# Constructing the electromagnetic tensor from a four-potential

1. Jul 8, 2014

### coleman123

*Edit: I noticed I may have posted this question on the wrong forum... if this is the case, could you please move it for me instead of deleting? thanks! :)

Hello, I am having problems on building my electromagnetic tensor from a four-potential. I suspect my calculations are not right. Here are the steps:

My metric is (coordinates are t,r,z,ø):

$\left( \begin{array}{cccc} e^{2 \psi (r,z)} & 0 & 0 & 0 \\ 0 & -e^{2 \gamma (r,z)-2 \psi (r,z)} & 0 & 0 \\ 0 & 0 & -e^{2 \gamma (r,z)-2 \psi (r,z)} & 0 \\ 0 & 0 & 0 & -e^{-2 \psi (r,z)} r^2 \\ \end{array} \right)$

Then I have a magnetic vector potential in spherical coordinates (the coordinates are r,θ,ø):

$\left\{0,0,\frac{4 \pi M \sin (\theta )}{3 r^2}\right\}$

I convert it to cylindrical coordinates r,z,ø:

$\left\{0,0,\frac{4 \pi M r}{3 \left(r^2+z^2\right)^{3/2}}\right\}$

I build my contravariant four-potential (the electric potential is zero):

$\left\{0,0,0,\frac{4 \pi M r}{3 \left(r^2+z^2\right)^{3/2}}\right\}$

I will use this formula to calculate the covariant electromagnetic tensor components:

(1)

So I need the covariant four-potential, which I obtain by multiplying the contravariant four-potential by the covariant metric tensor, and performing a summation I get:

$\left\{0,0,0,-\frac{4 \pi M r^3 e^{-2 \psi }}{3 \left(r^2+z^2\right)^{3/2}}\right\}$

Using (1) I calculate the covariant electromagnetic tensor, and then, raising indexes, the contravariant one is:

$\left( \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & \frac{4 e^{2 \psi (r,z)-2 \gamma (r,z)} M \pi \left(2 r \left(r^2+z^2\right) \psi ^{(1,0)}(r,z)-3 z^2\right)}{3 \left(r^2+z^2\right)^{5/2}} \\ 0 & 0 & 0 & \frac{4 e^{2 \psi (r,z)-2 \gamma (r,z)} M \pi r \left(3 z+2 \left(r^2+z^2\right) \psi ^{(0,1)}(r,z)\right)}{3 \left(r^2+z^2\right)^{5/2}} \\ 0 & -\frac{4 e^{2 \psi (r,z)-2 \gamma (r,z)} M \pi \left(2 r \left(r^2+z^2\right) \psi ^{(1,0)}(r,z)-3 z^2\right)}{3 \left(r^2+z^2\right)^{5/2}} & -\frac{4 e^{2 \psi (r,z)-2 \gamma (r,z)} M \pi r \left(3 z+2 \left(r^2+z^2\right) \psi ^{(0,1)}(r,z)\right)}{3 \left(r^2+z^2\right)^{5/2}} & 0 \\ \end{array} \right)$

The notation above, next to my metric functions $\psi$ and $\gamma$, indicate derivatives:

(1,0) is a derivative in r

(0,1) is a derivative in z

I was not expecting my metric functions $\psi$ and $\gamma$ to appear on my contravariant electrogmagnetic tensor. I would like a help if I did all the steps right.

If I just take the curl of the magnetic vector potential to find the magnetic field, and plug in the expected components on the electromagnetic tensor, I get a different result, and was expecting the same... that is what bothers me. I'll show this now.

Calculating the magnetic field from the magnetic vector potential in cylindrical coordinates results in:

$\left\{\frac{4 \pi M r z}{\left(r^2+z^2\right)^{5/2}},-\frac{4 \pi M \left(r^2-2 z^2\right)}{3 \left(r^2+z^2\right)^{5/2}},0\right\}$

I can then plug the Br and Bz components directly in the electromagnetic field tensor which is defined as (my c equals 1, and instead of x,y,z I have r,z,ø):

So when calculating the electromagnetic tensor this other way, I obtain a different result, without my metric functions $\psi$ and $\gamma$.

What is happening? Thanks in advance for any clarifications on this issue..

Last edited: Jul 8, 2014
2. Jul 10, 2014

### Matterwave

It's pretty hard to figure out exactly what's happening because the calculations are long and tedious. Even the results are long and tedious. I have a few suggestions on where one might start looking.

1) Check if when you raise an lower indices, that you get the matrix multiplication order correct, unless of course, you do everything manually by making the summations.

For example:
$$F^{\mu\nu}=g^{\mu\rho}g^{\nu\tau}F_{\rho\tau}$$

In matrix multiplication this would be $g^{-1}Fg^{-1}$ and not for example $g^{-1}g^{-1}F$.

2) Make sure your inverse metric tensor matrix is correct.

3) Make sure your change of coordinates matrices are correct.

4) As above, make sure your change of coordinates matrix multiplication is correct.

5) Make sure you are taking the curl in cylindrical coordinates, with the correct coefficients in front of the derivative operators.

These are some common mistakes. Have you tried checking these?

3. Jul 10, 2014

### coleman123

Hello, thanks for the help. I did everything in Mathematica, so I think the results are ok, unless I messed up with the indexes.

I heard that I must take the Curl using covariant derivatives... is this correct?

Last edited: Jul 10, 2014
4. Jul 10, 2014

### WannabeNewton

No you don't need to use covariant derivatives. In a freely falling frame the curl just takes on the usual form from vector calculus and electrodynamics. In an arbitrary coordinate system the curl of a vector field $\xi^{\alpha}$ is given by the quantity $\omega^{\alpha} = \epsilon^{\alpha\beta\gamma\delta}\xi_{\beta}\nabla_{\gamma}\xi_{\delta} = \epsilon^{\alpha\beta\gamma \delta}\xi_{\beta}\partial_{\gamma}\xi_{\delta}$ so you can use $\nabla_{\mu}$ or $\partial_{\mu}$ (this is assuming a torsion-free connection as is usual in GR). Here $\epsilon^{\alpha\beta\gamma\delta}$ is the 4-dimensional Levi-Civita tensor density (not symbol!). In an arbitrary Lorentz frame this same equation for the curl holds but since we are now working in an orthonormal basis, $\epsilon^{\hat{\alpha}\hat{\beta}\hat{\gamma}\hat{\delta}}$ is just the Levi-Civita symbol.

5. Jul 10, 2014

### coleman123

Alright, since the equation (1) is the Curl, I must do it in cylindrical coordinates. That is where I am making the mistake... If I follow (1) the way it is, it gives the Curl in Cartesian coordinates.

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6. Jul 10, 2014

### WannabeNewton

That paper calculates things in an orthonormal basis. Your original calculation involves a coordinate basis. You have to convert your results from one to the other depending on which you wish to work with. In your case this just involves a simple rescaling of the coordinate basis by appropriate factors of the metric in your original coordinates since the coordinate basis is mutually orthogonal. See my comment in post #4 about $\epsilon^{\alpha\beta\gamma\delta}$ as a tensor density vs. a symbol in the expression for the curl.

7. Jul 10, 2014

### coleman123

Do you have a book or site reference on that WannabeNewton? Thanks for the help.

8. Jul 10, 2014

### WannabeNewton

See Appendix J of Carroll "Space-time and Geometry", Appendix B of Wald "General Relativity", and especially most importantly section 8.4 of MTW.

9. Jul 19, 2014

### coleman123

I tried to apply:

$\omega^{\alpha} = \epsilon^{\alpha\beta\gamma\delta}\xi_{\beta}\nabla_{\gamma}\xi_{\delta} = \epsilon^{\alpha\beta\gamma \delta}\xi_{\beta}\partial_{\gamma}\xi_{\delta}$

But got zeroes... are you sure it is the correct expression? Maybe was my Levi-Civita tensor.