electromagnetism - Ultraviolet catastrophe in a classical world

In the real world, the ultraviolet catastrophe doesn't happen because the quantization of photons modifies the classical behavior of light at frequencies comparable to and higher than the temperature. But classical electromagnetism is a mathematically self-consistent theory, so we could imagine a world where $\hbar = 0$ and electromagnetism remains classical to arbitrarily high frequencies. How would the ultraviolet catastrophe work in such a world? Classically, all frequencies are equally populated at all temperatures and each has energy $\frac{1}{2} k_B T$, apparently leading to infinite energy radiation at any temperature, which doesn't seem compatible with conservation of energy. What would happen if you put a theoretical fully classical system in contact with a thermal bath (which seems like a physically reasonable set-up)?

My guess is that since this system has an infinite number of quadratic degrees of freedom, the usual canonical-ensemble derivation of thermal equilibrium breaks down. I think that if you couple a perfectly conducing cavity to any thermal bath, no matter how large, then the cavity will absorb an unboundedly large amount of energy from the bath. In the usual derivation, we assume that the bath has so much more energy than the system that its energy density is independent of the state of the system, but in this case that assumption will eventually be violated. The cavity's energy will become comparable to the bath's, so to find its equilibrium state, we will need to consider the details of the bath and treat the combined cavity-bath system in the microcanonical ensemble. So the Boltzmann distribution and the equipartition theorem will no longer apply, and the ultraviolet catastrophe will be avoided.

Answer

I think this is an interesting question. If one tries to couple the EM field to matter somewhat realistically and classically, he needs a model for matter. One possible model would be that of an assembly of uncorrelated dipoles. If we assume these dipoles to be point dipoles then the energy they emit is related to the wavelength as $\varepsilon_{\lambda} \sim \lambda^{-4}$. There is no discussion that this is correct as this corresponds to the Rayleigh scattering result and we more or less witness its effect everyday when we go outside.

So, it could be that the "ultraviolet catastrophe" attributed to the Rayleigh-Jeans formula, which incidentally has also an EM energy density that goes as $\sim \lambda^{-4}$, assumes somewhere that the scattering material objects are point-like.

At the very least for consistency reason, at equilibrium, the energy density emitted by an assembly of dipoles has to agree with that calculated independently for the EM field and the two above results have to agree.

Now, the point is that at high frequency, the wavelength becomes of the order of the size of the scatterers and we could get for instance something close to Mie scattering instead of Rayleigh scattering. The effect it would have is that the energy fraction emitted by a scatterer becomes roughly independent of the wavelength in this regime and so something will probably happen and prevent a fully diverging energy density at short wavelength.

So in effect, one would have to cut off the spectrum somewhere and make it at least saturate at some value.

Note that this is something ubiquitous in statistical mechanics (on a quite different note the partition function of a single hydrogen atom diverges and one needs often to put by hand a cutoff corresponding to an atom radius big enough to create an ambiguity between a pair of H atoms and an H2 molecule) and it is probable that people are often too eager to claim, a posteriori, that a new theory was definitely needed while something could possibly have been done with a better model.

This of course does not mean that quantum mechanics is not required in the end but it means that the dramatic failure of the Rayleigh-Jeans formula is a problem owing to all the assumptions leading to the said formula, not just one.

electromagnetism - Why is infrared radiation associated with heat?

I am little confused with infrared radiation. I understand that when an object is hotter, it radiates electromagnetic waves of a bigger frequency and this waves are also more energetic, that is why blue stars are hotter than red stars, so that means that objects that radiate infrared are cooler and their waves are less energetic than the objects that radiate visible light.

On the other hand I have heard from professors saying things like: "the reason of why you feel the heat of a wood fire as you get closer is because it radiates infrared radiation", nobody says the reason is because it radiates red radiation. Another example: "the old incandescent light bulb was inefficient because it radiates infrared light so it waste energy to radiate light that we don't even see, but at least it keeps the room warm", nobody says the modern lamps are better because it radiates only visible light and because is more energetic than infrared also keeps you warmer. Is this a general misconception in physics?

Answer

It sounds like you know some of the most important summary points about blackbody radiation, but here is a reference on the subject, since I will be talking almost entirely about blackbody radiation: https://en.wikipedia.org/wiki/Black-body_radiation

Given any temperature, there is a certain emission spectrum (see https://en.wikipedia.org/wiki/Planck%27s_law) describing the mixture of photons which a body at that temperature emits. As you already noted, the peak frequency of this spectrum increases as the temperature increases. It is also of interest to note that an increase in temperature increases photon emission at ALL frequencies, not just higher frequencies.

The short form answer to your question is that we live in on Earth, where temperatures tend to be in the 200-400K range. Even a campfire doesn't make it much above 1500K. At all of these temperatures (yes, including the campfire) the VAST majority of the energy radiated is in the infrared range. If you put a filter between yourself and a campfire which absorbed all visible light and transmitted all infrared light, you would feel just as warm. So it is natural for us to associate infrared radiation with heat. The sun is the only everyday example of something that warms us noticeably with visible light, and sunlight already holds its own unique place in the human experience. Sunlight feels warm. A physicist can get out sensitive instruments and observe that in fact all light warms us slightly, but as far as what we can feel with our own nerves goes, it is only infrared and sunlight that seem warm.

If we were plasma beings that inhabited the core of the sun, perhaps we would associate visible light (or some energetic subatomic particle or other...) with the transmission of thermal energy, but we aren't and we don't.

In the end, all light transmits energy, and heat is just energy in the form of atoms exercising their degrees of freedom. So there is no clear cut distinction between the way infrared radiation interacts with heat and the way any other radiation does. But over most of the wide range of commonly studied environments, heat is mostly ratiaded as infrared photons. Thus the association.

Regarding your thought about incandescent lights vs more efficient alternatives, we replaced our 60-100 Watt bulbs with 7-20 Watt bulbs, so they really don't warm us up anywhere near as much as the old ones. If we had replaced the bulbs with equivalent wattages, then your thought would be correct, but we would be blinded by our lamps!

Has anyone measure the strength of the force of gravity in relation to the molecular bonding?

It's clear to me that gravity it a function of mass. It is also clear to me that gasses are less affected by gravity. So I'm thinking that there exists a measurable minimum binding distance $d$ between molecules that determines whether the molecules will collectively experience a gravitational pull.

In other words, for the same element molecules $M_1$ and $M_2$ the gravitational force is normally represented by the addition of both masses ($M_1 + M_2$). But when $M_1$ and $M_2$ exceed a distance $d$ of separation, the mass used to determine the gravitational force is just $M_1$ or $M_2$. Then if there were a molecular bond that separates the molecules by a distance greater than $d$, gravity could be nullified.

Is this possible or am I totally off base?

classical mechanics - Principle of Least Action

Is the principle of least action actually a principle of least action or just one of stationary action? I think I read in Landau/Lifschitz that there are some examples where the action of an actual physical example is not minimal but only stationary. Does anybody of know such a physical example?

Answer

By convention we try to set things up so that it's usually a minimum, but we can't make any definition of the action that would make that hold in all cases.

In optics, consider a situation in two dimensions where you have an ellipsoidal cavity with reflecting walls. If you release a ray of light from the center, along the major axis, it gets reflected back to the center by following a path of maximum time. If you start a ray from the center, along the minor axis, it comes back after following a path of minimum time. You can choose the action to be the time or minus the time, but no matter what, one of these rays will be a minimum of the action and the other will be a maximum.

In special relativity, you can take the action for a particle to be the proper time $s=\int ds$ (with $ds$ positive), or you can take it to be $-s$ or other possibilities such as $-mcs$ (Landau and Lifschitz's choice). It doesn't matter which sign you choose, because the physical predictions are the same either way.

In both of these examples (optics and SR), you can make a choice of sign such that the action is minimized for infinitesimally short trajectories in free space. However, the example of the ellipsoidal cavity shows that you cannot in general make it a minimum for all paths of finite length.

In relativity, the metric only defines $ds$ in absolute value, $ds^2=g_{ab}dx^a dx^b$. Also, we'd like to be able to talk about timelike, lightlike, and spacelike geodesics. If we choose the action for timelike geodesics to be real, then it has to be imaginary for spacelike ones -- or we could define it as $\int\sqrt{|ds^2|}$, but the absolute value could be a nuisance because it isn't a smooth function.

For L&L's discussion of this, see Mechanics (3rd ed.), section 2; and Classical theory of fields (2nd ed.), sections 8, 53, and 87.

neutrinos - What distinguishes the behaviour of particle from its antiparticle: C violation or CP violation?

It is said that a CP violation would mean that the behaviour of the particle is different from the behaviour of antiparticle. Why is C violation not good/enough?

Answer

The operation that maps particles to antiparticles is just $C$. (This is somewhat of a simplification. A better thing to say is that in theories with $C$ symmetry, you can pair particle states with the same spacetime quantum numbers but the opposite internal quantum numbers. When $C$ is violated, there may exist no pairing that gets the quantum numbers right. In extreme cases, there may not be any way to define a $C$-like operator at all, no matter how you modify the quantum numbers; an example is a theory with a single Weyl spinor. In such cases you can still define a pairing using $CP$, if it exists, or failing that using $CPT$, which always exists and is conserved, but these pairings don't have the familiar properties you would expect. For much much more, see here and here.)

Why do people focus on $CP$ violation? The issue is that $C$ violation is ubiquitous in the Standard Model; in fact, in a certain sense it is as strong as possible in the charged current weak interactions. However, there are interesting phenomena that require both $C$ violation and $CP$ violation. So since $CP$ violation is the hard part, we talk about it a lot more.

One key example is the creation of a matter/antimatter imbalance in baryogenesis. For simplicity, suppose that $C$ and $CP$ are both defined, though they may not be obeyed. For any particle states $i$ and $f$, there are four related processes: $$i \to f, \quad \bar{i} \to \bar{f}, \quad i_P \to f_P, \quad \bar{i}_P \to \bar{f}_P$$ where a bar denotes the antiparticle, defined by the action of $C$. If these processes have rates $a$, $b$, $c$, and $d$, and the states have different baryon number, then the rate of baryon number violation is proportional to $$a - b + c - d.$$ If $C$ symmetry is obeyed, then $a = b$ and $c = d$, giving a rate of zero. If $CP$ symmetry is obeyed, then $a = d$ and $b = c$, again giving a rate of zero. One needs both $C$ and $CP$ violation to get baryogenesis.

Unfortunately, in popular science these statements are sometimes oversimplified to just "$CP$ distinguishes matter from antimatter", which is confusing.

general relativity - Is this analogy of Hawking Radiation correct?

Through reading of textbooks and other research papers, I have settled on the analogy of hawking radiation below (Written completely by myself)

Within the ergosphere of the black hole, virtual pairs of particles and anti-particles are constantly appearing due to vacuum fluctuations. Typically, the pair would then annihilate before this could be of any consequence. However, as one of these two particles will be closer to the hole than the other, it will experience a greater gravitational force than the other. Thus, there is the possibility that one particle will fall into the black hole while the other escapes to infinity. The particle that falls in will have negative energy, and subsequently negative mass, leaving the positive energy particle outside of the hole.

It is not the black hole that is directly emitting the particle, but from an external reference frame, this appears to be case. As tidal forces are greater for smaller black holes, the rate of emission of Hawking radiation increases as the hole decreases in size. In an isolated system, this would lead to an exponential decay. However, black holes are constantly accreting mass too. As such, any black hole with a sufficient mass will accrete mass faster than it loses it via Hawking radiation. Yet this is not the case with smaller holes, which would slowly ’evaporate’.

As this is all outside of my school curriculum and way above my current level of taught physics, I was hoping someone could either confirm that my analogy is correct or provide some constructive criticism and advice. Thanks!

Answer

As the comments have suggested, the problem is that your description of virtual particles appearing to vacuum fluctuations is wrong. Have a look at my answer to Black holes and positive/negative-energy particles for more on this.

There isn't an explanation of what is really going on that is accessible to the non-quantum field theory nerd (though I have attempted one in the answer I've linked above). No-one seems to know exactly where the pairs of virtual particles analogy came from because it doesn't correspond to the current day descriptions of the quantum vacuum. In any case virtual particles are more of a computational device than anything real. The best article I've seen on this is this one on matt Strassler's blog.

conservation laws - Is there a systematic way to obtain all conserved quantities of a system?

I'd like to know whether, given a system, there's a way to obtain all the conserved quantities. For instance if the system consists of electric and magnetic fields, the fields must satisfy Maxwell's equations. These equations are invariant under many transformations (Lorentz transformation, rotations, spatial and temporal translations, etc. By the way is there a way, maybe from group theory, to find all the possible transformations that leaves the equation(s) invariant?) which imply as many conserved quantities thanks to Noether's theorem. In wikipedia I can see an equation that seems to give all the conserved quantities (wiki's article) but it involves the Lagrangian and I'm not sure whether the formula is valid for all systems whose Lagrangian is possible to obtain.