 Solution

With $\tau$ the proper time of the massive test particle, the four-velocity has squared norm $-1$:The conserved quantities in part (i) giveMultiplying the normalization by $F$ and substituting gives the charged-particle radial equation in Reissner-Nordstrom spacetime:At a radial turning point $r_{\ast }>0$, the left side is zero, soThe parenthesis on the left is strictly positive. Hence $F(r_{\ast })\ge 0$. The factorization in part (b) now yieldsThere is no radial turning point between the horizons. Indeed, for $r_{-}<r<r_{+}$, the right-hand side of the radial equation is strictly positive, independently of the electric charge $e$ of the test particle. This is the charged-particle turning-point exclusion in Reissner-Nordstrom spacetime.

The inequalities are necessary, not a claim that a radial turning point exists for every $E,l,e$. At a regular horizon crossing a timelike vector cannot be tangent to a null hypersurface, so a massive test particle actually has $\dot{r}\ne 0$ there. Algebraic equality at $r_{\pm }$ would require $E=eQ/r_{\pm }$ and does not describe a regular finite-proper time radial turning point on the horizon. In the Schwarzschild spacetime limit $Q=0$, the positive-radius inner alternative disappears.