This account is temporarily suspended to cool down. The suspension period ends on Aug 28 '16 at 0:06.
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494215
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location New York City
age 41
visits member for 4 years
seen yesterday

I do not participate on this site any longer, except to respond to comments regarding my own text, if that text is unavailable in another form. I do not accept the political moderation atmosphere here, it is not compatible with open science. Unfortunately, this seems to be a recurring pattern on such sites--- they grow with promises of open participation, and then shut down in a phase transition of censorious moderatorship. Hopefully physicsoverflow.org will be the first exception to this rule, as the policies there were crafted specifically to avoid this phenomenon.


Aug
25
comment Can Black Holes be the Dark Matter?
@RobJeffries: Ok, I didn't consider black holes so big--- I imagined lots of proton sized black holes. I agree with most of your comments, I haven't thought about this.
Aug
7
comment Second Law of Black Hole Thermodynamics
@Blazej: The examples you give are too simple, the boosted Schwarzschild black hole is better, or using a slice which is wiggling like t(r)=t_0 + Acos(r). The area is independent of the wiggles. The area depends on the slice only if there is a divergence (positive expansion) of the null geodesics making up the horizon between the two slices you are comparing. Otherwise, you are free to slide the points up and down and the area doesn't depend on the slice. The principle of it is the Minkowsi triangle thing I said.
Aug
7
comment Second Law of Black Hole Thermodynamics
@Blazej: It isn't there! I was annoyed by this. You have to work it out for yourself, and I did that years ago when studying this stuff, and put the main unstated theorem here, it doesn't appear anywhere else.
Aug
6
comment How well is the $\rho$ and $\omega$ coupling universality measured?
@MikeV: Oh, I didn't know that the nucleon form factor calculations were able to distinguish such fine details. Thanks, I'll look at it.
Jul
28
comment Could we prove that neutrinos have mass by measuring their gravitational signature?
@Lehs: You're asking if the cosmological neutrinos are dense and cold enough to be close to forming a Fermi surface. The answer is probably no, the density and temperature are both known, so you can check explicitly whether there is about one neutrino per typical wavelength at the thermal momentum. I didn't check myself, I don't know the density off the top of my head.
Jul
19
comment Could we prove that neutrinos have mass by measuring their gravitational signature?
@Lehs: From the manner of their creation, and dissipation. Any beta decay process or supernova happens at scales of KeVs or MeV, and the neutrinos are weakly interacting enough that they don't cool down. Cosmologically produced neutrinos during the big bang are the coolest, and the number of these can be estimated the same way you estimate the nuclear density, from Big Bang models (which are very accurate today).
Jul
4
comment Would a solution to the Navier-Stokes Millennium Problem have any practical consequences?
@mike4ty4: I explained it two comments above--- the bits in a reversible computation can be arbitrarily complex, and there should be no way to compute a subset of them from an arbitrarily small fraction of them. In order to figure them out, you need to guess a sizable fraction of them, which means exponential search. It's actually stronger than "P!=NP", it's that NP complete problems require at least exponential in a fractional power of N search, and perhaps exponential in N search. P!=NP is too weak.
Jul
3
comment Would a solution to the Navier-Stokes Millennium Problem have any practical consequences?
@mike4ty4: I suppose you could say it that way. But heuristics are useful. If I ask you if you scan "pi" looking for places with two consecutive even digits, and find their density, it's going to be 1/4, even though nobody has proved it, because the digits are "random enough" for this to be true. This is the scientific standard of evidence, not mathematical proof. But P!=NP for me is effectively certain. On the other hand, this statement about Navier Stokes is not.
Jul
3
comment Would a solution to the Navier-Stokes Millennium Problem have any practical consequences?
@mike4ty4: Proof is mathematical certainty, but scientific certainty is about heuristics and common sense. The question P!=NP is equivalent to the statement that given the output of a computation of size N going forward, you can reverse it to guess the initial condition with only polynomial effort in N. But this computation can be embedded in a reversible computation, by introducing spectator waste bits, and the number of waste bits grows as N. To reverse the computation, you need to guess the waste bits, and they have arbitrary complexity, so you need exponential search. That's not a proof.
Jun
24
comment Escape velocity of a rocket standing on Ganymede (Moon of Jupiter)
@Nazaf: I neglected R_G where it is negligible.
Jun
24
comment Please explain me how the Higgs boson gives mass to other particles, more detail?
@Paganini: I never distinguished between the two until recently! I used the terms interchangeably until someone corrected me here. But since the massless fermions of given helicity are of a fixed helicity, it doesn't make much difference, so I hope you can forgive me.
Jun
12
comment What happens to matter in a standard model with zero Higgs VEV?
I made a wrong statement in the comments above regarding the coupling of pions. I said they don't couple to nucleons by direct Yukawa interaction but by gradients, which is entirely false. This falsehood was due to an intuitive idea--- shifting the pion field just moves to a different vacuum (this is true), so a coherent condensate of pions can't interact with anything (false). The shift in vacuum affects the Nucleon because it is a massive non-chirally invariant excitation and the correct statement of this sentiment is the Goldberger Trieman relation. I apologize, the rest is ok.
Jun
7
comment QM without complex numbers
@Argyll: The real dimension of H is twice the original complex dimension. The Hamiltonian is multiplied by "i" to make the eigenvalues imaginary. This version, which isn't really mine, is just the obvious way to turn complex vector spaces into real vector spaces, by doubling the dimension, and considering a real basis consisting of the original basis vectors e_1 ... e_n and i times these ie_1 .... ie_n, so 2n basis vectors. It is easy to do in both finite dimensions and infinite dimensions, and there are no subtleties. You can check that multiplication by i acts as the 2 by 2 matrix I gave.
Jun
6
comment Is the EmDrive, or “Relativity Drive” possible?
@BAR: Failure of conservation of momentum implies failure of conservation of energy in a moving frame, because energy and momentum mix up together under boosting even nonrelativistically.
Jun
6
comment QM without complex numbers
@Argyll: i is a 2 by 2 real matrix which squares to -1. Eigenvalues are complex even when the theory is real.
Jun
3
comment Is the EmDrive, or “Relativity Drive” possible?
@BAR: Solid scientific knowledge has this property. I don't need to pick up a pencil to prove the things I am saying, I understand the theory. The gullible folks at NASA don't, so one can only feel pity for them. The claim is theoretically unsound, but it is an experimental claim, so one must look at the experiments. The experiments are also unsound, so there is no basis for this claim, and it is simply fraudulent.
Jun
2
comment Is the EmDrive, or “Relativity Drive” possible?
@VolkerSiegel: No, NASA claims it does work, which means I have more smarts than all the NASA idiots put together.
Jun
1
comment Is the EmDrive, or “Relativity Drive” possible?
@BAR: I do not need to read anything to know that this claim is false. The vacuum is a unique state, and to produce thrust, you need to produce something going the other way, either air or radiation, and radiation is ruled out as I explained. The experiments are fraudulent, and your comment is gullible.
May
12
comment Chemical potential
@Timo: I mean the energy cost for adding a particle can be negative, whatever your sign convention for the chemical potential. I thought I was using the standard convention, where positive chemical potential means it costs you a positive energy to introduce a particle, and negative chemical potential means you want to introduce a particle, so it costs energy to remove one.
May
1
comment How do we know that nonperturbative canonical quantum gravity is wrong?
@user1247: Because 2d geometries are classifiable, you can sum over them, and if you look at 3-branes, they don't delocalize the same way, they stay where you put them, so that AdS/CFT on those doesn't have to sum over all topologies of these, rather only over the string exchanges between them, or over the asymptotic 4-d boundary theory in the low-energy limit. Points delocalize, but point-black-hole AdS/CFT gives matrix theory which is even better behaved than strings. The problem just never comes up, which is one reason why strings are so amazing.