I have a question that's half about a definition, half about computation. In fluid dynamics, if we have a velocity field $\mathbf u(\mathbf x,t)$ defined in some domain in $\mathbb R^N\times \mathbb R$ for $N=1,2,$ or $3$, then we define the strain rate tensor $(\varepsilon_{ij})_{ij}$ by
\begin{equation}
\varepsilon_{ij} = \frac12\left(\frac{\partial u_i}{\partial x_j}
+ \frac{\partial u_j}{\partial x_i}\right).
\end{equation}
Now, I am asked to derive the formula for $\varepsilon$ in cylindrical coordinates. Presumably though, we would have a different formula than this.
My background is mainly mathematical, and I'm familiar with tensor calculus from differential geometry. Expressed in coordinate invariant notation, it would seem to me that $\varepsilon$ is simply the symmetrization of $\nabla\mathbf u^\sharp$ considered as a contravariant $2$-tensor field.
However, this leads to a calculation that would appear to be erroneous, since I then get that $$\varepsilon_{\theta\theta} = \frac{1}{r^2}\partial_\theta u_\theta + \frac{1}{r^3}u_r$$ while I am asked to show that $$\varepsilon_{\theta\theta}=\frac{1}{r}\partial_\theta u_\theta + \frac{1}{r}u_r.$$
So is this the correct way of calculating $\varepsilon$?
Here is my work:
We already know the Christoffel symbols $\Gamma_{r\theta}^\theta=\Gamma_{\theta r}^\theta=\frac{1}{r}$, $\Gamma_{\theta\theta}^r = -r$, and that the rest are $0$. We may then calculate that
\begin{equation}
\nabla\mathbf u^\sharp = g^{ik}\left(
\frac{\partial u^j}{\partial x_k} + u^l\Gamma_{kl}^j
\right)
\frac{\partial}{\partial x_i}\otimes \frac{\partial}{\partial x_j}
\end{equation}
so that
$$
\varepsilon_{\theta\theta}=\frac{1}{r^2}\left(
\frac{\partial u_\theta}{\partial\theta}+\frac{u_r}{r}
\right)
$$
which is different from the desired answer. Where did I go wrong?
Best Answer
In a coordinate system with metric $g_{\mu\nu}$ the strain rate $\epsilon_{\mu\nu}$ is defined as one-half the Lie derivative of the metric tensor with respect to the velocity field $V^\mu$. The latter is $$ ({\mathcal L}g)_{\mu\nu}= V^{\alpha} \partial_\alpha g_{\mu\nu}+ g_{\alpha\nu} \partial_\mu V^\alpha+g_{\mu\alpha} \partial_\nu V^{\alpha}. $$ When the coordinate system is Cartesian, so $g_{\mu\nu}=\delta_{\mu\nu}$, this expression reduces to the definition in the original question. The definition
$$ \epsilon_{\mu\nu}\equiv \frac 12 ({\mathcal L}g)_{\mu\nu} $$ is therefore natural as the Lie derivative has the co-ordinate free interpretation of encoding the distortion of the angles and lengths in a volume of fluid as it flows.
Notice that Christoffel symbols are not really needed in the definition of the Lie derivative, although one can also write it as $$ ({\mathcal L}g)_{\mu\nu}= \nabla_\mu V_\nu + \nabla_\nu V_\mu $$ where the $\nabla$'s are the usual covariant derivatives. The two definitions are equal because both are tensors, and they agree in Cartesian co-ordiates. Alternatively some algebra with the explicit form of the Levi-Civita connection will show their equivalence.
My formula gives the strain rate in the co-ordinate basis. It is presumably equivalent to that given by Chester Miller in the dreibein basis of unit vectors.