Temperature of a black hole

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 3.7.

$\displaystyle S=\frac{8\pi^{2}GM^{2}}{hc}k \ \ \ \ \ (1)$

Taking ${U=Mc^{2}}$ as the energy of a black hole, we can write this as

$\displaystyle S=\frac{8\pi^{2}Gk}{hc^{5}}U^{2} \ \ \ \ \ (2)$

The temperature is therefore

$\displaystyle T=\left(\frac{\partial S}{\partial U}\right)^{-1}=\frac{hc^{5}}{16\pi^{2}GkU}=\frac{hc^{3}}{16\pi^{2}GkM} \ \ \ \ \ (3)$

which agrees with our earlier result from general relativity.

For a solar mass black hole, this gives a value of

 $\displaystyle T$ $\displaystyle =$ $\displaystyle \frac{\left(6.62\times10^{-34}\right)\left(3\times10^{8}\right)^{3}}{16\pi^{2}\left(6.67\times10^{-11}\right)\left(1.38\times10^{-23}\right)\left(2\times10^{30}\right)}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 6\times10^{-8}\mbox{ K} \ \ \ \ \ (5)$

Not only are black holes dark, they are also very cold.

The entropy-versus-energy curve 2 is a parabola so its slope ${\frac{\partial S}{\partial U}}$ increases as ${U}$ increases. As we’ve seen, this means that a black hole has negative heat capacity, and thus decreases in temperature as more energy (mass) is added. This is also clear from 3, since ${T\propto\frac{1}{M}}$.

Black hole with static charge; Reissner-Nordström solution

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 23; Problem 23.4.

The derivation of the Schwarzschild metric can be enhanced to include a source, such as a black hole, with a static electric charge ${Q}$. The resulting metric is known as the Reissner-Nordstöm solution. In order to include the effects of the charge, we have to realize that even if there is no mass outside the source, the electric field carries energy so its contribution to the stress-energy tensor must be included. We’ll begin with a quick review of the electromagnetic stress-energy tensor. The electromagnetic field tensor is

$\displaystyle F^{\mu\nu}=\left[\begin{array}{cccc} 0 & E_{x} & E_{y} & E_{z}\\ -E_{x} & 0 & B_{z} & -B_{y}\\ -E_{y} & -B_{z} & 0 & B_{x}\\ -E_{z} & B_{y} & -B_{x} & 0 \end{array}\right] \ \ \ \ \ (1)$

where ${E_{i}}$ and ${B_{i}}$ are the spatial components of the electric and magnetic fields. We can just as well write this in spherical coordinates as

$\displaystyle F^{\mu\nu}=\left[\begin{array}{cccc} 0 & E_{r} & E_{\theta} & E_{\phi}\\ -E_{r} & 0 & B_{\phi} & -B_{\theta}\\ -E_{\theta} & -B_{\phi} & 0 & B_{r}\\ -E_{\phi} & B_{\theta} & -B_{r} & 0 \end{array}\right] \ \ \ \ \ (2)$

The electromagnetic stress-energy tensor can be written in terms of ${F^{\mu\nu}}$ (generalized to non-flat space with a metric ${g^{\mu\nu}}$):

 $\displaystyle T^{\mu\nu}$ $\displaystyle =$ $\displaystyle -\frac{1}{8\pi k}\left(2F^{\mu\kappa}F_{\kappa\lambda}g^{\lambda\nu}+\frac{1}{2}F_{\lambda\kappa}F^{\lambda\kappa}g^{\mu\nu}\right)\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{1}{8\pi k}\left(2F^{\mu\kappa}F_{\kappa}^{\;\nu}+\frac{1}{2}F_{\lambda\kappa}F^{\lambda\kappa}g^{\mu\nu}\right)\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{1}{8\pi k}\left(2F^{\mu\kappa}g_{\kappa\alpha}F^{\alpha\nu}+\frac{1}{2}F_{\lambda\kappa}F^{\lambda\kappa}g^{\mu\nu}\right)\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{8\pi k}\left(2F^{\mu\kappa}g_{\kappa\alpha}F^{\nu\alpha}-\frac{1}{2}F_{\lambda\kappa}F^{\lambda\kappa}g^{\mu\nu}\right) \ \ \ \ \ (6)$

where to get the last line, we used the anti-symmetry of ${F^{\alpha\nu}=-F^{\nu\alpha}}$. The constant ${k=1/4\pi\epsilon_{0}}$ in more conventional notation.

The Einstein equation (with ${\Lambda=0}$) is:

$\displaystyle R^{\mu\nu}=8\pi G\left(T^{\mu\nu}-\frac{1}{2}g^{\mu\nu}T\right) \ \ \ \ \ (7)$

so to work out the components of ${R^{\mu\nu}}$ we need the scalar ${T=T_{\;\mu}^{\mu}}$. For any metric ${g^{\mu\nu}}$ we have from 3

 $\displaystyle T=T_{\;\mu}^{\mu}$ $\displaystyle =$ $\displaystyle g_{\mu\nu}T^{\mu\nu}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{8\pi k}\left(2F^{\mu\kappa}g_{\kappa\alpha}g_{\mu\nu}F^{\nu\alpha}-\frac{1}{2}F_{\lambda\kappa}F^{\lambda\kappa}g^{\mu\nu}g_{\mu\nu}\right)\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{8\pi k}\left(2F_{\nu\alpha}F^{\nu\alpha}-2F_{\lambda\kappa}F^{\lambda\kappa}\right)\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (11)$

The double contraction in the second line is ${g^{\mu\nu}g_{\mu\nu}=4}$ (as can be seen by working it out in a local inertial frame where ${g^{\mu\nu}=\eta^{\mu\nu}}$), and in the third line we lowered the two indices of ${F^{\mu\kappa}}$ in the first term.

To proceed, we need to make a few assumptions. First, we’ll assume that a charged black hole gives rise to a spherically symmetric field tensor with only an electric field (the magnetic field is zero). We’ll also assume that the metric obeys Birkhoff’s theorem, so that it is independent of time. In that case, the only non-zero component of ${F_{\mu\nu}}$ is the radial electric field, which is an unknown function of ${r}$ only. That is

$\displaystyle F_{\mu\nu}=\left[\begin{array}{cccc} 0 & -E\left(r\right) & 0 & 0\\ E\left(r\right) & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{array}\right] \ \ \ \ \ (12)$

Note that we’re looking at the components of ${F_{\mu\nu}}$ with both indices lowered, so that the electric field is negative for ${F_{tr}}$ and positive for ${F_{rt}}$. [I should add that the original derivation of this assumed flat space, so I’m a bit hazy on how we can make this assumption for non-flat space. I suppose, given that we’re not specifying ${E}$ at this point, we can make this assumption.]

At this stage, we’ll take the metric to be

$\displaystyle ds^{2}=-Adt^{2}+Bdr^{2}+r^{2}d\theta^{2}+r^{2}\sin^{2}\theta d\phi^{2} \ \ \ \ \ (13)$

with ${A}$ and ${B}$ to be determined. The components of ${F^{\mu\nu}}$ with raised indices are then since the metric is diagonal,

 $\displaystyle F_{\mu\nu}$ $\displaystyle =$ $\displaystyle g_{\mu\alpha}g_{\nu\beta}F^{\alpha\beta}\ \ \ \ \ (14)$ $\displaystyle F_{tr}$ $\displaystyle =$ $\displaystyle g_{tt}g_{rr}F^{tr}\ \ \ \ \ (15)$ $\displaystyle -E$ $\displaystyle =$ $\displaystyle -ABF^{tr}\ \ \ \ \ (16)$ $\displaystyle F^{tr}$ $\displaystyle =$ $\displaystyle \frac{E}{AB}=-F^{rt} \ \ \ \ \ (17)$

Given ${F^{\mu\nu}}$ and ${F_{\mu\nu}}$ we can now work out the RHS of 7, remembering that ${T=0}$ from 11:

 $\displaystyle 8\pi GT^{\mu\nu}$ $\displaystyle =$ $\displaystyle \frac{G}{k}\left(2F^{\mu\kappa}g_{\kappa\alpha}F^{\nu\alpha}-\frac{1}{2}F_{\lambda\kappa}F^{\lambda\kappa}g^{\mu\nu}\right)\ \ \ \ \ (18)$ $\displaystyle 8\pi GT_{\mu\nu}$ $\displaystyle =$ $\displaystyle \frac{G}{k}g_{\gamma\mu}g_{\delta\nu}\left(2F^{\gamma\kappa}g_{\kappa\alpha}F^{\delta\alpha}-\frac{1}{2}F_{\lambda\kappa}F^{\lambda\kappa}g^{\gamma\delta}\right) \ \ \ \ \ (19)$

From 12 and 17 we have

 $\displaystyle F_{\lambda\kappa}F^{\lambda\kappa}$ $\displaystyle =$ $\displaystyle -\frac{2E^{2}}{AB}\ \ \ \ \ (20)$ $\displaystyle 8\pi GT_{\mu\nu}$ $\displaystyle =$ $\displaystyle \frac{G}{k}g_{\gamma\mu}g_{\delta\nu}\left(2F^{\gamma\kappa}g_{\kappa\alpha}F^{\delta\alpha}+\frac{E^{2}}{AB}g^{\gamma\delta}\right)\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{G}{k}g_{\gamma\mu}g_{\delta\nu}2F^{\gamma\kappa}g_{\kappa\alpha}F^{\delta\alpha}+g_{\mu\nu}\frac{GE^{2}}{kAB} \ \ \ \ \ (22)$

For the individual components, we get, using ${g_{tt}=-A}$, ${g_{rr}=B}$ and ${g_{\theta\theta}=r^{2}}$

 $\displaystyle 8\pi GT_{tt}$ $\displaystyle =$ $\displaystyle \frac{G}{k}A^{2}\left(2F^{tr}BF^{tr}\right)-\frac{GE^{2}}{kB}\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{G}{k}\left[2A^{2}B\frac{E^{2}}{A^{2}B^{2}}-\frac{E^{2}}{B}\right]\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{GE^{2}}{kB}\ \ \ \ \ (25)$ $\displaystyle 8\pi GT_{rr}$ $\displaystyle =$ $\displaystyle \frac{G}{k}\left[2B^{2}F^{rt}\left(-A\right)F^{rt}+B\frac{E^{2}}{AB}\right]\ \ \ \ \ (26)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{G}{k}\left[-2AB^{2}\frac{E^{2}}{A^{2}B^{2}}+\frac{E^{2}}{A}\right]\ \ \ \ \ (27)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{GE^{2}}{kA}\ \ \ \ \ (28)$ $\displaystyle 8\pi GT_{\theta\theta}$ $\displaystyle =$ $\displaystyle 0+g_{\theta\theta}\frac{GE^{2}}{kAB}\ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{GE^{2}r^{2}}{kAB} \ \ \ \ \ (30)$

We can now plug these into 7 to get the equations that must be solved to find ${A}$ and ${B}$. We get

 $\displaystyle R_{tt}$ $\displaystyle =$ $\displaystyle \frac{GE^{2}}{kB}\ \ \ \ \ (31)$ $\displaystyle R_{rr}$ $\displaystyle =$ $\displaystyle -\frac{GE^{2}}{kA}\ \ \ \ \ (32)$ $\displaystyle BR_{tt}+AR_{rr}$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (33)$

When we worked out the Ricci tensor in terms of the metric, we got the equations

 $\displaystyle \frac{1}{2B}\left[\partial_{rr}^{2}A-\partial_{tt}^{2}B+\frac{\left(\partial_{t}B\right)^{2}}{2B}+\frac{\left(\partial_{t}A\right)\left(\partial_{t}B\right)-\left(\partial_{r}A\right)^{2}}{2A}-\frac{\left(\partial_{r}A\right)\left(\partial_{r}B\right)}{2B}+\frac{2\partial_{r}A}{r}\right]$ $\displaystyle =$ $\displaystyle R_{tt}\ \ \ \ \ (34)$ $\displaystyle \frac{1}{2A}\left[\partial_{tt}^{2}B-\partial_{rr}^{2}A+\frac{\left(\partial_{r}A\right)^{2}-\left(\partial_{t}A\right)\left(\partial_{t}B\right)}{2A}+\frac{\left(\partial_{r}A\right)\left(\partial_{r}B\right)-\left(\partial_{t}B\right)^{2}}{2B}+\frac{2A\partial_{r}B}{rB}\right]$ $\displaystyle =$ $\displaystyle R_{rr}\ \ \ \ \ (35)$ $\displaystyle -\frac{r\partial_{r}A}{2AB}+\frac{r\partial_{r}B}{2B^{2}}+1-\frac{1}{B}$ $\displaystyle =$ $\displaystyle R_{\theta\theta}\ \ \ \ \ (36)$ $\displaystyle \frac{\partial_{t}B}{rB}$ $\displaystyle =$ $\displaystyle R_{tr} \ \ \ \ \ (37)$

Because of Birkhoff’s theorem, all time derivatives are zero, so these equations simplify to

 $\displaystyle \frac{1}{2B}\left[\partial_{rr}^{2}A-\frac{\left(\partial_{r}A\right)^{2}}{2A}-\frac{\left(\partial_{r}A\right)\left(\partial_{r}B\right)}{2B}+\frac{2\partial_{r}A}{r}\right]$ $\displaystyle =$ $\displaystyle R_{tt}\ \ \ \ \ (38)$ $\displaystyle \frac{1}{2A}\left[-\partial_{rr}^{2}A+\frac{\left(\partial_{r}A\right)^{2}}{2A}+\frac{\left(\partial_{r}A\right)\left(\partial_{r}B\right)}{2B}+\frac{2A\partial_{r}B}{rB}\right]$ $\displaystyle =$ $\displaystyle R_{rr}\ \ \ \ \ (39)$ $\displaystyle -\frac{r\partial_{r}A}{2AB}+\frac{r\partial_{r}B}{2B^{2}}+1-\frac{1}{B}$ $\displaystyle =$ $\displaystyle R_{\theta\theta}\ \ \ \ \ (40)$ $\displaystyle 0$ $\displaystyle =$ $\displaystyle R_{tr} \ \ \ \ \ (41)$

Applying 33 we get

 $\displaystyle \frac{\partial_{r}A}{r}+\frac{A\partial_{r}B}{rB}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (42)$ $\displaystyle \frac{\partial_{r}A}{A}+\frac{\partial_{r}B}{B}$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (43)$

We can replace the partial derivatives by total derivatives since ${A}$ and ${B}$ depend only on ${r}$. Multiplying through by ${AB}$ we get

$\displaystyle B\frac{dA}{dr}+A\frac{dB}{dr}=\frac{d}{dr}\left(AB\right)=0 \ \ \ \ \ (44)$

so ${AB=\mbox{constant}}$. For very large ${r}$, the metric must reduce to ${\eta_{\mu\nu}}$ so both ${A\rightarrow1}$ and ${B\rightarrow1}$. Thus the product ${AB=1}$ everywhere, which means from 17 that ${F^{tr}=-F^{rt}=E}$ and thus ${F^{\mu\nu}=-F_{\mu\nu}}$.

So far, we have established that ${A=\frac{1}{B}}$ but to get the two components separately, we need to use the fact that we’re dealing a charged black hole. Maxwell’s equations can be written in tensor form as

$\displaystyle \nabla_{\nu}F^{\mu\nu}=4\pi kJ^{\mu} \ \ \ \ \ (45)$

where the absolute gradient is defined in terms of Christoffel symbols as

$\displaystyle \nabla_{\rho}F^{\mu\nu}=\partial_{\rho}F^{\mu\nu}+F^{\mu\alpha}\Gamma_{\alpha\rho}^{\nu}+F^{\alpha\nu}\Gamma_{\alpha\rho}^{\mu} \ \ \ \ \ (46)$

Contracting ${\rho}$ with ${\nu}$ we get

$\displaystyle \nabla_{\nu}F^{\mu\nu}=\partial_{\nu}F^{\mu\nu}+F^{\mu\alpha}\Gamma_{\alpha\nu}^{\nu}+F^{\alpha\nu}\Gamma_{\alpha\nu}^{\mu} \ \ \ \ \ (47)$

In empty space, ${J^{\mu}=0}$ since there is no charge or current, so for ${\mu=t}$ we have

$\displaystyle \nabla_{\nu}F^{t\nu}=\partial_{\nu}F^{t\nu}+F^{t\alpha}\Gamma_{\alpha\nu}^{\nu}+F^{\alpha\nu}\Gamma_{\alpha\nu}^{t} \ \ \ \ \ (48)$

Unfortunately, this means working out a few Christoffel symbols, but we can use the worksheet to make things easier. The only non-zero components of ${F^{\mu\nu}}$ are ${F^{tr}=-F^{rt}}$. Because ${\Gamma_{\alpha\nu}^{t}=\Gamma_{\nu\alpha}^{t}}$, the last term is zero after the sums are done, so

$\displaystyle \nabla_{\nu}F^{t\nu}=\partial_{r}F^{tr}+F^{tr}\Gamma_{r\nu}^{\nu} \ \ \ \ \ (49)$

In the notation of the worksheet, we have, using ${B=\frac{1}{A}}$, ${C=r^{2}}$ and ${D=r^{2}\sin^{2}\theta}$ and a subscript 1 means ‘take the derivative with respect to ${r}$‘:

 $\displaystyle \Gamma_{r\nu}^{\nu}$ $\displaystyle =$ $\displaystyle \Gamma_{1\nu}^{\nu}\ \ \ \ \ (50)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2A}A_{1}+\frac{1}{2B}B_{1}+\frac{1}{2C}C_{1}+\frac{1}{2D}D_{1}\ \ \ \ \ (51)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2A}A_{1}-\frac{1}{2A}A_{1}+\frac{1}{r}+\frac{1}{r}\ \ \ \ \ (52)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2}{r}\ \ \ \ \ (53)$ $\displaystyle \nabla_{\nu}F^{t\nu}$ $\displaystyle =$ $\displaystyle \partial_{r}E+\frac{2E}{r}=0\ \ \ \ \ (54)$ $\displaystyle r^{2}\partial_{r}E+2rE$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (55)$ $\displaystyle \frac{d}{dr}\left(r^{2}E\right)$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (56)$ $\displaystyle E\left(r\right)$ $\displaystyle =$ $\displaystyle \frac{b}{r^{2}} \ \ \ \ \ (57)$

where ${b}$ is a constant of integration. If this is to reduce to the Coulomb field at large ${r}$, then we require

$\displaystyle b=kQ=\frac{Q}{4\pi\epsilon_{0}} \ \ \ \ \ (58)$

From 40 and 30 with ${AB=1}$ we have

 $\displaystyle -\frac{r\partial_{r}A}{2}+\frac{r\partial_{r}B}{2B^{2}}+1-\frac{1}{B}$ $\displaystyle =$ $\displaystyle \frac{GE^{2}r^{2}}{k}\ \ \ \ \ (59)$ $\displaystyle -\frac{r\partial_{r}A}{2}+\frac{rA^{2}\partial_{r}\frac{1}{A}}{2}+1-A$ $\displaystyle =$ $\displaystyle \frac{GkQ^{2}}{r^{2}}\ \ \ \ \ (60)$ $\displaystyle -r\frac{dA}{dr}+1-A$ $\displaystyle =$ $\displaystyle \frac{GkQ^{2}}{r^{2}}\ \ \ \ \ (61)$ $\displaystyle -\frac{d\left(rA\right)}{dr}+1$ $\displaystyle =$ $\displaystyle \frac{GkQ^{2}}{r^{2}}\ \ \ \ \ (62)$ $\displaystyle \frac{d\left(rA\right)}{dr}$ $\displaystyle =$ $\displaystyle 1-\frac{GkQ^{2}}{r^{2}}\ \ \ \ \ (63)$ $\displaystyle A\left(r\right)$ $\displaystyle =$ $\displaystyle 1+\frac{GkQ^{2}}{r^{2}}+\frac{K}{r} \ \ \ \ \ (64)$

where ${K}$ is a constant of integration. In order for this to reduce to the Schwarzschild metric component ${-g_{tt}=\left(1-\frac{2GM}{r}\right)}$ when ${Q=0}$, we must have ${K=-2GM}$, so

 $\displaystyle A\left(r\right)$ $\displaystyle =$ $\displaystyle 1-\frac{2GM}{r}+\frac{GkQ^{2}}{r^{2}}\ \ \ \ \ (65)$ $\displaystyle B\left(r\right)$ $\displaystyle =$ $\displaystyle \left[1-\frac{2GM}{r}+\frac{GkQ^{2}}{r^{2}}\right]^{-1} \ \ \ \ \ (66)$

An event horizon occurs whenever ${g_{tt}=0}$. In this case, this gives rise to a quadratic equation in ${r}$:

 $\displaystyle r^{2}-2GMr+GkQ^{2}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (67)$ $\displaystyle r$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[2GM\pm\sqrt{4G^{2}M^{2}-4GkQ^{2}}\right]\ \ \ \ \ (68)$ $\displaystyle$ $\displaystyle =$ $\displaystyle GM\pm\sqrt{G^{2}M^{2}-GkQ^{2}} \ \ \ \ \ (69)$

For real solutions, we must have

 $\displaystyle G^{2}M^{2}$ $\displaystyle \ge$ $\displaystyle GkQ^{2}\ \ \ \ \ (70)$ $\displaystyle GM^{2}$ $\displaystyle \ge$ $\displaystyle kQ^{2} \ \ \ \ \ (71)$

If the charge ${Q}$ is large enough to violate this condition, there are no event horizons meaning that the singularity at ${r=0}$ becomes a naked singularity.

Falling into a black hole: tidal forces

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 19; Problem P19.9.

Using the Riemann tensor, we can get an idea of the force felt by someone falling into a black hole. Recall the original definition of the Riemann tensor was in terms of the equation of geodesic deviation:

$\displaystyle \ddot{\mathbf{n}}^{i}=-R_{\; j\ell m}^{i}u^{m}u^{j}n^{\ell} \ \ \ \ \ (1)$

where ${\mathbf{n}}$ is the four-vector separating two infinitesimally close geodesics, and ${u}$ is the four-velocity of an object in freefall along one of the geodesics.

For an object, such as a person, that has a large enough size that different parts of the object would, if they weren’t connected, follow different geodesics, a tension force is felt as the various geodesics that pass through different parts of the object diverge during the object’s journey. Suppose our unfortunate person is falling feet first into a black hole (we’ll assume that the person started at rest very far away from the black hole). If we set up a locally inertial frame (LIF) at the person’s centre of mass and align the person’s local ${z}$ axis with the radial direction in the Schwarzschild (S) metric, then we’ve seen that we can write the geodesic deviation as

$\displaystyle \frac{d^{2}n^{i}}{dt^{2}}=-R_{\; t\ell t}^{i}n^{\ell} \ \ \ \ \ (2)$

In the case of the falling person, we can look at the ${z}$ direction, since this is where most of the tidal effects will be felt. In that case, we get, using the person’s LIF as the reference frame:

$\displaystyle \ddot{n}^{z}=-R_{\; t\ell t}^{z}n^{\ell} \ \ \ \ \ (3)$

If we neglect separations in the ${x}$ and ${y}$ directions, this becomes

$\displaystyle \ddot{n}^{z}=-R_{\; tzt}^{z}n^{z} \ \ \ \ \ (4)$

To get the Riemann component, we can use the symmetry of the tensor:

$\displaystyle g_{za}R_{\; tzt}^{a}=R_{ztzt}=R_{tztz}=g_{ta}R_{\; ztz}^{a} \ \ \ \ \ (5)$

In the LIF, ${g_{ij}=\eta_{ij}}$ so this equation becomes

$\displaystyle R_{\; tzt}^{z}=-R_{\; ztz}^{t} \ \ \ \ \ (6)$

and we worked out the RHS in the last post, so we have

$\displaystyle R_{\; tzt}^{z}=-\frac{2GM}{r^{3}} \ \ \ \ \ (7)$

The acceleration of the ${z}$ separation of the two geodesics is then

$\displaystyle \ddot{n}^{z}=\frac{2GM}{r^{3}}n^{z} \ \ \ \ \ (8)$

which we can rewrite with ${r}$ as a function of the acceleration felt at that distance from the black hole:

$\displaystyle r=\left[\frac{2GMn^{z}}{\ddot{n}^{z}}\right]^{1/3} \ \ \ \ \ (9)$

To see how long it takes the person to fall from this distance to ${r=0}$, we can use the formula we derived earlier:

$\displaystyle \Delta\tau=\frac{\pi r^{3/2}}{\sqrt{8GM}} \ \ \ \ \ (10)$

So the time measured by the observer is

$\displaystyle \Delta\tau=\frac{\pi}{2}\sqrt{\frac{n^{z}}{\ddot{n}^{z}}} \ \ \ \ \ (11)$

To put this in practical terms, a typical person can handle up to about 5g (five times the acceleration due to gravity at the Earth’s surface) along their vertical direction before losing consciousness. If we take ${n^{z}\approx1\mbox{ m}}$ (about half a person’s height) and ${\ddot{n}^{z}\approx50\mbox{ m s}^{-2}}$ then the time from first experiencing this force to annihilation at the singularity at the centre of the black hole is about

$\displaystyle \Delta\tau\approx0.2\mbox{ sec} \ \ \ \ \ (12)$

Using Moore’s estimate of the speed of pain impulses (around 1 m/sec) any pain resulting from this probably wouldn’t be felt before the person gets annihilated.

Black hole in equilibrium with a thermal reservoir

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 16; Problem P16.8.

The entropy of a black hole is given by

$\displaystyle S=\frac{4\pi k_{B}G}{\hbar}M^{2} \ \ \ \ \ (1)$

Given that the entopy of a thermal reservoir with temperature ${T_{R}}$ and internal energy ${U}$ (another fact from thermodynamics) is

$\displaystyle S_{R}=\frac{U}{T_{R}}+C \ \ \ \ \ (2)$

where ${C}$ is a constant. If a black hole with mass ${M}$ is exchanging thermal radiation with the reservoir, we can say that the combined (and conserved) total energy of the system is ${U_{tot}=U+M}$ so

$\displaystyle S_{tot}=\frac{U_{tot}-M}{T_{R}}+C+\frac{4\pi k_{B}G}{\hbar}M^{2} \ \ \ \ \ (3)$

Also from thermodynamics, it is known that when a system is in thermal equilibrium (no net transfer of radiation between the black hole and the reservoir), then the entropy should be at a local maximum. In this case, we can find the extrema of ${S_{tot}}$ as a function of ${M}$, since the mass of the black hole varies as it exchanges radiation with the reservoir. We get

 $\displaystyle \frac{dS_{tot}}{dM}$ $\displaystyle =$ $\displaystyle -\frac{1}{T_{R}}+\frac{8\pi k_{B}G}{\hbar}M=0\ \ \ \ \ (4)$ $\displaystyle M$ $\displaystyle =$ $\displaystyle \frac{\hbar}{8\pi k_{B}GT_{R}} \ \ \ \ \ (5)$

However, from the second derivative:

$\displaystyle \frac{d^{2}S_{tot}}{dM^{2}}=\frac{8\pi k_{B}G}{\hbar}>0 \ \ \ \ \ (6)$

we see that the entropy is actually at a local minimum at this mass and, since the curve 3 is a parabola, the maxima of entropy occur at the two ends ${M=0}$ and ${M=U_{tot}}$. Which of these two points is the overall maximum depends on ${U_{tot}}$ and ${T_{R}}$, but it does seem to indicate that equilibrium is achieved only when either the black hole evaporates and sends all its energy (mass) into the reservoir, or else the black hole swallows up the reservoir completely.

Black hole temperatures at different distances

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 16; Problem P16.5.

We’ve seen that a temperature can be defined for a black hole that radiates particles (mainly photons) due to quantum pair creation:

$\displaystyle T_{\infty}=\frac{\hbar}{8\pi k_{B}GM} \ \ \ \ \ (1)$

This is the temperature as seen by an observer at infinity. If we take the temperature to be proportional to the observed energy of the emitted particles, we can get the temperature as measured by an observer at rest at some finite distance ${R}$ from the black hole. The energy of the photons is

$\displaystyle E=\frac{E_{\infty}}{\sqrt{1-2GM/R}} \ \ \ \ \ (2)$

where ${E_{\infty}}$ is the energy at infinity. Using the proportionality of ${T}$ and ${E}$, we get

$\displaystyle T=\frac{T_{\infty}}{\sqrt{1-2GM/R}} \ \ \ \ \ (3)$

The temperature thus becomes infinite at ${R=2GM}$. For a solar mass black hole, we have

$\displaystyle T_{\infty}=6.173\times10^{-8}\mbox{ K} \ \ \ \ \ (4)$

We have to get extraordinarily close to a black hole to see the temperature rise significantly. Solving 3 for ${R\left(T\right)}$ we get

$\displaystyle R\left(T\right)=2GM\left(1+\frac{T_{\infty}^{2}}{T^{2}-T_{\infty}^{2}}\right) \ \ \ \ \ (5)$

For example, if we want ${T=300\mbox{ K}}$ (around room temperature), then

$\displaystyle R\left(300\right)=2GM\left(1+4.23\times10^{-20}\right) \ \ \ \ \ (6)$

Since ${GM=1477\mbox{ m}}$, we need to approach to within ${1.25\times10^{-16}\mbox{ m}}$ of the event horizon. This is smaller than the radius of a proton (around ${8.8\times10^{-16}\mbox{ m}}$).

Black hole heat engine

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 16; Problem P16.4.

According to thermodynamics, a heat engine is a device that absorbs heat energy ${Q_{H}}$ from a reservoir at temperature ${T_{H}}$, uses some of this energy to do work ${W}$, and then expels the remaining energy to a colder reservoir at temperature ${T_{C}}$. The efficiency ${\epsilon}$ of the engine is defined as ${\epsilon\equiv W/Q_{H}}$, that is, the ratio of the work done to the total energy absorbed in the first place. The laws of thermodynamics state that the efficiency is limited by the condition

$\displaystyle \epsilon\le1-\frac{T_{C}}{T_{H}} \ \ \ \ \ (1)$

A proposal for a perfectly efficient engine is as follows. We collect some radiation inside a perfectly reflecting box from some hot object at temperature ${T_{H}}$ that is very far from a black hole. We take the box to the black hole and lower it so that it is just above the event horizon, where we hold it at rest. In this case, the angular momentum of the box is ${\ell=0}$ and since it is at rest, ${dr/d\tau=0}$, so from the equation

$\displaystyle \frac{1}{2}\left(\frac{dr}{d\tau}\right)^{2}+\frac{1}{2}\frac{l^{2}}{r^{2}}-GM\left(\frac{1}{r}+\frac{l^{2}}{r^{3}}\right)=\frac{1}{2}\left(e^{2}-1\right) \ \ \ \ \ (2)$

we get

$\displaystyle -\frac{GM}{r}=\frac{1}{2}\left(e^{2}-1\right) \ \ \ \ \ (3)$

If the box is held right at the event horizon, ${r=2GM}$ and thus ${e=0}$. If the energy of the radiation when it was added to the box is taken to be an addition to the box’s mass, then the energy per unit mass at the starting point is ${e=1}$. If we release the radiation from the box into the black hole at the event horizon, we are releasing zero energy, so in some sense we could say that all the original energy was used as work on the trip to the black hole. (Incidentally, although ${e}$ was originally defined as a constant of particle motion in Schwarzschild (S) space, that applied only for motion on a geodesic, and the current situation doesn’t satisfy this, since the box is pulled to a stop before it crosses the horizon. In such a case, ${e}$ will change with the motion.) The efficiency of the engine could thus be defined as

$\displaystyle \epsilon=1-\frac{e_{f}}{e_{0}} \ \ \ \ \ (4)$

where ${e_{0}}$ is the energy per unit mass absorbed initially and ${e_{f}}$ is the energy released into the black hole at the end. Since we’ve seen that we can define a non-zero temperature for a black hole, this appears to violate the laws of thermodynamics. (As a side note, it’s not entirely obvious how this process converts the radiation’s energy to work. The comparison with a heat engine seems to be from the fact that we are absorbing some radiation at one temperature and then depositing it into the black hole at a different temperature.)

However, this derivation assumes that the box can be brought right up to the event horizon, which isn’t really true. For a blackbody, Wien’s displacement law says that the peak wavelength emitted by a body at temperature ${T_{H}}$ is:

$\displaystyle \lambda_{max}=\frac{2\pi c\hbar}{4.965k_{B}T_{H}} \ \ \ \ \ (5)$

Since we’re interested in order of magnitude calculations in GR units we take ${c=1}$ and ignore the numerical constants to get

$\displaystyle \lambda_{max}\sim\frac{\hbar}{k_{B}T_{H}} \ \ \ \ \ (6)$

If the box is a cube, ${\lambda_{max}}$ is the minimum size of an edge if the box is to contain radiation of this wavelength, and since no part of the box can be lowered below the event horizon, the closest the centre of the box can approach the black hole is ${\lambda_{max}/2}$. Using the formula we derived earlier for the physical distance between a point outside the event horizon and the horizon itself, we can find the ${r}$ coordinate of the box’s centre at closest approach.

$\displaystyle \Delta s=r\sqrt{1-\frac{2GM}{r}}+2GM\tanh^{-1}\sqrt{1-\frac{2GM}{r}} \ \ \ \ \ (7)$

Since we expect ${r}$ to be close to ${2GM}$, the argument of the square root terms will be small, so we can use the approximation ${\tanh^{-1}x\approx x}$ for small ${x}$. We therefore get

 $\displaystyle \Delta s$ $\displaystyle =$ $\displaystyle \frac{\lambda_{max}}{2}\approx4GM\sqrt{1-\frac{2GM}{r}}\ \ \ \ \ (8)$ $\displaystyle r$ $\displaystyle =$ $\displaystyle 2GM\left(1-\frac{\lambda_{max}^{2}}{64\left(GM\right)^{2}}\right)^{-1} \ \ \ \ \ (9)$

The approximation of ${r}$ being close to ${2GM}$ is therefore valid provided ${\lambda_{max}\ll8GM}$.

We can rewrite 3 as

 $\displaystyle e^{2}$ $\displaystyle =$ $\displaystyle 1-\frac{2GM}{r}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\lambda_{max}^{2}}{\left(8GM\right)^{2}} \ \ \ \ \ (11)$

Thus if we stop lowering box when its centre is ${\lambda_{max}/2}$ away from the horizon, the energy is not zero, so we will be releasing some energy into the black hole when we release the radiation. The efficiency is therefore

 $\displaystyle \epsilon$ $\displaystyle =$ $\displaystyle 1-e\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1-\frac{\lambda_{max}}{8GM}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle \approx$ $\displaystyle 1-\frac{\hbar}{8k_{B}T_{H}GM} \ \ \ \ \ (14)$

This is consistent with the laws of thermodynamics if the black hole’s temperature is

$\displaystyle T_{C}=\frac{\hbar}{8k_{B}GM} \ \ \ \ \ (15)$

The actual temperature of a black hole is

$\displaystyle T=\frac{\hbar}{8\pi k_{B}GM} \ \ \ \ \ (16)$

so for an order of magnitude estimate, this calculation isn’t too bad.

Black hole radiation: energy of emitted particles

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 16; Box 16.1.

In Schwarzschild (S) space-time, the energy of a particle is given by

$\displaystyle e=\left(1-\frac{2GM}{r}\right)\frac{dt}{d\tau} \ \ \ \ \ (1)$

Outside the event horizon ${r>2GM}$ and since the S coordinate ${t}$ represents time, it must constantly increase as the proper time ${\tau}$ increases, so ${dt/d\tau>0}$, with the result that ${e>0}$. Inside the event horizon ${r<2GM}$, but time and space swap round so ${dt/d\tau}$ can be either positive or negative, since ${t}$ is now a space coordinate. The the energy can be positive or negative inside the event horizon. For a single particle moving along a geodesic, ${e}$ is a constant of the motion, so since ${1-\frac{2GM}{r}}$ becomes zero at the event horizon, ${dt/d\tau}$ becomes infinite which results from the fact that the S time becomes infinite at the horizon. After crossing the horizon, ${1-\frac{2GM}{r}<0}$ so ${dt/d\tau}$ must also be negative to keep ${e}$ constant. However, it is possible for the particle to interact with another particle inside the horizon, which could cause ${dt/d\tau}$ to change sign, resulting in a negative energy.

The idea behind black hole radiation is actually quite simple. Quantum field theory predicts that vacuum fluctuations can occur, in which a particle-antiparticle pair spontaneously appears, essentially out of nothing, provided that the energies of the two particles are equal and opposite; that is, one particle has positive energy ${E}$ while the other has negative energy ${-E}$. This phenomenon can occur only for a very short time interval of the order of ${\Delta t\sim\hbar/E}$ (at this stage, you can just accept all this as god-given, since we haven’t studied quantum field theory yet), after which the particle recombines with its antiparticle.

Now suppose this pair creation event occurs very close to the event horizon, and the negative energy particle crosses the horizon before it has a chance to recombine, and that the positive energy particle therefore escapes to infinity. The energy of the black hole is thus reduced by ${E}$ and the energy ${E}$ is radiated away to infinity. To see how this works in a simplified model, suppose the pair creation event occurs at some small distance ${\epsilon}$ above the event horizon. We need ${\epsilon}$ to be small enough that the negative energy particle can cross the horizon before it recombines, so we need to estimate how long it takes the particle to travel to the horizon. Since we’re outside the horizon, the S coordinate ${r}$ is still spatial, so we can use the following expression to estimate the proper time required:

$\displaystyle \frac{dr}{d\tau}=-\sqrt{e^{2}-\left(1-\frac{2GM}{r}\right)\left(1+\frac{\ell^{2}}{r^{2}}\right)} \ \ \ \ \ (2)$

We’ve used the minus sign to indicate that ${r}$ is decreasing. If the particle falls radially, the angular momentum is ${\ell=0}$, so the proper time required is

$\displaystyle \Delta\tau=-\int_{2GM+\epsilon}^{2GM}\frac{dr}{\sqrt{e^{2}-\left(1-\frac{2GM}{r}\right)}} \ \ \ \ \ (3)$

To go further, we need to know ${e}$. It’s unlikely that the pair is created at rest, but suppose we’re in a locally flat reference frame that is released from rest with its origin at ${r=2GM+\epsilon}$ and that the pair creation event occurs at this location. An object in the observer’s frame has an energy per unit mass of

$\displaystyle e=\sqrt{1-\frac{2GM}{2GM+\epsilon}} \ \ \ \ \ (4)$

so for such an object the integral becomes

 $\displaystyle \Delta\tau$ $\displaystyle =$ $\displaystyle -\int_{2GM+\epsilon}^{2GM}\frac{dr}{\sqrt{1-\frac{2GM}{2GM+\epsilon}-\left(1-\frac{2GM}{r}\right)}}\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\int_{2GM+\epsilon}^{2GM}\frac{dr}{\sqrt{\frac{2GM}{r}-\frac{2GM}{2GM+\epsilon}}} \ \ \ \ \ (6)$

This integral can be done using Maple although the result is a bit too complex to write out here, and then a series expansion of the result around ${\epsilon=0}$ gives the leading terms

$\displaystyle \Delta\tau=2\sqrt{2GM\epsilon}+\frac{5}{6}\sqrt{\frac{2}{GM}}\epsilon^{3/2}+\mathcal{O}\left(\epsilon^{5/2}\right) \ \ \ \ \ (7)$

If we want just the first term, we can transform the integral a bit using ${\rho\equiv r-2GM}$:

$\displaystyle \Delta\tau=\int_{0}^{\epsilon}\frac{d\rho}{\sqrt{\frac{2GM}{\rho+2GM}-\frac{2GM}{2GM+\epsilon}}} \ \ \ \ \ (8)$

If we expand each term in the square root in a series and keep terms up to first order in ${\rho}$ and ${\epsilon}$ we get

$\displaystyle \frac{2GM}{\rho+2GM}-\frac{2GM}{2GM+\epsilon}=\frac{1}{2GM}\left(\epsilon-\rho\right)+\mathcal{O}\left(\epsilon^{2}\right)+\mathcal{O}\left(\rho^{2}\right) \ \ \ \ \ (9)$

so to this order, we have

$\displaystyle \Delta\tau=\int_{0}^{\epsilon}\frac{d\rho}{\sqrt{\left(\epsilon-\rho\right)/2GM}} \ \ \ \ \ (10)$

We can use another substitution ${u\equiv\epsilon-\rho}$ to get

 $\displaystyle \Delta\tau$ $\displaystyle =$ $\displaystyle \int_{0}^{\epsilon}\frac{du}{\sqrt{u/2GM}}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2\sqrt{2GM\epsilon} \ \ \ \ \ (12)$

This is the time required for an object released from rest at ${r=2GM+\epsilon}$ to reach the event horizon. Presumably if the particle is not at rest when it is created, but is heading towards the event horizon, it will require less time than this to reach it.

Plugging this into the field theory estimate above, we can get an estimate of the energy of the radiated particle:

 $\displaystyle \Delta\tau$ $\displaystyle =$ $\displaystyle \frac{\hbar}{E}\ \ \ \ \ (13)$ $\displaystyle E$ $\displaystyle =$ $\displaystyle \frac{\hbar}{2\sqrt{2GM\epsilon}} \ \ \ \ \ (14)$

I have to admit I’m not particularly comfortable with this calculation, since ${\Delta\tau}$ is calculated for an object released at rest from ${r=2GM+\epsilon}$ whereas the pair of particles would probably not be produced at rest. However, as an order of magnitude estimate, I suppose it’s reasonable.

Kruskal-Szekeres diagrams: another space ship disaster

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 15; Problem 15.8.

In this example, we have a spaceship originally at some fixed distance from a black hole, but the engines fail and the ship starts to fall vertically downward (tail first) towards the black hole.

In the KS diagram, the tail of the ship follows the black world line and the front of the ship follows the parallel magenta world line. The diagonal lines with slope ${+1}$ indicate photons emitted by the tail end (with the green line being also the event horizon). The violet photon is emitted before the tail crosses the horizon and is received by the front before it crosses the horizon. A photon emitted just as the tail crosses the horizon follows the green world line, and since this line also corresponds to a constant ${r=2GM}$, the photon remains at the event horizon and is received by the front when the front crosses the horizon. The brown photon leaves the tail after it crosses the horizon and is received by the front after it too crosses the horizon. Thus there isn’t any time during the period where the ship is crossing the horizon that the front receives no photons from the tail, but any photons received by the front are emitted by the tail when both the front and the tail are on the same side of the horizon. When the front is outside the horizon and the tail is inside, the front still sees the tail, but it is seeing the tail as it was before it crossed the horizon.

Closer to the singularity at ${r=0}$, however, there is a cutoff point beyond which any photons (such as the short red line) emitted by the tail are absorbed at ${r=0}$ before reaching the front, so the front end will not be able to see the tail hit ${r=0}$.

Kruskal-Szekeres diagrams: saving a space shuttle

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 15; Problems 15.7, 15.9.

In this example (Moore, problem P15.9), we have a spaceship maintaining a constant distance ${R}$ from a black hole, such that the Kruskal-Szekeres (KS) coordinates satisfy

$\displaystyle u^{2}-v^{2}=\left(\frac{R}{2GM}-1\right)e^{R/2GM}=1 \ \ \ \ \ (1)$

At Schwarzschild (S) time ${t=0}$ (corresponding to ${v=0}$), a shuttle leaves the spaceship and gets pulled towards the black hole along the line ${u=1}$ in the KS diagram. If the mother ship is capable of speeds up to light speed, what is the latest time that it can leave its orbit to intercept the shuttle before the shuttle crosses the event horizon?

In the diagram, the shuttle’s world line is shown in yellow and the ship’s world line is the grey hyperbola. If the ship suddenly breaks out of its orbit and travels at light speed towards the black hole, and intercepts the shuttle just before it crosses the event horizon, the ship will have to follow the turquoise diagonal. (Remember that photon world lines have slopes of ${\pm1}$ on a KS diagram.) We therefore need to find the intersection of the turquoise line and the grey hyperbola, and then find the S time ${t_{r}}$ at which the ship starts on its rescue mission. On a KS diagram, curves of constant ${t}$ are straight lines through the origin, so the time is the thin green line in the diagram.

The turquoise line has slope ${-1}$ and passes through the point ${\left(1,1\right)}$ so its equation is

 $\displaystyle v-1$ $\displaystyle =$ $\displaystyle -\left(u-1\right)\ \ \ \ \ (2)$ $\displaystyle v$ $\displaystyle =$ $\displaystyle -u+2 \ \ \ \ \ (3)$

The intersection with the hyperbola is

 $\displaystyle u^{2}-\left(-u+2\right)^{2}$ $\displaystyle =$ $\displaystyle 1\ \ \ \ \ (4)$ $\displaystyle 4u-4$ $\displaystyle =$ $\displaystyle 1\ \ \ \ \ (5)$ $\displaystyle u$ $\displaystyle =$ $\displaystyle \frac{5}{4}\ \ \ \ \ (6)$ $\displaystyle v$ $\displaystyle =$ $\displaystyle \frac{3}{4} \ \ \ \ \ (7)$

The S time is

 $\displaystyle t_{r}$ $\displaystyle =$ $\displaystyle 2GM\ln\frac{u+v}{u-v}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2GM\ln\frac{8}{2}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 4GM\ln2\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle \approx$ $\displaystyle 2.77GM \ \ \ \ \ (11)$

As a slight variant on this problem (Moore, problem P15.7), suppose you are working on the outside of the spaceship hovering at radius ${R}$ and you drop a valuable piece of equipment, which then falls towards the black hole. Even though an observer at infinity would say that the object gets stuck at the event horizon (since the S ${t}$ coordinate becomes infinite there), the KS diagram shows that you have a limited time (as measured by your own proper time) to go after the object if you are to catch it before it crosses the horizon. In this case, the dropped object has a world line similar to the yellow line on the diagram (not quite the same line, since the dropped object would follow a geodesic, which isn’t a straight line here), and to catch it you’d have to leave your orbit at light speed along a turquoise diagonal that intercepts the object just before it crosses the horizon.

Kruskal-Szekeles metric: what can you see as you fall into a black hole?

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 15; Problem 15.4.

The Kruskal-Szekeres (KS) metric is an alternative to the Schwarzschild (S) metric that makes many phenomena easier to visualize by drawing a diagram in which world lines are plotted on a graph of the coordinate ${v}$ versus ${u}$. For radial motion, the KS metric is

$\displaystyle ds^{2}=\frac{32\left(GM\right)^{3}}{r}e^{-r/2GM}\left(du^{2}-dv^{2}\right) \ \ \ \ \ (1)$

and the relation between the KS coordinates ${u}$ and ${v}$ and the S coordinates ${r}$ and ${t}$ is

 $\displaystyle u^{2}-v^{2}$ $\displaystyle =$ $\displaystyle \left(\frac{r}{2GM}-1\right)e^{r/2GM}\ \ \ \ \ (2)$ $\displaystyle 2GM\ln\left|\frac{u+v}{u-v}\right|$ $\displaystyle =$ $\displaystyle t \ \ \ \ \ (3)$

As an example of using a KS diagram, suppose we have an observer in a space ship that is falling into a black hole. As we saw in the last post, any massive object’s world line must have a slope with a magnitude greater than 1 at all of its points in the KS diagram. One possible world line is shown as the heavy black curve in the diagram:

The motion of the observer is upwards in the diagram. In S coordinates, an observer at infinity sees the ship slow down as it approaches the event horizon, eventually freezing at the horizon, since the S ${t}$ coordinate becomes infinite there (as shown in the diagram by the heavy green line in the upper right quadrant). However, as measured by the observer in the ship, it takes a finite timeto move from a point outside the horizon to a point inside it; there is no divergence at ${r=2GM}$. This is shown in the KS diagram as the black curve simply continues across the green line without anything unusual happening.

Does the fact that the ship crosses the ${t=\infty}$ line mean that the observer in the ship can see the entire future of the outside world as he crosses ${r=2GM}$? To answer this, we first note that in order for the observer to see an event from the outside world, a photon from that event must be able to reach him before he gets to ${r=0}$, where his own world line ends. Since photons travel along lines with slopes of ${\pm1}$, the only photons that can reach him after he crosses the horizon are those between the two heavy red lines shown. Thus he can see into the distant future (that is, very large ${t}$ values) only for events that happen just outside the horizon (that is, events just to the lower right of the green line, between the two red lines). Events that occur at ${t=0}$ lie on the ${u}$ axis, so he can see only those events that lie between approximately ${u=0.75}$ and ${u=1.1}$. The ${r}$ values of these events are determined by the set of hyperbolas that cross the ${u}$ axis between these points (some of the hyperbolas of constant ${r}$ are shown as dashed grey curves).

Thus even though he is inside the event horizon, the observer can still see some events that happened outside the horizon, but he cannot see all events that happen at any time in the future.