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 [math]\displaystyle{ \mathbf S }[/math] is a surface bounded by a simple closed contour [math]\displaystyle{ C }[/math], then

[math]\displaystyle{ \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. }[/math]

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

[math]\displaystyle{ \int_{\mathbf S}(\mathbf x\times\mathbf n)\nabla\cdot\mathbf n\ dS =0, }[/math]

or, in index notation, we have

[math]\displaystyle{ \int_{\mathbf S}x_j\nabla\cdot\mathbf n\ dS_k = \int_{\mathbf S} x_k \nabla \cdot \mathbf n\ dS_j. }[/math]

That is to say the tensor

[math]\displaystyle{ T_{ij} = \int_{\mathbf S}x_j\nabla\cdot\mathbf n\ dS_i }[/math]

defined on a closed surface is always symmetric, i.e., [math]\displaystyle{ T_{ij}=T_{ji} }[/math].

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

[math]\displaystyle{ L_i = \oint_C [dx_i(n_ix_j+n_kx_k) + dx_j(-n_ix_j)+dx_k(-n_ix_k)]. }[/math]

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

[math]\displaystyle{ L_i = \int_{\mathbf S} \left\{n_i\left[\frac{\partial }{\partial x_j}(-n_ix_k) - \frac{\partial }{\partial x_k}(-n_ix_j)\right] + n_j \left[\frac{\partial }{\partial x_k}(n_jx_j+n_kx_k) - \frac{\partial }{\partial x_i}(-n_ix_k)\right] + n_k\left[\frac{\partial }{\partial x_i}(-n_ix_j) - \frac{\partial }{\partial x_j}(n_jx_j+n_kx_k)\right]\right\}\ dS. }[/math]

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

[math]\displaystyle{ 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_jx_k\left(\frac{\partial n_i}{\partial x_i} + \frac{\partial n_k}{\partial x_k}\right) - n_kx_j \left(\frac{\partial n_i}{\partial x_i} + \frac{\partial n_j}{\partial x_j}\right)\right]\ dS, }[/math]

or, in other words,

[math]\displaystyle{ 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_jn_k-x_kn_j)\nabla\cdot\mathbf n\right]\ dS. }[/math]

And since [math]\displaystyle{ |\mathbf n|^2=1 }[/math], we have

[math]\displaystyle{ L_i = - \int_{\mathbf S}(x_jn_k-x_kn_j)\nabla\cdot\mathbf n\ dS, }[/math]

thus proving the lemma.