# Jumping into a non-rotating black hole

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• cozycoz

#### 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?

## Answers and Replies

$$(1-\frac{r'}{r_0})^{\frac{1}{2}}≈1-\frac{r'}{r_0}$$ 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.

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

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##.

$$(1-\frac{r'}{r_0})^{\frac{1}{2}}≈1-\frac{r'}{r_0}$$ 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!

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