# Tensor derivates

1. Dec 3, 2013

### Niles

1. The problem statement, all variables and given/known data
Hi

I am reading about some fluid mechanics, when suddenly I read saw that someone took the derivate of a tensor. It is in this thesis, on page 26 eq. (70). It is the final equality I can't understand.

So the author is taking the derivate $\partial_{x_{\alpha}} P_{\alpha\beta}$ of the momentum flux tensor. How on earth does this end up giving $\rho u_{\alpha}\partial_{x_\alpha}u_{\beta} + \partial_{x_\alpha}p$?

2. Dec 3, 2013

### ShayanJ

You give us the definition of $P_{\alpha\beta}$ and we will answer your question!!! Deal?

3. Dec 4, 2013

### Niles

Sorry, here it is:

$$P_{\alpha\beta} = p\delta_{\alpha\beta} + (u_1^2, u_1u_2; u_2u_1, u_2^2)$$

Here p is a constant and and u a vector.

Deal!

Last edited: Dec 4, 2013
4. Dec 4, 2013

### ShayanJ

Your tensor can also be written as $P_{\alpha\beta}=p \delta_{\alpha\beta}+u_{\alpha}u_{\beta}$.

Let $\partial_{\alpha}=\partial_{x_{\alpha}}$.

Then we have $\partial_{\alpha}P_{\alpha\beta}=\partial_{\alpha} p \delta_{\alpha\beta}+u_{\beta}\partial_{\alpha}u_{\alpha}+ u_{\alpha} \partial_{\alpha} u_{\beta}$

This is all that can be said without adding other assumptions.Only that $\partial_{\alpha} p \delta_{\alpha\beta}=0$ because p is a constant!

So,you should see whether there are other assumptions too or not.For example $\partial_{\alpha}u_{\alpha}$ is the divergence of the vector u.It may be zero so we will have $\partial_{\alpha}P_{\alpha\beta}=u_{\alpha} \partial_{\alpha} u_{\beta}(+ \partial_{\alpha} p \delta_{\alpha\beta}=0)$ which is near to what you want.But I don't know where that $\rho$ comes from.Can you give the definition of u and also other equations involving them?

5. Dec 4, 2013

### Niles

I'll check it out, but it seems $p\propto \rho$ (from the thesis). It doesn't say anything about the gradient of u though.

Due to Einstein summation $\partial_{\alpha}u_\alpha$ is the gradient of u, but what is $u_{\alpha\partial_\alpha u\beta}$?