# How close does a charged particle have to be “attracted” to a charged black hole?

Let’s say see have a “mildly charged” black hole ($$r_q \ll r_s$$), and an electron or positive that has the same charge. While the charge to mass ratio of the black hole is orders of magnitude below the extremal limit, the particle still experiences an electrostatic repulsion orders of magnitude stronger than gravity.

Placed near the black hole, the particle will be repelled away. However once it’s inside the black hole, it will have to move towards the singularity.

So how close to the event horizon does the particle have to be before it will move towards the event horizon? Or is this not the case at all?

I tried having a look at the equations of motion for a charged black hole, but those just confused me, as they seem to imply that for an sufficiently light particle the exact mass of said particle won’t actually affect it’s trajectory to any non-negligible degree.

$$\ddot{r} = \frac{(r^2-2Mr+Q^2)(Q^2-Mr)\dot{t}^2}{r^5} + \frac{(Mr-Q^2)\dot{r}^2}{r(r^2-2Mr+Q^2)} + \frac{(r^2-2Mr+Q^2)\dot\theta^2}{r} + \frac{qQ(r^2-2mr+Q^2)}{r^4}\dot{t}$$

• Do you mean opposite charge rather than opposite mass there in your first line? Jul 29 at 19:09

For a particle initially at rest, so

$$\rm v=0 \to \dot{r}=\dot{\theta}=\dot{\phi}=0, \ \dot{t}=\sqrt{g^{tt}}$$

and in natural units of $$\text{G=M=c=}{\rm k_B}\text{=1}$$, with

$$\rm f=Q^2+r^2-2 \ r$$

the condition for the border between attraction and repulsion is

$$\rm \ddot{r}=\frac{f \ \left(q \ Q \ \sqrt{\frac{r^2}{f}}+\frac{Q^2 \ r-r^2}{f}\right)}{r^4}=0$$

Now we can solve for the specific charge $$\rm q$$ required to hover at a given $$\rm r$$:

$$\rm q=\frac{\left(r-Q^2\right) \sqrt{\frac{r^2}{f}}}{Q \ r}$$

or the stationary $$\rm r$$ for the given charges $$\rm Q$$ and $$\rm q$$:

$$\rm r= \frac{\pm \sqrt{q^4 \ Q^4-q^2 \ Q^4-q^4 \ Q^6+q^2 \ Q^6}-q^2 Q^2+ Q^2}{1-q^2 \ Q^2}$$

but the product of the charges needs to be larger than $$1$$ to give real solutions outside of the horizon (inside the horizon even neutral particles get repelled, see here).

If $$\rm q=1/Q$$ the stationary radius is at infinity (under Newton and Coulomb where $$\rm \ddot{r}=(Q \ q-1)/r^2$$ every $$\rm r$$ would do in that case).

blademan9999 wrote: "those just confused me, as they seem to imply that for an sufficiently light particle the exact mass of said particle won’t actually affect it’s trajectory to any non-negligible degree."

Reissner Nordström assumes that the mass $$\rm m$$ of the testparticle is neglible compared to the black hole's mass $$\rm M$$, see here for the experiment of the hammer and feather drop on the moon where the $$\rm m$$ of hammer and feather didn't matter either.

We only need the charge to mass ratio, so $$\rm m$$ is already included in the test particle's specific charge $$\rm q$$. The small $$\rm m$$ in your equation that you copied from Wikipedia is a typo and should be a large $$\rm M$$ by the way, the last stable version of that article is here, for an other source with the correct equations see here.

• Fantastic answer, Simon +1 Jul 30 at 3:32
• I get a deceptive website warning Error from that link Jul 30 at 6:17
• Applies to both mobile and desktop. Jul 30 at 11:58
• @blademan9999 I can't reproduce your problem, and since the safe browsing tests here, here, here, here or anywhere else also don't back up your claims I assume you must have some fake protection on your devices, otherwise you have to provide more information about what exactly happened on which site. Jul 31 at 6:03