# Would a particle the size of a neutron, if it had enough mass, collapse into a blackhole?

For example, a neutron is a particle that occupies a certain volume. If you pack enough mass into that volume, it would collapse into a black hole (I assume there is not enough mass now). At least if you don't consider quantum effects. Now, if QM is considered would that prevent a singularity from occurring?

Another example is the electron. It's a point particle so you'd predict a collapse. But it's also spread out due to quantum mechanics. So, no collapse.

The reason for this question is to create an analog to the prediction by classical EM that an electron should lose energy and spiral into the nucleus. Quantum mechanics comes to the rescue.

• You wouldn't be using neutrons but protons because you need an accelerator to get them to really high energies. If you could do that, theoretically microscopic black holes could form. LHC data analysis is looking for the signatures of such events, even though they are extremely unlikely to occur at the energies available at LHC (or at any accelerator we can hope to build for another century, or so). I would strongly urge you to give up the point particle picture. It's completely false and useless for any such speculation. – CuriousOne May 20 '16 at 22:05
• @CuriousOne: even theoretically, boosting a compact object to a high speed won't produce a black hole. The potential LHC black holes are the results of collision events, not rapidly moving objects. – Jerry Schirmer May 20 '16 at 22:09
• @JerrySchirmer: I didn't not say anywhere that the boosting will produce a black hole, I just didn't mention that the collisions will (or better... might), which I thought was pretty obvious since I was talking about a collider? – CuriousOne May 20 '16 at 22:15
• @aK1974: I don't think that talking about a radius would make much sense for the smallest of black holes. They evaporate as soon as they form and the tidal forces are probably so large that the quantum fluctuations of the horizon (if one can even call it that) is similar to its size. Apart from that it's not clear what the relevant energy scale is. At LHC black holes could only form if there are macroscopic compact dimensions which make the Planck scale completely obsolete. – CuriousOne May 20 '16 at 22:18

If you believe classical parameters for black holes down to a neutron's scale, then it turns out that a neutron has too much angular momentum to form a black hole, and would be best interpreted as a naked singularity.

The reason why is that the radius of a black hole's event horizon is predicted to be:

$$M \pm \sqrt{M^{2} - a^{2}}$$

where $M$ is the mass of the black hole, and $a$ is the angular momentum per unit mass of the black hole. If you put in the parameters for a neutron, $a > M$, and there are no real values for the horizon.

• The neutron is as definitely not a naked singularity. Don't put nonsensical ideas into the kids' minds. :-) – CuriousOne May 20 '16 at 22:20
• @CuriousOne: that's what the model says. This is why you can't apply general relativity to neutrons. And naked singularity is a better fit than black hole. – Jerry Schirmer May 21 '16 at 0:36
• Physics is not about what models say, it's about what the data says and there is no data whatsoever that links neutrons to naked singularities (which absolutely nobody has a reason to believe even exist... based on the model). – CuriousOne May 21 '16 at 1:56
• @CuriousOne: if you don't understand the models, you don't understand physics. If you think the model is wrong, you are rejecting the predictions of the models, but the question in the OP is a question about our understanding of neutrons and black holes. There is no way to answer that without back-predicting from our knowledge of general relativity. – Jerry Schirmer May 22 '16 at 15:03
• Curiously, I do understand both the correct models for a neutron and its phenomenology and none of that has anything to do with naked singularities, for which there is neither phenomenology nor models that can fit it. – CuriousOne May 22 '16 at 21:47

The general relativity model for an electron is the Kerr-Newman solution, that is a charged rotating black holes. Unfortunately the radius of the exterior horizon in geometric units is given by:

$$r_{+}=M + \sqrt{M^2-Q^2-P^2-\left(\frac{J}{M}\right)^2}$$

Here P is the magnetic charge, zero in this case. If you convert this relation to standard units ( for instance insert the mass in Kg and multiply for the conversion factor $G/c^2$) you will find that this radius is imaginary. That is, elementary particle are naked singularities from the point of view of general relativity.

As an aside, inspired from string theory and the fuzzball proposal it seems possible to find a completely regular (no singularity!) solution describing microstates of what you would call a naked singularity in general relativity. Basically, a solution of the equation of motion with the same asymptotics charges of a naked singularity, but without pathologies! This is still working in progress anyway.

This is the classical picture. Roughly, the quantum picture tells you that when the Compton wavelenght is equal to the Schwarzschild radius (that is, at the Planck mass) you have reached the mininum mass for a black hole. So from this point of view there is no possibility for a particle to collapse .

The full answer of course cannot be given without the full quantum gravity theory in hand.

A black hole in the Schwarzschild case has a radius $$r=\frac{2GM}{c^2}$$ Let us solve for the mass with a radius of about Fermi or $10^{-15}$m, or the radius of a baryon such as a neutron, $$M = \frac{rc^2}{2G}=\frac{10^{-15}m\times 9\times 10^{16}m^2/s^2}{2\times 6.67\times 10^{-11}Nm^2/kg^2}$$ $$=6.7\times 10^{11}kg.$$ This is close to a billion tons! This would be a quantum black hole, and a pretty hot one with Hawking radiation and is racing towards it final evaporation.