# Tensor Calculus: total differential

• Apr 24th 2012, 01:54 AM
linda5962
Tensor Calculus: total differential
I am having a problem understand a particular result in tensor calculus. It is quite a simple result. I shall set the scene.

It is possible to define a double contraction of two second order tensors $\displaystyle A, B$, which is given by

$\displaystyle A:B=trace(A^T B)$

Now we come onto the idea of taking the gradient of a scalar-valued function of a second order tensor. Consider $\displaystyle \phi(A)$ to be a scalar-valued function of second order tensors, $\displaystyle A$. It is a non-linear function, and we aim to approximate $\displaystyle \phi$ at $\displaystyle A$ by a linear function. Taylor expanding $\displaystyle \phi$ at $\displaystyle A$ apparently yields

$\displaystyle \phi(A+dA)=\phi(A)+d\phi + o(dA)$

where $\displaystyle d \phi$ is given by the double contraction:

$\displaystyle d\phi = \frac{\partial \phi(A)}{\partial A} : dA$

Now I have tried and tried to understand how this follows from some form of Taylor expansion, but I'm not getting anywhere. Can someone point me in the right direction?

Thanks,

Lind
• Apr 24th 2012, 06:04 AM
HallsofIvy
Re: Tensor Calculus: total differential
The Taylor's expansion of a function, f(x), around $\displaystyle x= x_0$ is $\displaystyle f(x_0)+ f'(x_0)(x- x_0)+$ terms of order [tex]x^2[/itex] and higher. In particular, if we write $\displaystyle dx= x- x_0$ that becomes $\displaystyle f(x_0)+ f'(x_0)dx+$ higher order terms in dA.

That looks like just what you have.
• Apr 24th 2012, 07:01 AM
linda5962
Re: Tensor Calculus: total differential
Thanks for clearing up the step about Taylor series expansion. My main question is, why does this term $\displaystyle \frac{\partial \phi(A)}{\partial A} dA$ end up being the double contraction given. I don't understand how these two things are related.