# Identities for covariant derivative

1. Oct 20, 2007

### CompuChip

Hi.

I'm considering the covariant derivative
$$\nabla_\mu V^\nu = \partial_\mu V^\nu + \Gamma_{\mu\nu}^\lambda V^\lambda$$
in spherical coordinates in flat 3D space (x = r cos sin, y = r sin sin, z = r cos; usual stuff).

Now I wrote down the gradient of a scalar function f, for which I got $\nabla_\mu f = \partial_\mu f$ (having constructed the derivative to reduce to the ordinary partial derivative on scalars) which is of course not correct, it should be something like
$\partial_r f, \frac{1}{r} \partial_\theta f \text{ and } \frac{1}{r \sin\theta} \partial_\phi f$ for the three components.

Same when I try to derive an expression for the divergence of a vector field, and then I need to show that it is the same as the familiar
$$\frac{1}{r^2} \frac{\partial}{\partial r} (r^2 V^1) + \frac{1}{r \sin\theta} \frac{\partial}{\partial\theta} (\sin\theta V^2) + \frac{1}{r \sin\theta} \frac{\partial V^3}{\partial \phi}$$
Of course,
$$\nabla_\mu V^\nu = \partial_\mu V^\mu + \Gamma_{\mu\mu}^\nu V^\nu$$
(that must be correct) but if plug in the connection coefficients I calculated, I get an otherwise correct result, except there are some factors 1/r and 1/r sin theta missing.

I have a feeling I'm mixing up my coordinate systems here, but how?

I did get the right result for the Laplacian $\nabla^2 f$ by the way ... isn't that odd?

2. Oct 20, 2007

### robphy

Can you show some of your calculations?

3. Oct 20, 2007

### CompuChip

The one for the gradient seems clear.
For the divergence:

Non-zero connection coefficients are
$$\begin{array}{rcl} \Gamma^\phi_{r \phi} = \Gamma^\phi_{\phi r} &=& \frac{1}{r} \\ \Gamma^r_{\theta\theta} &=& -r \\ \Gamma^\theta_{\phi\phi} &=& -\sin\theta \cos\theta \\ \Gamma^\theta_{r \theta} = \Gamma^\theta_{\theta r} &=& \frac{1}{r} \\ \Gamma^r_{\phi\phi} &=& -r \sin^2 \theta \\ \Gamma^\phi_{\theta\phi} = \Gamma^\phi_{\phi\theta} &=& \tan^{-1} \theta \\ \end{array}$$

Then I did (just filling in, I used Mathematica - forgive me for sinning ) $$\partial_\mu V^\mu + \Gamma_{\mu\nu}^\mu V^{\nu}$$ where $V^\mu(r, \theta, \phi)$ are the components of the vector field.
This gave me
$$\frac{2 V^1(r,\theta ,\phi )}{r}+\cot (\theta ) V^2(r,\theta ,\phi )+ \partial_\phi V^3(r,\theta ,\phi )+\partial_\theta V^2{}(r,\theta ,\phi )+\partial_r V^1(r,\theta,\phi )$$
which should be
$$\frac{2 V^1(r,\theta ,\phi )}{r}+\frac{\cot (\theta ) V^2(r,\theta ,\phi )}{r}+\frac{\csc (\theta ) \partial_\phi V^3(r,\theta ,\phi )}{r}+\frac{V^2(r,\theta ,\phi )}{r}+\partial_rV^1(r,\theta ,\phi )$$
Quite straightforward in principle, but wrong answer in practice.

Last edited: Oct 20, 2007
4. Oct 20, 2007

### CompuChip

But perhaps you can first explain it for the gradient, as that's the basic case. I get
$$\nabla_\mu f = \partial_\mu f$$
for $\mu = r, \theta, \phi$. Where do the prefactors (1, 1/r, 1/r sin$\theta$) come from?

Normally, you would find them by taking partials of x, y, z w.r.t. r, theta, phi. But in this case, I don't think x, y and z are even involved, since we just have a coordinate chart with coordinates r, theta, phi and a metric (which is induced by the flat metric, but we might as well have just defined it as it is).

And I am also curious why my approach - which is obviously wrong in some way - does give the right result for the Laplacian operator.

5. Oct 21, 2007

### CompuChip

OK, the problem was of course that my basis vectors (1, 0, 0), (0, 1, 0) and (0, 0, 1) are not the unit vectors that are in the familiar formula for the gradient.
Calculating the length of the basis vectors by multiplying by the metric diag(1, r^2, (r sin theta)^2) gives exactly the factors I was missing.

For the Laplacian, we plug in a scalar and get a scalar back, so this is not necessary, therefore I got the correct result.

Going to try if this solves my problem with the divergence as well now.