Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Divergence of tensor times vector

  1. Apr 1, 2013 #1
    (My question is simpler than it looks at first glance.)

    Here is Reynolds Transport Theorem:

    $$\frac{D}{Dt}\int \limits_{V(t)} \mathbf{F}(\vec{x}, t)\ dV = \int \limits_{V(t)} \left[ \frac{\partial \mathbf{F}}{\partial t} + \vec{\nabla} \cdot (\mathbf{F} \vec{u}) \right] \ dV$$

    where boldface means tensor and over-arrow means vector. I am trying to apply this to the expression:

    $$\frac{D}{Dt} \int \limits_{V(t)} (\vec{x} \times \rho \vec{u}) \ dV$$

    where x is the cross product. So

    $$\mathbf{F}(\vec{x}, t) = \epsilon_{ijk} x_{j} (\rho u)_{k} = \rho \epsilon_{ijk} x_{j} u_{k}$$


    $$\mathbf{F} \vec{u} = \rho \epsilon_{ijk} x_{j} u_{k} u_{l}$$

    Now my problem occurs why I try to figure out what

    $$\vec{\nabla} \cdot (\mathbf{F} \vec{u})$$

    is. Should it be

    $$\frac{\partial(\rho \epsilon_{ijk} x_{j} u_{k} u_{l})}{\partial x_{i}}$$ or $$\frac{\partial(\rho \epsilon_{ijk} x_{j} u_{k} u_{l})}{\partial x_{l}}$$?

    I am confused as to which index to take the derivative with. It's easy when you are taking the divergence of a vector; it's just the same index used in the vector. But when applied to a tensor, there's two indices (or more). So which is it?

  2. jcsd
  3. Apr 2, 2013 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I don't know the answer, as I'm not familiar with the notation ##\vec\nabla\cdot(\mathbf F \vec u)##. The only thought I have is that if your book proves the theorem, you should be able to figure out what they mean by examining the proof. If it's a good book, the notation should also be explained earlier.
  4. Apr 3, 2013 #3


    User Avatar
    Science Advisor
    Homework Helper

    A simple analysis of tensorial rank means that [itex] \nabla\cdot(\bf{F}\vec{u}) [/itex] means that the product of F and u must be free, that is a pure tensor product without contraction. That way the divergence of the resulting tensor has the same tensor rank as the time derivative of the tensor itself.
  5. Apr 4, 2013 #4
    From the wikipedia version
    it looks like what you want is
    [tex] \frac{D}{Dt} \int_V f^i dV = \int_V \left[ \frac{\partial f^i}{\partial t}
    + \frac{\partial}{\partial x_j}\left( v^j f^i \right)
    \right] dV. [/tex]
  6. Apr 4, 2013 #5
    Ah, that works well. Thanks!
  7. Apr 5, 2013 #6
    In your case the tensor F is really a vector.

    A handy vector identity is:

    [itex]\nabla \cdot \vec A \vec B = \vec B\nabla \cdot \vec A+ \vec A \cdot \nabla \vec B[/itex]

    Another identity that you might find useful is:

    [itex]\nabla \cdot \left(\vec A \times \vec B \right)= \vec B \cdot \nabla \times \vec A- \vec A \cdot \nabla \times \vec B[/itex]
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook