# Divergence theorem in curvilinear coordinates

Suppose I have a tensor

\begin{gather} \stackrel{\leftrightarrow}{A} = \begin{bmatrix} a_{11}(\vec{r}) & a_{12}(\vec{r}) & a_{13}(\vec{r})\\ a_{21}(\vec{r}) & a_{22}(\vec{r}) & a_{23}(\vec{r})\\ a_{31}(\vec{r}) & a_{32}(\vec{r}) & a_{33}(\vec{r}) \end{bmatrix} \end{gather}

where $\vec{r} = x_{1} \hat{e}_1 + x_{2} \hat{e}_2 + x_{3} \hat{e}_3$

The divergence of this tensor in general curvilinear coordinates is given by

\begin{gather} \nabla^{c} \cdot \stackrel{\leftrightarrow}{A} = \left[ \frac{\partial A_{ij}}{\partial x^{k}} - \Gamma_{ki}^{l} A_{lj} - \Gamma_{kj}^{l} A_{il} \right] g^{ik} \vec{b}^{j}\\ \vec{b}_{i} = \frac{\partial_{x_{i}} \vec{r}}{ \left| \partial_{x_{i}} \vec{r} \right| } \end{gather}

Using Mathematica, I computed the volume integral of the curvilinear divergence for cylindrical coordinates, giving \begin{gather} \iiint \nabla^{c} \cdot \stackrel{\leftrightarrow}{A} dV = \begin{bmatrix} r \int a_{11} d\theta dz + \int a_{12} dr dz + \int r a_{13} dr d\theta - \int a_{22} dr d\theta dz\\ r \int a_{21} d\theta dz + \int a_{22} dr dz + \int r a_{23} dr d\theta + \int a_{12} dr d\theta dz\\ r \int a_{31} d\theta dz + \int a_{32} dr dz + \int r a_{33} dr d\theta \end{bmatrix} \end{gather}

This does not match the traditional Divergence theorem I'm familiar with, or at least it doesn't appear so to me because of the extra triple integral terms $\vec{C}$:

\begin{gather} \iiint \nabla^{c} \cdot \stackrel{\leftrightarrow}{A} dV = \oint A_{ij} n_j \vec{b}_i dS + \vec{C}\\ \vec{C} \neq 0 \end{gather}

What is the correct transformation from the volume integral of the curvilinear divergence to some surface integral?

• The equation you wrote in cylindrical coordinates is not reasonable as the LHS is a vector and the RHS is a scalar! – H. R. Mar 10 '16 at 9:30
• @H.R. The intention is that the surface integral describes the i'th component of the resulting vector. I've added an indication by multiplying by $\vec{b}_i$, though I've almost certainly gotten the covariance/contravariance wrong (first time I'm really trying to figure out what the difference between these two are). I'm working on a more fundamental proof of the divergence theorem in curvilinear coordinates to see if I can't figure out why the answer I got from Mathematica is/isn't correct. For now, though, I need to sleep :P – helloworld922 Mar 10 '16 at 10:05
• The problem is that you cannot take ${\bf{g}}^i$ inside or outside the integral. and hence you cannot easily obtain the co-variant or contra-variant component by multiplying ${\bf{g}}_i$ or ${\bf{g}}^i$. Take a look at the equation $(6)$ below in my answer. – H. R. Mar 10 '16 at 11:07

## 1 Answer

Let us first consider the invariant form of classical divergence theorem in vector analysis

$$\int_{\Omega}\nabla \cdot {\bf{v}} dV = \int_{\partial \Omega} {\bf{n}} \cdot {\bf{v}}dS \tag{1}$$

For the sake of memorizing, they say that the gradient operator turns into the the unit normal vector. You can choose your vector ${\bf{v}}$ to be

$${\bf{v}} = {\bf{A}} \cdot {\bf{c}} \tag{2}$$

where ${\bf{A}}$ is a second order tensor and ${\bf{c}}$ is a constant vector. Then using $(1)$ and $(2)$ you can prove that

$$\int_{\Omega} \nabla \cdot {\bf{A}} dV = \int_{\partial \Omega} {\bf{n}} \cdot {\bf{A}} dS \tag{3}$$

Note that $(1)$ is a scalar equation while $(2)$ is a vector equation.

Now, you can use $(3)$ to write the divergence theorem in a curve-linear coordinate. So the next step is to compute the divergence of a second order tensor

\begin{align} \nabla \cdot {\bf{A}} &= {\bf{g}}^i \partial_{i} \cdot (A_{jk} {\bf{g}}^j \otimes {\bf{g}}^k) \\ &= {\bf{g}}^i \cdot \partial_{i} (A_{jk} {\bf{g}}^j \otimes {\bf{g}}^k) \\ &= \partial_i A_{jk} ({\bf{g}}^i \cdot {\bf{g}}^j) {\bf{g}}^k + A_{jk} ({\bf{g}}^i \cdot \partial_i{\bf{g}}^j) {\bf{g}}^k + A_{jk} ({\bf{g}}^i \cdot {\bf{g}}^j) \partial_i {\bf{g}}^k \\ &= \partial_i A_{jk} g^{ij} {\bf{g}}^k - \Gamma_{il}^{j} A_{jk} ({\bf{g}}^i \cdot {\bf{g}}^l) {\bf{g}}^k - \Gamma_{il}^{k} A_{jk} g^{ij} {\bf{g}}^l \\ &= \partial_i A_{jk} g^{ij} {\bf{g}}^k - \Gamma_{il}^{j} A_{jk} g^{il} {\bf{g}}^k - \Gamma_{il}^{k} A_{jk} g^{ij} {\bf{g}}^l \\ &= \partial_i A_{jk} g^{ij} {\bf{g}}^k - \Gamma_{ij}^{l} A_{lk} g^{ij} {\bf{g}}^k - \Gamma_{ik}^{l} A_{jl} g^{ij} {\bf{g}}^k \\ &= \left[ \partial_i A_{jk} - \Gamma_{ij}^{l} A_{lk} - \Gamma_{ik}^{l} A_{jl} \right] g^{ij} {\bf{g}}^k \\ &= \left[ \partial_i A_{kj} - \Gamma_{ik}^{l} A_{lj} - \Gamma_{ij}^{l} A_{kl} \right] g^{ik} {\bf{g}}^j \end{align} \tag{4}

and also we have

\begin{align} {\bf{n}} \cdot {\bf{A}} &= n_i {\bf{g}}^i \cdot ( A_{kj} {\bf{g}}^k \otimes {\bf{g}}^j ) \\ &= n_i A_{jk} ( {\bf{g}}^i \cdot {\bf{g}}^k ) {\bf{g}}^j \\ &= n_i A_{jk} g^{ik} {\bf{g}}^j \end{align} \tag{5}

and so the final result is

$$\boxed{\int_{\Omega} \left[ \partial_i A_{kj} - \Gamma_{ik}^{l} A_{lj} - \Gamma_{ij}^{l} A_{kl} \right] g^{ik} {\bf{g}}^j dV = \int_{\partial \Omega} n_{i} A_{jk} g^{ik} {\bf{g}}^{j} dS} \tag{6}$$

• The problem is that I'm not getting the invariant form by plugging in the full cylindrical divergence into the volume integral; I get some extra volume integral terms on the right hand side. – helloworld922 Mar 9 '16 at 12:12
• @helloworld922: Sorry! :) I don't get you! – H. R. Mar 9 '16 at 15:03
• @helloworld922: I don't know whether there is typo in wiki formulas or not so I computed the whole thing here from scratch. :) – H. R. Mar 9 '16 at 15:42