0$ is the energy taken by the string. Since we are interested in the tunnelling process of the string through the event horizon, and for that purpose, we only need to evaluate the imaginary part of $W(r,\theta)$. By inserting the above ansatz into the Hamilton-Jacobi equation, we obtain \be \frac{\partial W}{\partial r} = H_1^{\frac{1}{2}} F_1^{-1} \sqrt{(\delta E)^2 - \left( 2\pi T_s r \sin\theta \right)^2 \frac{H_2 F_1}{H_1} - \frac{F_1 p_\theta^2}{H_1 r^2}}, \ee for the outgoing string. Here we assume that the angular momentum $p_\theta$ is not divergent around the event horizon. Thus around the horizon, by using $F_1 \sim 0$, the function of $W(r,\theta)$ is approximately solved as \be W(r,\theta) \sim \delta E \int dr \, H_1^{\frac{1}{2}} F_1^{-1} = \frac{\delta E}{2\pi T} \int \frac{d\rho}{\rho}. \ee Here we made a coordinate transformation $d\rho = H_2^{\frac{1}{4}} F_1^{-\frac{1}{2}} dr$, which gives $H_1^{-1} H_2^{\frac{1}{2}} F_1 \sim \frac{1}{4} H_1^{-1} (\frac{dF_1}{dr})^2 \big|_{r_\text{horizon}} \rho^2 = 4\pi^2 T^2 \rho^2$. This procedure is similar to the derivation of the Hawking temperature $T$ in the eq.~(\ref{eq:Tsmear}), so we find it in the last equality. The region of the integral is chosen from inside to outside the horizon. The imaginary part of $W(r,\theta)$ gives the tunnelling probability of the string $P_s$, which is evaluated as \be P_s = e^{-2 \text{Im} W} = e^{-\frac{\delta E}{T}} = e^{-|\delta S|}. \ee We used the first law of the black hole thermodynamics for $\delta Q = 0$. $\delta S < 0$ is a difference of the entropy of the smeared quantum black 0-brane due to the radiation of the string. ]]>

0$ is the energy taken by the D0-brane. we consider the outgoing D0-brane through the event horizon. By inserting the above ansatz into the Hamilton-Jacobi equation, the function $W(r)$ is solved around the horizon as \be W(r) = \int dr \, H_1^{\frac{1}{2}} F_1^{-1} \sqrt{ \left( \delta E - T_0 C^{(1)}_t \right)^2 - T_0^2 H_1^{-1} H_2^{-1} F_1 } \sim \frac{\delta E - T_0 \Phi}{2\pi T} \int \frac{d\rho}{\rho},\qquad \ee where $\Phi$ is the electric potential. Here, as in the previous subsection, we used $d\rho = H_2^{\frac{1}{4}} F_1^{-\frac{1}{2}} dr$ and $H_1^{-1} H_2^{\frac{1}{2}} F_1 \sim \frac{1}{4} H_1^{-1} (\frac{dF_1}{dr})^2 \big|_{r_\text{horizon}} \rho^2 = 4\pi^2 T^2 \rho^2$. The Hawking temperature is given by the eq.~(\ref{eq:Tsmear}). The region of the integral is chosen from inside to outside the horizon. Finally WKB approximation of the tunneling probability through the even horizon is estimated by the imaginary part of $W(r)$, and we obtain \be P = e^{-2\,\text{Im}\,W} = e^{- \frac{\delta E - T_0 \Phi}{T}} = e^{-|\delta S|}. \ee Note that $\delta Q = T_0$ for the radiation of the test D0-brane. We used the first law of the black hole thermodynamics in the last equality. $\delta S < 0$ is a difference of the entropy of the smeared quantum black 0-brane due to the radiation of the D0-brane. For the near horizon limit, $\delta S \sim N$ and the tunnelling process is quite suppressed for large $N$. ]]>