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Chapter 15

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)

Step-by-Step

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.