Ex 2 solved in NR assignment 2.

Assignments/Assignment_2/Assignment_2.tex

@@ -68,23 +68,31 @@ Therefore,~$\nabla_a \nabla^a \phi = 0$ gives
 \begin{align*}
 - (1+2H) \p_t^2 \Phi + 4H \p_r \p_t \Phi + (1-2H) \p_r^2 \Phi - \frac{2H}{r} \p_t \Phi + \frac{2H}{r} \p_r \Phi - \frac{2 H}{r^2} \Phi = 0 \, .
 \end{align*}
-Taking~$A \equiv - (1+2H)$,~$B \equiv 4H$ and~$C \equiv (1-2H)$, we get~$AC - B^2 = - (1+2H) (1-2H) - (4H)^2 = - 1 < 0$ for all~$r$. Therefore, this equation iss hyperbolic both inside and outside the horizon at~$2H = 1$, or~$r = 2M$.
+Taking~$A \equiv - (1+2H)$,~$B \equiv 4H$ and~$C \equiv (1-2H)$, we get~$AC - B^2 = - (1+2H) (1-2H) - (4H)^2 = - 1 < 0$ for all~$r$. Therefore, this equation is hyperbolic both inside and outside the horizon at~$2H = 1$, or~$r = 2M$.
 
 Let,~$\p_t \Phi \equiv - \Pi$ and~$\p_r \Phi \equiv \Psi$. This gives the following system of eqations
 \begin{align*}
 & \p_t \Phi = - \Pi \, , \\
 & \p_t \Psi + \p_r \Pi = 0 \, , \\
-& \p_t \Pi + \frac{4H}{1+2H} \p_r \Pi - \frac{1-2H}{1+2H} \p_r \Psi = \frac{2H}{r} (\Pi + \Psi) - \frac{2H}{r^2} \Phi \, .
+& \p_t \Pi - \frac{4H}{1+2H} \p_r \Pi + \frac{1-2H}{1+2H} \p_r \Psi = - \frac{2H}{r} (\Pi + \Psi) + \frac{2H}{r^2} \Phi \, .
 \end{align*}
 This gives
 \begin{align*}
 \mathbf{A} = \left( \begin{array}{ccc}
 0 & 0 & 0 \\
 0 & 0 & 1 \\
-0 & - \frac{1-2H}{1+2H} & \frac{4H}{1+2H}
-\end{array} \right)
+0 & \frac{1-2H}{1+2H} & - \frac{4H}{1+2H}
+\end{array} \right) \, .
+\end{align*}
+Eigenvalues of this matrix are~$0$,~$\frac{- 4H \pm \sqrt{16 H^2 + 4(1-4H^2)}}{2(1 + 2H)}$, or~$0$,~$-1$ and~$\frac{1-2H}{1 + 2H}$, and so the nontrivial characteristic speeds are~$c_1 = -1$ and~$c_2 = \frac{1-2H}{1+2H}$.
+
+A radial null geodesic in the Kerr-Schild metric given above is given by~$- (1 - 2H) dt^2 + 4 H dt dr + (1 + 2H) dr^2 = 0$, or~$- (1 - 2H) + 4 H u + (1 + 2H) u^2 = 0$, where~$u \equiv dr/dt$. Solutions of this quadratic equation are the characteristic speeds in Kerr-Schild coordinates~$u = -1, \frac{1-2H}{1+2H}$. Integrating~$dr/dt = -1$ gives~$t + r = constant$.
+
+Integrating~$\frac{dr}{dt} = \frac{1-2H}{1+2H}$ gives
+\begin{align*}
+t = \int \frac{r+2M}{r-2M} dr + const = r + 4M \int \frac{dr}{r-2M} + const = r + 4M \ln |r - 2M| + const \, ,
 \end{align*}
-Eigenvalues of this matrix are~$0$,~$\frac{4H \pm \sqrt{16 H^2 - 4(1-4H^2)}}{2(1 + 2H)}$, or~$0$,~$-1$ and~$\frac{1-2H}{1 + 2H}$.
+which gives~$t - r = 4M \ln |r - 2M| + constant$. The name outgoing is misleading as both characteristic speeds are negative for~$r < 2M$.~$\blacksquare$
 
 \item Consider the equation $a u_{xx} + 2b u_{xy} + c _{yy} = f(u, \p u, x,y)$, where $u = u(x,y)$ and the subscripts denote partial derivatives. For what value of $a$, $b$ and $c$ is this equation elliptic, parabolic and elliptic PDEs, give reason: (a). $a = b = c = 1$, (b). $a = c = 1$, $b = 0$, (c) $a = 1$, $b = 0$, $c = -1$. (3 marks) \\