Solving for the Missing Factor: Jumping into a Non-Rotating Black Hole

[cozycoz]

I'm reading Lambourne's <Relativity, Gravitation and Cosmology>, and I cannot get a result the book describes. It's on equation (6.7) in 173p.

When a person free-falls into a non-rotating black hole from $r=r_0$ to some position $r=r'$, the proper time becomes
$$τ(r')-τ(r_0)=\frac{r_0}{c}\sqrt{\frac{r_0}{r_S}}[\frac{π}{2}+\sqrt{\frac{r'}{r_0}(1-\frac{r'}{r_0})}+\arctan(-\sqrt{\frac{r'}{r_0-r'}})].$$

Now we can simplify the above by taking the limits of $r_0>>r'$.
First, using $\arctan{x}=x+O(x^3)$,
$$\arctan(-\sqrt{\frac{r'}{r_0-r'}})=-\sqrt{\frac{r'}{r_0-r'}}+O[(-\sqrt{\frac{r'}{r_0-r'}})^3],$$
and here, for
$$\sqrt{\frac{r'}{r_0-r'}}=\sqrt{\frac{\frac{r'}{r_0}}{1-\frac{r'}{r_0}}} << 1,$$
we can ignore $O(x^3)$. Then
$$\sqrt{\frac{r'}{r_0}(1-\frac{r'}{r_0})}+\arctan(-\sqrt{\frac{r'}{r_0-r'}})$$ $$≈\frac{r_0}{c}\sqrt{\frac{r_0}{r_S}}[\frac{π}{2}+\sqrt{\frac{r'}{r_0}}(1-\frac{r'}{r_0})^{\frac{1}{2}}-\sqrt{\frac{r'}{r_0}}(1-\frac{r'}{r_0})^{-\frac{1}{2}}].$$

By using
$$(1-\frac{r'}{r_0})^{\frac{1}{2}}≈1-\frac{r'}{r_0}$$ $$(1-\frac{r'}{r_0})^{-\frac{1}{2}}≈1+\frac{r'}{r_0},$$
I get
$$τ(r')-τ(r_0)=\frac{r_0}{c}\sqrt{\frac{r_0}{r_S}}[\frac{π}{2}-(\frac{r'}{r_0})^{\frac{3}{2}}].$$
This is my result, but the book says

$$τ(r')-τ(r_0)=\frac{r_0}{c}\sqrt{\frac{r_0}{r_S}}[\frac{π}{2}-\frac{2}{3}(\frac{r'}{r_0})^{\frac{3}{2}}].$$

Where did the factor $\frac{2}{3}$ come from?

 

[mfb]

$$(1-\frac{r'}{r_0})^{\frac{1}{2}}≈1-\frac{r'}{r_0}$$ This is missing a factor 1/2 for the fraction on the right side. Same for the following equation.

You probably have to consider the x³ term of the arctan. The leading term of the argument to the third power has the same power as your result.

 

[stevendaryl]

$$τ(r')-τ(r_0)=\frac{r_0}{c}\sqrt{\frac{r_0}{r_S}}[\frac{π}{2}-\frac{2}{3}(\frac{r'}{r_0})^{\frac{3}{2}}].$$

There's something strange about that result. An alternative derivation of the $r_0 \gg r_S$ case is as follows:

Start with the equation:

$(\frac{dr}{d\tau})^2 = c^2 [ \frac{E}{mc^2} - 1 + \frac{r_s}{r}]$

$E$ turns out to be the constant value of the expression $mc^2 (1 - \frac{r_s}{r}) \frac{dt}{d\tau}$. If the object drops from $r_0 \gg r_S$, then $E \approx mc^2$. So the equation simplifies to:

$(\frac{dr}{d\tau})^2 = c^2 [ \frac{r_s}{r}]$

This has the exact solution:

$\tau = \frac{r_0}{c} \sqrt{\frac{r_0}{r_S}} (1 - (\frac{r}{r_0})^{\frac{3}{2}})$

There is no $\pi/2$. With the choice $E = mc^2$, there is no trigonometric functions involved, it's just rational powers of $r$.

One way to see that this equation works is that in the limit $r \rightarrow r_0$, $\tau \rightarrow 0$.

 

[cozycoz]

$$(1-\frac{r'}{r_0})^{\frac{1}{2}}≈1-\frac{r'}{r_0}$$ This is missing a factor 1/2 for the fraction on the right side. Same for the following equation.

You probably have to consider the x³ term of the arctan. The leading term of the argument to the third power has the same power as your result.

I omitted $\frac{1}{2}$ only here. I actually calculated with it so it changes nothing, sorry for the mistake!

But I've done considering $x^3$ term and got $\frac{2}{3}$ factor. Thank you!