# Divergence of tensor times vector

1. Apr 1, 2013

### Runner 1

(My question is simpler than it looks at first glance.)

Here is Reynolds Transport Theorem:

$$\frac{D}{Dt}\int \limits_{V(t)} \mathbf{F}(\vec{x}, t)\ dV = \int \limits_{V(t)} \left[ \frac{\partial \mathbf{F}}{\partial t} + \vec{\nabla} \cdot (\mathbf{F} \vec{u}) \right] \ dV$$

where boldface means tensor and over-arrow means vector. I am trying to apply this to the expression:

$$\frac{D}{Dt} \int \limits_{V(t)} (\vec{x} \times \rho \vec{u}) \ dV$$

where x is the cross product. So

$$\mathbf{F}(\vec{x}, t) = \epsilon_{ijk} x_{j} (\rho u)_{k} = \rho \epsilon_{ijk} x_{j} u_{k}$$

and

$$\mathbf{F} \vec{u} = \rho \epsilon_{ijk} x_{j} u_{k} u_{l}$$

Now my problem occurs why I try to figure out what

$$\vec{\nabla} \cdot (\mathbf{F} \vec{u})$$

is. Should it be

$$\frac{\partial(\rho \epsilon_{ijk} x_{j} u_{k} u_{l})}{\partial x_{i}}$$ or $$\frac{\partial(\rho \epsilon_{ijk} x_{j} u_{k} u_{l})}{\partial x_{l}}$$?

I am confused as to which index to take the derivative with. It's easy when you are taking the divergence of a vector; it's just the same index used in the vector. But when applied to a tensor, there's two indices (or more). So which is it?

Thanks

2. Apr 2, 2013

### Fredrik

Staff Emeritus
I don't know the answer, as I'm not familiar with the notation $\vec\nabla\cdot(\mathbf F \vec u)$. The only thought I have is that if your book proves the theorem, you should be able to figure out what they mean by examining the proof. If it's a good book, the notation should also be explained earlier.

3. Apr 3, 2013

### dextercioby

A simple analysis of tensorial rank means that $\nabla\cdot(\bf{F}\vec{u})$ means that the product of F and u must be free, that is a pure tensor product without contraction. That way the divergence of the resulting tensor has the same tensor rank as the time derivative of the tensor itself.

4. Apr 4, 2013

### qbert

From the wikipedia version
it looks like what you want is
$$\frac{D}{Dt} \int_V f^i dV = \int_V \left[ \frac{\partial f^i}{\partial t} + \frac{\partial}{\partial x_j}\left( v^j f^i \right) \right] dV.$$

5. Apr 4, 2013

### Runner 1

Ah, that works well. Thanks!

6. Apr 5, 2013

### the_wolfman

In your case the tensor F is really a vector.

A handy vector identity is:

$\nabla \cdot \vec A \vec B = \vec B\nabla \cdot \vec A+ \vec A \cdot \nabla \vec B$

Another identity that you might find useful is:

$\nabla \cdot \left(\vec A \times \vec B \right)= \vec B \cdot \nabla \times \vec A- \vec A \cdot \nabla \times \vec B$