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bio website motls.blogspot.com
location Czech Republic
age 41
visits member for 4 years, 4 months
seen 47 mins ago

Hi, I am a string theorist and a publicist.


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.
Jan
29
comment What is the mass of individual components in a gravitationally bound system?
Well, I wouldn't say that the reason why it's OK with the conservation law is the nonlinearity. The reason is the red shift, i.e. that the locally measured mass/energy in a gravitational field isn't the same thing as the contribution of this mass/energy to the total mass/energy as seen from infinity. Roughly speaking, the two quantities differ by the multiplicative constant $\sqrt{g_{00}}$, related to the gravitational potential, but studying motion of objects in general relativity is "harder" than just adding simple rescaling factors such as $\sqrt{g_{00}}$.
Jan
29
comment Does a charging capacitor emit an electromagnetic wave?
Yup! If there is a discontinuity, the function of time may be decomposed into all frequencies - pretty much all frequencies are represented, including the high ones. In practice, it may be hard to observe this flash but it's there.
Jan
29
comment Where does the energy come from when a current heats a wire (resistor)?
Don't worry about doing more! ;-)
Jan
28
comment Where does the energy come from when a current heats a wire (resistor)?
Imagine that you have two definitions of $\vec S$ such that both of them obey the continuity equation with the divergence above but the values of $\vec S$ are different. These two configurations of $\vec S$ may describe the very same physical situation. This means that $\vec S$ can't be "directly measured"; the role of $\vec S$ is to have a vector to be used in "accounting" to prove that the energy is locally conserved. Concerning the ambiguity, see en.wikipedia.org/wiki/…
Jan
28
comment Where does the energy come from when a current heats a wire (resistor)?
Oh, one more addition. In your comment (3 above me), you may have referred to my remark about the various ambiguous definitions of the Poynting vector. I meant that one may redefine the Poynting vector in such a way that the conservation law above will still hold (and the total energy will be unchanged) but the flows (circulation of energy) will be different. But this freedom to redefine $\vec S$ really means that the energy flux density vector, when loosely defined, is ambiguous and unphysical. That makes it obvious that such subtle changes of $\vec S$ can't be used to send any signals.
Jan
28
comment Where does the energy come from when a current heats a wire (resistor)?
In general relativity, by the way, the energy conservation law becomes problematic and when we want to include energy carried by gravitational waves etc., it cannot really be written locally in any meaningful way. But in non-gravitational physics, we can do it. When I said that energy was an abstract quantity, I meant that even in the vacuum, $\vec E, \vec B$ may be nonzero and generic, and therefore $\vec S$ is probably nonzero, too. Energy is flowing even though there are no "particles" over there. Nothing wrong about it. My focus was the word "abstract", like "numerical".
Jan
28
comment Where does the energy come from when a current heats a wire (resistor)?
You're welcome, bitmask. Maxwell's equations may be explicitly proven to respect relativity, including the condition that signals never propagate faster than light. I didn't really say that you may subtract energy from any point and return it anywhere. On the contrary, I quoted the differential equation with ${\rm div}\,\,\vec S$ which is manifestly local: if energy disappears from some place, it is because it is flowing to a nearby point as shown by the $\vec S$ vector. The conservation is local. I just wanted to say that energy doesn't have to have a "material carrier".
Jan
28
revised Where does the energy come from when a current heats a wire (resistor)?
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