## Three dimensional divergence theorem with generally parametrized volume element.

Posted by peeterjoot on April 10, 2011

# Obsolete with potential errors.

This post may be in error. I wrote this before understanding that the gradient used in Stokes Theorem must be projected onto the tangent space of the parameterized surface, as detailed in Alan MacDonald’s Vector and Geometric Calculus.

See the post ‘stokes theorem in geometric algebra‘ [PDF], where this topic has been revisited with this in mind.

# Original Post:

# Motivation.

With the divergence of the energy momentum tensor converted from a volume to a surface integral given by

I got to wondering what a closed form algebraic expression for this curious (and foreign seeming) quantity was. It obviously must be related to the normal to the surface. It seemed to me that a natural way to answer this question was to consider this divergence integral over an arbitrarily parametrized volume. This turns out to be overkill, but a useful seeming digression.

## A generally parametrized parallelepiped volume element.

Suppose we parametrize a volume by specifying that all the points in that volume are covered by the position vector from the origin, given by

At any point in the volume of interest, we can create a level curve, holding two of the parameters constant, and varying the remaining one. In particular, we can construct three direction vectors along these level curves, one for each parameter not held constant

The span of these vectors, provided they are non-degenerate, forms a parallelepiped, the volume of which is

This volume element can be expanded in a number of ways

where the Jacobian determinant is given by

Provided we are interested in a volume for which the sign of this Jacobian determinant does not change sign, our task is to evaluate and reduce the integral

to a set (and sum of) two dimensional integrals.

## On the geometry of the surfaces.

Suppose that we integrate over the ranges , , . Observe that the outwards normals along the face is . This is

Similarly our normal on the face is

and on the face the outward normal is

Along the faces these are just negated. We can summarize these as

## Expansion of the Jacobian determinant

Suppose, to start with, our divergence volume integral 1.8 has just the following term

The specifics of how the scalar is indexed will not matter yet, so let’s suppress it. The Jacobian determinant can be expanded along the column for

Performing the same task (really just performing cyclic permutation of indexes) we can now construct the whole divergence integral

Regrouping we have

Observe that we can factor these sums utilizing the normals for the parallelepiped volume element

Let’s look at the first of these integrals in more detail. We integrate the values of the evaluated on the points of the surface for which . To perform this integral we dot against the outward normal area element . We do the same, but subtract the integral where is evaluated on the surface , where we dot with the area element that has the inwards normal direction on that surface. This is then done for each of the surfaces of the parallelepiped that we are integrating over.

In terms of the outwards (area scaled) normals on the and surfaces respectively we can write

This can be written more concisely in index form with

so that the divergence integral is just

latex a_{1+}$, , }} d^2 \sigma^\beta T^{\beta \alpha}-\int_{\text{over level surfaces , , }} d^2 \sigma^\beta T^{\beta \alpha}\end{aligned} \hspace{\stretch{1}}(1.16)$

In each case, for the surfaces, our negated inwards normal form can be redefined so that we integrate over only the outwards normal directions, and we can use the oriented integral notation

To encode (or imply) whether we require a positive or negative sign on the area element tensor of 1.15 for the surface in question.

## A look back, and looking forward.

Now, having performed this long winded calculation, the meaning of becomes clear. What’s also clear is how this could have been arrived at directly utilizing the divergence theorem in its normal vector form. We had only to re-write our equation as a vector equation in terms of the gradient

From this we see directly that .

Despite there being an easier way to find the form of , I still consider this a worthwhile exercise. It hints how one could generalize the arguments to the higher dimensional cases. The main task would be to construct the normals to the hypersurfaces bounding the hypervolume, and how to do this algebraically utilizing determinants may not be too hard (since we want a Jacobian determinant as the hypervolume element in the “volume” integral). We also got more than the normal physics text book proof of the divergence theorem for Cartesian coordinates, and did it here for a general parametrization. This wasn’t a complete argument since we didn’t consider a general surface, broken down into a triangular mesh. We really want volume elements with triangular sides instead of parallelograms.

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