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I was playing with the Hawking temperature formula $$T_H = \frac{\hbar\ c^3}{8 \pi\ G\ M\ k_B}$$ and I thought it would be interesting to associate a velocity to this temperature: $$k_B\ T_H = \frac{1}{2} M\ v^2$$ (I didn't use the relativistic kinetic energy because both $T_H$ and $k_B$ are quite small.)

This becomes: $$v = \sqrt{\frac{\hbar\ c^3}{4 \pi\ G\ M^2}}$$ and $$p = \sqrt{\frac{\hbar\ c^3}{4 \pi\ G}}$$ where $p = M\ v$.

I tried to give an interpretation to this momentum. It's obviously not the black hole's macroscopic momentum (its mass times the speed at which the event horizon moves through space), but it may be something like its internal momentum, tied to the movement of the collapsed mass inside the gravity well.

If that's the case, I can consider this momentum as $\sigma_p$, the uncertainty on momentum used in quantum mechanics, and thus calculate the uncertainty in position $\sigma_r$: $$\sigma_r\ \sigma_p = \frac{\hbar}{2}$$ (The $=$ sign comes from the fact that in such an extreme situation the uncertainty must be minimal.)

And thus: $$\sigma_r = \sqrt{\frac{\pi\ \hbar\ G}{c^3}} = \sqrt{\pi}\ \ell_P$$ I've always found it hard to believe that the mass inside a black hole is a 0-sized object inside an infinite gravity well. Wherever I see infinities, I look for an approximation that caused it. So I was really happy to interpret this number as the actual size of the black hole's collapsed mass inside the event horizon. The fact that this number is just slightly bigger than the Planck length further convinced me.

But then I reasoned a bit more and found some possible problems with all this calculations:

  1. The Hawking temperature may not be tied to actual thermal motion of the black hole's mass.
  2. The momentum I found may not be interpreted as an uncertainty.
  3. The extreme gravity may modify Heisenberg's uncertainty principle.
  4. The uncertainty in position may not be interpreted as the black hole's mass size.

Sadly I can't give an answer to all of this by myself, so I came here looking for help.

If this question is too broad, I can split it in up to four questions, depending on where (and if) I'm wrong, so in case we can concentrate just on the first point.

Edit: I had the time today to try a better approach (at least that's what I hope).

The second equation becomes: $$k_B\ T_H = \left(\gamma - 1 \right) M\ c^2$$ where $\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}$ is the Lorentz-factor.

It follows then that: $$v = \frac{\sqrt{\hbar\ c^3 \left(16\ \pi\ G\ M^2 + \hbar\ c \right)}}{8\ \pi\ G\ M^2 + \hbar\ c} = M_p\ c\ \frac{\sqrt{16\ \pi\ M^2 + M^2_P}}{8\ \pi\ M^2 + M^2_P} = c\ \frac{\sqrt{16\ \pi\ \mathfrak{M}^2 + 1}}{8\ \pi\ \mathfrak{M}^2 + 1}$$ where $M_P = \sqrt{\frac{\hbar\ c}{G}}$ is the Planck mass and $\mathfrak{M} = \frac{M}{M_P}$ is the mass measured in units of the Planck mass.

From $p = \gamma\ M\ v$ I derived the expression: $$p = M_p\ c\ \frac{\sqrt{16\ \pi\ M^2 + M^2_P}}{8\ \pi\ M} = M_p\ c\ \frac{\sqrt{16\ \pi\ \mathfrak{M}^2 + 1}}{8\ \pi\ \mathfrak{M}}$$

After putting everything into Heisenberg's uncertainty principle: $$\sigma_r = \frac{4\ \pi\ \ell_P\ M}{\sqrt{16\ \pi\ M^2 + M^2_P}} = \frac{4\ \pi\ \ell_P\ \mathfrak{M}}{\sqrt{16\ \pi\ \mathfrak{M}^2 + 1}}$$

Note that in the case where $M \gg M_P$, or $\mathfrak{M} \gg 1$, the formulas become the same as the ones given above.

This should answer BySimmetry's comment about the absence of mass in the formula: there is a dependence, but in the limit $M \gg M_P$, the one I implicitly and involuntarily considered in my first, classical approach, the dependence is so weak it's negligible. Now however, whatever the meaning of all this is (if any), it should work for smaller masses too.

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    $\begingroup$ By multiplying by $M$ you have removed all characteristics of the original black hole from your equations and are just left with a jumble of fundermental constants. Consequently your $l_P$ is the Planck length as it is the only length you can make out of those constants. $\endgroup$ – By Symmetry Jul 10 at 12:37
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    $\begingroup$ To add to my previous comment. This means that your "size of the collapsed mass" does not depend on the amount of collapsed mass. In particular it has the same value for a black hole with no mass at all, that is empty space. For the reasons you listed (among others) I don't think your calculation means much, unfortunately. $\endgroup$ – By Symmetry Jul 10 at 13:06
  • $\begingroup$ @BySymmetry That's an interesting point. I have to say though that I didn't really remove the mass, I just hid it inside the momentum. I'm not sure a zero-mass black hole makes any sense, at ~$2 \times 10^{-8}$ kg the Schwarzschild radius is $\sqrt{\pi}\ \ell_P$. Relativistic momentum should probably be used at this point instead of $p = M\ v$. $\endgroup$ – GRB Jul 10 at 15:10

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