# Why $\mathbf n\cdot(\mathbf L^T\mathbf n)=\mathbf n\cdot(\mathbf D\mathbf n)$?

The deformation gradient $\mathbf F$ describes how material line elements change their length and orientation during deformation; the velocity gradient $\mathbf L$ describes the rate of these changes.

We can write $\mathbf L=\mathbf D+\mathbf W$, where $\mathbf D=(\mathbf L+\mathbf L^T)/2$ (thus symmetric) and $\mathbf W=(\mathbf L-\mathbf L^T)/2$ (thus skewsymmetric).

In words, $\mathbf D$ is the Eulerian strain-rate tensor and $\mathbf W$ is the body spin.

Now suppose $\mathbf n$ is a unit vector field.

We have $\mathbf n\cdot(\mathbf L^T\mathbf n)=\mathbf n\cdot(\mathbf D^T\mathbf n)+\mathbf n\cdot(\mathbf W^T\mathbf n)=\mathbf n\cdot[(\mathbf D^T+\mathbf W^T)\mathbf n]=\mathbf n\cdot(\mathbf D\mathbf n)$.

But I do not understand why the last equation is true.

I would appreciate any help or hint. Thank you.

It's because $$\mathbf W$$ is skew symmetric:

$$\mathbf n·\mathbf W^T\mathbf n=-\mathbf n·\mathbf W^T\mathbf n=0$$

But $$\mathbf D$$ is symmetric, so is $$\mathbf D=\mathbf D^T$$, and it follows.

$$\mathbf m·\mathbf W^T\mathbf n=\mathbf n·\mathbf W\mathbf m$$
$$\mathbf n·\mathbf W\mathbf m=-\mathbf n·\mathbf W^T\mathbf m$$
With $$\mathbf n=\mathbf m$$, $$2\mathbf n·\mathbf W^T\mathbf n=0$$
• Do you mean $\mathbf W^T=-\mathbf W$? – johnny09 Apr 13 '17 at 10:28