(My question is simpler than it looks at first glance.)(adsbygoogle = window.adsbygoogle || []).push({});

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

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Divergence of tensor times vector

**Physics Forums | Science Articles, Homework Help, Discussion**