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Page $190$ of “Tensors, Relativity, and Cosmology”:

Consider a particle falling radially into a black hole with a radial velocity $u^1=dr/ds$. As the particle is falling radially, we have $u^2=u^3=0$. The motion of the particle is described by the geodesic equation

$$\frac{du^n}{ds}+\Gamma^n_{lk}u^l u^k=0 \tag{21.60}$$ Using $u^2=u^3=0$ and the result for the Christoffel symbols of the second kind, we may write the temporal equation of motion as follows

$$\frac{du^0}{ds}=-\Gamma^0_{lk}u^lu^k=-2\Gamma^0_{10}u^1u^0 \tag{21.61}$$ or $$\frac{du^0}{ds}=-2\frac{dv}{dr}\frac{dr}{ds}u^0=-2\frac{dv}{ds}u^0 \tag {21.62}$$

where the factor of two appears because of the term proportional to $\Gamma^0_{10}=\Gamma^0_{01}$ appear twice in the sum. Thus we obtain

$$\frac{du^0}{ds}+2\frac{dv}{ds}u^0=\exp(-2v)\frac{d}{ds}\left(exp(2v)u^0\right)=0 \tag{21.63}$$

which can be integrated twice to give $$\exp(2v)u^0=g_{00}u^0=K=Constant \tag{21.64}$$, where K is the integration constant which is equal to the value of $g_{00}$ at the point where the particle starts to fall towards the black hole. Using the identity $g_{kn}u^ku^n=1$, we may also write $$1=g_{kn}u^ku^n=g_{00}(u^0)^2+g_{11}(u^1)^2 \tag{21.65}$$

Multiplying by $g_00$ and using (21.64) as well as $g_{00}g_{11}=-1$ $$g_{00}=(g_{00})^2(u^0)^2+g_{00}g_{11}(u^1)^2=K^2-(u^1)^2 \tag{21.66}$$ or $$(u^1)^2=K^2-g_{00}=K^2-1+\frac{r_G}{r} \tag{21.67}$$ where $r_G$ is the gravitational radius of the blackhole. For the radially falling body, $u^1<0$, thus

$$u^1=-\left(K^2-1+\frac{r_G}{r}\right)^{-\frac{1}{2}} \tag {21.68}$$ Using (21.64) and (21.68)

$$\frac{dt}{dr}=\frac{u^0}{cu^1}=-\frac{K}{c g_{00}}\left(K^2-1+\frac{r_G}{r}\right)^{-\frac{1}{2}} \tag{21.69} $$ or $$\frac{dt}{dr}=-\frac{K}{c}\left(1-\frac{r_G}{r}\right)^{-1} \left(K^2-1+\frac{r_G}{r}\right)^{-\frac{1}{2}} \tag{21.70}$$

Assume tat the particle falling radially into a black hole is close to the gravitational radius $r_G$, rewrite $r=r_G+\epsilon$ with $\epsilon<<r_G$. (21.70) then becomes $$ \frac{dt}{dr}=-\frac{K}{c}\left(1-\left(1+\frac{\epsilon}{r_G}\right)\right)^{-1} \left(K^2-1+\left(1+\frac{\epsilon}{r_G}\right)^{-1}\right)^{-\frac{1}{2}} \tag{21.71}$$ Using the approximation $$\left(1+\frac{\epsilon}{r_G}\right)^{-1}\approx 1-\frac{\epsilon}{r_G} \tag{21.72}$$ $$\frac{dt}{dr} \approx -\frac{Kr_G}{c\epsilon}\left(K^2-\frac{\epsilon}{r_G}\right)^{-\frac{1}{2}} \approx \frac{-r_G}{c\epsilon} \tag{21.73}$$

  1. “where K is the integration constant which is equal to the value of $g_{00}$ at the point where the particle starts to fall towards the black hole.” Does this imply that $$u^0=\frac{c\,\mathrm dt}{\mathrm ds}=1$$ and hence $$\mathrm ds=c\,\mathrm d\tau= c\,\mathrm dt$$ at this point?

  2. I do not see how in (21.73) can be approximated to give $$-\frac{Kr_G}{c\epsilon}\left(K^2-\frac{\epsilon}{r_G}\right)^{-\frac{1}{2}} \approx \frac{-r_G}{c\epsilon}$$

and doesn't this approximation assume that $K$ is very large?

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1)

Think of $K$ as the value of (assuming metric signature (+---) )

$$g_{00}=1-\frac{r_G}{r}=\exp(2v)$$

at the $r$ distance from which the particle starts falling $\left(r_G=\frac{2G_{N}M}{c^2}\right.$ is the so called Schwarzschild radius, with $M$ the mass of the black hole). Note that $c dt$ and such only makes sense in a flat space.

The integration leads to $g_{00}u^0=K$, where $K$ is an integration constant. Since it is constant, one can evaluate it at any point in time, so if we choose the initial time (before the particle starts moving toward the black hole), then $u^0=1$ and so $g_{00}=K$ (evaluated at the initial value of $r$).

2)

It seems that the expansion is assuming that $\frac{\epsilon}{r_G}\ll1$, so that

$$\left(K^2-\frac{\epsilon}{r_G}\right)^{-1/2}\approx \frac{1}{\sqrt{K^2}}+\frac{\sqrt{K^2}}{2K^4 r_G}\epsilon=\frac{1}{K}+\frac{1}{2K^3 r_G}\epsilon \approx \frac{1}{K}$$

The goal in this calculation is to get the leading behavior for small $\epsilon$. One can keep subleading terms in $\epsilon$ along the way and then get rid of everything other than the leading one at the end, which is fine and consistent.

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  • $\begingroup$ Shouldn't $g_{00}$ be $(1-\frac{r_G}{r})$ instead and why should $K$ ever be equal to $g_{00}$? $\endgroup$ – Chern-Simons Jun 25 '19 at 18:28
  • $\begingroup$ Doesn't 2) imply that $\frac{\epsilon}{r_G} \approx 0$ then (which is inconsistent with the first expansion)? $\endgroup$ – Chern-Simons Jun 25 '19 at 18:31
  • $\begingroup$ Apologies for bombarding you with questions.... $\endgroup$ – Chern-Simons Jun 25 '19 at 19:11
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    $\begingroup$ @EXINT I made some edits, see updated reply. You are right, if the metric signature is chosen +---, then $g_{00}$ has opposite sign, which is more appropriate here. $\endgroup$ – Kagaratsch Jun 25 '19 at 19:16

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