# Why is the Schwarzschild radius associated with the tiniest micro black hole formed by a Planck mass twice the Plank length?

If one calculates the Schwarzschild radius, $r_s$, of a Plank mass $m_p=2,18*10^{-8} (m)$ one gets:

$$r_s=2{\frac{G{m_p}}{c^2}}=1,48*10^{-27}*2,18*10^{-8}=3,22*10^{-35}(m)$$

Now the Planck length $m_p$ is $1,61*10^{-35}(m)$, which is exactly half the Schwarzschild radius of the black hole associated with a Plank mass. Thus $l_p=\frac{G{m_p}}{c^2}$. Filling in the expression for $m_p=\sqrt{\frac{\hbar c} G}$ in the expression for $l_p$ gives the more familiair form for $l_p$, namely $\sqrt{\frac{\hbar G} {c^3}}$.

Why is it that the Schwarzschild radius of a Planck mass micro black hole (the tiniest that exists) is exactly twice the Planck length?

You can of course answer this question by saying that the formulae tell us that this is the case. But is the factor two by which the two lengths are connected just a coincidence, or has this connection some deeper significance?

Things like the Planck mass, the Planck length, etc. are order of magnitude estimates of when certain effects should be important. You shouldn't take it literally that Quantum gravity becomes important at EXACTLY $\ell_{p}$, and is completely irrelevant at any longer distance. It's a heuristic to help us think. In reality, just like in electromagnetism, we get a factor of $\frac{1}{4\pi}$ in front of the equation, we can expect some sort of pure number factor to modify the various relevant quantum gravity equations.
• I gave you a +1, but shouldn't the reduced Planck mass be divided by $\sqrt{8 \pi}$? Not that it's thát important... – descheleschilder Jun 26 '17 at 22:40