Where does a radially falling test particle cross the inner horizon of RN BH?

[LAHLH]

Hi,

If a particle falls radially into a Reissner-Nordstrom BH with Q<M (i.e the case where you have an inner and and out horizon), then where in the conformal diagram does the test particle cross the inner horizon $$r_{-}$$?

See for example the righthand Penrose diagram of http://tinyurl.com/6e54fu2 to see what I'm talking about. This Penrose diagram shows the particle crossing to the right of where the horizon crossing occurs, not directly through the cross X nor to the left of it. I would like to able to see why it doesn't go directly through the centre of the X or somewhere else.

I believe one can analyse this in Schwarzschild type coords in the region between the outer and inner horizons, for radial timelike geodesics of RN we have:

$$\frac{1}{2} \dot{r}^2+V(r)=\frac{E^2}{2}$$

with $$V(r)=\left[\frac{1}{2}-\frac{M}{r}+\frac{Q^2}{2r^2}\right]$$

This potential goes to zero at both horizons and tends to positive infinity at r goes to zero.

It's also true that $$\frac{dt}{d\tau}=\frac{E}{1-\frac{2M}{r}+\frac{Q^2}{r^2}}$$, is going to go to infinity as you approach the horizons (just like it does when you approach r=2M in Schw), but this doesn't mean you can't cross the horizon, it's an illusion and just like in Schw if you transform into the equivalent of ingoing Eddington Finkelstein there is no problem crossing the horizon.

But I have no idea how any of this tells me where it crosses the inner horizon in the sense I mentioned above?

 

[PeterDonis]

I have no idea how any of this tells me where it crosses the inner horizon in the sense I mentioned above?

That's because it doesn't. A radially infalling particle can cross anywhere on the inner horizon; there are radial timelike geodesics through every point on the horizon. Where a given particle actually crosses depends on the initial conditions of its motion, i.e., the conditions that determine which particular geodesic is its worldline.