## Chapter 3

**On p. 43,** I wrote, "Time runs slower at the surface of the earth than it does in outer space. The difference is slight: it's a little less than one part in a billion." Here's how to calculate the amount of gravitational time dilation. The Schwarzschild metric, including all factors of $G$ and $c$, is

$$ds^2 = -f(r) c^2 dt^2 + {dr^2 \over f(r)} + r^2 (d\theta^2 + \sin^2\theta d\phi^2)$$

where

$$f(r) = 1 - {2GM \over rc^2} \,.$$

The time dilation at a radius $r$ is $\sqrt{f(r)}$. We now calculate

$$\eqalign{1 - \sqrt{f(r)} &\approx {GM \over rc^2} = {(6.67 \times 10^{-11} {\rm m^3} \, {\rm kg}^{-1} \, {\rm s}^{-2}) \times (6.0 \times 10^{24} \, {\rm kg}) \over (6370\,{\rm km}) \times (3 \times 10^8\,{\rm m}/{\rm s})^2} {1\,{\rm km} \over 1000\,{\rm m}} \cr &= 7.0 \times 10^{-10} \,.}$$

In the first step I used a binomial expansion of the square root. In the second step I plugged in earth's mass and radius for $M$ and $r$. The final result is a bit less than $10^{-9}$, which is a part in a billion.

It's correct to calculate this gravitational redshift using the Schwarzschild metric because of Birkhoff's theorem: outside a spherically symmetric object (like the earth), the metric can only be Schwarzschild, provided you don't have appreciable rotation or electric fields.

**On p. 37,** in the figure caption, I wrote "The black hole is much smaller than the size depicted here, and so is the region around it empty of stars". Later, on p. 40, I say "As the star approaches the black hole horizon, it eventually gets shredded to bits," and I mention that it's simpler to ignore the rotational motion for simplicity. It's interesting to quantify better these statements. Roughly, the size of a black hole of a million solar masses is a millionth of the distance from us to the sun. So a black hole like the one at the center of the galaxy, huge as it is, would fit easily in our solar system. It would enormously change solar system dynamics, though. For starters, earth's orbit would be about $1000$ times faster. Next, the earth would be shredded to bits by the tidal forces. To see this, we can estimate the Roche radius (or Roche limit) where the tidal forces on an object exceed its own gravitational self-attraction. It comes out to be about the distance from the earth to the sun for densities comparable to water.

**On p. 45,** I wrote "-273.15 degrees Celsius (which is -459.67 degrees Fahrenheit)." Here's how the conversion works between a temperature $T_C$ in Celsius and $T_F$ in Fahrenheit:

$$T_C = {5 \over 9} (T_F - 32) \qquad\qquad T_F = 32 + {9 \over 5} T_C \,.$$

Plugging $T_C = -273.15$ into the second formula immediately gives the claimed result.

**On p. 46,** I wrote "For example, at the melting point of ice, this formula [$E = k_B T$] says that the typical energy of the thermal vibrations of a single water molecule is a fortieth of an electron volt." To get this, we just plug in $T = 273.15 \, {\rm K}$ into the formula

$$E = k_B T = (1.38 \times 10^{-23} \, {\rm J} / {\rm K}) \times (273.15 \, {\rm K}) \times {1\,{\rm eV} \over 1.60 \times 10^{-19} \, {\rm J}} = 2.35 \times 10^{-2} \, {\rm eV} \,.$$

This is close to a fortieth of an electron volt, which is $2.5 \times 10^{-2} \, {\rm eV}$.

**On p. 47,** I said that "a black hole formed in ... gravitational collapse ... might contain a few times as much mass as teh sun. Its temperature would then be about twenty billionths of a Kelvin, or $2 \times 10^{-8}\,{\rm Kelvin}$. To calculate this, we need the formula for the black holes temperature (the Hawking temperature):

$$\eqalign{T_{\rm Hawking} &= {\hbar c^3 \over 8\pi G M k_B} = {(1.05 \times 10^{-34} \, {\rm m}^2 \, {\rm kg} / {\rm s}) (3 \times 10^8\,{\rm m}/{\rm s})^3 \over 8\pi (6.67 \times 10^{-11}\,{\rm m}^3 \, {\rm kg}^{-1} \, {\rm s}^{-2}) (6.0 \times 10^{30} \, {\rm kg}) (1.38 \times 10^{-23} \, {\rm J}/{\rm K})} \cr &\approx 2.0 \times 10^{-8} \, {\rm K} \,,}$$

where to get numbers I plugged a value for $M$ equal to three times the solar mass. If instead we used $M$ equal to five million solar masses, then the same formula would give $T_{\rm Hawking} = 1.2 \times 10^{-14} \, {\rm K}$.