If I use the spacetime geometry for the Reissner-Nordström case and try to see the black hole's effect on an accelerated charged observer nearby, can I use the covariant equation for a charge moving in a general geometry and EM field? i.e.

$$\frac{d^2 x^i}{ds^2} + \Gamma^i_{kl} \frac{dx^k}{ds} \frac{dx^l}{ds} = \frac{e}{m_o c^2} F^i_k \frac{dx^k}{ds}$$

I have a little hunch about it since the RN Black hole would already "have" the $F_{ik}$ (EM filed tensor) taken account of and I am kind of using it twice.

Should I be replacing the given $x^i$ coordinates $(ct, r)$ with the Rindler coordinates using the usual transformation?


1 Answer 1


The equation of motion for a charged particle is (in natural units)

$$ \rm \frac{d^2 x^{\mu}}{d \tau^2} = -\sum_{\alpha, \beta} \ \left( \Gamma^{\mu}_{\alpha \beta} \ u^{\alpha}u^{\beta} +q \ F^{\mu \beta} \ u^{\alpha} \ g_{\alpha \beta} \right) $$

see here, so if you have a neutral test particle with charge $ \rm q=0 $ you only need the metric tensor, but if $ \rm q \neq 0 $ you also need the Maxwell Tensor $ \rm F $, see here for the covariant and contravariant versions.

The components of the proper acceleration

$$ \rm a = \surd \ |\sum_{\alpha, \beta} g_{\alpha \beta} \ a^{\alpha} a^{\beta}| $$

felt by the particle or measured with an accelerometer are given by

$$ \rm a^{\mu}= \frac{d^2 x^{\mu}}{d \tau^2}+\sum_{\alpha, \beta} \ \Gamma^{\mu}_{\alpha \beta} \ u^{\alpha}u^{\beta} $$

  • $\begingroup$ Yeah thanks for the correction in my Equation above. But the question still remains, my test particle is charged + accelerated. How should I go about taking acceleration into account? $\endgroup$
    – mnuizhre
    Jan 12 at 6:20
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    $\begingroup$ I updated the answer, the force F is simply ma with m being the test particle's rest mass. $\endgroup$
    – Yukterez
    Jan 12 at 18:07
  • $\begingroup$ While your equation is correct but to look it, its the same what I have written in my question. what I need to know is how can I modify this equation for a charged particle which is accelerating to begin with! $\endgroup$
    – mnuizhre
    Jan 16 at 13:51
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    $\begingroup$ If the force is only due to the charge of the particle and the black hole it is simply F=ma from equation 2, and if you also have a rocket propulsion or some other additional force that accelerates your particle besides gravity and the electromagnetic force you sum up the four-components of the additional acceleration and add it to the sum in equation 1 to get the new d²xᵘ/dτ² $\endgroup$
    – Yukterez
    Jan 17 at 3:32
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    $\begingroup$ For example, if you want your particle to hover at a constant distance over the black hole you choose it's charge q so that the second term in the sum of equation 1 cancels out the first term, and then calculate the force that particle would feel. That is also the same force an uncharged particle would feel if the force needed to hover would come from a rocket instead of the charge repulsion. If you have more forces you just add up their four-components like you add the term qFᵘᵇuᵃgₐᵦ to get d²xᵘ/dτ², or the other way around, depending on whether your input is the force or the trajectory. $\endgroup$
    – Yukterez
    Jan 17 at 3:49

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