112,191 reputation
7164327
bio website motls.blogspot.com
location Czech Republic
age 41
visits member for 4 years, 4 months
seen 13 hours ago

Hi, I am a string theorist and a publicist.


Feb
9
comment The quantum state can be interpreted statistically, again
I think that these discussions don't really belong to a physics forum. All these papers – whether they have "can" or "cannot" in the title – implicitly and sometimes explicitly assume that there is an objective and in this sense classical "model" that underlies the probabilistic predictions. The absence of such a model can't be "proved quite rigorously" which is why additional research about such proofs is a hobby for mathematical nitpickers but the physics observations surely do imply, in the physics sense, that the positivist/Copenhagen/probabilistic/Born interpretation is the only one.
Feb
7
comment About the decay of Higgs into 2 Z bosons
It's very strange how you're canceling mistakes against correct factors. The SM Lagrangian at two places is either equivalent up to a field redefinition or one of the places are wrong. Moreover, the decay width is proportional to the squared amplitude, so if the amplitude is doubled by a factor of 2 error at the beginning, then the probabilities and decay rates will be quadrupled ie changed by a factor of 4 which doesn't cancel against 2.
Feb
7
comment Redshifting of Light and the expansion of the universe
Dear @Anixx, it is a law of physics but it only holds, via Noether's theorem, for systems whose laws of physics explicitly obey the time-translational symmetry. Cosmology doesn't belong to this list. Lawrence's confused presentation of cosmology as "photons in the Pound-Rebka experiment" i.e. in Earth's gravity field makes no sense because of the cosmological principle which guarantees that the photon energy can't depend on the location. At any rate, its energy is just getting lost in cosmology and there's no "non-vacuous" way to account for it. The law breaks down, much like the symmetry.
Feb
7
comment Redshifting of Light and the expansion of the universe
Dear @Lawrence, the cosmological principle guarantees that the Universe is uniform. So if an object, i.e. a photon, had a potential energy, it would be independent of its position. The size of the Universe is the only genuine observable in the FRW equations and it's clear that the photon's "potential energy", whatever you may mean by that, could only be a function of this size. But the total content of photon-carried energy obviously isn't a function of the size only. It follows that your confusion/analogy between Earth's gravity field and cosmology is totally invalid.
Feb
7
comment Why does maximal entropy imply equilibrium?
Dear @Nick, I don't understand your question. How am I supposed to answer such questions about thermodynamics without the laws of thermodynamics? Moreover, $U$ is conserved: that's the first law of thermodynamics. What I said is that $S$ increases, it's the second law of thermodynamics. It follows from the two laws that $U-TS$, i.e. free energy, decreases.
Feb
7
answered Is the long range neutron-antineutron interaction repulsive?
Feb
6
awarded  Enlightened
Feb
6
awarded  Nice Answer
Feb
6
awarded  Nice Answer
Feb
5
comment Different kinds of S-matrices?
Dear @Squark, it depends on whether there are lines of marginal stability in the theory (values of couplings at which some stable external states cease to exist) etc. If the "adiabatic" prescription for the S-matrix works, then the $g\to 0$ and $g\to \infty$ Hilbert spaces of free particles are isomorphic and there is a simple isomorphism. Of course, if there exist asymptotic states but this existence depends on the finiteness of $g$, then the isomorphism of the two "free" Hilbert spaces breaks down: but your original adiabatic method won't work, anyway.
Feb
5
answered Different kinds of S-matrices?
Feb
2
comment Why does maximal entropy imply equilibrium?
Dear @Nick, the second law of thermodynamics is about time evolution. It says that the entropy goes up in time, or at least doesn't go macroscopically down. Of course, one must know the detailed dynamical laws of a particular physical system to know how quickly the entropy will go up but he may be sure it will go up if it is not maximized yet.
Feb
2
comment Why doesn't a bike/bicycle fall if going with a high speed?
Dear @John, I apologize but I don't agree with the statement you make about him. The page is completely illogical. It pretends that if you weaken or reverse the angular momentum of the wheels, the gyroscopic effect won't work. But it works even with the opposite sign. The angular momentum is big enough - as everyone who was holding and tried to "precess" a rotating wheel knows - and the changes of $\vec L$ needed for the bike to fall are in the vertical direction, exactly what the contact with the road can't provide because the contact point is vertically down from the center of mass.
Feb
2
comment Why doesn't a bike/bicycle fall if going with a high speed?
The angular momentum from the rotating bikes is rather small in absolute sense but it doesn't matter because the vertical torque that can be obtained from the road - the lowest point of the tires - is even smaller, essentially zero. By moving the center of mass of the person, one may change what is the allowed "quasivertical" direction in which the conditions above hold and in which stability is maintained. At any rate, the stability does increase with the speed of the bike, exactly as the "angular momentum based" theory predicts.
Feb
2
comment Why doesn't a bike/bicycle fall if going with a high speed?
I've made about 30,000 kilometers on my bikes throughout the years, most of which was in a hands-free model, so I assure you that the existence of the handles doesn't have much to do with the stability of my bike when I am riding it. The angular momentum goes in the left horizontal direction; when the bike is falling, it would go slightly up or down. So the change of the angular momentum or torque would have to be in the vertical direction. But that's exactly the direction of torque that the contact with the road can't offer us because it's down from the center-of-mass.
Feb
1
comment Nonextensive statistical mechanics
Also, what I find misleading about the non-additive formula for $S(P\times Q)$ is that it tries to pretend that $S(P\times Q\times R)$ have to be determined after that. Effectively, the formula introduces some "two-body-interaction-like" modifications of entropy but pretends that there isn't any three-body or higher-body interaction. This is just meaningless. And the more complex interactions - which are really relevant for the very-many-body entropy - do matter. At most, this business may inspire one to get some new probability distributions but there are infinity of others, too.
Feb
1
comment Nonextensive statistical mechanics
I know that the papers with the "nonextensive/Tsallis entropy" keyword say the same things that you do but it makes no sense, @propaganda. There isn't any "new kind of statistical mechanics" here. The entropy is exactly additive in particles if they're perfectly non-interacting (ideal gas) but in almost all systems, they are interacting (real gases, liquids, or anything else) and this is normally taken into account when calculating the proper good old logarithm-based entropy. Various additivities break down due to the interactions; however, no evidence for the Shannon entropy to break down...
Jan
31
comment How fast do electrons travel in an atomic orbital?
It's an order-of-magnitude estimate, so I mean any bound state - any state for which the probability is nearly 100% at all times for the electron to be less than $r<r_0$ for some large but fixed $r_0$. Mathematically, they're arbitrary linear superpositions of the bound (discrete spectrum) energy eigenstates: the estimate above works for them.
Jan
31
comment Do Maxwell's Equations overdetermine the electric and magnetic fields?
Dear @Nick, Lubos or Lubosh is surely easier to write and I don't get insulted. Many people, even those outside Central and Eastern Europe where š may be typed on keyboard, are actually able to write the character in a second or so, by copy-and-paste etc., so it's not a huge sacrifice if they write it correctly. But yes, š is pronounced as sh.
Jan
31
comment Do Maxwell's Equations overdetermine the electric and magnetic fields?
Dear @Vladimir, ${\rm div}\,\vec D=\rho$ is a constraint for $\vec D$, surely in the technical sense. It is not a purely algebraic constraint; if it were, then it would be solvable and one could just erase some components of $\vec D$ immediately. Instead, it contains spatial derivatives. But this difference only allows the overall $\vec D$ in space to move by constant: independent of space. At individual points of the initial slice, the presence of derivatives is irrelevant for the counting and there's 1 constraint per point (except for one point in space) just like if it were an algebraic one