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bio website motls.blogspot.com
location Czech Republic
age 40
visits member for 3 years, 6 months
seen 10 hours ago

Hi, I am a string theorist and a publicist.


Jul
10
comment Is entropy in quantum mechanics emergent or fundamental?
Dear @user53081, I don't know how this comment differs from the original question, and if I would write something, it would probably not differ from what I already wrote. Just one extra comment: a large(r) number of degrees of freedom – if not the "thermodynamic limit" (infinitely many degrees of freedom, or atoms, if you wish) – is necessary for the entropy to be any accurate. One may formally define entropy even for the elementary particles or tiny systems but its exact value will depend on conventions and subjective knowledge too much. What's important is that these errors may average out.
Jul
10
awarded  entropy
Jul
8
answered Is entropy in quantum mechanics emergent or fundamental?
Jul
7
comment Can special relativity be extended such that the frame of a photon makes sense?
Density matrix isn't a modification of quantum mechanics. It's just a way to describe incomplete knowledge about the state vector, a straightforward combination of quantum mechanics of pure vectors and the usual classical ways to consider probabilistic weighted averages (without any information about the relative phase). Quantum mechanics cannot be modified by an epsilon without ruining it completely, and it is really the best example of the important principle I was trying to convey.
Jul
6
comment Can special relativity be extended such that the frame of a photon makes sense?
You can modify any important theory you want but if you modify a theory that is correct and essential, like relativity, you get a wrong and worthless theory. Why do you want to "modify" it? It's exactly like asking whether one may modify Darwin's theory so that animals never try to eat each other or compete with each other. Yes, you can "modify" it but the modification is wrong. The equations of relativity make clear conclusions about the value $v=c$ as well - and especially about it - and you can't cherry-pick or modify selective conclusions without destroying the whole structure.
Jul
6
answered Why is scattering vector $\vec{q}$ called vector of 'momentum transfer'?
Jul
4
answered Did “big bang” radiate light?
Jul
3
awarded  Nice Answer
Jul
3
awarded  Talkative
Jul
3
comment Electromagnetic radiation and black body radiation
The correct, quantum explanation is effectively a theory of particles because it implies that the energy carried by the field mode at frequency $f$ is always an integer multiple of the photon energy $E=hf$, i.e. $Nhf$. Because the energy is an integer multiple, the integer itself may be interpreted as the number of particles - photons - in the state.
Jul
3
comment Electromagnetic radiation and black body radiation
The classical wave theory makes wrong predictions because it's a wrong theory. The prediction for the total thermal energy is infinite because the classical theory of waves infinitely overestimates the amount of "degrees of freedom" - dynamical variables that oscillate and that can store energy. Quantum mechanics modifies these predictions so that the results for measurable quantities become finite. The frequency at which the radiation is maximized is $f\sim kT/h$ but it's just an order-of-magnitude estimate - actually all frequencies are present in the black body radiation.
Jul
3
comment Electromagnetic radiation and black body radiation
Dear @Gummybears, the classical theory of waves or electromagnetism is predicting what the field is doing at temperature $T$, but the prediction is wrong. It is self-evidently wrong because it predicts that the total energy stored in the field at a finite temperature is infinite. It's called the "ultraviolet catastrophe". Quantum mechanics modifies the prediction at frequencies $f$ such that $hf\sim kT$ or $hf\gt kT$, adds an exponential decrease of the energy with frequency, and the resulting total energy in the field at temperature $T$ is then finite and dominated by a certain range of $f$.
Jul
3
comment Electromagnetic radiation and black body radiation
Dear @Gummybears, you are misreading. The text - and its pages 36-37 - are a flawless introduction to the Planck radiation. None of the sentences over there say that the black body radiation only occurs at a single frequency. Still, for each temperature, some range of frequencies - around a maximum-intensity frequency - dominates. All these things are calculable. I am afraid that every answer to your question will say exactly what your good textbook does, and you will misunderstand just like you misunderstand the textbook.
Jul
2
comment Could Legolas actually see that far?
Dear Ross, right, the distance may be a multiple of the wavelength but it's not the only condition for this trick to work. The information about the relative phase must still be maintained and it's simply not possible for non-monochromatic visible light at distances 9 cm. ... You can't see "3D" by one eye. Seeing "3D" by definition means to have the information from two different angles. One eye may see that some objects are blurry at some focusing, and estimate their distance, but that's not called "3D vision".
Jul
2
revised Could Legolas actually see that far?
added 457 characters in body
Jul
2
comment Could Legolas actually see that far?
Dear @tpg2114, $D$ is the diameter of the aperture for a simple reason. It is a diffraction limit - limitation caused by the interference of light. Light rays coming into the two eyes - which are really, really far from each other, relatively to the wavelength - don't interfere with each other. The phases of the light waves in the two eyes are completely random and uncorrelated. So the distance between the two eyes doesn't affect diffraction (wave optics). It only affects 3D vision (which is geometric optics), and 3D vision loses all beef at distance 24 km which is effectively infinity.
Jul
2
answered Could Legolas actually see that far?
Jul
2
awarded  Curious
Jul
2
awarded  quantum-electrodynamics
Jul
1
comment Massless $\lambda \phi^4$ QFT
Hi, I already wrote it, but if $m=0$ is imposed at one scale, it will be violated at all other scales, anyway. The $m=0$ condition randomly occurs at one scale where the $\phi$ starts to get a vev because there is one minimum on one side and two minima on the other side. Otherwise, nothing special happens at that scale and there are no IR divergences analogous to those with the soft photons. Have you read my answer at all?