- #1
JoutlawPhysics
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Hi PF,
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.
I'm trying to derive the rate-of-strain tensor in cylindrical coords, starting with the Christoffel symbols.
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}
\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
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.
Homework Statement
I'm trying to derive the rate-of-strain tensor in cylindrical coords, starting with the Christoffel symbols.
Homework 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}
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