Chandrasekhar–Wentzel lemma

In vector calculus, Chandrasekhar–Wentzel lemma was derived by Subrahmanyan Chandrasekhar and Gregor Wentzel in 1965, while studying the stability of rotating liquid drop. The lemma states that if $𝐒$ is a surface bounded by a simple closed contour $C$, then

$𝐋={\oint }_C𝐱\times (d𝐱\times 𝐧)=-\underset{𝐒}{\int }(𝐱\times 𝐧)\mathrm{\nabla }\cdot 𝐧dS.$

Here $𝐱$ is the position vector and $𝐧$ is the unit normal on the surface. An immediate consequence is that if $𝐒$ is a closed surface, then the line integral tends to zero, leading to the result,

$\underset{𝐒}{\int }(𝐱\times 𝐧)\mathrm{\nabla }\cdot 𝐧dS=0,$

or, in index notation, we have

$\underset{𝐒}{\int }{x}_j\mathrm{\nabla }\cdot 𝐧d{S}_k=\underset{𝐒}{\int }{x}_k\mathrm{\nabla }\cdot 𝐧d{S}_j.$

That is to say the tensor

$T_{ij}=\underset{𝐒}{\int }{x}_j\mathrm{\nabla }\cdot 𝐧d{S}_i$

defined on a closed surface is always symmetric, i.e., $T_{ij}=T_{ji}$.

Proof

Let us write the vector in index notation, but summation convention will be avoided throughout the proof. Then the left hand side can be written as

$L_i={\oint }_C[d{x}_i(n_j{x}_j+n_k{x}_k)+d{x}_j(-n_i{x}_j)+d{x}_k(-n_i{x}_k)].$

Converting the line integral to surface integral using Stokes's theorem, we get

$L_i=\underset{𝐒}{\int }\{n_i[\frac{\partial }{\partial x_j}(-n_i{x}_k)-\frac{\partial }{\partial x_k}(-n_i{x}_j)]+n_j[\frac{\partial }{\partial x_k}(n_j{x}_j+n_k{x}_k)-\frac{\partial }{\partial x_i}(-n_i{x}_k)]+n_k[\frac{\partial }{\partial x_i}(-n_i{x}_j)-\frac{\partial }{\partial x_j}(n_j{x}_j+n_k{x}_k)]\}dS.$

Carrying out the requisite differentiation and after some rearrangement, we get

$L_i=\underset{𝐒}{\int }[-\frac{1}{2}{x}_k\frac{\partial }{\partial x_j}(n_i^2+n_k^2)+\frac{1}{2}{x}_j\frac{\partial }{\partial x_k}(n_i^2+n_j^2)+n_j{x}_k(\frac{\partial n_i}{\partial x_i}+\frac{\partial n_k}{\partial x_k})-n_k{x}_j(\frac{\partial n_i}{\partial x_i}+\frac{\partial n_j}{\partial x_j})]dS,$

or, in other words,

$L_i=\underset{𝐒}{\int }[\frac{1}{2}(x_j\frac{\partial }{\partial x_k}-x_k\frac{\partial }{\partial x_j})|𝐧{|}^2-(x_j{n}_k-x_k{n}_j)\mathrm{\nabla }\cdot 𝐧]dS.$

And since $|𝐧{|}^2=1$, we have

$L_i=-\underset{𝐒}{\int }(x_j{n}_k-x_k{n}_j)\mathrm{\nabla }\cdot 𝐧dS,$

thus proving the lemma.