In 2002, Lubos Motl wrote an interesting paper based on Dreyer’s work on quasinormal modes and black hole entropy. Carl Brannen has now observed that Lubos makes some intriguing remarks in section 4, after finding that the asymptotics of transmission amplitudes look like

$\tau (\omega) = \frac{1}{e^{\beta \omega} + 3}$

when $j \in 2 \mathbb{Z}$, where $\beta$ is the inverse of the Hawking temperature. Lubos notes that the plus in the denominator is indicative of Fermi statistics, but the strange 3 is something new, which he calls Tripled Pauli Statistics. He says: does it mean that scalar quanta and gravitons near the black hole become (or interact with) J=1 links (triplets) in a spin network that happens to follow Pauli’s principle?

Well Lubos, as kneemo has already noted, there might just be something important going on here with our idempotent swapping parity cubes. What do you think?

### Like this:

Like Loading...

*Related*

## CarlBrannen said,

April 12, 2007 @ 2:28 am

I would add that Lubos’ result is important in that it verifies a bunch of numerical studies into the question, and also, when j is an odd integer or a half integer, the result of the calculation is the usual Bose and Fermi statistics, respectively. Thus the spin-j for j even result is particularly striking.

The natural conclusion from all this is that the fundamental spin-0 and spin-2 particles do not obey the spin statistics theorem. These are the Higgs and the graviton, which just happen to be the parts of the standard model that have not yet been observed and for which no statistics are known, other than what theory predicts.

This calls into question the spin statistics theorem. This theorem is a complicated theorem that depends on a lot of assumptions. The one I suspect is defective is Poincare invariance.

## Kea said,

April 12, 2007 @ 2:39 am

Thanks, Carl. Yes, your conclusion seems perfectly reasonable. I would like to play around a bit more with ribbon diagrams to try and understand this better.

## kneemo said,

April 12, 2007 @ 4:29 pm

Michael Duff and Sergio Ferrara posted a new paper on April 4th entitled E6 and the bipartite entanglement of three qutrits. You might want to show this to your skeptical friends over in NZ.

## L. Riofrio said,

April 12, 2007 @ 5:08 pm

Interesting result Lubos has come up with. It shows that one can do good work while blogging. At first people may ignore his result, but if true it will someday seem obvious. Kea, this might make another good ribbon diagram.

## nige said,

April 12, 2007 @ 6:18 pm

Hawking’s theory does have the problem that it assumes that pair production is occurring all the time in the vacuum, and that pairs of fermions can become separated, with one on either side of the event horizon.

Quantum field theory says that pair production in the vacuum requires either an extremely intense electric field, or an extremely high frequency oscillatory field.

You get radiation (Casimir force effects for example) in the vacuum without accompanying pair production.

To get pairs to appear in the first place, to create Hawking radiation, you basically need an electric field exceeding the IR cutoff energy which is equivalent to 10^18 volts/metre field gradient. That electric field occurs about 1 fm from an electron.

You get this by putting the IR cutoff energy for pair production/vacuum polarization (required to make QFT work correctly) into Coulomb’s law, F = qQ/(4*Pi*Permittivity*r^2). Since force F = qE where q is charge and E is field strength (v/m),

E = F/q = Q/(4*Pi*Permittivity*r^2)

we get distance r from the closest approach when two electrons with IR cutoff energy 0.511 Mev each are scattered.

When the electrons are at closest approach distance r, all their kinetic energy (0.511 MeV each, i.e. 1.022 MeV in total) is converted into electrostatic potential energy because they have been working against a repulsive electric field while approaching:

1.022 MeV = QQ/(4*Pi*Permittivity*r),

hence: r = 1.44 x 10^-15 m.

Putting this value of r into E = F/q = Q/(4*Pi*Permittivity*r^2) gives 7*10^20 v/m.

This is the minimum electric field strength needed to get polarizable pairs of particles in the vacuum.

QFT gives a threshold for pair production of E = (m^2)(c^3)/(e*h bar) = 1.3*10^18 volts/metre

This equation, E = (m^2)(c^3)/(e*h bar), is equation 359 in Freeman Dyson’s lectures http://arxiv.org/abs/quant-ph/0608140 and is also equation 8.20 in Luis Alvarez-Gaume and Miguel A. Vazquez-Mozo, http://arxiv.org/abs/hep-th/0510040

*********

Now, what is Hawking assuming? Is he assuming that pair production occurs near the event horizon of a black hole due to intense electric fields of the magnitudes estimated above?

Or some other case?

Pair production can’t occur in weaker electric fields, because if they did, there would be no electric charges in the universe. The role of any such dielectric is to polarize around charges, shielding them! This is why the electric charge increases within the IR cutoff zone (i.e., in higher energy collisions).

I don’t believe that in general black holes will radiate anything, because that contradicts experimentally known facts of QFT as given above. The renormalization of charge is an experimentally shown necessity in the correct calculation of phenomena like the magnetic moment of leptons, known to many more decimal places than any other physical constant in the whole of science.

Renormalization requires taking effective charges which correspond to the polarization and pair production cutoffs mentioned.

QFT wouldn’t work if the vacuum contained polarizable (i.e. movable) pairs of charges everywhere. It works only because the vacuum doesn’t. Hawking is assuming otherwise.

Hawking radiation from electrons (treated as black holes) is however possible and likely, see my discussion at http://nige.wordpress.com/2007/03/08/hawking-radiation-from-black-hole-electrons-causes-electromagnetic-forces-it-is-the-exchange-radiation/

## Doug said,

April 13, 2007 @ 12:46 am

Hi Kea,

I found this article and these statements attributed to Hawking in ‘CosmicLog’ by Alan Boyle:

1 – “The History of the Universe Backwards”

2 – “Information is not lost, but it is not returned in a useful way,” he said. “It is like burning an encyclopedia. Information is not lost, but it is very hard to read.”

http://cosmiclog.msnbc.msn.com/archive/2007/04/10/129538.aspx

1 – Quote #1 is how game theory often solves problems.

2 – Quote #2 perhaps suggests that ‘Black Holes’ should be treated as “Information Transformers”.

In such an approach, the gastrointestinal tract is an information transformer of food into both nutrients and waste, with information not lost but difficult to reconstitute – a biophysical ‘black hole’?

## Kea said,

April 13, 2007 @ 12:53 am

Hi Doug. Yeah, that ‘burning the encyclopedia’ idea is exactly what I used in a discussion here a few years ago. I was looking at it also from the point of view of

recreating history. Is there a way we could read what was in a book that was burnt in 1450? Sure, if we could gather the information!## nige said,

April 13, 2007 @ 8:44 am

Professor Smolin has a discussion of this entropy argument at page 90 in TTWP:

“The first crucial result connecting quantum theory to black holes was made in 1973 by Jacob Bekenstein … He made the amazing discovery that black holes have entropy. Entropy is a measure of disorder, and there is a famous law, called the second law of thermodynamics, holding that the entropy of a closed system can never decrease. [

Notice he says “closed system” conveniently without defining it, and if the universe is a closed system then the 2nd law of thermodynamics is wrong: at 300,000 years after the big bang the temperature of the universe was a uniform 4000 K with extremely little variation, whereas today space is at 2.7 K and the centre of the sun is at 15,000,000 K. Entropy for the whole universe has been.] Bekenstein worried that if he took a box filled with a hot gas – which would have a lot of entropy, because the motion of the gas molecules was random and disordered – and threw it into a black hole, the entropy of the universe would seem to decrease, because the gas would never be recovered. [falling, in contradiction to the laboratory based (chemical experiments) basis of thermodynamics. The reason for this is the role ofgravitationin lumping matter together, organizing it into hot stars and empty space. This is insignificant for the chemical experiments in labs which the laws of entropy were based upon, but itissignificant generally in physics, where gravity lumps things together over time, reducing entropy and increasing order. There is no inclusion of gravitational effects in thermodynamic laws, so they’re plain pseudoscience when applied to gravitational situations.This is nonsense because gravity in general works] To save the second law, Bekenstein proposed that the black hole must itself have an entropy, which would increase when the box of gas fell in, so that the total entropy of the universe would never decrease. [againstrising entropy; it causes entropy to fall! Hence the big bang went from uniform temperature and maximum entropy (disorder, randomness of particle motions and locations) at early times to very low entropy today, with a lot of order. The ignorance of the role of gravitation on entropy by these people is amazing.But the entropy of the universe is decreasing due to gravitational effects anyway. At early times the universe was a hot fireball of disorganised hydrogen gas at highly uniform temperature. Today space is at 2.7 K and the centres of stars are at tens of millions of Kelvin. So order has increased with time, and entropy – disorder – has fallen with time.]”On page 91, Smolin makes clear the errors stemming from Hawking’s treatment:

“Because a black hole has a temperature, it will radiate, like any hot body.”

This isn’t in general correct either, because the mechanism Hawking suggested for black hole radiation requires pair production to occur near the event horizon, so that one of the pair of particles can fall into the black hole and the other particle can escape. This required displacement of charges is the same as the condition for polarization of the vacuum, which can’t occur unless the electric field is above a threshold/cutoff of about 10^18 v/m.

In general, a black hole will not have a net electric field at all because neutral atoms fall into it to give it mass. Certainly there is unlikely to be an electric field strength of 10^18 v/m at the event horizon of the black hole. Hence there are no particles escaping. Hawking’s mechanism is that the escaping particles outside the event horizon annihilate into gamma rays which constitute the “Hawking radiation”.

Because of the electric field threshold required for pair production, there will be no Hawking radiation emitted from large black holes in the universe: there is no mechanism because the electric field at the event horizon will be too small.

The only way you can get Hawking radiation is where the condition is satisfied that the event horizon radius of the black hole, R = 2Gm/c^2, corresponds to an electric field strength exceeding the QFT pair production threshold of E = (m^2)(c^3)/(e*h bar) = 1.3*10^18 volts/metre, where e is the electron’s charge.

Since E = F/q = Q/(4*Pi*Permittivity*r^2) v/m, the threshold net electric charge Q that a black hole must carry in order to radiate Hawking radiation is

E = (m^2)(c^3)/(e*h bar)

= Q/(4*Pi*Permittivity*r^2)

= Q/(4*Pi*Permittivity*{2Gm/c^2}^2)

Hence, the minimum net electric charge a black hole must have before it can radiate is

Q = 16*Pi*(m^4)(G^2)*(Permittivity of free space)/(c*e*h har)

Notice the fourth power dependence on the mass of the black hole! The more massive the black hole, the more electric charge it requires before Hawking radiation emission is possible.

## kneemo said,

April 13, 2007 @ 6:03 pm

nige,

Concerning black hole radiation, you may find the recent paper by Yuan K. Ha to be of interest.