B "Gravitational Compression in Neutron Stars"

[sevenperforce]

If Hawking's predictions are correct all the way down to the Planck scale, then there would have to be a minimum-mass black hole not based on the Planck mass, but based on the point at which the Hawking radiation particles would have an energy equal to half the mass-energy of the black hole.

I don't know where you are getting this from. If you have a peer-reviewed paper that shows how this result is derived, or if you can give the derivation yourself, then please show your work. Otherwise, as above, you're just handwaving.

Certainly. I was speaking qualitatively for the sake of brevity up until now, but the math is definitely where things get interesting.

Extrapolating Hawking's predictions all the way down to the Planck scale for very, very small black holes will run you into nonsense pretty fast because the power output goes to infinity as the mass goes to zero. It may be, of course, that Hawking's equations simply no longer hold well before this point is reached. But if they do hold true all the way down, then the very last step of black hole evaporation must necessarily involve the entire remaining mass-energy of the black hole being released at once. However, Hawking's model requires that the radiation match a blackbody curve corresponding to the black hole temperature. Since the wavelength is also going to zero as the mass goes to zero, the particle energy is necessarily going to go to infinity, and so the minimum possible mass will correspond to the point where the peak wavelength of the Hawking radiation corresponds to a particle energy E_p = hc/λ(T_H) matching the total mass-energy of the black hole E = Mc²:

hc/λ(T_H) = Mc² or M = h/λc(T_H)

For the sake of momentum conservation, however, this has to be adjusted; you actually need two particles, each with half the mass-energy of the black hole. I did the math in the other thread here.

 

[sevenperforce]

As an example of such a calculation, consider that the Hawking temperature of a black hole is about $6 \times 10^{-8} \times M_\text{S} / M$ K, where $M_\text{S}$ is one solar mass. If we suppose that a "hypernova" of some sort requires a temperature of, say, $6 \times 10^8$ K (probably a significant underestimate) in order to blast the upper layers of a collapsing star away and prevent them from falling in with the core, then a hole mass of about $10^{-16} M_\text{S}$ would be required, or about $10^{14}$ kg.

What density does this correspond to? We can't just calculate the Euclidean volume of a sphere with radius equal to the Schwarzschild radius for the above mass, because space in this case is not Euclidean. But we can estimate the density using the math of the Oppenheimer Snyder model, at the instant when collapsing matter of this mass would just be falling through its Schwarzschild radius. The key is that the matter region in this model looks like a portion of a closed universe that is collapsing toward a "Big Crunch", and we can calculate the 3-volume of such a closed universe if we know the scale factor and the range of spatial coordinates that it occupies.

This gives $V \approx 9.96 M^3$, but $M$ here is in geometric units, so we need to add a factor of $(G / c^2)^3$ to obtain $V = 4.07 \times 10^{-81} M^3$, where $M$ now is in kilograms. So the density will be $M / V = 10^{81} / 4.07 M^2 \approx 10^{52}$ kg per cubic meter.

I'll leave it to you to compare this to typical neutron star or even quark-gluon plasma densities to figure out how likely it is that any kind of shock wave could compress matter to that density without being successfully resisted by the pressure of the matter.

Perfect! This is exactly what I was trying to get a feel for.

I would presume that neutron-degenerate matter (and, subsequently, quark-gluon plasma) will resist up to a certain failure pressure, and then they will essentially "collapse" and stop resisting altogether. Is that correct, or will degeneracy pressure continue to provide resistance even after they have been compressed past their failure point?

 

[PeterDonis]

Hawking's model requires that the radiation match a blackbody curve corresponding to the black hole temperature

More precisely, Hawking's model requires that the radiation match a black body curve for macroscopic black holes, i.e., holes for which the corresponding "quantum number" (the number of "area quanta" or whatever other microstate turns out to be the right one for doing statistical mechanics) is very large. For small quantum numbers, the concept of "black body curve" has no meaning. That's not just true of black holes; it's true of any system whatsoever. Go and look at how the black body formula is derived for the case of an ordinary object: you will see that it assumes that the number of atoms in the object (here atoms are the microstates on which statistical mechanics is based) is very large. If it isn't, the derivation doesn't work.

The rest of your reasoning is based on the mistake I just described, so it isn't valid.

 

[PeterDonis]

the very last step of black hole evaporation must necessarily involve the entire remaining mass-energy of the black hole being released at once.

Yes, because one Planck mass (more or less--actually, as I said in a previous post, the quantum is 1/4 of the Planck area, so the minimum mass for a hole would be the mass corresponding to that area) is the smallest possible mass, so when a one Planck mass black hole evaporates, it does it in one step, so to speak. This has nothing to do with black body radiation (as my previous post showed, the concept of "black body radiation" has no meaning for systems with small quantum numbers). It's just a simple consequence of the fact that black hole masses are quantized in units of (more or less) one Planck mass (if it is a fact--bear in mind, again, that all this is speculation based on a heuristic guess about quantum gravity, with no experimental evidence to back it up).

 

[sevenperforce]

Yes, because one Planck mass (more or less--actually, as I said in a previous post, the quantum is 1/4 of the Planck area, so the minimum mass for a hole would be the mass corresponding to that area) is the smallest possible mass, so when a one Planck mass black hole evaporates, it does it in one step, so to speak. It's just a simple consequence of the fact that black hole masses are quantized in units of (more or less) one Planck mass (if it is a fact--bear in mind, again, that all this is speculation based on a heuristic guess about quantum gravity, with no experimental evidence to back it up).

Right; I was supposing that perhaps this speculation and heuristic guess are incorrect, and wondering if there was another way to come up with a minimum mass that could provide predictions we could actually test.

More precisely, Hawking's model requires that the radiation match a black body curve for macroscopic black holes, i.e., holes for which the corresponding "quantum number" (the number of "area quanta" or whatever other microstate turns out to be the right one for doing statistical mechanics) is very large. For small quantum numbers, the concept of "black body curve" has no meaning. That's not just true of black holes; it's true of any system whatsoever. Go and look at how the black body formula is derived for the case of an ordinary object: you will see that it assumes that the number of atoms in the object (here atoms are the microstates on which statistical mechanics is based) is very large. If it isn't, the derivation doesn't work.

Does emission at the peak of the wavelength function satisfy the conditions of the blackbody curve, even if the derivation itself wouldn't be valid?

The way I see it, since we can't probe the Planck scale directly (at least not now), the only way to test any quantization models would be to relate some quantum-level phenomena (like tunneling decay of quanta) to larger, measurable things (like black hole blackbody spectra) in a predictable way. Einstein didn't have a way of showing anybody individual photons, but he proved that light was composed of photons by using quantization to explain the photoelectric effect.

 

[PeterDonis]

Does emission at the peak of the wavelength function satisfy the conditions of the blackbody curve, even if the derivation itself wouldn't be valid?

Of course not. Emission at one wavelength is not the same as emission over a whole range of wavelengths, which is what the black body curve describes.

The way I see it, since we can't probe the Planck scale directly (at least not now), the only way to test any quantization models would be to relate some quantum-level phenomena (like tunneling decay of quanta) to larger, measurable things (like black hole blackbody spectra) in a predictable way.

Yes. Now go read up on the quantum gravity literature to see all the proposals that have already been made for trying to do that. As I said before, this area of research is several decades old. You would be well advised to familiarize yourself with what has already been done.

 

[PeterDonis]

Thread closed for moderation.