# How to compute the strain rate tensor in non-Euclidean coordinates

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 $$\varepsilon_{ij} = \frac12\left(\frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i}\right).$$

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 $$\nabla\mathbf u^\sharp = g^{ik}\left( \frac{\partial u^j}{\partial x_k} + u^l\Gamma_{kl}^j \right) \frac{\partial}{\partial x_i}\otimes \frac{\partial}{\partial x_j}$$ 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?

• This is the component represented in terms of the unit vectors in the theta direction. The coordinate basis vector in the circumferential direction is related to the unit vector in the theta direction by a factor of r. – Chet Miller Nov 29 '17 at 21:45
• But what change of variables do we use for tensors? Since these are tensors we're working with, rather than vectors. – Chikyuujin Nov 29 '17 at 21:48
• I thought you derived it directly, not by transforming. Are you familiar with dyadic tensor notation? – Chet Miller Nov 29 '17 at 23:15
• No, I am not, unless I have learnt it by another name. – Chikyuujin Nov 29 '17 at 23:47

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.

• What is a dreibein basis? – Chikyuujin Nov 30 '17 at 20:52
• @Chikyuujin It's a a three-frame of orthonormal vectors --- i.e a three dimensional verision of the General Relativist's vierbein (ein, zwei, drie, vier, = 1,2,3,4 in German). the ${\bf i}_r, {\bf i}_\theta, {\bf i}_z$ in Miller's answer constitute am orthonormal dreibein. – mike stone Nov 30 '17 at 21:38

Well, for whatever it's worth, here is how it would be done within the framework of dyadic notation.

In cylindrical coordinates, the velocity vector is given by: $$\mathbf{u}=u_r\mathbf{i}_r+u_{\theta}\mathbf{i}_{\theta}+u_z\mathbf{i}_z$$and the gradient vector operator is given by:$$\mathbf{\nabla}=\mathbf{i}_r\frac{\partial}{\partial r}+\mathbf{i}_{\theta}\frac{1}{r}\frac{\partial}{\partial \theta}+\mathbf{i}_z\frac{\partial}{\partial z}$$So the gradient of the velocity vector is given by:$$\mathbf{\nabla}\mathbf{u}=\mathbf{i}_r\frac{\partial\mathbf{u}}{\partial r}+\mathbf{i}_{\theta}\frac{1}{r}\frac{\partial \mathbf{u}}{\partial \theta}+\mathbf{i}_z\frac{\partial \mathbf{u}}{\partial z}$$ From this, we see that we need to evaluate the partial derivatives of the velocity vector with respect to r, $\theta$, and z: $$\frac{\partial\mathbf{u}}{\partial r}=\mathbf{i}_r\frac{\partial u_r}{\partial r}+\mathbf{i}_{\theta}\frac{\partial u_{\theta}}{\partial r}+\mathbf{i}_z\frac{\partial u_z}{\partial r}$$

$$\frac{\partial\mathbf{u}}{\partial \theta}=\mathbf{i}_r\frac{\partial u_r}{\partial \theta}+u_r\mathbf{i}_{\theta}+\mathbf{i}_{\theta}\frac{\partial u_{\theta}}{\partial \theta}-u_{\theta}\mathbf{i}_r+\mathbf{i}_z\frac{\partial u_z}{\partial \theta}$$

$$\frac{\partial\mathbf{u}}{\partial z}=\mathbf{i}_r\frac{\partial u_r}{\partial z}+\mathbf{i}_{\theta}\frac{\partial u_{\theta}}{\partial z}+\mathbf{i}_z\frac{\partial u_z}{\partial z}$$ So, $$\mathbf{\nabla}\mathbf{u}=\mathbf{i}_r\mathbf{i}_r\frac{\partial u_r}{\partial r}+\mathbf{i}_r\mathbf{i}_{\theta}\frac{\partial u_{\theta}}{\partial r}+\mathbf{i}_r\mathbf{i}_z\frac{\partial u_z}{\partial r}+\mathbf{i}_{\theta}\mathbf{i}_r\left(\frac{1}{r}\frac{\partial u_r}{\partial \theta}-\frac{u_{\theta}}{r}\right)+\mathbf{i}_{\theta}\mathbf{i}_{\theta}\left(\frac{1}{r}\frac{\partial u_{\theta}}{\partial \theta}+\frac{u_r}{r}\right)+\mathbf{i}_{\theta}\mathbf{i}_z\frac{1}{r}\frac{\partial u_z}{\partial \theta}+\mathbf{i}_z\mathbf{i}_r\frac{\partial u_r}{\partial z}+\mathbf{i}_z\mathbf{i}_{\theta}\frac{\partial u_{\theta}}{\partial z}+\mathbf{i}_z\mathbf{i}_z\frac{\partial u_z}{\partial z}$$ The transpose of this velocity gradient tensor is obtained by interchanging the two unit vectors in each term. The rate of deformation tensor is obtained by adding the velocity gradient tensor to its transpose and dividing by 2.

I know this is rather old question, but since I had to go through this myself I figured out an answer - or rather I managed to dress up @Chester Miller 's answer with some formal differential geometry. Since I'm more of a physicist than mathematician I'll use index notation rather than musical isomorphisms. The key misunderstanding that leads to mismatch between OP's formula for $\epsilon_{\theta\theta}$ and desired one is the definition of velocity, or rather the chosen basis in the tangent space. While OP uses (as usually in differential geometry) coordinate basis, i.e. $$e_i = \partial_{x_i}$$ in which (for the case of polar coordinate system) the basis vectors are orthogonal but not normalised, @Chester seems to be using orthonormal basis (called sometimes physical basis) in which the metric has a canonical form $$g_{IJ} = \delta_{IJ}$$ (from now on I'll use convention where capital indices denote component in physical basis and the other ones -- in coordinate basis)

So, when OP defined his velocity field it was $u^ie_i$, while @Chester had $u^I e_I$. Also, components of OP's $\epsilon$ tensor were $\epsilon_{ij}$, while Chester presented $\epsilon_{IJ}$. To relate one to another we need to know the transformation matrix $$e_A = e_A^b e_b$$ between coordinate and physical basis. This is easy in our case, since the coordinate basis is already orthogonal, so it's enough to normalise the vectors: $$e_I = |e_i|^{-1}e_i,~ I=i$$ in other words $$e_A^b = \frac{\delta_A^b}{|e_b|}$$ (both notations are somehow obscure, but I hope understandable). To obtain Chester's result using OP's method one has to compute $$\epsilon_{IJ}=e_I^a e_J^b \nabla_{\left(a\right.}(e_{\left.b\right)}^C u_C)$$ where $e_b^C$ is an inverse of $e^b_C$ in the following sense: $e_b^C e_C^d = \delta_b^d$ and the covariant derivative $\nabla$ contains normal Levi-Civita connection in coordinate basis (ones mentioned by the OP).

The whole procedure described here is similar to Einstein gravity vielbein formulation, where one uses locally orthonormal field of basis on a tangent space, and so the metric is always of Lorentzian shape diag$[-1, 1, 1...]$ depending on the dimension and the dynamical gravity field is the (local) transition matrix $e$ -- and then the covariant derivative is based on a (functionally) different spin connection