How do I properly construct the electromagnetic tensor in curved space-time? I have my curved spacetime metric $(+,-,-,-)$ and my magnetic vector potential $A$. I tried two ways but not sure which is right (if there is one).

First way:

  1. Compute the magnetic field $B$ from the curl of the magnetic vector potential $A$: $$ \mathbf{B} = \nabla \times \mathbf{A}. $$

  2. Place the resulting components directly in the contravariant electromagnetic tensor definition in cylindrical coordinates: $$ F^{\mu\nu} = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & B^z \\ 0 & 0 & 0 & -B^r \\ 0 & -B^z & B^r & 0 \end{pmatrix}. $$

Second way:

  1. Define the electromagnetic four-potential ($\phi$ is zero in my problem): $$ A^\alpha = (\phi, \mathbf{A}). $$

  2. Lower the four-potential index by contracting it with my covariant metric tensor.

  3. Compute the electromagnetic field components with the formula $$ F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu. $$ I replaced the ordinary derivatives with covariant derivatives.

  4. Raise the indexes of this covariant electromagnetic field tensor to compare with the first way.

The problem is that I can't seem to have the same results with both methods, which says pretty clearly that I am doing something wrong. Is there something fundamentally wrong in taking these steps?


1 Answer 1


Your second method is correct.

To compare, say, the magnetic field with what you find in Jackson, you really need to realize that there's an assumption that you have unit basis vectors there, and that the cross product is actually a hodge dual (which will invoke factors of the square root of the determinant of the metric). These will make direct comparisons a bit tricky when going from one notation to the other.

Of course, ultimately, both methods will work. The second method is far more error proof and is also coordinate (and the only metric dependent step is the lowering of the index of $A^{\mu}$) independent.

(note that what I"m saying above is that it doesn't matter if you replace the ordinary derivatives with covariant derivatives, since:

$$\nabla_{a}v_{b} = \partial_{a}v_{b} - \Gamma_{ab}{}^{c}v_{c}$$,

which means that

$$\nabla_{a}v_{b} - \nabla_{b}v_{a} = \partial_{a}v_{b} - \partial_{b}v_{a} - \Gamma_{ab}{}^{c}v_{c} + \Gamma_{ba}{}^{c}v_{c} = \partial_{a}v_{b} - \partial_{b}v_{a}$$,

so they really are the same thing.)

  • $\begingroup$ So even if I am using cylindrical coordinates (my metric is Weyl), I can directly apply those partial derivatives of the second method e carry on with the calculations? Is there anything that I must do if I am not working with cartesian coordinates? $\endgroup$ Jul 16, 2014 at 20:00
  • $\begingroup$ @Giovanni Yes, that is the way to do it. But take care with raising and lowering when e.g. finding the source of the field - that's the part where the extra metric stuff enters. $\endgroup$
    – Void
    Jul 16, 2014 at 20:13
  • $\begingroup$ @Giovanni: yes, but see above about the hodge dual. Cross product doesn't generalize trivially to curved space the way that dot product does. It's all a matter of how you want to present the components. I would suggest starting with ordinary cartesian space in three coordinates and seeing what's necessary to reproduce the canned formulae in Griifiths/Jackson. $\endgroup$ Jul 16, 2014 at 20:13
  • $\begingroup$ I am not familiarized with the hodge dual yet, but I read somewhere that I could also apply the Curl using a the four-dimensional Levi-Civita tensor density. $\endgroup$ Jul 17, 2014 at 15:16

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