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Rate-of-strain tensor in cylindrical coords.

  1. Apr 28, 2014 #1
    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.

    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
     
  2. jcsd
  3. May 4, 2014 #2
    I'm sorry you are not generating any responses at the moment. Is there any additional information you can share with us? Any new findings?
     
  4. May 6, 2014 #3
    Nothing yet. Would still appreciate some insight into my error.
     
  5. May 6, 2014 #4
    Well, for one thing, in your specific tensor components in question, the terms are not dimensionally consistent. The units of each end every term in the rate of deformation tensor must be s-1. This is not the case with one of your vø terms.

    Here are some questions:

    Are you doing the covariant differentiation correctly, including representing the del operator properly?

    Are you taking into account that you might be using the contravariant components of the velocity vector, while Bachelor is using the physical components of the velocity vector (i.e., in terms of the unit vectors, rather than in terms of the coordinate basis vectors).

    What about the rate of deformation tensor? Bachelor is expressing the components in terms of the unit vectors. What are you using?

    I'm guessing that, when you obtained the velocity gradient tensor, it was in terms of mixed covariant/contravariant components. If that was the case, then you can't simply switch the rows and columns to get the transpose.

    Chet
     
    Last edited: May 6, 2014
  6. May 13, 2014 #5
    Chet,

    Thanks for the reply. I will follow up on your suggestions. Meanwhile, let me offer one bit of discovery. According to MathWorld, http://mathworld.wolfram.com/CylindricalCoordinates.html, there are two representations of the Christoffel symbols in cylindrical coordinates, one promulgated by Afrken (1985) and one by Misner et al. (1973). I had been using the Arfken representation, since that was what Batchelor presents. However, using the Misner et al. representation does yield the correct form.

    My suspicion is that there are features of covariance/contravariance that explain the difference. Perhaps someone with more literacy in these matters can take up the challenge of explaining it. :-)

    Thanks,
    Jeff
     
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