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 $\mathbf{\nabla}\cdot P(r)$.