# Physics 616

• Prof. Andrew W. Steiner
(or Andrew or "Dr. Steiner")
• Office hour: 103 South College, Thursday 11am
• Email: awsteiner@utk.edu
• Homework: Electronically as .pdf
• You may work with each other on the homework, but you must write the solution in your own words

Use down and up arrows to proceed to the next or previous slide.

## Black Holes and Effacement

• Effacement: ("process of eliminating something", in this case, eliminating complexity) black holes evolve towards a state which can be fully characterizied by $M$, $J$, $Q$
• Also may have other properties, depending on quantum theories of gravity
• Q likely small
• Total area of all horizons cannot decrease in time "area theorem" (violated by Hawking radiation)
• Curvature becomes infinite at the center ("the singularity")
• Area proportional to entropy

## No-Hair Theorem

• Kerr metric is the unique time independent solution for a black hole
• This means that a black hole (assuming no electric field at large distances) in GR is completely specified by its mass and spin
• At large distances, we get the Schwarzchild metric
• This simplifies black holes considerably: we need not know anything about QCD, for example
• Adding E&M gives the "Einstein-Maxwell" equations $$T^{\alpha \beta} = - \frac{1}{\mu_0} \left(F^{\alpha \psi} {F_{\psi}}^{\beta} + \frac{1}{4} g^{\alpha \beta} F_{\psi \tau} F^{\psi \tau}\right)$$ and then the black hole may have an electric charge (Kerr-Newman metric)
• No magnetic moment if there are no magnetic monopoles
• This gives the no-hair theorem: black holes only have $M$, $J$ and $Q$
• Large values of $Q$ not expected
• Not related to the hairy ball theorem

## Other Metrics

• Kerr-Newman $\Rightarrow$ Kerr if $Q=0$
• Kerr-Newman $\Rightarrow$ Reissner-Nordström metric if $J=0$
• Kerr-Newman $\Rightarrow$ Schwarzchild if $Q=J=0$
• Kerr-Newman $\Rightarrow$ Minkowski space if $Q=J=M=0$
•

## No-Hair and Neutron Stars

 However, no-hair theorem doesn't tell us what's going on inside the event horizon It is expected that GR breaks down before density becomes infinite in the center Neutron stars are more complicated Recently, several have suggested that there are no-hair like results which simply neutron star structure Tidal deformability is called "Love number" Yagi and Yunes (2013)

## Titles

• Four-Hair Relations for Differentially Rotating Neutron Stars in the Weak-Field Limit
• I-Love-Q anisotropically: Universal relations for compact stars with scalar pressure anisotropy
• Relating follicly-challenged compact stars to bald black holes: A link between two no-hair properties
• Why I-Love-Q: Explaining why universality emerges in compact objects
• Three-Hair Relations for Rotating Stars: Nonrelativistic Limit
• Love can be Tough to Measure

## Black Hole Properties, Etc.

• Any horizon will eventually become stationary (but not necessarily non-rotating)
• Non-spherical properties radiated away in gravitational waves
• Area theorem: Total area of all horizons cannot decrease in time (violated by quantum effects)
• Curvature becomes infinite at singularities, but all singularities are separated from spacetime by horizons. No "naked singularities" (cosmic censorship conjecture).
• Birkhoff's Theorem: the Schwarzchild metric is the only spherically symmetric metric (you get time independence for free)

## Black Holes in X-ray Binaries

• Kepler's third law gives the "mass function" $$f(M) = \frac{(M \sin i)^3}{(M + M_C)^2} = \frac{P K^3}{2 \pi G}$$ where $P$ and $K$ are the period and semi-amplitude of the radial velocity, $i$ is the inclination angle and $M_C$ is the mass of the companion and $M$ is the mass of the black hole.
• Given $P$ and $K$ $$\sin^3 i = \frac{P K^3(M+M_C)^2}{2 \pi G M^3} \leq 1$$ thus $$\frac{1}{M} \lt \frac{(M+M_C)^2}{M^3} \leq \frac{2 \pi G}{P K^3}$$ or $M\gt P K^3/(2 \pi G)$

## Black Hole Mass Distribution

• Supernovae believed to create low-mass black holes
• Almost all galaxies thought to have black holes, $10^6 - 10^{10}~\mathrm{M}_{\odot}$
• Neutron star maximum mass less than 3 $\mathrm{M}_{\odot}$
• Very few black holes with masses less than 5 $\mathrm{M}_{\odot}$ (why?)
• Intermediate mass black holes $\left(100-10^6~\mathrm{M}_{\odot}\right)$ are difficult to create
• Maybe Pop III stars?

## Milky Way BH

Guidry ch. 18
• Observe proper motion of stars near galaxy's central BH
• Motion provides mass distribution, $M \sim 2.6 \times 10^{6}~\mathrm{M}_{\odot}$
• Why is it just one? Why not two orbiting BHs?

• To explain: start with observer inside the horizon with $p_0 = 0 \Rightarrow U^{0}=0$ and no angular momentum, then since $\vec{U}\cdot\vec{U}=-1$, $$U^{r} = - \left( \frac{2 M}{r} -1\right)^{1/2}$$
• Now consider a radially moving photon with $E=\pm p^{r}$
• Energy of photon relative to observer is $-\vec{p}\cdot\vec{U}$: $$-\vec{p}\cdot\vec{U}= - p^{r} U^{r} g_{rr} = - \left( \frac{2 M}{r} -1\right)^{-1/2} p^{r}$$ which is positive only if $p^{r} \lt 0$, but if $E=p^{r}$, then a negative energy photon travels inside the event horizon (in a finite proper time)
• Since the negative energy photon is allowed to travel in, the positive energy photon can travel out to infinity.

• What is the energy of the outgoing photon? Now consider an observer freely falling in to $r=2M$ from a finite radius $R=2M+\varepsilon$
• The observer has $$\tilde{E} = \left(2 - \frac{2 M}{2 M + \varepsilon}\right)^{1/2} \approx(\varepsilon/2M)^{1/2}$$
• Remember that $$d \tau = - \frac{dr}{\left(\tilde{E}^2-1+2 M/r\right)^{1/2}}$$
• Integrate this $$\Delta \tau = - \int^{2M}_{2 M+\varepsilon} \left(\frac{2 M}{r} - \frac{2 M}{2 M + \varepsilon}\right)^{-1/2} dr$$ which to lowest order in $\varepsilon$ gives $$\Delta \tau = 2 \left(2 M \varepsilon\right)^{1/2}$$

• The energy of the outgoing photon in this frame is limited by the uncertainty principle $\Delta \tau = \hbar/2/{\cal E}$, thus $${\cal E} = \frac{\hbar}{4} \left(2 M \varepsilon\right)^{-1/2}$$
• But the dot product $-\vec{p} \cdot \vec{U}$ is invariant, so $-\vec{p} \cdot \vec{U}$ is also the photon as observed at infinity
• At infinity, the observer is effectively stationary, so $U^{0}$ is the only nonzero component, $$U_0 = - \left(1 - \frac{2 M}{r} \right)^{1/2} \quad \mathrm{and} \quad g^{00} {U_{0}}^2 = -1$$ but this is just equal to $-\tilde{E} \approx [\varepsilon/(2M)]^{1/2}$, so $${\cal E} = - g^{00} p_0 U_0 = g^{00} U_0 E = E \left( \frac{\varepsilon}{2 M} \right)^{-1/2}$$ thus $E=\hbar/(8 M)$

• In fact, photons are emitted with a black body spectrum with a temperature $$T_H = \frac{\hbar}{8 \pi k_B M}$$ where the peak temperature is $E \approx 1.58 \hbar/(8 M)$.
• Negative energy photons decrease the black hole mass and thus increase its temperature
• Because the temperature is proportional to $1/M$ and the area is proportional to $M^2$, the luminosity goes like $M^{-2}$ and the lifetime like $M^3$.
• Not yet directly detected

## Area theorem and Entropy

• From thermodynamics, $dE = T dS$, but the black hole's energy is just its mass and it's temperature is just it's inverse mass
• Thus the entropy is proportional to the area, and the full result is $S=k_B A/(4 \hbar)$
• Since $dA/dt \geq 0$, and the black hole's entropy always increases
• In usual units, $$S = \frac{k_B A}{4 \ell_P^2}$$ where $\ell_P$ is the Planck length $\ell_P = \sqrt{G \hbar/c^3}$