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 \begin{equation} V = \frac{Q}{a_0}\left(1 - \sum(n-1)\iint \frac{F_n^2}{4\pi}d\sigma\right), \end{equation} 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 \begin{equation} F_n = \sum_{m=-n}^n Y_n^m(\theta, \phi). \end{equation} 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.

Lord Rayleigh's paper

  • $\begingroup$ The expressions for the Laplace series are a bir more complicated: mathworld.wolfram.com/LaplaceSeries.html $\endgroup$ Sep 6, 2019 at 16:53
  • 1
    $\begingroup$ 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. $\endgroup$
    – Amey Joshi
    Sep 6, 2019 at 16:58
  • $\begingroup$ 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. $\endgroup$
    – Amey Joshi
    Sep 6, 2019 at 16:59

2 Answers 2


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 \begin{equation}\tag{1}\label{e1} \vec{r} = r(\theta, \phi)\hat{e}_r. \end{equation} 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 \begin{equation}\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. \end{equation} The surface area of the deformed sphere is \begin{equation}\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. \end{equation} Simplifying the integrand using the binomial theorem and retaining only the lowest ordered derivatives, \begin{equation}\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. \end{equation} We now express $r$ as a Laplace series \begin{equation}\tag{9}\label{e9} r(\theta, \phi) = a_0 + \sum_{n \ge 1}a_n Y_n(\theta, \phi). \end{equation} where $Y_n$ is the Laplace function. It is a solution of the partial differential equation \begin{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. \end{equation} 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 \begin{equation}\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}. \end{equation} Similarly \begin{equation}\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}. \end{equation} We now express the surface area of the deformed sphere as \begin{equation}\tag{13}\label{e13} A = A_1 + A_2 + A_3, \end{equation} 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", \begin{equation}\tag{17}\label{e17} \int_0^{2\pi}Y_n d\phi = 2\pi P_n(\cos\theta)P_n\cos(\theta^\prime) \end{equation} 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 \begin{equation}\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, \end{equation} 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 \begin{equation}\tag{19}\label{e19} \iint_S Y_n Y_m \sin\theta d\theta d\phi = \frac{4\pi}{2n+1}\delta_{mn}. \end{equation} Using equations \eqref{e18} and \eqref{e19} in the expression to get $A_1$ we get \begin{equation}\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. \end{equation} We now evaluate the sum \begin{equation}\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. \end{equation} 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 \begin{equation}\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. \end{equation} 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 \begin{equation}\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. \end{equation} From equations \eqref{e19}, \eqref{e20} and \eqref{e23}, we get \begin{equation}\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. \end{equation} From equations \eqref{e13}, \eqref{e20} and \eqref{e24}, \begin{equation}\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). \end{equation}

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 \begin{equation}\tag{26}\label{e26} d\tau = r^2\sin\theta drd\theta d\phi. \end{equation} The volume of the deformed sphere is \begin{equation}\tag{27}\label{e27} \tau = \iint_S \frac{r^3}{3}\sin\theta d\theta d\phi. \end{equation} 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 \begin{equation}\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). \end{equation}

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 \begin{equation}\tag{29}\label{e29} V = V_1 + V_2 + V_3, \end{equation} 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 \begin{equation}\tag{33}\label{e33} V_1 = 4\pi\sigma a_0. \end{equation} 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 $a_n$, \begin{equation}\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}. \end{equation} Similarly, \begin{equation}\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}. \end{equation} Using a procedure similar to the one used to get equation \eqref{e24}, we get \begin{equation}\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. \end{equation} Therefore, \begin{equation}\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). \end{equation} Substituting $Q/A$ for $\sigma$, where $A$ is given by equation \eqref{e25} we get \begin{equation}\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). \end{equation} 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 \begin{equation}\tag{39}\label{e39} r = a_0\left(1 + \sum_{n \ge 1}F_n\right), \end{equation} where \begin{equation}\tag{40}\label{e40} F_n = \frac{a_n}{a_0}\frac{Y_n}{\sqrt{n-1}} \end{equation} then equation \eqref{e38} becomes \begin{equation}\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). \end{equation}

  • $\begingroup$ If the rest of the development is sound, I need help in justifying equation (40). $\endgroup$
    – Amey Joshi
    Sep 16, 2019 at 5:41
  • $\begingroup$ 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. $\endgroup$
    – Amey Joshi
    Sep 23, 2019 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.


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