Top of last page in German version of Jackson |

### Question

You are familiar with the operations of gradient (##\nabla\phi##), divergence (##\nabla\bullet\mathbf{V}##) and curl (##\nabla\times\mathbf{V}##) in ordinary vector analysis in three-dimensional Euclidean space. Using covariant derivatives, derive formulae for these operations in spherical polar coordinates ##\left\{r,\theta,\phi\right\}## defined by\begin{align}

x&=r\sin{\theta}\cos{\phi}&\phantom {10000}(1)\nonumber\\

y&=r\sin{\theta}\sin{\phi}&\phantom {10000}(2)\nonumber\\

z&=r\cos{\theta}&\phantom {10000}(3)\nonumber

\end{align}Compare your results to those in Jackson (1999) or an equivalent text. Are they identical? Should they be? (JD Jackson, Classical Electrodynamics, Wiley 1999)

### Answer

One of the most difficult things about this exercise was finding a copy of Jackson (1999). It is a 'classic' text, so, unlike most English mathematical text books in the Berlin library, the only copies available were translation into German. Luckily it wasn't too hard to find the required results. They were on the very last page. The heading is 'Description of vector operations in various coordinate systems'. Simple really!

Full answer at Ex 3.02 Spherical gradient divergence curl as covariant derivatives.pdf (7 pages)

**Analyse Fonctionnelle left a comment on 7 March 2021**. Here it is again with ## signs round latex expressions, so it looks nicer. I also made some replies in

*italics*.

Hello, I wonder if you are still working on the exercises of Carroll's book. Continue the good work, I think it's very helpful.

*Thank you! I'm kind of running out of exercises, so there are not many new ones now.*First of all, all the Christoffell symbols are correct, but a few things,

1. When working with Christoffell symbols, it's best not to contract, lower or raise indices because they are not tensors. So when you calculate the covariant form of the divergence, you should do ##g^{ij}\nabla_i V_j## first, instead of carrying the contraction of ##i## into the calculation. And then do the covariant derivative of ##V_j## (there'll be a minus sign in front of the ##\Gamma## 's).

*Quite right!*2. The covariant form of curl should be ##\epsilon^{ijk}\nabla_j V_k \partial_i## and the whole thing divided by the square root of the determinant of the metric. The way you wrote in the pdf will give you a number, not a vector. And the square root of ##det(g)## is because ##\epsilon## is not a tensor but a tensor density.

*Sounds plausible.*3. As you said at the end of the pdf, you'll need to convert everything to an orthonormal basis in order to get the form in Jackson's book. Basically, the new basis is ##\hat{\partial_i} = \partial_i/\sqrt{g_{ii}}## (no summation) and correspondingly, the new component is ##\hat{A^i} = A^i \sqrt{g_{ii}}## (no summation).

*Sounds plausible.*And there will be places where you need to change back and forth between ##V_i## and ##V^i##. It's not straightforward in any means.

Thank you for putting in the time and effort to try to work out the exercises. I really appreciate it. I'm sure many other people do too.

*Thanks very much for the thoughtful feedback.*
Hello, I wonder if you are still working on the exercises of Carroll's book. Continue the good work, I think it's very helpful.

ReplyDeleteFirst of all, all the Christoffell symbols are correct, but a few things,

1. When working with Christoffell symbols, it's best not to contract, lower or raise indices because they are not tensors. So when you calculate the covariant form of the divergence, you should do g^{ij}\nabla_i V_j first, instead of carrying the contraction of i into the calculation. And then do the covariant derivative of V_j (there'll be a minus sign in front of the \Gamma 's).

2. The covariant form of curl should be \epsilon^{ijk}\nabla_j V_k \partial_i and the whole thing divided by the square root of the determinant of the metric. The way you wrote in the pdf will give you a number, not a vector. And the square root of det(g) is because \epsilon is not a tensor but a tensor density.

3. As you said at the end of the pdf, you'll need to convert everything to an orthonormal basis in order to get the form in Jackson's book. Basically, the new basis is \hat{\partial_i} = \partial_i/\sqrt{g_{ii}} (no summation) and correspondingly, the new component is \hat{A^i} = A^i * sqrt{g_{ii}} (no summation).

And there will be places where you need to change back and forth between V_i and V^i. It's not straightforward in any means.

Thank you for putting in the time and effort to try to work out the exercises. I really appreciate it. I'm sure many other people do too.

A comment or two on the curl. (a) when we move to a general coordinate system, we probably want to replace the Levi-Civita symbol by the Levi-Civita tensor. (b) curl is in fact a two-form disguised as a vector. A more compact way to write it is simply $$dV_\mu$$.

ReplyDeleteI believe that the solution to the problem of why the expressions are not exactly the same as Jackson's is because the escalation factors are not taken into account.

ReplyDeleteh_r = 1, \quad h_\theta = r, \quad h_\phi = r \sin \theta

You do an excellent job I like what you do, keep it up, congratulations!!!

ReplyDelete

ReplyDeleteI think the correct calculation would be using this formula

\nabla \times \mathbf{V} = \left( \frac{1}{h_j h_k} \epsilon_{ijk} \left( \frac{\partial}{\partial x^j} (h_k V_k) + \Gamma^l_{jk} V_l \right) \right) \hat{x}^i

where

h_j h_k

Are the scale factors (also known as "metric coefficients") in curvilinear coordinates. In spherical coordinates, for example, they account for the fact that distances along the coordinate directions change differently as you move through space. Remenber h_r = 1, \quad h_\theta = r, \quad h_\phi = r \sin \theta