Tensor Notation and derivatives

[emirates]

Hi folks.

Hope that you can help me.

I have an equation, that has been rewritten, and i don't see how:

has been rewritten to:

Can someone explain me how?

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


σ_ij,jζu_i = (σ_ijζu_i)_,j

really hope, that someone can help me

 

[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.

 

[emirates]

i didnt get that.

Is it possible for you to help me a bit?

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

 

[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.

 

[emirates]

Hi.

I do understand this part.

What I'm not understanding is, how to get this expression:
(σ_ijδu_i)_,j FROM σ_ij,jδu_i

because I earlier asked if the (σ_ijδu_i)_,j = σ_ij,jδu_i

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

 

[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}.

 

[emirates]

okay. I see this now.
But I still don't see where is expression comes from (σ_ijδu_i)_,j

 

[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.