Question

A black hole is a region of space where gravity is so strong that nothing, not even light, can escape. Throwing something into a black hole is therefore an irreversible process, at least in the everyday sense of the word. In fact, it is irreversible in the thermodynamic sense as well: Adding mass to a black hole increases the black hole's entropy. It turns out that there's no way to tell (at least from outside) what kind of matter has gone into making a black hole. Therefore, the entropy of a black hole must be greater than the entropy of any conceivable type of matter that could have been used to create it. Knowing this, it's not hard to estimate the entropy of a black hole.
$\left(a\right)$Use dimensional analysis to show that a black hole of mass $M$should have a radius of order $GM/c^2$, where $G$is Newton's gravitational constant and $c$is the speed of light. Calculate the approximate radius of a one-solar-mass black hole$\left(M=2\times {10}^{30}kg\right)$ .
$\left(b\right)$In the spirit of Problem $2.36$, explain why the entropy of a black hole, in fundamental units, should be of the order of the maximum number of particles that could have been used to make it.

$\left(c\right)$To make a black hole out of the maximum possible number of particles, you should use particles with the lowest possible energy: long-wavelength photons (or other massless particles). But the wavelength can't be any longer than the size of the black hole. By setting the total energy of the photons equal to$M{c}^2$ , estimate the maximum number of photons that could be used to make a black hole of mass $M$. Aside from a factor of $8{\mathrm{\pi }}^2$, your result should agree with the exact formula for the entropy of a black hole, obtained* through a much more difficult calculation:

$S_{\text{b.h.}}=\frac{8{\pi }^{2}G{M}^2}{hc}k$

$\left(d\right)$ Calculate the entropy of a one-solar-mass black hole, and comment on the result.

Short answer

Part $\left(a\right)$

$\left(a\right)$The radiusof black hole is $r=1483m.$

Part $\left(b\right)$

role="math" localid="1650336523710" $\left(b\right)$The proton wavelength is equal to radius of black hole.

Part role="math" localid="1650336528503" $\left(c\right)$

role="math" localid="1650336526138" $\left(c\right)$the maximum number of photons make a black hole of mas is $S~\frac{G{M}^{2}k}{hc}.$

Part role="math" localid="1650336531208" $\left(d\right)$

role="math" localid="1650336533559" $\left(d\right)$The entropy of a one-solar-mass black hole is$S_{b.h.}=1.463\times {10}^{54}\mathrm{J}\times {\mathrm{K}}^{-1}.$

Step-by-step solution

Step: 1 Finding radius of black hole: (part a)

Using Newton's physics as,

$\begin{array}{l}\frac{GM}{r^2}=\frac{c^2}{r}\\ r=\frac{GM}{c^2}\end{array}$

For solar mass black hole is $M=2\times {10}^{30}kg.$the radius of black hole is

$$\begin{gathered} r=\frac{6.674\times {10}^{-11}\times 2\times {10}^{30}}{{\left(3\times {10}^8\right)}^2} \\ r=1483\mathrm{m}. \end{gathered}$$

Step: 2 About proton wavelength: (part b)

If we believe that a black hole's entropy is $Nk$multiplied by some logarithm, we can use the logic presented earlier to say that an order of magnitude approximation of the entropy is:

$S~NK$

As a result, we'll require an estimate of how many particles went into making the black hole. The following is the argument. We'd prefer to employ as many particles as possible because there's no way to identify how much mass or energy went into the creation of the black hole (assuming no charge or angular momentum is involved). To put it another way, we'd want to identify the lowest energy $m{c}^2$of a large particle, o r$h$for a photon, that can be employed in experiments.The proton wavelength is equal to radius of black hole.

Step: 3 The max number of photons in black hole: (part c)

We have,

$\begin{array}{c}\lambda =r=\frac{GM}{c^2}\end{array}$

Energy as

$\begin{array}{l}E=hf=\frac{hc}{\lambda }\\ E=\frac{h{c}^3}{GM}\end{array}$

The number of protons as

role="math" localid="1650337568639" $$\begin{gathered} N=\frac{\text{Total energy of the black hole}}{\text{Energy of the photon}} \\ N=\frac{M{c}^2}{E} \\ N=\frac{G{M}^2}{hc} \end{gathered}$$

Entropy as

$S~\frac{G{M}^{2}k}{hc}$

The actual value as

$S_{\text{b.h.}}=\frac{8{\pi }^{2}G{M}^2}{hc}k.$

Step: 4 Entropy of one-solar-mass black hole: (part d)

The actual black hole entropy as

$S_{b.h.}=\frac{8{\pi }^{2}G{M}^{2}k}{hc}$

Substituting the values of solar mass hole as $M=2\times {10}^{30}kg$,we get

$\begin{array}{l}{S}_{b.h.}=\frac{8{\pi }^2\times \left(6.674\times {10}^{-11}\right){\left(2\times {10}^{30}\right)}^2\times 1.38\times {10}^{-23}}{6.626\times {10}^{-34}\times 3\times {10}^8}\\ S_{b.h.}=1.463\times {10}^{54}\mathrm{J}\times {\mathrm{K}}^{-1.}\end{array}$

This is far bigger than our estimate of the sun's entropy, which is $1.66\times {10}^{34}\mathrm{J}\times {\mathrm{K}}^{-1}.$order of magnitude.