Solving a Tensor Field: Divergence of P = 0

[Päällikkö]

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
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)

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

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
{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}

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

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.

 

[gabbagabbahey]

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
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)

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

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
{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}

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

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

\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

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

Of course, this is utter nonsense.

Since you are given P(r) 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):

\mathbf{\nabla}\cdot (\mathbf{A}\mathbf{B})=\mathbf{B}(\mathbf{\nabla}\cdot \mathbf{A})+\mathbf{A}\cdot(\mathbf{\nabla}\mathbf{B})

\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]

+ \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]

The desired result then follows by inspection of the radial component of \mathbf{\nabla}\cdot P(r).