- #1

- 519

- 11

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.