# Tensor Notation and derivatives

1. Dec 14, 2013

### emirates

Hi folks.

Hope that you can help me.

I have an equation, that has been rewritten, and i dont see how:

has been rewritten to:

Can someone explain me how?

Or can someone just tell me if this is correct in tensor notation:

σij,jζui = (σijζui),j

really hope, that someone can help me

2. Dec 14, 2013

### DeIdeal

Try to apply the Leibniz rule to the derivative ($\partial_j$) of the product $\sigma_{ij}\delta u_i$. It should be fairly obvious after that.

3. Dec 14, 2013

### emirates

i didnt get that.

Is it possible for you to help me a bit?

was the expression valid?: σij,jζui = (σijζui),j

4. Dec 14, 2013

### DeIdeal

Sure, I was a bit reluctant to give out the full answer, but I guess this isn't really a homework question.

First of all, no, it's not valid unless δu is a constant (note, I guess you mean δ, delta, by ζ).

Secondly, to get the expression just expand

$(\sigma_{ij,j}+p_i)\delta u_i=\sigma_{ij,j}\delta u_i+p_i\delta u_i$

And, as I suggested, use the Leibniz rule (the derivative of a product)

$(\sigma_{ij}\delta u_i)_{,j}\equiv \partial_j(\sigma_{ij}\delta u_i)= (\partial_j\sigma_{ij})\delta u_i +\sigma_{ij}(\partial_j\delta u_i )\equiv\sigma_{ij,j}\delta u_i+\sigma_{ij}\delta u_{i,j}$

(Just in case it's the comma notation that's confusing you, I wrote the derivative operators out explicitly)

Rearrange to get

$\sigma_{ij,j}\delta u_i=(\sigma_{ij}\delta u_i)_{,j}-\sigma_{ij}\delta u_{i,j}$

So

$\sigma_{ij,j}\delta u_i+p_i\delta u_i=(\sigma_{ij}\delta u_i)_{,j}-\sigma_{ij}\delta u_{i,j}+p_i\delta u_i$

Which is the expression you were given.

Last edited: Dec 14, 2013
5. Dec 14, 2013

### emirates

Hi.

I do understand this part.

What I'm not understanding is, how to get this expression:
ijδui),j FROM σij,jδui

because I earlier asked if the (σijδui),j = σij,jδui

where you said no, but as i see from you calculations, then the expression is valid? :) Or?

6. Dec 14, 2013

### DeIdeal

Well, no. As per the previous post,

$$\sigma_{ij,j}\delta u_i=(\sigma_{ij}\delta u_i)_{,j}-\sigma_{ij}\delta u_{i,j}$$

So clearly $\sigma_{ij,j}\delta u_i \neq (\sigma_{ij}\delta u_i)_{,j}$ unless $\sigma_{ij}\delta u_{i,j}=0$.

Notice that you have the same additional term $\sigma_{ij}\delta u_{i,j}$ in the expression you posted in the expression you posted in post #1, it doesn't say anywhere that $\sigma_{ij,j}\delta u_i =(\sigma_{ij}\delta u_i)_{,j}$ but instead that, exactly as I said, $\sigma_{ij,j}\delta u_i=(\sigma_{ij}\delta u_i)_{,j}-\sigma_{ij}\delta u_{i,j}$.

7. Dec 14, 2013

### emirates

okay. I see this now.
But I still dont see where is expression comes from (σijδui),j

8. Dec 14, 2013

### DeIdeal

Hmm. I wonder whether I'm able to explain this properly, but I'll try. It is sometimes convenient to express a product of an object and the derivative of another object, for example $\sigma_{ij,j}\delta u_i$, in a form that contains the derivative of their product (This might allow the use of eg the divergence theorem when the product appears under an integral). Then you just notice that you can indeed do that, by applying the Leibniz rule, to the product of the two objects, in this case to $\sigma_{ij}\delta u_i$ by writing what I did above. It might also be that the derivative of the second object (so $\delta u_{i,j}$) is more desirable and easier to handle for one reason or another.