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 $\mathbf {S}$ is a surface bounded by a simple closed contour $C$, then

$\mathbf {L} =\oint _{C}\mathbf {x} \times (d\mathbf {x} \times \mathbf {n} )=-\int _{\mathbf {S} }(\mathbf {x} \times \mathbf {n} )\nabla \cdot \mathbf {n} \ dS.$

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

$\int _{\mathbf {S} }(\mathbf {x} \times \mathbf {n} )\nabla \cdot \mathbf {n} \ dS=0,$

or, in index notation, we have

$\int _{\mathbf {S} }x_{j}\nabla \cdot \mathbf {n} \ dS_{k}=\int _{\mathbf {S} }x_{k}\nabla \cdot \mathbf {n} \ dS_{j}.$

That is to say the tensor

$T_{ij}=\int _{\mathbf {S} }x_{j}\nabla \cdot \mathbf {n} \ dS_{i}$

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

$L_{i}=\oint _{C}[dx_{i}(n_{i}x_{j}+n_{k}x_{k})+dx_{j}(-n_{i}x_{j})+dx_{k}(-n_{i}x_{k})].$

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

$L_{i}=\int _{\mathbf {S} }\left\{n_{i}\left[{\frac {\partial }{\partial x_{j}}}(-n_{i}x_{k})-{\frac {\partial }{\partial x_{k}}}(-n_{i}x_{j})\right]+n_{j}\left[{\frac {\partial }{\partial x_{k}}}(n_{j}x_{j}+n_{k}x_{k})-{\frac {\partial }{\partial x_{i}}}(-n_{i}x_{k})\right]+n_{k}\left[{\frac {\partial }{\partial x_{i}}}(-n_{i}x_{j})-{\frac {\partial }{\partial x_{j}}}(n_{j}x_{j}+n_{k}x_{k})\right]\right\}\ dS.$

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

$L_{i}=\int _{\mathbf {S} }\left[-{\frac {1}{2}}x_{k}{\frac {\partial }{\partial x_{j}}}(n_{i}^{2}+n_{k}^{2})+{\frac {1}{2}}x_{j}{\frac {\partial }{\partial x_{k}}}(n_{i}^{2}+n_{j}^{2})+n_{j}x_{k}\left({\frac {\partial n_{i}}{\partial x_{i}}}+{\frac {\partial n_{k}}{\partial x_{k}}}\right)-n_{k}x_{j}\left({\frac {\partial n_{i}}{\partial x_{i}}}+{\frac {\partial n_{j}}{\partial x_{j}}}\right)\right]\ dS,$

or, in other words,

$L_{i}=\int _{\mathbf {S} }\left[{\frac {1}{2}}\left(x_{j}{\frac {\partial }{\partial x_{k}}}-x_{k}{\frac {\partial }{\partial x_{j}}}\right)|\mathbf {n} |^{2}-(x_{j}n_{k}-x_{k}n_{j})\nabla \cdot \mathbf {n} \right]\ dS.$

And since $|\mathbf {n} |^{2}=1$, we have

$L_{i}=-\int _{\mathbf {S} }(x_{j}n_{k}-x_{k}n_{j})\nabla \cdot \mathbf {n} \ dS,$

thus proving the lemma.