The displacement gradient tensor transformation rule

Question

The transformation rule of a 2nd rank tensor expresssed in a given basis is often written as follow:

$$F' = P^T FP $$

where $F$ is the matrix representation of the tensor in the old basis B, $F'$ its representation in the new basis B', $P$ is the transformation matrix and finally $P^T$ its transpose.

I'm currently trying to proove this using the displacement gradient tensor as an example. Its elements in a given basis can be defined from the derivatives of the displacement field $\overrightarrow u$ with respect to the coordinates $(x_i, i = 1,2,3)$:

$$ u_{ij} = \frac{\partial{u_i}}{\partial{x_j}} $$

I've tried to express the tensor components in a new basis $u_{ij}^{'} = \frac{\partial{u_{i}^{'}}}{\partial{x_{j}^{'}}}$ as a function of the $u_{ij}$ using the conventionnal chain rule and vector decomposition and I end up with $F' = P^{-1}FP$ which is correct only if the old and new basis are related by a rotation. In the most general case it's not true because $P^{T}\neq P^{-1}$ and I can't figure out where is my mistake. Is my reasoning wrong or did I make a calculus error ?

Any help in solving that issue would be greatly appreciated, thanks a lot.

Answer 1

The $P^TFP$ rule is for tensors that have both indices downstairs such as the strain tensor $e_{ij}$. Remembering that a displacement is a contravariant vector, the displacement gradient tensor $$ {e^i}_j = \frac{\partial u^{i}}{\partial x^j} $$ has one index upstairs and one downstairs. It therefore transforms as $P^{-1} FP$ which is what you found. The two transformation rules coincide if your restrict to orthogonal transformations for which $P^T=P^{-1}$. That's why intro elasticity usually resricts to cartesian coordinates. In curvilinear coordinates, the strain tensor is much more complicated than the symmetrised displacement gradient.

If you want an genuine tensor under more than orthogonal transformations you need a Lie derivative. If the metric is $$ ds^2 = g_{\mu\nu}(x) dx^\mu dx^\nu $$ then the strain tensor due to an infinitesimal displacement $x^\mu \to x^\mu+\eta^\mu$ is given by one-half of the Lie derivative of the metric with respect to the displacement: $$ e_{\mu\nu}= \frac 12 ({\mathcal L}_\eta g)_{\mu\nu} \stackrel{\rm def}{=} \frac 12 (\eta^\lambda \partial_\lambda g_{\mu\nu} + g_{\mu\lambda}\partial_\nu \eta^\lambda+ g_{\lambda\nu} \partial_\mu \eta^\lambda). $$
This reduces to the orthogonal cartesian expression $$ e_{\mu\nu}= \frac 12 \left(\partial_\mu \eta_\nu + \partial_\nu \eta_\mu\right) $$ when $g_{\mu\nu}(x)= \delta_{\mu\nu}$ so there is no need to make a distinction between $\eta^\mu$ and $\eta_\mu = g_{\mu\lambda} \eta^\lambda$. When $g_{\mu\nu}$ is constant but not orthonormal, the first term in the Lie derivative vanishishes, but your still need the metric to lower the indices on the $\eta^\mu $to get your $P^T FP$.

For a discussion of the Lie derivative and its connection with the strain tensor see pages 433 and 435 in our book.