# Radially falling into a Black Hole

• Confused Physicist

#### Confused Physicist

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.

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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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).