quantum mechanics - Mass or no mass?

Do all forms of energy have a mass? We know by $E=mc^2$ that mass and energy are directly proportional, but there are massless forms of energy such as electro-magnetic waves. I am also told that there are different forms of mass, such as invariant mass, virtual mass, and relativistic mass. This electro-magnetic wave seems to be a kind of quantum mechanical wave, that collapses into a particle dubbed a photon. When it interacts with matter, it gives that matter energy, which gives it more mass. Particles with an invariant mass also can be described by probabilistic waves, and behave in a similar way as photons. That is, when its wave function is disturbed in some way, it too collapses into a particle of some form, depending on what it is. The wave function must not be able to exceed the speed of light, because it has an invariant mass.

So my question is, what happens to the energy of an electromagnetic wave, when it transforms into an energy that has a form of mass, and where is that mass from, and what is the difference between the wave function of an invariant mass, verses the wave function of a photon without mass? Okay lots of question here, sorry.

Answer

To answer your question let me start with the most basic constituents of the universe, i.e. elementary particles.

Standard model of particle physics contains matter particles (quarks & leptons), force carriers (W/Z, photon, gluon) and the Higgs particle. Photon and gluon are massless, and the rest of the particles have non-zero masses. In a non-interacting situation, all those particles have their masses fixed, independent of their speed (So no relativistic mass which is an old way of thinking about relativistic kinematics anyways).

When there are interactions between particles and enough energy to produce new particles (through $E=mc^2$), weird things happen. At this point we need quantum field theory to fully understand what's going on. Without going much into details, the way we think about the interactions is through force carrying particles I mentioned in the previous paragraph. When two particles interact, they can simply scatter off each other or the interaction can create a complete different set of particles. We understand the process of going from an "initial" set of particles to a "final" set of particles in terms of all the "paths" connecting them with the interactions allowed by the Standard Model. This situation is very similar to what happens in the double slit experiment. During this process virtual particles (from the above set I mentioned) are created. These intermediate particles have the exact same properties as the real ones, except their mass (or invariant mass) can be different than their actual mass. As in the double slit experiment, nature takes all the allowed paths between initial and final states.

At this point we can talk about the energy of our particles. According to special relativity there is a rest frame for any massive particle and in this special frame even though our particle is at rest it will have an energy given by $E_0 = mc^2$. So for a massless particle like photon there is no rest frame and hence no rest energy. But all the particles (massive or massless) have momentum, and a total energy given by $E = \sqrt{(mc^2)^2 + (pc)^2}$. Note that this reduces to $E = pc$ for a massless particle like photon.

If you think in terms quantum physics, an electromagnetic wave contains photons. So the energy of the electromagnetic wave comes from the energy of the photons i.e. the momentum carried by the photons (which is also related to their frequency or wavelength).

So far we talked about mass and energy of fundamental particles. What happens when we have bound states like protons/neutrons or nuclei or atoms or molecules? The mass of these compound objects depend on the bounding energy (or potential energy) that keeps them together. So for example the mass of a proton is not equal to the total mass of the quarks that makes up the proton (two up, one down quark) but is mostly created by the interaction between these quarks explained by quantum chromodynamics (QCD). Since there is more than one fundamental particle making up our bound states, there can also be excited states with slightly different masses. An atom for example can absorb a photon and switch to an excited state for a very brief time which has a different mass. This mass difference is usually very small. This only happens when the energy of the photon matches the difference between the discrete energy levels of the atom. Most of the time they will only scatter like billiard balls.

For an even larger bound system like a solid object, an absorbed electromagnetic wave usually changes the temperature of the object. In this larger system, we can think of atoms organized in a geometric pattern. On the average they sit at fixed points on a 3D lattice but they all individually vibrate around their equilibrium points. So an electromagnetic wave absorbed by the object usually changes the amplitude of these vibrations or the average vibration energy i.e. the temperature of the object.

yang mills - Field strength vanishes iff $A_{mu}$ is pure gauge

Is it true that the field strength $F_{\mu\nu}$ in a non-Abelian gauge theory with gauge group $G$ vanishes if, and only if, the gauge field $A_{\mu}$ is a pure gauge?

I can show one implication.

If $A_{\mu}=\frac{i}{g}U\partial_{\mu}U^{\dagger}$ where $U \in G$, then the field strength vanishes, but I am struggling with the other implication.

Answer

I) Vanishing field-strength $F=0$ does not imply that the gauge potential $A$ is pure gauge. It only holds locally. There could be global obstructions. In fact, topological obstructions could happen even if the gauge group $G$ is Abelian.

II) Let us sketched the proof of the local statement in a sufficiently small neighborhood $\Omega\subseteq M$ of a point $x_{0}\in M$.

1. 
   
   

   For a point $x\in \Omega$ choose a path/curve $C$ from $x_0$ to $x$.

   
   


2. 
   

   Define group element via a Wilson line $$\tag {1} U(x)~:=~P e^{\int_{C} \!A},$$ where $P$ denotes path ordering.

   
   


3. 
   

   Next use the non-Abelian Stokes' theorem to argue that this definition (1) does not depend on the curve $C$, because $F=0$.

   
   


4. 
   

   Finally, use the group-valued section (1) to gauge transform the gauge potential $A$ to be zero.

   
   
   

acoustics - Relation of power and intensity of sound/point source wave with distance travelled

I have two conceptual doubts about 3D waves while self-studying,they are:

(1) For a 3D wave,we know that $$ A \propto 1/r $$ Relations for intensity are as follows $$ I \propto A^2 \ and \ I \propto 1/r^2 $$ Shouldn't this imply that $ I \propto 1/r^4$? Please tell me where am I messing up, as I can't figure out whats wrong here.

P.S.- I know the $ A\propto 1/r$ comes from the intensity relations only and I have a feeling thats where im getting my logic wrong.

(2) By same relation $$A \propto 1/r$$ shouldn't the power ($p\propto A$) of a sound/point source wave decrease with distance?

classical mechanics - How does a lever magnify force?

I understand that energy is conserved when a force is applied to the end of a lever and magnified closer to the pivot point. However, I would like to know how it is the force is transferred and magnified between the atoms of the lever.

computer puzzle - What phrase does this program rebus represent?

A lazy software engineer was recently browsing over the source code in his company's code repository (Because he was bored of reading puzzles on Puzzling.SE) when he found the following source code:

class Egg {
public:
    Egg() : hatched_{ false } { }
    bool hatched() { return hatched_; }
protected:
    Egg(bool hatched) : hatched_{ hatched } { }

private:
    bool hatched_;
};

struct Chicken : public Egg {
    Chicken() : Egg{ true } { } 
};

void count(Egg* (&eggs)[5]) {
    for (const auto& egg : eggs)

        if (!egg->hatched())
            throw "Exception";
}

void hatch(Egg*& egg) {
    delete egg;
    egg = new Chicken;
}

int main() {

    Egg* eggs[5];

    for (auto& egg : eggs)
        egg = new Egg;

    hatch(eggs[0]);

    // count(eggs);

    hatch(eggs[1]);

    hatch(eggs[2]);
    hatch(eggs[3]);

    // count(eggs);

    hatch(eggs[4]);

    count(eggs);

    for (const auto& egg : eggs) // Resources cleaned up!

        delete egg;
}

Looking over the source code, the lazy engineer couldn't find anything meaningful. Running the code produced no output. The only (shocking) thing he noticed was the memory leak in the case of the thrown exception (Someone needs to teach this guy RAII)

What phrase does the source code above represent?

Very Big Hint:

The phrase is a (fairly common) idiom.

Answer

The phrase is:

Don't count your chickens before they're hatched

Because

The hatch function turns an Egg into a Chicken. The user has an array of eggs but hatches all of them before counting the chickens. It looks like he tried counting them before they were hatched (resulting in the exception which prevented him from doing that) so he commented those out. Finally he was able to count them after they had all been hatched.

electromagnetic radiation - Why is glass much more transparent than water?

There is a related question (Why glass is transparent?) but I am coming at it only from Maxwell's equations. One can determine the skin depth $δ$ for poor conductors like (pure) water and glass using (see Wikipedia)

$$δ =2ρ \sqrt{\frac{ϵ}{μ_0}}$$

If I ignore the frequency dependence of the permittivity (only to get a board range for the skin depth of glass), using appropriate values for the resistivity ρ (water = $2.5×10^5$ Ω∙m and glass = $10^{10}−10^{14}$ Ω∙m), electric permittivity ($ϵ=ϵ_0ϵ_r$) and magnetic permeability ($μ ≈ μ_0$), I calculate that

$$δ(water) =10^4m$$ $$δ(glass) =10^8-10^{12} m$$

Maxwell’s equations determine the behavior of electromagnetic waves in conductors (as well as poor conductors), so if glass and water have such larger skin depths, then this is the reason why light is transparent for these two mediums – right? If so, I then have two related questions:

1. 
   

   Mathematically, it’s fairly straight forward to show that the skin depth is independent of frequency. However, is there a physical explanation why the skin depth is independent of frequency for poor conductors but not for good conductors?

   
   


2. 
   

   At least at optical frequencies, the skin depth is mainly dependent on the resistivity of the material. Since glass has a higher resistivity (is a poorer conductor) than water, electromagnetic waves penetrate farther through glass. So the key to understanding why glass is more transparent than water is physically understanding why δ ∝ ρ?

   
   

I have looked through the books of Griffiths and Jackson for help on this, and found nothing. Thank you in advance for any help on these questions.

Correction and edit due to Johannes’s comment below for question 2

Answer

I disagree with the premise of this question. Using DC permittivity and DC resistivity is an awful starting point if you want to understand anything about visible-light response. [Update: I should say that it's not that bad a starting point for metals specifically. Much worse for other materials.] When electrons move back and forth at 60 Hz, they usually move in a totally different way than when they move back and forth at 1 quadrillion Hz.

For example, in an n-type semiconductor, at 60 Hz, the conductance comes from electrons in the conduction band getting shifted within the band and traveling and sometimes bumping into defects. The conductance at 1 quadrillion Hz comes from electrons in the valence band being pulled into a quantum superposition state between valence and conduction band states. The superposition state happens to jiggle back and forth (by atomic-scale distances) at 1 quadrillion Hz, because of the energy difference between the two states and the laws of quantum mechanics. Soon the superposition is disturbed and you get an electron-hole pair.

For example, rubber has a very high resistivity but is not transparent. Indium-tin-oxide has a low resistivity but is transparent.

To understand visible absorption, you need to be thinking about energy levels and modes, not DC resistivity.

Water absorbs visible light because of various weak (harmonic) vibrational modes. Normally, vibration modes are only in the infrared, but water has unusually high-frequency vibration modes that reach just a bit into the visible. (Because hydrogen is light and bonds very tightly to oxygen. Just like a taut thin string on a guitar will vibrate at a higher frequency than a loose thick string.) Glass does not have that property.

Glass can be much more transparent than water: For example, fiber optics are glass strands through which light can travel many kilometers with negligible absorption. Fiber optics are manufactured very carefully to reduce absorption; if you made ordinary window glass that was 1km thick, it would certainly be opaque.

everyday life - Why friction is zero when wheel slip is zero?

From most graph, when the tire doesn't slip, then the friction is zero, for example see the below image

Why there is no friction when there is no slip? As the car stand still on slope, there is no slip but still there is friction holding the car against gravity

Answer

For an (idealized) perfectly round wheel on a perfectly smooth road, there is only a single point of contact between the wheel and the road at any given time. If you were to plot the motion of a single point on the wheel's surface as it goes around and then touches the ground, you would see that it follows a curve called a cycloid. The picture in that wikipedia article explains it better than I possibly could.(*) As you can see from the image, the point on the wheel's surface is actually changing directions as it touches the road, so at that point in time its instantaneous velocity is zero. Because it is stationary relative to the road, there is no kinetic friction.

However, there can still be static friction, such as if you're driving the car around a curve. In that case, it's static friction on the wheel that prevents you from slipping and keeps you following the curved path. (Or static friction plus a contribution from gravity if the curve is banked.)

There is also static friction between the wheels and the road that causes the car to accelerate in the first place. (I'm assuming it starts from rest.) If friction between wheels and ground were zero, the wheels would spin in place but the car would never go anywhere.

(*) The picture makes it very clear, but if you prefer a verbal explanation: The wheel as a whole is moving forward (relative to the road), but when the point on the wheel's surface is at the bottom of its rotation, it's moving backward relative to the center of the wheel. The result is that the point on the surface of the wheel is stationary (relative to the road) when it's at the bottom of its rotation.