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''

:$\mathbf L = \oint_C \mathbf x\times(d\mathbf x\times\mathbf n) = -\int_(\mathbf x\times\mathbf n)\nabla\cdot\mathbf n\ dS.$

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,

:$\int_(\mathbf x\times\mathbf n)\nabla\cdot\mathbf n\ dS =0,$

or, in index notation, we have

:$\int_x_j\nabla\cdot\mathbf n\ dS_k = \int_ x_k \nabla \cdot \mathbf n\ dS_j.$

That is to say the tensor

:$T_ = \int_x_j\nabla\cdot\mathbf n\ dS_i$

defined on a closed surface is always symmetric, i.e., $T_=T_$.

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 x_i(n_jx_j+n_kx_k) + dx_j(-n_ix_j)+dx_k(-n_ix_k)

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

:$L_i = \int_ \left\\ dS.$

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

:L_i = \int_ \left \fracx_k\frac(n_i^2+n_k^2) + \frac x_j\frac(n_i^2+n_j^2)+n_jx_k\left(\frac + \frac\right) - n_kx_j \left(\frac + \frac\right)\right dS,

or, in other words,

:L_i = \int_ \left \mathbf n, ^2 - (x_jn_k-x_kn_j)\nabla\cdot\mathbf n\right dS.

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

:$L_i = - \int_(x_jn_k-x_kn_j)\nabla\cdot\mathbf n\ dS,$

thus proving the lemma.

References

{{DEFAULTSORT:Chandrasekhar-Wentzel lemma
Vector calculus