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## Q. 15.3

Determine how much time a black hole having 3 solar masses will take to radiate its energy.

Strategy We use Equation (15.9) to determine how long it takes a blackbody of mass-energy $M c^{2}$ to radiate all its energy.

$P(T)=4 \pi \sigma r_{ S }^{2}\left(\frac{\hbar c^{3}}{8 \pi k G M}\right)^{4}$ (15.9)

## Verified Solution

We use the mass-energy $M c^{2}$ to find that it loses its energy at the rate $P(T)=-d\left(M c^{2}\right) / d t=-c^{2}(d M / d t)$, where we have taken the time derivative. We set this result equal to Equation (15.9) and find

\begin{aligned}-c^{2} \frac{d M}{d t} &=4 \pi \sigma r_{ S }^{2}\left(\frac{\hbar c^{3}}{8 \pi k G M}\right)^{4} \\&=4 \pi \sigma\left(\frac{2 G M}{c^{2}}\right)^{2}\left(\frac{\hbar c^{3}}{8 \pi k G M}\right)^{4}=\frac{2 \sigma \hbar^{4} c^{8}}{(8 \pi)^{3} k^{4} G^{2}} \frac{1}{M^{2}}\end{aligned}

and the rate of mass loss becomes

$\frac{d M}{d t}=-\frac{2 \sigma \hbar^{4} c^{6}}{(8 \pi)^{3} k^{4} G^{2}} \frac{1}{M^{2}}=-\alpha \frac{1}{M^{2}}$ (14.10)

where we have collected the constants into α. Evaluation of the constant $\alpha=3.96 \times 10^{15} kg ^{3} / s$ is left to Problem 18.

We rewrite Equation (15.10) and integrate to find

\begin{aligned}M^{2} d M &=-\alpha d t \\\int M^{2} d M &=-\alpha \int d t \\\frac{M^{3}}{3} &=-\alpha t+C^{\prime} \\M^{3} &=-3 \alpha t+C=-3 \alpha t+M_{0}^{3}\end{aligned}

where the new constant $C=3 C^{\prime}=M_{0}^{3}$ is related to the mass $M_{0} \text { at } t=0$, The time for all the mass to radiate can be found by letting M = 0, and we find

\begin{aligned}M_{0}^{3} &=3 \alpha t \\t &=\frac{M_{0}^{3}}{3 \alpha}\end{aligned}

Now we find the time for the 3-solar-mass black hole to evaporate.

$t=\frac{\left[3\left(1.99 \times 10^{30} kg \right)\right]^{3}}{3\left(3.96 \times 10^{15} kg ^{3} / s \right)}=1.8 \times 10^{76} s =5.7 \times 10^{68} y$

This is obviously many times the lifetime of our universe, which is 13.7 billion years.