I'm running into trouble understanding what is going wrong, fooling around with the maxwell equations in tensor form, and playing with Cartesian and spherical coordinates.

To be more clear, I'm trying to follow the derivation of the Reissner Nordstrom metric which I found here (starting page 25) The main differences are that I'm working in a (-+++) signature opposed to the article's (+---) signature. And the fact that I'm trying to do the derivation without ruling out magnetic monopoles. All is well until the beginning of page 29. Let me give some more context:

The electromagnetic tensor in the (-+++) metric signature. $$ F_{\mu \nu} = \begin{pmatrix} 0&-Ex&-Ey&-Ez\\ Ex&0&Bz&-By\\ Ey&-Bz&0&Bx\\ Ez&By&-Bx&0 \end{pmatrix} $$ Now in order to change these tensor components into the spherical ones we do the following new metric is $$ ds^2 = -dt^2 + dr^2 + r^2d\theta^2 + r^2 \sin^2(\theta) d\phi^2 $$ with $r \in [0,\infty)$, $\theta \in [0,\pi)$ and $\phi \in [0,2\pi)$

Then by using the following we should be able to rewrite $F_{\mu \nu}$'s components into spherical components $$ F_{\mu' \nu'} = \frac{dx^\mu}{dy^{\mu'}}\frac{dx^\nu}{dy^{\nu'}}F_{\mu \nu} $$ where $x^\mu$ are the cartesian coordinates $(t,x,y,z)$ and $y^\mu$ are the spherical coordinates $(t,r,\theta,\phi)$. According to my calculations this results in the following: $$ F_{\mu\nu}= \begin{pmatrix} 0 & -E_r & -rE_\theta & -r\sin{\theta}E_\phi \\ E_r & 0 & rB_\phi & -r\sin{\theta}B_\theta \\ rE_\theta & -rB_\phi & 0 & r^2\sin{\theta}B_r \\ r\sin{\theta}E_\phi & r\sin{\theta}B_\theta & -r^2\sin{\theta}B_r & 0 \end{pmatrix} $$ Again we maintain a (-+++) signature. This is where my problem starts to arise. The divergence in spherical coordinates is given by: $$ \nabla \cdotp \vec{B} = \frac{1}{r^2}\partial_r(r^2B_r) + \frac{1}{r\sin(\theta)}\partial_\theta(\sin(\theta)B_{\theta}) + \frac{1}{r\sin(\theta)}\partial_\phi B_\phi $$

And the following of the two Maxwell equations $$ F_{\alpha \beta ; \gamma} + F_{\beta \gamma ; \alpha} + F_{\gamma \alpha ; \beta} $$

Now in cartesian coordinates picking $\alpha = x$, $\beta = y$ and $\gamma = z$ this ofcourse yields the divergence of the B field $$ F_{xy; z} + F_{yz; x} + F_{zx; y} = \partial_z B_z + \partial_x B_x + \partial_y B_y = \nabla \cdotp \vec{B} $$ The same should happen when taking the components of $F_{\mu \nu}$ in spherical coordinates. Picking $\alpha = r$, $\beta = \theta$ and $\gamma = \phi$ \begin{align} F_{r \theta; \phi} + F_{\theta \phi; r} + F_{\phi r; \theta} &= \partial_\phi F_{r\theta} - \Gamma^m_{\phi r}F_{m\theta} - \Gamma^m_{\phi \theta} F_{r m}\\ &+ \partial_r F_{\theta \phi} - \Gamma^m_{r \theta}F_{m \phi} - \Gamma^m_{r \phi}F_{\theta m} \\ &+ \partial_\theta F_{\phi r} - \Gamma^m_{\theta \phi}F_{m r} - \Gamma^m_{\theta r}F_{\phi m} \\ &= \partial_\phi (r B_\phi) + \partial_r (r^2 \sin(\theta) B_r) + \partial_\theta (r \sin (\theta) B_\theta) \\ &=\sin(\theta)\partial_r (r^2 B_r) + r \partial_\theta (\sin(\theta)B_\theta) + r \partial_\phi B_\phi \\ &= r^2 \sin (\theta) (\frac{1}{r^2}\partial_r (r^2 B_r) + \frac{1}{r\sin(\theta)} \partial_\theta (\sin(\theta)B_\theta) + \frac{1}{r\sin(\theta)} \partial_\phi B_\phi ) \\ &= r^2 \sin (\theta) \nabla \cdotp \vec{B} \end{align} But I was just expecting to get $\nabla \cdot \vec{B}$ without the factor of $r^2 \sin(\theta)$. Somewhere along this thought process something is going wrong probably, but I really can't find where. Anyone who'd like to help? If my problem is in any way unclear, please let me know! Help would be greatly appreciated!

  • $\begingroup$ That is true, but if you calculate the christoffel symbols for this metric and put these in the equation, all terms with christoffel symbols in them cancel each other out. Should be modified to what? Or did you end that sentence on purpose? $\endgroup$ Commented Jan 23, 2017 at 21:19
  • $\begingroup$ Edited my original post to include the Christoffel symbols. But these cancel each other out. If I'm not mistaken $\endgroup$ Commented Jan 23, 2017 at 21:30
  • $\begingroup$ When you define your spacetime metric should you have some unknown functions for the metric tensor components? $\endgroup$ Commented Jan 23, 2017 at 22:22

2 Answers 2


Pushing forward along the projection to a constant time slice (this gets rid of $dt$ factors), the E-field is a 1-form while the B-field is a 2-form (in physics lingo, the B-field is a pseudovector). For 1-forms, the divergence is defined as $* d *$ with $*$ the Hodge star associated to the metric, but for 2-forms, the divergence is $* d$. In coordinates, $*$ on $p$-forms looks like $\sqrt{g}\cdot \epsilon^{i_1 \ldots i_p}_{i_{p+1}\ldots i_{n}}$, hence your extra factor of $\sqrt{g}$ you get by treating $B$ as a 1-form (the Maxwell equation you use).


I have not done the algebra you have, but it seems to me that you are confusing physical components of ${\bf B}$ and the co-ordinate components of ${\bf B}$. What I mean is more easily explained with the ${\bf E}$ field. The physical spherical components $E_r, E_\theta, E_\phi$ all have units of volts-per-meter. The co-ordinate components of ${\bf E}$ have the same names and symbols but $E_r$ has units of Volts-per-meter, while $E_\theta$ and $E_\phi$ have units of volts-per-radian (dimensionally the same as volts).

The spherical components you obtain from the tensor tranformation formula are the co-ordinate components --- but the formula you quote for the divergence is the one for the physical components. Books are rather bad at explaining the difference. It is touched on in section 8.1.1 of Stone and Goldbart "Mathematics for Physics."

If $$ ds^2= h_1^2 (dx^1)^2 + h_2^2 (dx^2)^2+h_3^2 (dx^3)^2$$ for example, the co-ordinate gradient components of an ${\bf E}$ field derived from a potential are $$ E_i = -\frac{\partial V}{\partial x^i} $$ while the physical components are $$ E_i= -\frac 1{h_i} \frac{\partial V}{\partial x^i}. $$ I'd have to think how this works for the magnetic field whose physical components all have dimension webers per meter squared as opposed to the coordinate components that presumably have units of webers per co-ordinate square.

  • $\begingroup$ It's noticible that the extra factor you have is precisely $\sqrt{g}$ for the spherical polars. So it part of converting co-ordiate measure $d^3x$ to the physical $ \sqrt{g} d^3x$. $\endgroup$
    – mike stone
    Commented Jan 24, 2017 at 23:07

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