113,109 reputation
7168334
bio website motls.blogspot.com
location Czech Republic
age 41
visits member for 4 years, 5 months
seen 13 hours ago

Hi, I am a string theorist and a publicist.


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
Jan
30
comment Can a photon see ghosts?
Let me also mention that it is not true that gauge-fixing is "necessary", not even in non-Abelian gauge theories. The physical states upon the vacuum are cohomologies of the BRST operators. These cohomologies form a specific Hilbert space and to describe this Hilbert space, one doesn't have to gauge fix in any specific way. Gauge fixing is more "paramount" to describe classical backgrounds etc. but not physical multiparticle states.
Jan
30
comment Why doesn't a bike/bicycle fall if going with a high speed?
Dear @David, I just gave you +1, after more than a year when you wrote it. I don't believe a letter about the comments that the bicycle stability has nothing to do with angular momentum. Your answer is perfectly valid and I am also going to downvote all the nonsensical pop-science "alternative answers" although it's fighting windmills in this case.
Jan
30
comment Why does maximal entropy imply equilibrium?
Dear Nick, the increase of entropy is the second law of thermodynamics. When the entropy isn't maximized, the second law may really be interpreted as a "strict inequality": the entropy strictly increases, so something has to be changing about the system because $S$, a function of the variables, goes up. That's why it's not an equilibrium. The actual rate by which $S$ increases depends on the situation but as I indicated, the force trying to move the system in the higher-entropy direction is really given by $F=-\nabla (E-TS)$, a gradient of free energy. It includes $+T\cdot\nabla S$.
Jan
30
comment Why does maximal entropy imply equilibrium?
Dear Nick, as I said, I think that your statement that "something will change" is only true if the entropy is (locally) non-maximal among states with the same energy. Then it is guaranteed that the system will evolve in the direction in which the entropy increases - and free energy decreases. The force is really $-\nabla \Phi$, a gradient of the free energy, when one does it right, so this increases with the gradient of the entropy.
Jan
29
comment Can a photon see ghosts?
If you want specific states that one encounters in gauge-fixing, you have to be gauge-fixing. But the simplification in Abelian theories is that there won't actually be any loop diagrams with FP ghosts, no nontrivial Jacobians from the gauge fixing etc. Those things are non-issues for U(1) which is why the FP ghosts are redundant in this case.
Jan
29
answered Why does maximal entropy imply equilibrium?
Jan
29
answered Can a photon see ghosts?
Jan
29
answered How fast do electrons travel in an atomic orbital?
Jan
29
comment How can constants… change?
Just to be sure that the main point of your question/confusion is answered: the constant isn't actually changing in Nature. It's just our (physicist's) opinion about the value that is changing as they're using newer, and hopefully more accurate, experiments to measure such constants. The right value, whatever it is, should be somewhere in the mean value from the books, plus minus the error margin (or twice error margin, but very unlikely for the gap to be much greater than the announced uncertainty), and the uncertainty should be shrinking with time.
Jan
29
comment What is the mass of individual components in a gravitationally bound system?
In particular, all the components of the metric, $g_{\mu\nu}$, and not just $g_{00}$, matter for the reactions of electrons to themselves, external fields, etc. BTW when I said "redshift", I used the relationship between energy and frequency, $E=hf$: when an object "climbs out" of the gravitational field, its frequency is decreasing if it is a photon, and its kinetic (and total) energy is also decreasing as the gravity slows it down. This is really why the electron closer to the horizon is counted as "lighter" from the viewpoint of observers at infinity.