Does this theorem have a name?

[techmologist]

For a surface S bounded by a curve C,

$$\int_{C} \mathbf{t} \times \mathbf{n}ds = -2\int_{S} H \mathbf{n}da$$

t is the unit tangent to C, and n is the unit normal to the surface S. H is the mean curvature of S. It can be derived from Stokes' theorem, but it seems like the kind of result that might have been known earlier.

 

[techmologist]

For any constant vector c,

$$\mathbf{c} \cdot \int_C \mathbf{t} \times \mathbf{n} ds = \int_C (\mathbf{n} \times \mathbf{c}) \cdot \mathbf{t} ds =\int_S \nabla \times (\mathbf{n} \times \mathbf{c}) \cdot \mathbf{n}da$$

Using the vector identity

$$\nabla \times (\mathbf{a} \times \mathbf{b}) = \mathbf{b} \cdot \nabla \mathbf{a} - \mathbf{a} \cdot \nabla \mathbf{b} - (\nabla \cdot \mathbf{a} ) \mathbf{b} + (\nabla \cdot \mathbf{b} )\mathbf{a}$$

and the fact that c is constant, the above integral becomes

$$\int_S [\mathbf{c}\cdot \nabla \mathbf{n} - (\nabla \cdot \mathbf{n})\mathbf{c} ] \cdot \mathbf{n} da = \mathbf{c} \cdot \int_S (\nabla \mathbf{n} ) \cdot \mathbf{n} - (\nabla \cdot \mathbf{n }) \mathbf{n} da$$

The first term in the integrand is zero, since it is just the gradient of the length of the unit normal. Also, since the vector c was arbitrary, its dot product with the integral can be dropped:


$$\int_C \mathbf{t} \times \mathbf{n} ds = -\int_S (\nabla \cdot \mathbf{n} ) \mathbf{n} da$$

Depending on how you define the normal (outward or inward), div n = -2H up to a sign^*. So that gives the result

$$\int_C \mathbf{t} \times \mathbf{n} ds = \int_S 2H \mathbf{n} da$$

I had an extra minus sign in the first post.


* The mean curvature H of a surface is equal to half the trace of the second fundamental form B: 2H = g^jkb_jk. The components of the second fundamental form can be defined as

$$b_{jk} = -\mathbf{x}_j \cdot \mathbf{n}_{,k} = -\mathbf{x}_j \cdot \mathbf{n}_{,k} = - n^i_{{ };k}\mathbf{x}_j \cdot \mathbf{x}_i = -n^i_{{ };k}g_{ij}$$

Multiplying both sides by the inverse of g_ij and then summing the diagonal elements gives

$$g^{jk}b_{jk} = - n^k_{{ };k} = -\nabla \cdot \mathbf{n}$$