Hi PF,(adsbygoogle = window.adsbygoogle || []).push({});

I posted this in HW a week ago and got no response. Might be a bit beyond the typical HW forum troller. So, please excuse the double-post.

1. The problem statement, all variables and given/known data

I'm trying to derive the rate-of-strain tensor in cylindrical coords, starting with the Christoffel symbols.

2. Relevant equations

The cylindrical coordinate Christoffel matrices:

\begin{equation}\Gamma^r=\left(

\def\arraystretch{1.5}\begin{array}{ccc}

0&0&0\\

0&-r&0\\

0&0&0

\end{array}\right)

\qquad

\Gamma^\phi=\left(

\def\arraystretch{1.5}\begin{array}{ccc}

0&\dfrac{1}{r}&0\\

\dfrac{1}{r}&0&0\\

0&0&0

\end{array}\right)

\qquad

\Gamma^z=\left(

\def\arraystretch{1.5}\begin{array}{ccc}

0&0&0\\

0&0&0\\

0&0&0

\end{array}\right).

\end{equation}

The gradient of the basis vectors in cylindrical coordinates is

defined in terms of the Christoffel symbols, \[\Gamma^k_{ij}\] such that

\begin{equation}

\nabla_ie_j=\Gamma^k_{ij}e_k.

\end{equation}

The only non-zero terms are

\begin{equation}

\nabla_r e_\phi = \frac{1}{r}e_\phi,\qquad \nabla_\phi e_r =

\frac{1}{r}e_\phi,\qquad \nabla_\phi e_\phi = -r e_r

\end{equation}

3. The attempt at a solution

\begin{eqnarray}

\underline{\underline{\dot{\gamma}}}&=&\nabla\underline{v}= \left(

\def\arraystretch{1.2}\begin{array}{c}

\nabla_r(\underline{v})\\

\nabla_\phi(\underline{v})\\

\nabla_z(\underline{v})

\end{array}\right)=\left(

\def\arraystretch{1.2}\begin{array}{c}

\nabla_r(v_r e_r+v_\phi e_\phi+v_z e_z)\\

\nabla_\phi(v_r e_r+v_\phi e_\phi+v_z e_z)\\

\nabla_z(v_r e_r+v_\phi e_\phi+v_z e_z)

\end{array}\right)\\

&=& \left(

\def\arraystretch{2.2}\begin{array}{c}

(\nabla_rv_r)e_r+\left(\nabla_rv_\phi +v_\phi\dfrac{1}{r}\right)e_\phi+(\nabla_rv_z)e_z\\

(\nabla_\phi v_r-v_\phi r)e_r+\left(\nabla_\phi v_\phi +v_r\dfrac{1}{r}\right )e_\phi +(\nabla_\phi v_z)e_z\\

(\nabla_zv_r)e_r+(\nabla_zv_\phi) e_\phi+(\nabla_zv_z) e_z

\end{array}\right)\nonumber.

\end{eqnarray}

Grouping by basis vectors into individual columns,

\begin{eqnarray}

\underline{\underline{\dot{\gamma}}}

&=&\left(

\def\arraystretch{2.2}\begin{array}{ccc}

\dfrac{\partial v_r }{\partial r}& \dfrac{\partial v_\phi}{\partial r}+\dfrac{v_\phi}{r} & \dfrac{\partial v_z}{\partial r}\\

\dfrac{1}{r}\dfrac{\partial v_r}{\partial\phi}-v_\phi r & \dfrac{1}{r}\dfrac{\partial v_\phi}{\partial\phi}

+\dfrac{v_r}{r} & \dfrac{1}{r}\dfrac{\partial v_z}{\partial\phi} \\

\dfrac{\partial v_r}{\partial z} & \dfrac{\partial v_\phi}{\partial z}

& \dfrac{\partial v_z}{\partial z}

\end{array}\right)

\end{eqnarray}

Symmetrizing,

\begin{eqnarray}

\underline{\underline{\dot{\gamma}}}&=&\frac{\nabla\underline{v}}{2}+\frac{(\nabla\underline{v})^T}{2}\\

&=&\frac{1}{2}\left(

\def\arraystretch{2.2}\begin{array}{ccc}

2\dfrac{\partial v_r }{\partial r}& \dfrac{1}{r}\dfrac{\partial v_r}{\partial\phi}-v_\phi r+\dfrac{\partial v_\phi}{\partial r}+\dfrac{v_\phi}{r} & \dfrac{\partial v_r}{\partial z}+\dfrac{\partial v_z}{\partial r}\\

\dfrac{1}{r}\dfrac{\partial v_r}{\partial\phi}-v_\phi r+\dfrac{\partial v_\phi}{\partial r}+\dfrac{v_\phi}{r} & 2\left(\dfrac{1}{r}\dfrac{\partial v_\phi}{\partial\phi}

+\dfrac{v_r}{r}\right) & \dfrac{\partial v_\phi}{\partial z}+ \dfrac{1}{r}\dfrac{\partial v_z}{\partial\phi} \\

\dfrac{\partial v_r}{\partial z}+\dfrac{\partial v_z}{\partial r} & \dfrac{\partial v_\phi}{\partial z}+\dfrac{1}{r}\dfrac{\partial v_z}{\partial\phi}

& 2\dfrac{\partial v_z}{\partial z}.\nonumber

\end{array}\right)

\end{eqnarray}

To check, I compared this to the rate-of-strain

tensor reported in Batchelor (1967, pg. 602):

\begin{eqnarray}

\underline{\underline{\dot{\gamma}}}&=&\frac{1}{2}\left(

\def\arraystretch{2.2}\begin{array}{ccc}

2\dfrac{\partial v_r }{\partial r}& \dfrac{1}{r}\dfrac{\partial v_r}{\partial\phi}+r\dfrac{\partial}{\partial r}\left(\dfrac{v_\phi}{r}\right) & \dfrac{\partial v_r}{\partial z}+\dfrac{\partial v_z}{\partial r}\\

\dfrac{1}{r}\dfrac{\partial v_r}{\partial\phi}+r\dfrac{\partial}{\partial r}\left(\dfrac{v_\phi}{r}\right) & 2\left(\dfrac{1}{r}\dfrac{\partial v_\phi}{\partial\phi}

+\dfrac{v_r}{r}\right) & \dfrac{\partial v_\phi}{\partial z}+ \dfrac{1}{r}\dfrac{\partial v_z}{\partial\phi} \\

\dfrac{\partial v_r}{\partial z}+\dfrac{\partial v_z}{\partial r} & \dfrac{\partial v_\phi}{\partial z}+\dfrac{1}{r}\dfrac{\partial v_z}{\partial\phi}

& 2\dfrac{\partial v_z}{\partial z}

\end{array}\right)\\

&=&\frac{1}{2}\left(

\def\arraystretch{2.2}\begin{array}{ccc}

2\dfrac{\partial v_r }{\partial r}& \dfrac{1}{r}\dfrac{\partial

v_r}{\partial\phi}+\dfrac{\partial v_\phi}{\partial r}-\dfrac{v_\phi}{r} & \dfrac{\partial v_r}{\partial z}+\dfrac{\partial v_z}{\partial r}\\

\dfrac{1}{r}\dfrac{\partial

v_r}{\partial\phi}+\dfrac{\partial v_\phi}{\partial r}-\dfrac{v_\phi}{r} & 2\left(\dfrac{1}{r}\dfrac{\partial v_\phi}{\partial\phi}

+\dfrac{v_r}{r}\right) & \dfrac{\partial v_\phi}{\partial z}+ \dfrac{1}{r}\dfrac{\partial v_z}{\partial\phi} \\

\dfrac{\partial v_r}{\partial z}+\dfrac{\partial v_z}{\partial r} & \dfrac{\partial v_\phi}{\partial z}+\dfrac{1}{r}\dfrac{\partial v_z}{\partial\phi}

& 2\dfrac{\partial v_z}{\partial z}

\end{array}\right) .\nonumber

\end{eqnarray}

As you can see, the (2,1),(1,2) elements are very different. What am I missing?

Thanks in advance,

Jeff

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# Rate-of-strain tensor in cylindrical coords.

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