# Strain Rate Tensor

I was going through Schaum's outline for Continuum Mechanics and came across the strain rate tensor. As the strain rate tensor, when represented geometrically, shows the deformation that is happening, is there any equation that correlates the strain rate tensor with the metric tensor? The line element between initial and final configurations keeps changing.So can we express the strain rate tensor in terms of the metric tensor?

Yes! Good insight! If you model the deformation as a body flowing through rectangular coordinates, the metric tensor is fixed and there's no relation to strain. However, an alternative point of view (one that fits better with a differential form model of flow) is that the axes are 'body-fixed' curvilinear coordinates that flow with the body; then the metric tensor evolves and can be related to the strain tensor. Specifically (Stone and Goldbart, Maths for Physics, p389) you can "define the strain tensor to be ½ of the Lie derivative of the metric".

• What about the dimensions? Strain rate tensor has 1/ length dimension but in General Relativity, the metric tensor is said to dimensionless? Sep 20, 2015 at 13:19

No. Any symmetric tensor, of the form $$E=\begin{pmatrix} E_{11}& E_{12}&E_{13}\\ E_{12}& E_{22}&E_{23}\\ E_{13}& E_{23}&E_{33}\\ \end{pmatrix}$$ can be the strain rate tensor of a body.

The metric tensor, on the other hand, is simply the identity matrix. There is no way to obtain tensors with off-diagonal elements in a natural, invariant way from de identity matrix, and the same holds for tensors with non-constant diagonals.

The only part of the strain rate tensor that you can express in terms of the metric tensor $g$, then, is its constant-diagonal part. This is the rate-of-expansion tensor, as explained in Wikipedia, but it leaves out the rate-of-shear tensor which is equally important.

More or less. The strain rate tensor is in fact not directly related to the metric tensor, although this tensor is involved in its computation. The right Cauchy-Green tensor can be interpreted as the pullback of the metric tensor:

$$\mathbf{C} = \phi^* \mathbf{g}$$

where $$\phi$$ is the deformation map. On the other hand, the strain rate tensor is given by:

$$\mathbf{D} = \frac{1}{2}\frac{\partial \mathbf{C}}{\partial t}$$

In components:

$$D_{AB} = \frac{1}{2}\frac{\partial}{\partial t}\left( g_{ab}\frac{\partial x^a}{\partial X^A}\frac{\partial x^b}{\partial X^B} \right)$$

where the final and initial coordinates are related by the deformation map $$x^a(t) = \phi(X^A,t)$$