LOG#174. Basic black hole physics.

Posted on by July 30, 2015

hidden_blackhole_lg BH

In this year, 2015, we celebrate the Light Of Year (LOY) and one century of general relativity (GR), a misnomer for something that should be called relativistic theory of gravity.

Let me celebrate the centenary of general relativity with some (not-ordered) posts about different topics, ultimately related to black holes.

What are black holes? Well, simply put, black holes are objects so dense that the own light can not, at least in classical terms, escape from them! In fact, a naive simple calculation in classical physics (that should be taken with care and not as complete or rigorous in GR or even in the framework of quantum theories) allow us to compute the escape velocity of light! The escape velocity is the speed at which the sum of an object’s kinetic energy and its gravitational potential energy is equal to zero. Mathematically you get it as follows

(1)   \begin{equation*} \dfrac{1}{2}mv_e^2-\dfrac{GMm}{R}=0 \end{equation*}

and thus

(2)   \begin{equation*} \boxed{v_e=\sqrt{\dfrac{2GM}{R}}} \end{equation*}

In fact, escape velocity IS the speed something/someone needs to become freedom from the gravitational attraction of a massive body or material system (with mass M), without further propulsion, i.e., without spending more propellant. When you plug the escape velocity equals to the speed of light, i.e., inserting v_e=c in the previous formulae, you get the so-called Schwarzschild radius R_S, the radius of a non-rotating, uncharged black hole without cosmological constant:

(3)   \begin{equation*} \boxed{R_S=\dfrac{2GM}{c^2}} \end{equation*}

This results are completely valid for a black hole of mass M (energy E=Mc^2), non-rotating, uncharged and in 4 space-time dimensions without cosmological constant. Thus, we have a non-rotating uncharged black hole in 4 dimensions with escape velocity v_e, mass M, energy E and the above Schwarzschild radius. Are there any other magnitudes we should note or calculate in black hole physics and thermodynamics! Yes, there are. Assuming spherical symmetry, we can calculate the Schwarzschild area or event horizon/surface area of the Schwarzschild’s black hole…

(4)   \begin{equation*} \boxed{A_S=4\pi R_S^2=\dfrac{16\pi G^2 M^2}{c^4}} \end{equation*}

We can also calculate the surface gravity \kappa, if the gravitational field of the black hole reads

    \[g=\dfrac{GM}{R^2}\]

then, at the Schwarzschild radius it becomes the mentioned surface gravity \kappa=g(R=R_S):

(5)   \begin{equation*} \boxed{\kappa=\dfrac{c^4}{4GM}} \end{equation*}

Interestingly, this surface gravity is 1/M the maximal force c^4/4G allowed by natural units…What else? Surface tides, or more precisely, the tidal acceleration at the black hole surface (sometimes called event horizon). The tidal acceleration is calculated with

(6)   \begin{equation*} a_T=\dfrac{2GMd}{R^3} \end{equation*}

If it is evaluated at R_S you get

(7)   \begin{equation*} \boxed{a_T(BH)=a_T(R=d=R_S)=d\kappa=R_S\kappa=\dfrac{c^6}{4G^2M^2}} \end{equation*}

Blackbody physics…A blackbody with temperature T has a luminosity given by the Stefan-Boltzman law

(8)   \begin{equation*} \boxed{L=A\sigma T^4=\dfrac{\hbar c^2}{3840\pi R^2}} \end{equation*}

The radiation flux is

(9)   \begin{equation*} \phi =\dfrac{L}{4\pi R^2}=\dfrac{\hbar c^6}{61440\pi^2 G^2}\left(\dfrac{1}{M^2 R^2}\right) \end{equation*}

and where L is the black body luminosity. This luminosity gives, evaluated at the Schwarzschild radius

(10)   \begin{equation*} \boxed{L_{BH}=L(R=R_S)=\dfrac{\hbar c^6}{15360\pi GM^2}} \end{equation*}

From this equation, you also have

(11)   \begin{equation*} -\dfrac{dE}{dt}=L=\dfrac{\hbar c^6}{15360\pi GM^2} \end{equation*}

and using E=Mc^2

(12)   \begin{equation*} -\dfrac{dM}{dt}=L=\dfrac{\hbar c^4}{15360\pi GM^2} \end{equation*}

and you can integrate this last equation the evaporation time of a black hole with initial mass M and final mass equal to zero, in order to get the evaporation time of this Schwarzschild’s BH t_{ev}

(13)   \begin{equation*} \boxed{t_{ev}(BH)=\dfrac{5120\pi G M^3}{\hbar c^4}=8.407\times 10^{-17} M(kg)^3\dfrac{s}{kg^3}=\left(\dfrac{M}{M_\odot}\right)^3\cdot 2.099\cdot 10^{67}yrs} \end{equation*}

and where M_\odot is the solar mass, M the mass in kilograms and yr indicates years. Finally, the gold medal: Hawking’s discovery! S. W. Hawking discovered that BH has a temperature, despite the fact he originally thought on the contrary! Jacob Bekenstein works on black hole physics produced a powerful set of analogies between thermodynamics and black holes. Bekenstein suggested that black hole entropy was like an area…Much to the disgust of Hawking, applying his knowledge of Quantum Field Theories (QFT) on curved spacetimes, and basic ideas of Quantum Mechanics, he finished probing that black holes are not black at all, they are hot and radiate like a blackbody with a temperature, for even the classical non-rotating Schwarzschild black hole we are studying here, given by

(14)   \begin{equation*} \boxed{T_{BH}=\dfrac{\hbar c^3}{8\pi k_BGM}} \end{equation*}

This formidable discovery allowed to complete and prove the thermodynamical analogy (identity or duality?) the Bekenstein suggested. In fact, Bekenstein did know that he could only state that black hole entropy was proportional to the event horizon area (or surface area) up to a multiplicative number. Assuming the QFT calculation of Hawking is right (and it is, it is solid and checked), and that

    \[S_{BH}=\alpha A_{BH}\]

You differentiate this to get

    \[dS_{BH}=\alpha k_B dA_{BH}\]

But

    \[A_{BH}=A_S=4\pi R_S^2\]

so

    \[dA_{BH}=8\pi R_S dR_S\]

Inserting the value of the Schwarzschild radius in terms of mass M, the gravitational constant G and the speed of light c, we get

    \[dA_{BH}=\dfrac{32\pi G^2}{c^4} M dM\]

The energy/mass of the black hole is

    \[dE=dMc^2=T_{BH}dS_{BH}\]

but according to our results

    \[T_{BH}dS_{BH}=\dfrac{\hbar c^3}{8\pi k_B GM} dS_{BH}=\dfrac{\hbar c^3}{8\pi k_B GM} \alpha k_B dA_{BH}=\dfrac{\hbar c^3}{8\pi k_B GM} \alpha k_B \dfrac{32\pi G^2}{c^4} M dM\]

or equivalently

    \[dE=dMc^2=TdS=\dfrac{\alpha 4\hbar G}{c} dM\]

and from here we can determine the numerical constant \alpha in order to fully match the two expressions…It yields

    \[\alpha=\dfrac{c^3}{4G\hbar}\]

and then, the BH entropy of the Schwarzschild solution of Einstein’s Field Equations becomes

(15)   \begin{equation*} \boxed{S_{BH}^{S}=\dfrac{k_B c^3}{4G\hbar} A=\dfrac{1}{4} k_B\left(\dfrac{A}{L_P^2}\right)=\dfrac{4\pi k_BG M^2}{\hbar c}=\dfrac{4\pi k_BM^2}{M_P^2}=4\pi k_B\left(\dfrac{M}{M_P}\right)^2=\dfrac{\pi k_Bc^3 A}{2Gh}} \end{equation*}

You got it! The equation that Hawking himself has asked to be in his own tombstone…I thought to title this blog entry as Hawking’s tombstone equation, but it was very creepy so I changed my mind. You can easily check that result by yourself…That is, I suggest you to calculate

    \[ dE=dM c^2=T_{BH}dS_{BH}\]

for this particular case. Many other black holes are similar, but only they have more complex formulae for the BH temperature and the BH entropy.

Remark: Generally, black hole entropy units are in the international system and in base e (or natural), so they are written in terms of nats . There are many other cool entropy units:  hartleys (previously called bans, ban in singular) or shannons, where

    \[ 1 \mbox{nat}=\dfrac{1}{\ln 10} \mbox{ban}=\dfrac{1}{\ln 10}\mbox{hartley}=\dfrac{1}{\ln 2} \mbox{shannon}\]

or

    \[ 1 \mbox{ban}= 1 \mbox{hartley} = \log_2 10 \mbox{bit}= \ln 10 \mbox{nat}\]

You can calculate all these wonderful black hole magnitudes and many others in the site http://xaonon.dyndns.org/hawking/ but you must be aware that the entropy units are bans/hartleys and not nats!

See you in my next blog post!

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Comments

LOG#174. Basic black hole physics. — 3 Comments

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