Radial Infall

Why do (shooting) stars fall down from the sky?

From the previous post, we know that $$\tau = \int \frac{dr}{\pm\sqrt{\tilde{E}^2 - \left(1-\frac{2M}{r}\right)\left(1+\frac{\tilde{L}^2}{r^2}\right)}}.$$

We study the simplified case with \(\dot{\phi}\) such that \(\tilde{L} = 0\), and \(\tilde{E} = 1\). Then, the proper time is \begin{align} \tau(r) &= \pm\int \frac{dr}{\sqrt{1-\left(1-\frac{2M}{r}\right)}}\\ &= \pm\int \sqrt{\frac{r}{2M}} dr \\ &= \pm\frac{2}{3\sqrt{2M}}r^{3/2} + \text{const} \end{align} Let \(r = r_0\) when \(\tau=0\). $$\bbox[5px,border:2px solid #666]{ \tau(r) = \frac{\pm 2}{3\sqrt{2M}}\left(r^{3/2} - r_0^{3/2}\right) }$$ The \(\pm\) sign is due to taking square root on the previous equation. If the testing particle is falling in, \(r\) decreases with time, then we should take a minus sign.

Up until now we have been using the proper time of the testing particle. Now we would like to see as an observer at infinity, i.e. time is equivalent to the flat spacetime. As we are describing infall, we take the negative sign and obtain $$\frac{d\tau}{dr} = -\sqrt{\frac{r}{2M}},$$ which implies $$\frac{d\tau}{dr} = \frac{dt}{dr}\frac{d\tau}{dt} = -\sqrt{\frac{r}{2M}}.$$ Substituting \(1 = \tilde{E} = \frac{dt}{d\tau}\left(1-2M/r\right)\), we have $$\frac{dt}{dr} = \sqrt{\frac{r}{2M}}\left(\frac{r}{2M - r}\right).$$ We can now do the integration \begin{align} \sqrt{2M}\frac{dt}{dr} &= \frac{r^{3/2}}{2M-r} = -r^{1/2} + \frac{2Mr^{1/2}}{2M-r}\\ \sqrt{2M} t(r) &= -\frac{2}{3}r^{3/2} - 4Mr^{1/2} + (2M)^{3/2}\ln\left|\frac{1+\sqrt{r/2M}}{1-\sqrt{r/2M}}\right| + \text{const}. \end{align} Let \(r=r_0\) when \(t=0\). $$\bbox[5px,border:2px solid #666]{ t(r) = \frac{1}{\sqrt{2M}}\left[-\frac{2}{3}(r^{3/2}-r_0^{3/2}) - 4M(r^{1/2}-r_0^{1/2}) + (2M)^{3/2}\left(\ln\left|\frac{1+\sqrt{r/2M}}{1-\sqrt{r/2M}}\right|-\ln\left|\frac{1+\sqrt{r_0/2M}}{1-\sqrt{r_0/2M}}\right|\right)\right] }$$

Plotting \(r\) against time, we can see that the testing particle is falling towards the source as time goes. From the point of view of a far away observer, the testing particle takes forever to fall through \(r=2M\), but actually, the testing particle has already fallen in. Clearly \(r=2M\) represents something special. We will see that the testing particle has taken a one-way journey and can never come back once it has cross the line \(r=2M\) later.