# Equation of motion for a charged particle

To compute the equations of motion of a neutral testparticle in the graviational field, one needs the metric tensor $g_{\mu \nu}$ and $g^{\mu \nu}$ to compute the Christoffel-symbols

$${\Gamma^{\rm i}_{\rm j k} = \sum _{\rm s=1}^4 \ \frac{{{g}}^{\rm i s}}{2} \left(\frac{\partial {g}_{\rm s j}}{\partial {\rm x^k}}+\frac{\partial {g}_{\rm s k}}{\partial {\rm x^j}}-\frac{\partial {g}_{\rm j k}}{\partial {\rm x^s}}\right)}$$

and get the coordinate acceleration of the test particle:

$${{\rm \ddot x^{i} = -\sum _{j=1}^4 \sum _{k=1}^4 \ \dot x^j \ \dot x^k \ \Gamma^{i}_{j k}}}$$

But in the vicinity of a charge, there is not only the metric tensor, which might look like this

$$g_{\mu \nu}=\left( \begin{array}{cccc} g_{\rm t t} & 0 & 0 & g_{\rm t \phi} \\ 0 & g_{\rm r r} & 0 & 0 \\ 0 & 0 & g_{\rm \theta \theta} & 0 \\ g_{\rm t \phi} & 0 & 0 & g_{\rm \phi \phi} \\ \end{array} \right)$$

where the $g$-components are functions of the coordinates and mass, charge and spin, but also a Coulomb-potential, which looks like

$$\rm A_{\alpha}=(Q/r,0,0,0)$$

How is the Coulomb-potential plugged into the geodesic equation? The metric tensor is a $4 \times 4$ matrix, but the Coulomb potential seems to be $1 \times 4$, how is this added together to find the geodesics $\rm \ddot x^i$ of a charged particle of charge $\rm q$ in the vicinity of a dominant mass of charge $\rm Q$?

On Wikipedia I found the equation

$${{\rm \ddot x^i = - \Gamma^i_{j k}{\dot x^j}{\dot x^k}\ +\frac{q}{m} \ {F^{i k}} \ {\dot x^j}} \ {g_{\rm j k}}}$$

but there is no definition of ${\rm F^{i k}}$, which is, as Chrisoph and Eddy mentioned, the electromagnetic tensor. But what is the electromagnetic tensor for the Kerr-Newman metric in spherical (Boyer Lindquist) coordinates?

Edit: thanks for the answears, that's what I got so far: Screenshot

• I edited the title: the equation of motion is not a geodesics when an EM field is present and the test particle is charged – magma Sep 21 '17 at 17:30
• Since you seem to be a Mathematica user, I suggest you download the [xAct]( xact.es) suite of packages for tensor calculi . – magma Sep 21 '17 at 17:35

The resulting equation of motion is as follows:

$${d^2 x^\mu \over ds^2} =- \Gamma^\mu {}_{\alpha \beta}{d x^\alpha \over ds}{d x^\beta \over ds}\ +{q \over m} {F^{\mu \beta}} {d x^\alpha \over ds}{g_{\alpha \beta}}.$$

with

$${g_{\alpha \beta}}{d x^\alpha \over ds}{d x^\beta \over ds}=-1.$$

Here, $F^{\mu\nu}$ denotes the electromagnetic field tensor given by

$$F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$$

in it's incarnation with lowered indices.

Regarding your edit, in Boyer-Lindquist coordinates, $A$ is given by

$$A = \frac{Qr}{\rho^2}(dt-a\sin^2\theta d\phi) + \frac{1}{\rho^2}P\cos\theta \left(a dt -(r^2+a^2)d\phi\right)$$

where

$$\rho^2(r,\theta) = r^2+a^2 \cos^2 \theta$$

according to this paper I just googled.

• the first part of this answer has become obsolete as OP has sneakily edited his question while I was writing my post... – Christoph Sep 20 '17 at 9:43
• I mentioned you comment, thanks so far but how does the electromagnetic tensor look in spherical coords, Wikipedia just gives the Minkowski form – Yukterez Sep 20 '17 at 9:47
• I think there is an error on Wikipedia, the indicies of F and g need to be shifted in the equation for x" hen they give the Coulomb law at far distances,otherwise you ge some nasty Csc[2θ] term in the equation för φ" (whis is infinite at the equator) – Yukterez Sep 24 '17 at 17:12

$F$ is the Electromagnetic tensor

Edit: this answer was posted in answer to the original question which can be paraphrased as 'I found this equation on Wikipedia, what does this term mean,' and it is clear from the OPs subsequent edit that this helped narrow down what they actually wanted to know.

• This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review – stafusa Sep 20 '17 at 10:07
• It was an answer to the question when I posted it. The question was nicely summed up at the end with 'I found this thing on wiki, but I don't know what this bit is', so I told him what that bit was. Definitely answered the question present at the time. If that had solved his problem, then had I posted it in the comments the question would have remained unanswered and no longer needed. – Eddy Sep 20 '17 at 10:10
• Hi Eddy, indeed, I'm considering the current question. – stafusa Sep 20 '17 at 10:12
• It's unreasonable for you to expect me to answer a question which is not present... If you want to take down my answer now the question has changed be my guest. – Eddy Sep 20 '17 at 10:14
• I agree. Someone else flagged you answer and I voted based on the current situation (question & answer). I'm no moderator, so I don't decide anything. If you'd like to increase the chances of your answer not being deleted, you might consider editing it, either to clarify that there was a change, or in order to address the "new" question. – stafusa Sep 20 '17 at 10:19