Gradient Tensor of a vector field

[stuxk]

Hi, I'm trying to compute the gradient tensor of a vector field and I must say I'm quite confused. In other words I have a vector field which is given in spherical coordinates as:
$$\vec{F}=\begin{bmatrix} 0 \\ \frac{1}{\sin\theta}A \\ -B \end{bmatrix}$$, with A,B some scalar potentials and I want to compute the gradient of the vector.

I found that I have to use the covariant derivative which is given by: $$\nabla\vec{B}= \frac{\partial{B_p}}{\partial{u^q}} - B_k\Gamma_{pq}^{k}$$

For some reason I can't get the right results. For instance the tensor element with p=2, q=1 is given as $$\frac{1}{\sin\theta}\frac{\partial A}{\partial r} = \frac{\partial F_\theta}{\partial r}$$ whereas according the covariant derivative formula should be: $$\frac{\partial F_\theta}{\partial r} - \frac{1}{r} F_\theta$$.

Probably I'm doing a big mistake and I think I haven't really understood how I can calculate the gradient tensor. Any help will be appriciated.

 

[espen180]

$$\Gamma^a_{bc}=\frac12 g^{ad}\left(\partial_bg_{cd}+\partial_cg_{bd}-\partial_dg_{bc}\right)$$

Aren't vectors usually contravariant? Does the problem state anything about this? Covariant vectors are usually gradiants or scalar-valued functions?

It turned out that your covariant vector is really contravariant. That may have caused the error.

Below this line is the calculation, so don't look unless you're fine with getting the key.

___________________________________

$$\nabla_qF^p=\partial_qF^p-\Gamma^p_{qk}F^k$$

setting p=2, q=1:

$$\nabla_rF^{\theta}=\partial_rF^{\theta}-\Gamma^{\theta}_{rk}F^k$$

$$=\frac{1}{\sin\,\theta}\frac{\partial A}{\partial r}-\Gamma^{\theta}_{rr}F^r-\Gamma^{\theta}_{r\theta}F^{\theta}-\Gamma^{\theta}_{r\phi}F^{\phi}$$

The metric is $$g_{rr}=1$$ , $$g_{\theta\theta}=r^2$$ , $$g_{\phi\phi}=r^2\sin^2\theta$$, others zero , so the only nonzero derivatives of the metric are

$$\partial_{r}g_{\theta\theta}=2r$$
$$\partial_{r}g_{\phi\phi}=2r\sin^2\theta$$
$$\partial_{\theta}g_{\phi\phi}=2r^2\sin\,\theta\,\cos\,\theta$$

Which means that $$-\Gamma^{\theta}_{rr}F^r-\Gamma^{\theta}_{r\theta}F^{\theta}-\Gamma^{\theta}_{r\phi}F^{\phi}=-\frac12g^{\theta\theta}\partial_rg_{\theta\theta}F^{\theta}=-\frac12 r^{-2}\cdot2r\cdot\frac{1}{\sin\,\theta}A=-\frac{A}{r\sin\,\theta}$$

so

$$\nabla_rF^{\theta}=\frac{1}{\sin\,\theta}\frac{\partial A}{\partial r}-\frac{A}{r\sin\,\theta}=\frac{\partial F^{\theta}}{\partial r}-\frac{1}{r}F^{\theta}$$

 

[stuxk]

espen180 thanks for the reply. I'm reading a paper and the thing I'm trying to do is to understand how the gradient tensor of the before mentioned vector field is derived. That means that I have the solution for the gradient tensor but not the derivation...

I must say I'm still a bit confused. The vector field $$\vec{F}$$ represents the magnetic field due to some currents in the atmosphere in spherical coordinates. Therefore I think you are right that it should be a contravariant vector. On the other hand, consulting the book "Mathematical Methods for Physicists" by Arfken the covariant derivative of a contravariant tensor is given by:

$$\nabla_qF^p=\partial_qF^p+\Gamma^p_{qk}F^k$$​

So my first question is why do you have $$\nabla_qF^p=\partial_qF^p-\Gamma^p_{qk}F^k$$ ?

Furthermore, If I calculate the element with p=2, q=1 as you nicely showed I get:

$$\nabla_rF^{\theta}=\frac{1}{\sin\,\theta}\frac{\partial A}{\partial r}+\frac{A}{r\sin\,\theta}=\frac{\partial F^{\theta}}{\partial r}+\frac{1}{r}F^{\theta}$$​

But, checking with the already calculated gradient tensor in that paper, for this element I have:

$$\frac{1}{\sin\theta}\frac{\partial A}{\partial r} = \frac{\partial F^\theta}{\partial r}$$​

So I don't get why I have this extra term $$\frac{1}{r}F^{\theta}$$ or $$-\frac{1}{r}F^{\theta}$$ in your case.

Note that in that paper in order to calculate the gradient tensor he uses the formula:

$$\nabla_qF^p=\nabla x^q(\frac{\partial F_q}{\partial x^p}-\Gamma^k_{pq}F_k)\nabla x^p$$​

in which I'm not sure whether I understand why we multiply with the basis $$\nabla x^q , \nabla x^p$$. Is it because we use the covariant vector instead of the contravariant vector ?.

 

[espen180]

espen180 thanks for the reply. I'm reading a paper and the thing I'm trying to do is to understand how the gradient tensor of the before mentioned vector field is derived. That means that I have the solution for the gradient tensor but not the derivation...

I must say I'm still a bit confused. The vector field $$\vec{F}$$ represents the magnetic field due to some currents in the atmosphere in spherical coordinates. Therefore I think you are right that it should be a contravariant vector. On the other hand, consulting the book "Mathematical Methods for Physicists" by Arfken the covariant derivative of a contravariant tensor is given by:

$$\nabla_qF^p=\partial_qF^p+\Gamma^p_{qk}F^k$$​

So my first question is why do you have $$\nabla_qF^p=\partial_qF^p-\Gamma^p_{qk}F^k$$ ?

Haha. It turns out that I started with the wrong formula but made a sign error, so the ruesult is still valid.

The term $$-\frac{1}{r}F^{\theta}$$ comes from this mess:

$$-\Gamma^{\theta}_{rr}F^r-\Gamma^{\theta}_{r\theta}F^{\theta}-\Gamma^{\theta_{r\phi}F^{\phi}$$ which is actually the sum $$-\Gamma^{\theta}_{rk}F^k$$ written out using the Einstein summation convention. Anyway, there is a lone nonzero term in there, inside the middle Christoffel symbol:

$$\frac12g^{\thetax}\left( \partial_r g_{\theta x} + \partial_{\theta}g_{rx} - \partial_xg_{r\theta}\right)$$

which a whole 'nouther sum itself.

In my previous post I listed all the nonzero derivatives of the metric for spherical coordinates. Just by looking at the this expression you can see the the only nonzero term will reveal itself when x=θ (the inverse metric multiplying the whole thing, plus that only metric elements on the diagonal have nonzero derivatives are two things which can lead to this conclusion), so plug in x=θ. Now, the only nonzero term inside the parenthesis will be the left one, so we are left with

$$-\Gamma^{\theta}_{rk}F^k=-\frac12g^{\theta\theta}\partial_rg_{\theta\theta}F^{\theta}=-\frac{A}{r\sin\,\theta}=-\frac{1}{r}F^{\theta}$$


Regarding the other covariant derivative formula you posted, I have never seen it before. The one I used is the one found in Shaum's Outline Series: Vector Analysis & Introduction to Tensor Analysis.