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bio website motls.blogspot.com
location Czech Republic
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visits member for 4 years, 7 months
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Hi, I am a string theorist and a publicist.


Oct
22
comment Why the Principle of Least Action?
A philosophical answer but a good one. ;-)
Oct
22
comment Why the Principle of Least Action?
A fresh article trying to answer such questions: motls.blogspot.com/2011/10/…
Oct
22
comment Non-unitarity of wave function collapse
The "calculation" you wrote is completely nonsensical, Jonathan. $\delta(x)$ is still a function of one variable – or, in your completely needlessly nitpicking terminology (it's the point of such distributions that one CAN treat them as functions), a distribution acting on the space of functions of a single variable. You are confusing $[\delta(x)]^2$ which I proved is something else than $\delta(x)$ and ill-behaved even though both are functions/distributions of one variable that vanish outside $x=0$ with $\delta^{(2)}(x,y)=\delta(x)\delta(y)$ which waits to be integrated over two variables.
Oct
21
comment Air flow coming out of a fan feels much stronger than air flow coming in. Why?
All dissipating processes such as friction and viscosity break the time-reversal symmetry of the effective laws of physics.
Oct
21
comment Air flow coming out of a fan feels much stronger than air flow coming in. Why?
Well, right. When the air goes out of the fan, it's directed, so the air molecules are hitting your hand. On the other hand, when the air goes into a fan, it may go from any direction to fill the pressure gap and the solid angle is therefore much greater, reducing the velocity. I think that your intuition about a "natural symmetry" is a good one and if there were no turbulence, I believe that the velocities would be symmetric and had the same absolute values. So the symmetry violation is a consequence of complicated turbulent phenomena.
Oct
21
comment Non-unitarity of wave function collapse
Dear @Jonathan, your claim is obviously incorrect. $\delta^2(x)$ is much greater than $\delta(x)$. For example, you may write $\delta(x)=1/2\epsilon$ in the interval $(-\epsilon,+\epsilon)$. Its square is then $1/4\epsilon^2$ in the same interval and it doesn't integrate to one but to $1/2\epsilon$. That's no coincidence: $\int_{-\infty}^\infty dx\,f(x)\delta(x) = f(0)$, so if you substitute $f(x)=\delta(x)$ to integrate $\delta^2$, you get $\delta(0)$ on the right hand side which is definitely not one as it should be for the integral of $\delta(x)$ but a divergent constant.
Oct
21
revised Why higher frequencies in Fourier series are more suppressed than lower frequencies?
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Oct
21
revised Why higher frequencies in Fourier series are more suppressed than lower frequencies?
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Oct
21
revised Why higher frequencies in Fourier series are more suppressed than lower frequencies?
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Oct
21
answered Why higher frequencies in Fourier series are more suppressed than lower frequencies?
Oct
20
answered Why does charge conservation due to gauge symmetry only hold on-shell?
Oct
19
comment Modern avatar of Englert's solution?
I see, thanks for the clarrified connections etc., @José.
Oct
19
comment Non-unitarity of wave function collapse
Dear @Jonathan, in some modest sense, you may be able to "define" the square root of the delta-function but you won't be able to calculate with it. For example, you won't know what it evolves to. It's easy to see why. An ordinary delta-function evolves to the well-known nonzero functions at time $t$, calculable from the Green's functions. However, the square root of the delta-functions is proportional to delta-function but infinitely times smaller. So it will evolve into "zero" and unitarity will be violated, anyway. This is just not how the calcuations may be done.
Oct
18
comment Is there a theortical limit to the amount of sound-energy air can contain?
I actually think it's a fun question. It depends what configurations of the air you're ready to call sound. For example, you may allow sound waves that make the air pressure drop everywhere except for $1/N$ of the volume. I guess that the work this system may perform while switching to the uniform pressure goes like $C.p_{\rm average}V_{\rm total}\ln(N)$ or so. Of course, if you may compress all the air to very small regions, you will get more energy. Then there are lots of questions whether you allow heat (hot air), phase transitions, hot air, nuclear reactions etc.
Oct
18
comment What is a simple intuitive way to see the relation between imaginary time (periodic) and temperature relation?
Nice, Moshe! Qmechanic: fun terminology. I guess the opposite is not true: the Heisenberg representation isn't an important representation of the Schrödinger algebra. However, what is true is that Wigner's friend is an important representation of Schrödinger's cat and Wigner's friend could even be Heisenberg himself, see e.g. this picture en.wikipedia.org/wiki/File:Heisenberg,W._Wigner,E._1928.jpg :-D
Oct
18
comment Modern avatar of Englert's solution?
Dear @José, sorry for being redundant then. I also added (into my answer) a link to a seemingly related paper on the octonionic membrane by Duff et al. Conical singularities in general don't solve Einstein's equations at the singular point except that sometimes they can when the UV physics governing the singular loci is favorable. After all, orbifolds are allowed and some of them may also be viewed as deficit angles etc. But I am afraid that if a full-fledged embedding of Englert's solution to a UV complete theory exists, it's not known.
Oct
18
answered Modern avatar of Englert's solution?
Oct
18
comment Non-unitarity of wave function collapse
This is a meaningless statement, lurscher. Of course that if you want to imagine that there exists a non-existent, non-local, inconsistent, non-unitary operation that messes up the wave function, you follow this pathological procedure at least by another procedure that rescales the wave function so that it has the same norm as before. But this is by additional fudging: there's no inherent preservation of the norm in the operation. It's wrong to say it's by "definition". By the most natural definition, the pathological non-unitary non-local operation you propose also modifies the norm.
Oct
18
comment Non-unitarity of wave function collapse
"measurements in the most basic formulation given in QM courses always assume that you are able to measure the state in a non-local way" - Nope, there is no contradiction between locality and the postulates of quantum mechanics, as demonstrated by the important example of local quantum field theories such as the Standard Model which are perfectly local. They also describe reality: all physical processes in this Universe are demonstrably local, which proves that all mechanisms that are nonlocal are unphysical.
Oct
17
comment Non-unitarity of wave function collapse
To show that the "decoherence map" (elimination of off-diagonal entries) isn't an isometry on the space of matrices, just consider what this map does with matrices $((a,b),(b,c))$ for different values of $b$. These matrices are clearly very far from each other in the natural metric on the space of matrices, especially if you pick a large $b$. But all these matrices get mapped to $((a,0),(0,b))$ so the distance of the values of the map is zero. ;-)