How many black holes can fit on the head of a pin?

Classical Mechanics Level 5

Black holes are regions of space from which nothing can escape. If you consider a spherical object of mass M and radius R and set the escape velocity from the object to be $c$ , the speed of light, you can determine a relationship between R and M, $R=2 GM/c^2$ , where $G$ is Newton's gravitational constant. This radius is called the Schwarzschild radius, denoted by $R_s$ . If mass M is concentrated into a region with a radius smaller than $R_s$ then you have a black hole, and if not, there is no black hole.

From the above relation you can determine the minimum mass of a black hole, as roughly speaking the Schwarzschild radius must be larger or equal to the Compton wavelength - the minimum size of the region in which an object at rest can be localized.

Find the minimum mass of a black hole in $\mu g$ .

Finally, a bonus thing to think about. What does this result mean for the masses of the particles that we see in nature?

Details and assumptions

The value of the gravitational constant is $G=6.67 \times 10^{-11} \text{ m}^3\text{/kg s}^2$ .
The speed of light is $c=3 \times 10^8\text{ m/s}$ .
Planck's constant is $h=6.63 \times 10^{-34} \text{ kgm}^2\text{/s}$ .
$1 ~\mu g = 10^{-6} \text{ g} = 10^{-9} \text{ kg}$

The answer is 38.6.

3 solutions

Saurabh Dubey
May 20, 2014

The Compton wavelength of any particle is given by, L= $\frac{h}{mc}$ where L =Compton wavelength, h =Planck's constant, m =Rest mass of the particle, and c =speed of light. Given in the question, R= $\frac {2Gm}{c^2}$ As stated in the problem, R $\geq$ L For minimum mass,the radius should be minimum (as R is directly proportional to mass) i.e. R = L . Therefore, $\frac{2Gm}{c^2}=\frac{h}{mc}$ . Solving the above equality, we get, m= $\sqrt {\frac {hc}{2G}}$ . Substituting the values of h,c and G, we find that m= $3.86 \times10^-8$ kg , or 38.6 $\mu g$ .

Pravin Venkatraman
May 20, 2014

Compton wavelength = h / (M * c) So, 2GM / c^2 >= h / (M * c) Thus, M^2 >= hc / 2G

Solving M >= 38.6 * 10^ (-9)

David Mattingly Staff
May 13, 2014

The smallest black hole must have $r_s=\lambda_c$ , where $\lambda_c$ is the Compton wavelength $h/mc$ so:

$\frac{2Gm_{cr}}{c^2}=\frac{h}{m_{cr}c} \Rightarrow m_{cr}=\sqrt{\frac{hc}{2G}} = 3.86 \times 10^{-8} (kg)=38.6(\mu g)$

How many black holes can fit on the head of a pin?

The answer is 38.6.

3 solutions

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