# Charge of a particle is running when getting closed to charged black hole horizon

Given some particle with a mass $$m$$ and charge $$q$$ in the background of some charged black hole with a mass $$M$$ and charge $$Q$$ given for example by the Reissner Nordström metric. Is it reasonable that the particle's charge $$q$$ will flow as the location of the particle will reach the outer horizon $$r_+$$ of the charged black hole, even in tree-level calculation? So in particular, $$q$$ will be a running coupling constant as a function of $$r$$, $$q \equiv q(r)$$

At the level of effective action, we can think for simplicity on complex scalar field $$\chi$$ with mass $$m$$ and charge $$q$$ under the background metric: $$ds^2= -f(r)dt^2+f^{-1}(r)dr^2+r^2d\theta^2+r^2sin^2\theta d\phi^2$$ with $$f(r)=1-\frac{2M}{r}+\frac{Q^2}{r^2}$$ so the resulting action would be: $$S=\int d^4x\sqrt{-g}\left[R-\frac{1}{4}g^{\alpha\mu}g^{\beta\nu}F_{\alpha \beta }F_{\mu\nu}+g^{\mu\nu}\partial_{\mu}\chi\partial_{\nu}\chi^*+iq(r)g^{\mu \nu}\partial_{\mu}\chi A_{\nu}\chi^*-iq(r)g^{\mu \nu}\partial_{\mu}\chi^* A_{\nu}\chi + q^2(r)g^{\mu \nu }A_\mu A_\nu \chi \chi^* \right]$$

One reason I thought it possible for this to happen is that in the case of an extremal black hole $$Q=2M$$, and when $$q>2m$$, there is a chance that near the horizon the particle will fall into the black hole, causing it to absorb the particle and now to have a higher charge: $$Q_{new}= Q + q > 2M_{new} = 2M + 2m$$ which is a contradiction to the physics of black holes. This is why in particular: $$q(r_+)=0$$ So $$q$$ has to flow between its value on the horizon and its value in flat infinity $$q_{\infty}=\lim_{r\to\infty}q(r)$$

• Charge is conserved in the black hole physics. Your question describes a personal theory contradicting the conservation of charge. Commented Dec 26, 2023 at 16:37
• in your example the electric repulsion is stronger than the gravitational attraction, so the particle simply flies away. Commented Dec 27, 2023 at 9:23