Tensor Derivatives Homework Help

[Niles]

Homework Statement

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 <br /> \rho u_{\alpha}\partial_{x_\alpha}u_{\beta} + \partial_{x_\alpha}p<br />?


Thanks in advance for hints/help.

 

[ShayanJ]

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

 

[Niles]

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

Sorry, here it is:

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

Here p is a constant and and u a vector.

Deal!

 

[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?

 

[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}?