# Radially falling into a Black Hole

Hi, I have the following problem:

Given the 5-D generalization of the Schwarszschild solution with line element:

$$ds^2=-\Bigg(1-\frac{r^2_+}{r^2}\Bigg)dt^2+\Bigg(1-\frac{r^2_+}{r^2}\Bigg)^{-1}dr^2+r^2[d\chi^2+\sin^2(\chi)(d\theta^2+\sin^2(\theta)d\phi^2)]$$

where $r_+$ is a positive constant. An observer falls radially starting from rest at $r=10r_+$. How much time elapses on their clock before they hit the singularity at $r=0$?

MY ATTEMPT HAS BEEN:

Using the Lemaitre coordinates $\tau$, $\rho$ to eliminate the singularity of $ds^2$ at $r_+$:

$$d\tau=dt+\frac{r_+}{r}\frac{dr}{1-\frac{r^2_+}{r^2}}$$

$$d\rho=dt+\frac{r}{r_+}\frac{dr}{1-\frac{r_+^2}{r^2}}\quad\quad\quad (1)$$

we have the following line element where the singularity at $r_+$ is removed:

$$ds^2=d\tau^2-\frac{r_+}{r}d\rho^2-r^2(d\theta^2+\sin^2(\theta)d\phi^2)$$

where $r=\sqrt{2(\rho-\tau)r_+}$, which is obtained by integrating $$d\rho-d\tau=\frac{r}{r_+}dr$$.

For a free falling body, $d\rho=0$, and equation (1) gives:

$$dt=-\frac{r}{r_+}\frac{1}{1-\frac{r_+}{r}}dr$$

Integrating this equation from $r=10r_+$ to $r=0$ should give me the time the problem asks for:

$$\Delta\tau=-\int_{10r_+}^0 \frac{r}{r_+}\frac{1}{1-\frac{r_+}{r}}dr$$

Is this correct?

Thanks!!!

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I have the feeling the last step, where I integrate $dt$ and get the time the observer takes to fall into the black hole, is not quite correct. But I'm not really sure. Could someone help me out? Thanks, I really appreciate it.

TSny
Homework Helper
Gold Member
Using the Lemaitre coordinates $\tau$, $\rho$ to eliminate the singularity of $ds^2$ at $r_+$:

$$d\tau=dt+\frac{r_+}{r}\frac{dr}{1-\frac{r^2_+}{r^2}}$$

$$d\rho=dt+\frac{r}{r_+}\frac{dr}{1-\frac{r_+^2}{r^2}}\quad\quad\quad (1)$$
Your expressions here don't appear to correspond to the usual definition of Lemaitre coordinates. You are missing some square roots. Also, should the $r$'s be squared in the denominators? See https://en.wikipedia.org/wiki/Lemaître_coordinates

[EDIT: Sorry. I overlooked the fact that the $r$'s are squared in the 5-D version of the metric. I'm not familiar with this. So, your expressions for the Lemaitre coordinates are probably OK.
However, I think there is a typo in your equation

$$ds^2=d\tau^2-\frac{r_+}{r}d\rho^2-r^2(d\theta^2+\sin^2(\theta)d\phi^2)$$
Shouldn't the coefficient of $d\rho^2$ on the right be $\left(\frac{r_+}{r}\right)^2$ instead of $\frac{r_+}{r}$? You have changed the overall sign of $ds^2$ when going from the Schwarzschild coordinates to the Lemaitre coordinates. Also, here your angular part is for 4D rather than 5D.

For a free falling body, $d\rho=0$,
$d\rho=0$ only holds for radial free-fall if the initial condition corresponds to starting at rest at infinity. For a particle that starts at rest at a finite value of $r$, $d\rho \neq 0$.

You can solve the problem sticking with the Schwarzschild coordinates. In standard texts, it is shown how to derive an expression for $\frac{dt}{d \tau}$ in terms of $r$ for any free-fall trajectory. Here, $\tau$ is proper time for the freely falling particle. See the first equation here: http://grwiki.physics.ncsu.edu/wiki/Schwarzschild_Black_Hole#Geodesic_Motion
I think this equation should still be valid in the 5-D spacetime of your problem.

[EDIT: The equation for $\frac{dt}{d \tau}$ will probably change for 5-D. I will try to find the time to work it out. The derivation should be similar to the 4-D derivation.]

You can use this equation along with the form of the metric to deduce an expression for $\frac{dr}{d \tau}$ as a function of $r$ for radial motion.

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TSny
Homework Helper
Gold Member
The 5-D case does appear to work out just as easily as the 4-D case. You should find that the expression for $\frac{dt}{d\tau}$ in the 5D case differs from the 4D case by just replacing $\frac{r_+}{r}$ in the 4D case by $\left(\frac{r_+}{r}\right)^2$. Then, working out an expression for $\frac{dr}{d\tau}$ from the 5D metric is very similar to working it out in the 4D case. The final integration to obtain the total infall time $\tau$ is actually easier in the 5D case (if I didn't make any mistakes).