thermodynamics - How does a gas of particles with uniform speed reach the Maxwell-Boltzmann distribution?

Take an empty container and fill it with $N$ gas particles (ideally a monoatomic gas), each having the same kinetic energy $E$, then isolate the container. Since initially the speeds don't follow the Maxwell-Boltzmann distribution, such a system cannot be in thermodynamic equilibrium. On the other hand, assuming perfectly elastic collisions (and there is no reason to assume otherwise, since the only form of energy the particles can possibly have is kinetic), I see no way such a system could spontaneously evolve to equilibrium: elastic collisions among equal masses keep speeds unchanged! What gives?

I have no background in non-quasistatic processes, but I tried nonetheless to work out a solution taking into account the container, which necessarily has a certain heat capacity, a certain initial temperature, and whose walls are not necessarily perfectly elastic, etc. Knowing the number of particles and their individual speed, it's possible to compute the system's total heat content (but is this exactly $NE$, or less?) and thus derive it's equilibrium temperature. Since the system is isolated, I take it the quantity that has to change must be entropy (namely increase, as the uniform speed state seems less likely; the change can probably be arrived at from a strictly combinatorial point of view). At any rate, the process I imagined goes like this: initially, the particles bombarding the wall transfer some amount of heat to it while slowing down; in turn the wall, now heated up, will transfer back some heat to the gas; eventually, the system will reach the expected equilibrium.

Is my assumption of perfectly elastic collisions wrong, and if so, where does the dissipated energy go?

Is there an increase in temperature that accompanies the increase in entropy?

Can someone point me to the rigorous mathematical framework for analyzing the problem?

Is there direct experimental evidence that the speed of gas particles attains the Maxwell-Boltzmann distribution, or is it just a theoretical result that everyone is just happy to work with?

Thanks for any suggestion.

cosmology - $Omega_{r}$ from WMAP results?

To do some Friedmann-Lemaître cosmology calculations, I would like to know an estimation of $\Omega_{r_0}$ ($\Omega$ radiation today). WMAP 7 give estimation of $\Omega_{b}$, $\Omega_{c}$ and $\Omega_{\Lambda}$ but nothing about $\Omega_{r}$ : http://lambda.gsfc.nasa.gov/product/map/dr4/best_params.cfm Where can I find the best value of $\Omega_{r}$ available today ?

Thank you very much.

Answer

Have you tried looking at these: http://arxiv.org/find/astro-ph/1/au:+Larson_D/0/1/0/all/0/1 (Seven year analysis of WMAP data)?

optics - Why do I see better under water using swimming goggles?

I am myopic (I don't really know if this is relevant or not) and I usually swim without contact lenses. My vision is clearly better underwater when I am using swimming goggles.

I have tried to understand why this happens and I think that it is probably due to the presence of just air (which should be more or less "still") between the googles and my eyes avoiding this way the turbulent flow of water that should somehow affect my vision, due to the fact that air has a different refraction index ($n_1$) than the one of water ($n_2$) and that the material of the goggles is not relevant. But it also could be that the material of the goggles is making the difference.

What is the actual physical explanation of this fact?

Answer

Is blurred effect due to turbulence?

No, it is not. The turbulence has a little effect here. Even if there is no turbulence, one see everything blurred underwater. The reason is explained below.

An eye is a natural lens. A clear shot of something you see depends on how well the image is focused on your eye. The most of the refraction in the eye occurs at the cornea and a little bit at the lens of the eye. The image is then focused on the retina.

But when you are under water, the optical density of the cornea and water are almost the same (or say both have similar refractive indices-1.376 for cornea and 1.333 for water)). If you open your eyes underwater, there will hardly be any refraction because now the light is going from a medium (water) to another medium (cornea) with the same density, so refraction never takes place. If light entering the cornea is not properly refracted, it will not be focused on the retina to give you a clear image. This is why you see everything blurred underwater.

The use of swimming goggles can overcome this defect. Your eyes work perfectly if light enters your eye from air. That principle is made use of in swimming goggles. When you use goggles, you have some air in-between the cornea and the glass of the goggle. So even if the light is coming from underwater it first passes through the air and then only it reaches the eye. So you feel exactly like on ground and see things well. This is for a normal person.

You are myopic. The air-glass interface in front of your eye helps you like you have your contact lenses while on ground by providing better refraction. It may work for every myopic persons. Try use goggles while on ground. This may not help. It's because the light is now entering your eye from air. So there will be a slight difference in focusing. Same effect is when you use your contact lenses underwater.

collision - $e^+e^-to qbar{q}$: Reconstructing $qbar{q}$ energy and momentum

Question

In a real collider experiment e.g. LHC / LEP how can one reconstruct the energy and momentum of the resultant $q\bar{q}$ pair produced from the process $e^+e^-\to q\bar{q}$?

Specifically, how can we relate the kinematic variables before ($q\bar{q}$) and after hadronisation (observed jets)?

Detail

I was working on a question that asserted as a side comment that one could reconstruct the final state energy and momentum of the resultant $q\bar{q}$ pair produces from the process $e^+e^-\to q\bar{q}$ observed experimentally in a collider.

The question posed itself was easy (and not relevant here) however, the assertion of the ability to reconstruct the energy and momentum of the quark and antiquark caused me some confusion.

Thoughts

I have studied Thompson Modern Particle Physics and the only explanation I have is rather hand wavy and I'm not sure of its legitimacy.

I figured that we could observe the cross section of the hadron jet process and compare at specific centres of mass $\sqrt{s}$ to known experimental results (in plot below) form the ratio,

$$ R= \frac{e^+e^- \to hadrons} {e^+e^- \to \mu^+\mu^-} $$

If we were clever about picking $\sqrt{s}$ then perhaps results could peak at a meson production process that we would be aware of and then hence, we could determine the final state energy and momentum (this is the really hand wavy bit as I don't actually know how one might do that!)

Answer

Hadronization still doesn't break conservation of energy and momentum. So getting the total energy, or the total momentum, of the quark-antiquark pair is easy: just add up the total energy and momentum of all the reaction products.

To get the individual energy or momentum of one particle, i.e. just the quark (or antiquark), we rely on the fact that they come out of the detector in opposite directions. That means it's easy to separate the outgoing particles into two clusters: one came from the quark and one came from the antiquark. You can then add up the energies and momenta within each jet to find the momentum of the quark or antiquark that produced it.

This image, which I found on Quantum Diaries, shows the kind of separation I'm talking about. Fair warning, though: this particular image came from a proton-proton collision, which means there's a bit of extra "junk" emitted in all directions, as you can see by looking between the jets. That doesn't happen much in electron-positron collisions. Unfortunately I couldn't find a better picture to illustrate it.

Another option for identifying the momentum of a progenitor quark (or antiquark, or even gluon) is to look at only one outgoing particle; usually the highest-momentum hadron. Suppose that particle has a fraction $z$ of its parent quark's momentum, and is moving in the same direction. We can apply some quantum field theory and a lot of data analysis to figure out the probability distribution for different values of $z$. This distribution (or something roughly like it) is called a fragmentation function, denoted $D(z)$ or something like $D_{\pi^+/u}(z)$ if you want to limit yourself to a specific kind of hadron and/or quark.

Fragmentation functions are more useful in proton-proton collisions or ion collisions, where the "junk" I mentioned gets in the way of cleanly grouping the outgoing particles into two jets.

quantum mechanics - Why are Cooper pairs formed by electrons of opposite momentum and spin?

I understand that Cooper pair in low-temperature superconductivity are formed by electron-phonon interaction. Normally one then assumes that electrons of opposite momentum and spin are paired. This is what i dont understand. In my understanding two electrons with opposite momentum should seperate and thus not feel an attractive force due to electron-phonon interaction.

Answer

It is indeed a counterintuitive fact. Let us go slowly through the argument.

You start with a degenerate gaz of fermions. They are piled-up in the momentum space, up to the so-called Fermi level. It means that the last electrons entering the game have the Fermi energy, which is pretty huge ($\sim 10\;000$ Kelvin). Also, these electrons are the only accessible ones, since the ones deep inside the Fermi sea are frozen up by the other ones.

On top of this stable situation (this is called a metal after all), you allow coupling of two-electrons through phonon exchange. This happens locally in space, since this is the simpler hypothesis. This is the usual naive picture: an electron creates a vibration of the lattice (i.e. the electron releases a phonon) which perturbs the next electron flowing (i.e. the next electron catches up the phonon), but this picture is confusing and you should not bother yourself with I believe.

Let us come back to our two-electrons interaction mediated locally by a phonon. It turns out that the interaction is slightly attractive for some parameters. You can verify that this small attraction leads to a huge effect : two electrons which were previously at the Fermi level collapse (so to say) to the zero-energy state (picturesquely said, they lost roughly $10\;000$ Kelvin in the process, but this also is a naive picture). This is called the Cooper instability of the Fermi sea. In fact, the electrons do not collapse, they bound to each other, like in a covalent bound if you wish. And so the kinetic energy of the bound state is zero, although the two electrons forming the bound state were previously at the Fermi energy... so how to reconcile these two ideas ? Well, you say that the former electrons were in states $k_{F}$ and $-k_{F}$, the Fermi momentum, such that the bound state has momentum $k_{F}+(-k_{F})=0$ is at rest ! So you indeed couple two electrons with opposite momenta into a bound state with no momentum.

Since the interaction is local in space, the bounding can only happens for singlet pairing: the former independent electrons were in opposite spin states, thanks to Pauli's principle.

Now, it is customary to call this state a Cooper pair, and to note it as $\left|k_{F},\uparrow;-k_{F},\downarrow\right\rangle $.

So far so good for two electrons and one phonon, what about the condensate ? You can verify that the condensate of paired fermions minimised the energy of the system of interacting electrons. Bardeen, Cooper and Schrieffer first demonstrated it, and so is the BCS explanation for superconductivity.

So to conclude: yes the two electrons forming the Cooper pair have opposite momenta (and spin, but I think your question was mostly about the momentum) but the resulting momentum of the bound state is still zero. The trick is that a Cooper pair is a novel object, which has not so much to do with the properties of the two electrons it's made with, and that's why the condensate is at rest of course ! It's even better to not see a Cooper pair as the pairing of 2 electrons at all, since it is a new state. Say differently, the Cooper instability absorbs two particles (picturesquely called electrons) at momentum $\pm k_{F}$ and with opposite spins and create a new particle (picturesquely coined Cooper pair) with zero momentum and zero spin. Then spin and momentum are conserved (hourra !). To conserve the charge, the new particle (the Cooper pair) must have twice the charge of the former ones (the electrons).

Has the concept of non-integer $(n+m)$-dimensional spacetime ever been investigated by theoretical physicists?

The following image serves to aid the reader in understanding the "privileged character" of $3+1$-spacetime.

The wikipedia article on spacetime, and the sub-article "The priveleged character of $3+1$-dimensional spacetime" in particular, made me think a bit about the possibility that we might live in a non-integer amount of spatial and/or time dimensions.

The notion of attaching a non-negative real number to a metric space has at least mathematically already been described by such concepts as "Hausdorff Dimension" and "Minkowski-Bouligand Dimension".

This may sound silly/ignorant/absurd to professional practicing theoretical physicists. To me (a layman), however, it doesn't sound much stranger than the idea of wrapping up six extra dimensions (which is, from what I understand, considered to be a serious possibility by those who study $10$-dimensional String Theory) into intricate shapes called "Calabi-Yau Manifolds".

Has any research on $(a +b)$-dimensional spacetime (where $a,b \in \mathbb{R}_{\geq 0} $) ever been done? If so, what where the findings? If not, why not?

Answer

One example of such an approach is Ambjorn and Loll's Causal Dynamical Triangulations, which is very similar in many ways to the very old idea of Regge calculus, whereby spacetime is discretized. At small scales, non integer dimensions can emerge. For an introductory article , see

Jan Ambjørn, Jerzy Jurkiewicz and Renate Loll. The Self-Organizing Quantum Universe. Scientific American (July 2008), 299, pp. 42-49. doi:10.1038/scientificamerican0708-42, available here.

quantum field theory - Instantons and Borel Resummation

As explained in Weinberg's The Quantum Theory of Fields, Volume 2, Chapter 20.7 Renormalons, instantons are a known source of poles in the Borel transform of the perturbative series. These poles are on the negative real axis, and the series remains Borel-summable as long as the coupling constant is not too large.

However, instantons are objects in the Euclidean version of QFT. What's the significance of the above Borel resummation in the Minkowski theory?