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Tensor calculus

  1. May 21, 2009 #1

    Päällikkö

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    The following isn't actually a homework problem, but this seems to be the natural place to ask questions of this sort. Without further ado,

    I have a spherically symmetric tensor field, which is written with the help of dyadics as
    [tex]P(r) = P_n(r)\mathbf{e}_r\mathbf{e}_r + P_t(r)(\mathbf{e}_\theta\mathbf{e}_\theta+\mathbf{e}_\varphi\mathbf{e}_\varphi)[/tex]

    From the condition that the divergence of P vanishes, I am to deduce that
    [tex]\frac{d}{dr}(r^2 P_n(r)) = 2r P_t(r)[/tex]

    As I've got little to no experience with tensors (especially when they're in dyadic form), I'm at a loss here.


    For a vector field A one would calculate the divergence as
    [tex]{1 \over r^2}{\partial \left( r^2 A_r \right) \over \partial r} + {1 \over r\sin\theta}{\partial \over \partial \theta} \left( A_\theta\sin\theta \right) + {1 \over r\sin\theta}{\partial A_\phi \over \partial \phi}[/tex]

    I tried throwing stuff into here, and ended up with nonsense:
    [tex]\frac{d}{dr}(r^2 P_n(r)) + \cot(\theta) r P_t(r) = 0[/tex]

    I suppose the divergence could be calculated, in tensor terminology, as the contraction of P with the covariant derivative. This would lead to three equations, which I couldn't get to give the wanted result.
     
  2. jcsd
  3. May 22, 2009 #2

    gabbagabbahey

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    First off, it's important to notice that the divergence of a second rank tensor will be a vector.

    What you seem to have done is say that

    [tex]\mathbf{\nabla}\cdot P(r)=\mathbf{\nabla}\cdot\left(P_n(r)\mathbf{e}_r + P_t(r)(\mathbf{e}_\theta+\mathbf{ e}_\varphi)\right)=0[/tex]

    (from which you obtain [itex]\frac{d}{dr}(r^2 P_n(r)) + \cot(\theta) r P_t(r) = 0[/itex])

    Of course, this is utter nonsense.

    Since you are given [itex]P(r)[/itex] in dyadic form, the easiest way to compute its divergence is probably to continue working in dyadic form and make use of the following product rule (which is a natural extension for the usual vector calculus rule involving the divergence of the product of a scalar and a vector):

    [tex]\mathbf{\nabla}\cdot (\mathbf{A}\mathbf{B})=\mathbf{B}(\mathbf{\nabla}\cdot \mathbf{A})+\mathbf{A}\cdot(\mathbf{\nabla}\mathbf{B})[/tex]

    [tex]\implies \mathbf{\nabla}\cdot \left[ P_n(r)\mathbf{e}_r\mathbf{e}_r + P_t(r)(\mathbf{e}_{\theta}\mathbf{e}_{\theta}+\mathbf{e}_{\varphi}\mathbf{e}_{\varphi}) \right] = \left[\mathbf{e}_r (\mathbf{\nabla}\cdot \mathbf{P_n(r)\mathbf{e}_r})+\mathbf{P_n(r)\mathbf{e}_r}\cdot(\mathbf{\nabla}\mathbf{e}_r)\right] [/tex]

    [tex]+ \left[\mathbf{e}_\theta (\mathbf{\nabla}\cdot \mathbf{P_t(r)\mathbf{e}_\theta})+\mathbf{P_t(r)\mathbf{e}_\theta}\cdot(\mathbf{\nabla}\mathbf{e}_\theta)\right] + \left[\mathbf{e}_\varphi (\mathbf{\nabla}\cdot \mathbf{P_t(r)\mathbf{e}_\varphi})+\mathbf{P_t(r)\mathbf{e}_\varphi}\cdot(\mathbf{\nabla}\mathbf{e}_\varphi)\right][/tex]

    The desired result then follows by inspection of the radial component of [itex]\mathbf{\nabla}\cdot P(r)[/itex].
     
    Last edited: May 22, 2009
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