**Black things: big and small**

Atish started with some basic introduction to black hole thermodynamics in string theory. The Bekenstein-Hawking entropy is "A/4G" where "A" is the horizon area and "G" is Newton's constant. In string theory, this value can be checked for many BPS objects by microscopic calculations. One usually obtains the result in the form

- S = 2.pi.sqrt(N1.N2.N3)

where "N1,N2,N3" are three kinds of charges expressed as multiples of the elementary charge - i.e. as integers - for example the number of D1-branes, the number of D5-branes, and the momentum (or some of their dual quantities). When you consider four-charge black holes, you obtain a remarkably similar formula

- S = 2.pi.sqrt(N1.N2.N3.N4)

to the result for three charges. The cubic and quartic expressions can actually be generalized to more general values of many types of charges as long as you replace the product of three or four factors by the corresponding cubic or quartic invariant of the duality group - an invariant expressed in the representation in which the charges transform. Also, one can consider rotating black holes with angular momentum "J". In these cases, the formulae above are typically modified to have "n.w-J" instead of just "n.w" - in an example with a momentum "n" and a winding "w". Better-than-expected agreement emerges for near-extremal black holes and even some classes of non-extremal ones. In all cases, if you express these nice results - 2.pi times a square root of an integer - in terms of the geometric parameters describing the black hole, you get an agreement with the Bekenstein-Hawking entropy.

Note that in the first "Strominger-Vafa" example with three charges, you obtain zero if one of the charges is zero i.e. if there are only two charges. Indeed, the classical horizon area shrinks to zero. However, this is only the case if you neglect the stringy corrections to the Einstein-Hilbert action. Once you include higher-derivative (four-derivative) corrections, your equations of motion change and the black hole develops a horizon even for a vanishing value of the third charge. The classical horizon area is nonzero but the entropy is no longer "A/4G". You must use the full Wald's formula that extracts the entropy as a function of the Lagrangian. The formula is essentially

- Entropy = int (horizon) 4.pi. partial(Lagrangian)/partial(R_{rtrt})

where "R_{rtrt}" is a component of the Riemann tensor in the r-t plane, a plane transverse to the (D-2)-dimensional horizon; imagine that you choose coordinates at each point of the horizon that put the metric to the standard Minkowski form. If you only have the Einstein-Hilbert Lagrangian "R/16.pi.G", the integral above gives you "A/4G". If you have "R^2" corrections, you will integrate not only a constant over the horizon but also a multiple of "R".

Typically, when you do such things in "R^2" theories, the black hole entropy becomes "A/2G": one half of the result is the old Bekenstein-Hawking entropy and the other half arises from the new term of the Wald formula. However, you should keep in mind that only the final result for the entropy is physical i.e. convention-independent. When you redefine the metric "g_{mn}" by adding a small multiple of "R_{mn}", for example, this field redefinition won't be visible at infinity but it will affect the way in which the entropy "A/2G" is divided between the Bekenstein-Hawking and the subleading Wald contributions.

**Heterotic black holes**

Atish's name is also associated with the Dabholkar-Harvey states, supersymmetric perturbative excitations of the heterotic string. These states only have the excitations on the left-moving, bosonic side, while the right-moving, supersymmetric sector of the string is kept in its ground state which guarantees that SUSY is unbroken. Atish decided to study the CHL string. The CHL string is a "Z2" orbifold of the "E8 x E8" heterotic string in which the "Z2" also exchanges the two gauge groups, together with a shift of a circular coordinate by one half of its circumference. Visually, in the Hořava-Witten strongly coupled picture, such a procedure makes the CHL string clearly equivalent to M-theory on the Möbius strip.

Atish decided to look at the CHL string because it can still preserve the same supersymmetry as the heterotic string while allowing one to investigate more discrete choices and see new mathematical structures. In all these heterotic cases, one only encounters the vector multiplets among the states of the black hole. This means that all contributions to the index are positive, and the index can therefore be identified with the exponentiated entropy. When he switches from the heterotic string to the CHL string, the resulting formula for the degeneracy of the states changes a bit. For the heterotic string, you would get a Bessel result "I_{13}" for some argument. The CHL operation effectively identifies two groups of 8 bosons which reduces "I_{13}" to "I_{9}". However, there are different rules and normalizations for the states charged under the gauge symmetry, so that the result must be multiplied by a theta function describing "E8" current algebra at level two; that should not be surprising because by identifying two "E8" groups, we give the CHL string its level two.

The group "Sp(2,Z)" of symplectic "4 x 4" symplectic matrices plays a rather important role in these calculations, much like the analysis of its modular functions. Mathematical results from the 19th century due to Jacobi and many others are relevant in this enterprise. One can show that the resulting formula corresponds to the partition sum of the bosonic string's CFT on a genus 2 surface, as recently argued by Davide Gaiotto. Indeed, the "Sp(2,Z)" is the group of large diffeomorphisms of a genus 2 surface that rearranges its four one-cycles in a way that preserves the intersections numbers. While the occurence of a genus two Riemann surface could look like a bizarre mathematical coincidence, there is actually a chain of arguments to map the black hole states into a string network and then a genus two M2-brane.

At any rate, these heterotic examples work perfectly and the correct number of the dyonic states in vacua with 16 supercharges was counted, re-counted, and re-recounted. In a similar way, the Calabi-Yau examples based on type II string theory (OSV etc.) also work pretty well.

**Natural unit of entropy**

Incidentally, the general stringy microscopic formulae for the black hole entropy have the form of 2.pi times a square root of an integer, as argued above. In all those naive attempts based on "loop quantum gravity", many people wanted to obtain a brutally discrete spectrum of entropy, something like "ln(3) times integer". First of all, the "integer" is replaced by "sqrt(integer)" in the correct theory. But you may argue that this difference can be suppressed for a proper choice of a rotating black hole. However, you also see a difference in the prefactor. The chaotic, confused and inconsistent Immirzi-like prefactors are replaced by a nice number "2.pi" in string theory - this number technically arises from the application of Cardy's formula.

The numerologists among the readers could ask: is there a relation between "2.pi" and "ln(integer)"? The answer is Yes. Most people interested in mathematical physics already know that

- 1 + 2 + 3 + 4 + 5 + ... = -1/12.

This result is easily expressed as "zeta(-1)" where "zeta" is Riemann's zeta function. But can we also calculate the product of all integers?

- 1 x 2 x 3 x 4 x 5 x ... = ?

Yes. Take the logarithm of the product. You obtain

- ln(1) + ln(2) + ln(3) + ... = ?

Can we regularize this mildly divergent sum? Yes. These logarithms of integers occur in the *derivative* of "zeta(s)" with respect to "s", namely in "zeta'(s)". If you want to get rid of the power law factors from "zeta(s)", you choose "s=0". Don't forget that the logarithms appear with a minus sign if you differentiate. In summary,

- ln(1) + ln(2) + ln(3) + ... = -zeta'(0) = ln(sqrt(2.pi))

This means that

- 1 x 2 x 3 x 4 x 5 x ... = sqrt(2.pi)

Whether you can combine different versions of loop quantum gravity in such a way that you obtain the correct entropy is up to your imagination. ;-)

"1 x 2 x 3 x 4 x 5 x ... = sqrt(2.pi)

ReplyDeleteWhether you can combine different versions of loop quantum gravity in such a way that you obtain the correct entropy is up to your imagination. ;-)" - Lumos

Lumos, this may lead to the final theory! I'm thinking about the Dirac sea and the way different particles are associated with their neighbors, as modelled by the product of integers you give. Of course the product looks infinite, but isn't. It's just great how maths works :-)

sqrt(2.Pi) = 2.5066... = 1 + 1.5066...

Now: 1.5066 x 137.03597 x 0.510999 = 105.503 MeV

The muon mass is 105.658 MeV. 0.1% higher than the value above. However we know that the correction factor for the magnetic moment of an electron or muon to the first vacuum interaction correction is 1 + 1/(2 x Pi x 137.0397) ~ 1.001, so assuming this also factor for some reason applies here as well,

Mass of muon ~ [mass of electron],[sqrt(2.pi)].[1/alpha].[1 + (alpha)/(2.pi)] = 105.63 MeV

(Above, the value 137.03597 = 1/alpha is the well known Sommerfield dimensionless number, and 0.510999 MeV is the rest-mass energy of the electron.)

Quantoken's grand unified theory is something along these lines, which were also used in a grand unified theory of Sir Arthur Stanley Eddington ("gematria" is where you take Hebrew words which also have numerical meanings, and add up the value of the letters, getting 666 for the Hebrew version of "string theorist" for instance, but I suppose this stuff is just called "numerology").

It is fairly easy to work out that the 137 number can be interpreted as the ratio of electron core charge to the weaker strength beyond the veil of polarised vacuum around the core (see my page).

What is harder to work out is the correct heuristic justification for the 1.5 number, and of course the exact value. I don't recall what Quankoken uses.

A.O. Barut came up with an empirical formula for lepton masses in the Physical Review Letters, v42, p1251 (1979).

I've also seen a paper on arXiv by A. Rivero which studies some kind of relationship of this variety.

Of course to Lumos, it is all crackpot unless it instantly finds a place within string theory.

I also recall a remark on Peter Woit's blog by Carl Brennen which says that if you add up the individual square roots of the masses of the three leptons, and square that sum, then divide that number by the ordinary sum of the three lepton masses, you get the dimensionless number 1.5. In fact, I've just done that calculation and it only gives 1.4991, far too low for this problem.

It is easy to dismiss ideas this this as crackpot. But at least until 1979 the Physical Review Letters editor thought it might be wise to keep an eye on empirical relationships between coupling constants and elementary particle masses.

The problem is that whatever the truth is, whether string theory or LQG, some kind of connection with reality of these numbers is needed. You have three leptons and three families of quarks. The quark masses are not "real" in the sense that you can never in principle observe a free quark (the energy needed to break a couple or traid of quarks apart is enough to form new pairs of quarks).

So the real problem is explaining the observable facts relating to masses: the three lepton masses (electron, muon, tauon, respectively about 0.511, 105.66 and 1784.2 MeV, or 1, 1.5 and 25.5 respectively if you take the electron mass as the unit), and a large amount of hadron data on meson (2 quarks each) and baryon (3 quarks each) masses.

When you multiply the masses of the hadrons by alpha (1/137.0...) and divide by the electron mass, you get, at least for the long-lived hadrons (half lives above 10^-23 second) pretty quantized (near-integer) sized masses:

Mesons

Charged pions = 1.99

Neutral Pions = 1.93

Charged kaons = 7.05

Neutral kaons = 7.11

Eta = 7.84

Hyperons

Lambda = 15.9

Sigma+ = 17.0

Sigma0 = 17.0

Sigma- = 17.1

Xi(0) = 18.8

Xi(-) = 18.9

Omega- = 23.9

Of course the exceptions are nucleons, neutrons and protons, which have both have masses on this scale of around 13.4.

I think it is a clue to why they are relatively stable compared to all the other hadrons, which all have half lives of a tiny fraction of a second (after the neutron, the next most stable hadron found in nature is the pion of course, which has a half life of 2.6 shakes (1 shake = 10^-8 second).

The general formula for at least hadron masses (excluding nucleons) is given on my home page:

This idea predicts that a particle core with n fundamental particles (n=1 for leptons, n = 2 for mesons, and obviously n=3 for baryons) coupling to N virtual vacuum particles (N is an integer) will have an associative inertial mass of Higgs bosons of:

(0.511 Mev).(137/2)n(N + 1) = 35n(N + 1) Mev,

where 0.511 Mev is the electron mass. Thus we get everything from this one mass plus integers 1,2,3 etc, with a mechanism. We test this below against data for mass of muon and all ‘long-lived’ hadrons.

The problem is that people are used to looking to abstruse theory due to the success of QFT in some areas, and looking at the data is out of fashion. If you look at history of chemistry there were particle masses of atoms and it took school teachers like Dalton and a Russian to work out periodicity, because the bigwigs were obsessed with vortex atom maths, the ‘string theory’ of that age. Eventually, the obscure school teachers won out over the mathematicians, because the vortex atom (or string theory equivalent) did nothing, but empirical analysis did stuff. It was eventually explained theoretically!

Clearly what's physically happening is that the true force is 137.036 times Coulomb's law, so the real charge is 137.036. This is reduced by the correction factor 1/137.036 because most of the charge is screened out by polarised charges in the vacuum around the electron core:

"... we find that the electromagnetic coupling grows with energy. This can be explained heuristically by remembering that the effect of the polarization of the vacuum ... amounts to the creation of a plethora of electron-positron pairs around the location of the charge. These virtual pairs behave as dipoles that, as in a dielectric medium, tend to screen this charge, decreasing its value at long distances (i.e. lower energies)." - arxiv hep-th/0510040, p 71.

‘All charges are surrounded by clouds of virtual photons, which spend part of their existence dissociated into fermion-antifermion pairs. The virtual fermions with charges opposite to the bare charge will be, on average, closer to the bare charge than those virtual particles of like sign. Thus, at large distances, we observe a reduced bare charge due to this screening effect.’ – I. Levine, D. Koltick, et al., Physical Review Letters, v.78, 1997, no.3, p.424.

Koltick found a 7% increase in the strength of Coulomb's/Gauss' force field law when hitting colliding electrons at an energy of 80 GeV or so. The coupling constant for electromagnetism is 1/137 at low energies but was found to be 1/128.5 at 80 GeV or so. This rise is due to the polarised vacuum being broken through. We have to understand Maxwell's equations in terms of the gauge boson exchange process for causing forces and the polarised vacuum shielding process for unifying forces into a unified force at very high energy.

The minimal SUSY Standard Model shows electromagnetic force coupling increasing from alpha of 1/137 to alpha of 1/25 at 10^16 GeV, and the strong force falling from 1 to 1/25 at the same energy, hence unification.

The reason why the unification superforce strength is not 137 times electromagnetism but only 137/25 or about 5.5 times electromagnetism, is heuristically explicable in terms of potential energy for the various force gauge bosons.

If you have one force (electromagnetism) increase, more energy is carried by virtual photons at the expense of something else, say gluons. So the strong nuclear force will lose strength as the electromagnetic force gains strength. Thus simple conservation of energy will explain and allow predictions to be made on the correct variation of force strengths mediated by different gauge bosons. When you do this properly, you may learn that SUSY just isn't needed or is plain wrong, or else you will get a better grip on what is real and make some testable predictions as a result.

Heisenberg's uncertainty says

pd = h/(2.Pi)

where p is uncertainty in momentum, d is uncertainty in distance.

This comes from his imaginary gamma ray microscope, and is usually written as a minimum (instead of with "=" as above), since there will be other sources of uncertainty in the measurement process.

For light wave momentum p = mc,

pd = (mc)(ct) = Et where E is uncertainty in energy (E=mc2), and t is uncertainty in time.

Hence, Et = h/(2.Pi)

t = h/(2.Pi.E)

d/c = h/(2.Pi.E)

d = hc/(2.Pi.E)

This result is used to show that a 80 GeV energy W or Z gauge boson will have a range of 10^-17 m. So it's OK.

Now, E = Fd implies

d = hc/(2.Pi.E) = hc/(2.Pi.Fd)

Hence

F = hc/(2.Pi.d^2)

This force is 137.036 times higher than Coulomb's law for unit fundamental charges.

Notice that in the last sentence I've suddenly gone from thinking of d as an uncertainty in distance, to thinking of it as actual distance between two charges; but the gauge boson has to go that distance to cause the force anyway.

Clearly what's physically happening is that the true force is 137.036 times Coulomb's law, so the real charge is 137.036e.

Best wishes,

'Crackpot' (according to stringers)

Nigel:

ReplyDeleteThe only thing correct in your above comment is the fact you used electron mass times 1/alpha. I am the first one to point out that the fundamental mass unit is such that alpha times it gives the correct electron mass. The rest of your comment, I consider all nonsense. I do not have an interest in numerology. Numerology coincidences, without a reasoning, are crackpots.

The correct muon mass does not have anything to do with sqrt(2*PI), like many other particles, it's related to its mean half life time. With the best known muon half lifetime figure of 2.19703 uS, the derived muon mass is 206.760 MeV, versus the measured 206.768 MeV. I can not explain the remaining discrepancy. Maybe when they measure the moun half lifetime, they have not considered the fact that a muon in motion lives just a slight bit longer, because of special relativity time dilation. Should the correct muon half lifetime be 2.187 uS instead fo 2.197 uS, then I will be able to calculate the exactly correct muon mass.

It is still a mistery exactly why muon has the half lifetime that it has. Certainly the standard model doesn't do any better, it just convert the question of half lifetime, to the question of coupling constant. No one has an answer why various coupling constants are the values they are.

Finally, Lubos, if you obtained an entropy which contains PI as a multiplier, then it tells you it's a wrong result. The entropy is the natural logarithm of all possible microscopic state a system can be in. so entropy will necessarily be equal to something like ln(n), with n being a certain

integer. Since PI is an irrational number so you can never have an entropy which equal to PI times some formula. It does NOT work. In an entropy formula you can have ln(2), ln(3), but you can never have PI.Quantoken

Well Nigel, Quantokin... it is nice to see both you cranks getting along so well.

ReplyDeleteQuantokin, since you are a crackpot, and reside in your own compactified dementia (http://users.pandora.be/vdmoortel/dirk/Physics/Gems/CompactifiedDementia.html)

I cannot expect your knowledge of mathematics to include the interesting and beautiful interrelations between the natural logarithm and Pi.

Pi shows up in the most interesting places.

As a minor example:

ln(-1) = i pi

ln(i) = = 1/2 pi i

Goodness, I guess you don't consider -1 an integer, yet there is a connection between an integer and pi. Are imaginary numbers allowed in crackpot mathematics like yours?

I am not sure if a crank like you understands that prime numbers are integers, but there is a relationship between prime numbers, pi and the natural logarithm. Lets see if you have enough brains to find it.

You don't even have to do your own thinking, for the relationship has been known for nearly 150 years. Can you use google?