}{4\pi} \sin\theta d\theta d\phi \right).$$$$ Substituting $$Q/A$$ for $$\sigma$$, where $$A$$ is given by equation \eqref{e25} we get $$$$\tag{38}\label{e38} V = \frac{Q}{a_0}\left(1 - \sum_{n \ge 1}\frac{a_n^2}{a_0^2}\iint_S\frac{Y_n^2}{4\pi} \sin\theta d\theta d\phi + O(a_n^4)\right).$$$$ To get this equation we have used the expression $$\frac{1}{A} = \frac{1}{4\pi a_0^2}\left(1 - \sum_{n \ge 1}\frac{a_n^2}{a_0^2}\left(1 + \frac{n(n + 1)}{2}\right)\iint_S\frac{Y_n^2}{4\pi}\sin\theta d\theta d\phi\right).$$ If we write equation \eqref{e9} as $$$$\tag{39}\label{e39} r = a_0\left(1 + \sum_{n \ge 1}F_n\right),$$$$ where $$$$\tag{40}\label{e40} F_n = \frac{a_n}{a_0}\frac{Y_n}{\sqrt{n-1}}$$$$ then equation \eqref{e38} becomes $$$$\tag{41}\label{e41} V = \frac{Q}{a_0}\left(1 - \sum_{n \ge 1}(n - 1)\iint_S\frac{F_n^2}{4\pi} \sin\theta d\theta d\phi\right).$$$$

• If the rest of the development is sound, I need help in justifying equation (40). – Amey Joshi Sep 16 '19 at 5:41
• The normalization choice of equation (40) is not important for the rest of the development in the paper. In fact, we can continue with equations containing $Y_n$ and get the correct expression for Rayleigh's criterion of break up of a charged drop of liquid. – Amey Joshi Sep 23 '19 at 16:58

This a bit too long for a comment, but this is just what it is. I suggest a simplified (and less rigorous) derivation based on the great material provided by @Amey Joshi 's answer.

Using Gauss' law and the expression for $$A_{1}$$:

$$\phi \approx E(r=a_{0}).A_{1}= Q/ \epsilon_{0}$$

So that $$E(r=a_{0})= Q / (4\pi\epsilon_{0} a_0^2 (1 + \sum_{n \ge 1}\frac{a_n^2}{a_{0}^2} \iint Y_n^2 \frac{d\sigma}{4\pi})$$

So to first order, the slightly deformed configuration is equivalent to considering the modified charge $$Q^{'}=Q(1 - \sum_{n \ge 1}\frac{a_n^2}{a_{0}^2} \iint Y_n^2 \frac{d\sigma}{4\pi})$$. If (40) holds then the correct expression for the potential may be derived. I really have no idea why Rayleigh would use the normalisation of its expansion functions as provided by (40). Note also that Rayleigh seems to be using $$4 \pi \epsilon_{0} =1$$ for simplification.