electromagnetism - Does the wavelength always decrease in a medium?

I was studying a GRE Physics Test problem where optical light with a wavelength of 500 nm travels through a gas with refractive index $n$.

If we look at the equations for wave motion and index of refraction

$$c=\lambda_0\nu\quad\text{(in vacuum)}$$

$$v = \lambda\nu\quad\text{(in medium)}$$

$$n = c/v$$

we see that, if the frequency is constant, the wavelength decreases in the medium compared to vacuum. Is this a consistent property at all frequencies and for all mediums with refractive index real and greater than 1?

Are there dielectrics which change the frequency (still for n > 1), and is there an example of that?

Answer

The index of refraction of a material can be less than 1 at high frequency, this is called "anomalous dispersion" and it happens as you cross an energy level of certain materials. It means that the phase velocity of light of a certain frequency is higher than c.

If the index of refraction is constant, as it is for long wavelengths, n has to be bigger than 1 to avoid superlumimal communication.

The principle of energy conservation in a static environment forbids a frequency shift for a photon, since this would add energy or take away energy, and nothing in the medium is changing with the right frequency to do that. But light entering a moving medium shifts frequency. photons can combine to make one of double the frequency in a strong light beam in a nonlinear medium, and this corresponds to making higher harmonics of the classical field.

rotational kinematics - Why is the moment of inertia (wrt. the center) for a hollow sphere higher than a solid sphere (with same radius and mass)?

Why is the moment of inertia (wrt. the center) for a hollow sphere higher than a solid sphere (with same radius and mass)? I have completely no idea and I am inquiring about this as it is an interesting question that popped in my head while doing physics homework.

Answer

The key is... the closer the mass to the axis of rotation, the easier it is to add angular velocity to the body.

For instance a figure skater rotates faster when she puts her limbs closer to her body.

Let's see how it works in a more intuitive fashion:

For instance, in the figure bellow, trying to lift up table (A) would be easier than table (B).

In both cases the mass of each individual box is the same, but in (A) you have a better lever because of the distance from the border where the force is being applied, to each box.

Therefore, table (B) would be harder to lift up, even when R (length of the table) and M (total mass of the four boxes) are the same.

Now let's see how it works in the case with the spheres:

1. Let's make the sphere a disk, and then divide it in pieces.


2. Make the center of mass of the disc fixed, and move all the pieces to one side.


3. Now we have a similar scenerario to the one with the tables.

Both spheres, the solid and hollow one, rotate around their center of mass in the same way that the table rotates around the legs at the opposite side to where the force is being applied.

To make sense of step 2, where the mass of the all pieces is collapsed, think on a Merry-Go-Round where all the kids move to one side keeping their distance to the axis of rotation fixed.

electromagnetism - We know that the protons in a nucleus are positively charged. So why does the nucleus stay intact?

We know that the protons in a nucleus are positively charged, whereas the neutrons do not possess a charge; we also know that unlike charges attract. So why does the nucleus stay intact, even though there isn't supposed to be any electrostatic force of attraction within it? Instead, why isn't there any repulsion between the protons which are like-charged?

Answer

Short answer: the strong nuclear force.

The strong nuclear force binds nucleons (protons and neutrons) together. It is a very short-range force, which is why it only acts over distances on the scale of atomic nuclei. There is repulsion between the protons, which is why, as the number of protons goes up, more and more neutrons are required to stabilize the nucleus (look at how atomic mass goes up relative to atomic number on a periodic table). More nucleons (protons and neutrons) means more strong force, and more neutrons means more space between the protons, reducing the repulsion. Together, these effects can produce a stable nucleus.

Stable here is a relative term, since nuclear decay occurs for many isotopes in which the particular number of protons and neutrons isn't stable in the long term (sometimes very long). All elements above lead on the periodic table have too many protons and are all radioactive, since no number of neutrons is able to fully stabilize the atoms against nuclear decay. This is partly a result of the proton-proton repulsion you refer to.

general relativity - Conserved quantity of Spactime Diffeomorphism Group

So it's my understanding that the underlying symmetry of GR is the Diffeomorphism Group of 3+1 spacetime.

It's also my understanding that a symmetry implies a corresponding conserved quantity in a physical theory.

Then, if my understanding is correct, there should be a conserved quantity associated with the Diffeomorphism Group of 3+1 spacetime. What is it?

electromagnetism - Gravititonal fields compared to electromagnetic fields - are they infinite in range?

me and my friend has a discussion last night, and he argued that both an electromagnetic field and gravititonal field are infinite in their area of effect, but with diminishing effects as you get farther away from the object.

I argued that only gravity field is infinite in range.

So which one of us is correct, and please explain in laymen's terms, as we're not physicists :)

Answer

Actually, your friend is right! The electromagnetic field has an infinite range, for example for a charged point particle the electric field is proportional to $\frac 1{r^2}$, where $r$ is the distance to the particle. However in very special circumstances, like inside a superconductor, the electromagnetic field will be short-range because the photons will effectively be massive.

See this table, where you can see both the electromagnetic and gravitational fields have infinite range of interaction, but the Weak doesn't. It's the famous Higgs boson which renders the weak field short range (by making the corresponding gauge bosons massive), very similar to what happens inside a superconductor.

quantum mechanics - Bound states, scattering states and infinite potentials

I am doing my first semester of Quantum Mechanics and we're using Griffith's Introduction to Quantum Mechanics. As he is introducing the Dirac delta function potential he explains bound and scattering states, and I understand that a system is considered bound if the energy of the system is less than the potential at infinity, that is

\begin{align} \text{Bound state: } E &< \lim_{|x| \to \infty} V(x)\\ \text{Scattering state: } E &> \lim_{|x| \to \infty} V(x). \end{align}

That makes sense, and he then continues by saying that this implies that $E<0$ for bound states and $E>0$ for scattering states, as you can always add a constant to the potential energy to make it zero at infinity.

He also explains how the solution to the Schrödinger equation for bound states is a discrete linear combination, and the solution for scattering states is an integral which cannot be normalized, and therefore does not exist. He then continues with the Dirac delta function potential unabated.

The problem I have is how to reconcile this with the previous chapter, in which he treated the harmonic oscilator - a bound system - and found the energy levels to be

$$E_n = \hbar \omega \left(\frac{1}{2}+n\right),$$

which is positive even though the potential goes to infinity at infinity. I suppose you could "subtract infinity" and get an infinitely negative energy (and 0 potential at infinity), but that's a bit weird at best. Part 1 of the question: Is that all there is to it? "Subtract infinity" and then the second inequality ($E<0$) works?

Part 2 of the qustion: since infinite potentials are just approximations and do not really exist (or do they?), how can bound states ever exist (Griffith remarks that finite potentials can be overcome by tunneling)? Additionally, scattering states also do not exists as their wavefunctions are non-normalizable. So the conclusion is that nothing really exists according to Quantum Mechanics... which can't be right, surely?

Answer

The distinction to be made here is that, for the quantum harmonic oscillator system, there are no unbound states, only bound states thus, there is no benefit to insisting the states have negative energy, no reason to 'subtract infinity' in order to zero the potential at infinity.

However, in systems that permit both bound and unbound states, it is reasonable to zero the potential at infinity for the same reason that we do this classically.

For example, in the classical central force problem, there is a state in which particle can 'escape to infinity' where it will have zero kinetic energy (more precisely, the kinetic energy of the particle asymptotically approaches zero). If we set the potential energy to be zero at infinity, then the total energy 'at infinity' is zero. Thus, the particle with zero total energy 'sits on the boundary' between those particles with not enough energy to 'reach' infinity and those that do.

But, for the classical harmonic oscillator potential, no particle can escape to infinity. The kinetic energy of the particle will periodically and instantaneously be zero. In this case, it is reasonable that the state where the total energy is always equal to the potential energy (the state where the kinetic energy is always zero) be the zero total energy state; all other states having positive total energy.

So the conclusion is that nothing really exists according to Quantum Mechanics... which can't be right, surely?

That's not remotely the correct conclusion to draw. One might conclude instead that

(1) The conception of bound state must be modified in the passage from classical mechanics to quantum mechanics and

(2) the physical (normalizable) unbound states are not eigenstates of the Hamiltonian, i.e., the physical unbound states are not states of definite energy but are, instead, a distribution of energy eigenstates, e.g., a wavepacket.

rotational kinematics - Why is angular velocity a vector quantity?

Angular velocity is $$\omega= \frac{dƟ}{dt},$$ here $\theta$ and $t$ are scalar quantities. But $\omega$ is a vector quantity. Why is it such?

So far I know the direction of $\omega$ is along the axis of rotation. If so, why?

newtonian mechanics - Kepler problem in time: how do two gravitationally attracted particles move?

Two particles with initial positions and velocities $r_1,v_1$ and $r_2,v_2$ are interacting by the inverse square law (with G=1), so that

$$ {d^2r_1\over dt^2} = - { m_2(r_1-r_2)\over |r_1-r_2|^3} $$ $$ {d^2r_2\over dt^2} = - { m_1(r_2-r_1)\over |r_1-r_2|^3} $$

(the inverse square law along the line of separation). What is the complete solution of these differential equations? What is the position of the two objects as a function of time?

After reading a lot on Wikipedia, I've come to the definition of center of mass and relative coordinates:

$$R(t) ~=~ \frac{m_1 r_1 + m_2 r_2}{m_1+m_2}$$

$$\ddot{r}(t) r(t)^2 ~=~ (m_1+m_2)G$$

Where $R$ is the center of mass, and $r$ is the displacement between the particles... Is this correct? How do I proceed to solve the differential equation?

Answer

Textbook solution

The first thing to do is to define the center of mass and relative coordinates:

$$ R(t) = {m_1 r_1 + m_2 r_2 \over m_1 + m_2} $$

$$ r(t) = r_2 - r_1 $$

You invert this to find

$$ r_1= R - {m_2\over M} r$$ $$r_2=R+ {m_1\over M} r$$

The equation of motion for R is trivial, since center of mass is a conservation law:

$$ {d^2 R \over dt^2 } = 0 $$

and it is solved by $$ R(t) = V_0 t + R_0 $$ where $V_0$ and $R_0$ are the initial center of mass velocity and position respectively (which are calculated from the given initial conditions).

The nontrivial equation is for the relative coordinate: $$ m{d^2 r \over dt^2} = - {m_1 m_2 r\over |r|^3}$$ where m is the reduced mass: ${1\over m} ={1\over m_1} + {1\over m_2}$.

Or: $$ {d^2 r \over dt^2} = - {M r\over |r|^3}$$

Where $M=m_1 + m_2$ is the total mass.

The problem is reduced to solving the Kepler motion in a 1/r potential. From now on, I will rescale time to make the mass parameter in the r equation 1.

You can choose the x axis to lie along the initial r, and the y-axis to lie along the component of the initial $\dot{r}$ perpendicular to the initial r. Another way of saying this is that you rotate the coordinates to make the angular momentum vector $r\times p$ where $p=m\dot{r}$ to lie along the z-axis. This rotation reduces the problem to a plane, and the rotation matrix columns are given by the normalized initial r (now along the x-axis), the component of the initial velocity perpendicular to r, normalized (along the y-axis), and normalized L along the z-axis.

You then use units to set the total over reduced mass to 1, and use polar coordinates in the x-y plane of the motion, and note that the angular momentum is constant:

$$ r^2 {d\theta\over dt} = L $$

This tells you that equal areas are swept out by the r-vector in equal times. The equation of motion for r(t) (no longer a vector, now a scalar radial coordinate) is:

$$ {d^2 r \over dt^2} = {L^2 \over r^3} - {1 \over r^2} $$

Then you change time out for $\theta$, expressing everything in terms of $r(\theta)$, which you can do using the equal area law, whenever the anguar momentum is nonzero (if the initial angular momentum is zero, or very close to zero, this is a one-dimensional two-body problem which can be solved directly by more elementary means). The equation of motion for $r(\theta)$ simplifies when you make a coordinate transformation to $u={1\over r}$:

$$ {d^2 u \over d\theta^2} = C - u $$

Where C is some unimportant constant, and this is solved by

$$u(\theta) = {1\over A} (1 + a \cos(\theta -\theta_0)) $$

Where A is the semi-major axis of the ellipse (if the orbit is an ellipse), $\theta_0$ determines the orientation in the x-y plane, and a is the eccentricity of the ellipse (if a<1), or determines the angle of the hyperbola (if a>1) or tells you the orbit is a parabola (a=1).

the only result you need is that

$$ r(\theta) = {A\over 1+ a\cos(\theta-\theta_0)} $$

This gives you the solution of r as a function of $\theta$, which gives the shape of the orbit. This is where textbooks stop.

Finding $\theta$ as a function of $t$

But you then want the solution for $\theta$ as a function of time, to get the r and $\theta$ as functions of time. This is determined from the area law, conceptually:

$$ r^2 {d\theta\over dt} = L$$

$$ {d\theta \over (1+ a \cos(\theta-\theta_0))^2 }= {L \over A^2} dt $$

and integrating this from time 0 to time t, tells you in principle what $\theta(t)$ is. The result can be written as:

$$ F(a,\theta) = F(a,\theta_0) + {L\over A^2} t $$

Where $F(a,\theta)$ is the special function that gives you the area of a conic section of parameter a, in a wedge from the focus where one half-line is along the major axis, and the other half-line makes an angle $\theta$ with the first. This special function is not expressible in terms of elementary functions.

This function is defined by the integral above, and you can calculate it numerically using any numerical integration method. Finding this function, and inverting it, is the only difficult part of this problem. There are three limits which are necessary for perturbations:

• for a=0, $ F(0,\theta) = \theta$


• for a=1, $ F(1,\theta) = {y\over 4} + {y^3\over 12} $ where $y=r\sin{\theta} = {\sin(\theta)\over 1+\cos(\theta)}$


• for $a=\infty$, $ F(a,\theta) \approx {1\over a} \tan(\theta) $

Each of these are elementary degenerations: the first is the circle, the second is the parabola, and the third is a straight line. The important thing is that each of these degenerations gives you x(t) and y(t) which are simple, and further, you can perturb around each of these three limits in a nice way. In the following, the parameter t is rescaled to absorb ${L\over A^2}$

• circle: $x(t) = \cos(t)$ $y(t) = \sin(t)$


• parabola: $y(t)= (\sqrt{1+36t^2} + 6t)^{1\over 3} - (\sqrt{1+36t^2} - 6t)^{1\over 3} $ $x(t) = {1\over 2} - {y^2\over 2} $


• line: $x(t) = {1\over a}$, $y(t) = t$

The line and circle are obvious, the parabola is found by inverting the cubic for y as a function of t using the cubic equation.

Near the circle, time is periodic with the orbital period, which is the area inside the ellipse $aA^2$ divided by the area sweep-rate $L/2$. So you have once-winding function from a circle to a circle, which can always be written as a Fourier series with a linear term, which is found from the power series of the integrand in a, integrated term by term. Near the straight-line hyperbola, you can similarly perturb in a series, and the only interesting degeneration is the parabola. Near the parabola, the perturbation theory is a little more complicated.

newtonian mechanics - Why are springs shaped the way they are?

Why are springs coiled the way they are? Why not some other shape? Is the shape due to its elasticity or something?

Answer

The shape is due to the underlying principle of what a coil-spring actually does.

A spring is originally a long (often metal) wire. When you coil it, you are changing the way the forces can be applied to the spring.

For coil springs, you are making two surfaces that are often flat or have connections, that can be coupled with other systems.

The interesting thing is what is actually happening to the wire when you stretch or compress a coil spring. Because of the helical nature of the winding, when you push/pull on the spring, most of the effort is not going into bending or stretching the wire. It is twisting it.

You can visualize this by thinking of a coil from the side. The twisting forces all the coils closer together or further apart, leading to the motion desired.

You can consider it as a long bar experiencing torsion, allowing it to twist. As long as the bar has a constant cross section this relationship will also be linear, which may also have something to do with the shape.

Essentially, my understanding is that the helical shape allows you to use vertical motions to create twisting in the horizontal planes.

Helical shapes seem to do similar things in other applications. Perhaps someone with a better understanding of the math could give a clear description of why helix shapes can change the nature of motion in general.

(for example, torsional springs actually create a spring that resists twisting, but the spring itself is actually acting like a long bent bar; basically the opposite scenario of a linear helical spring)

newtonian mechanics - Why is the centripetal force vector on a banked turn horizontal rather than parallel to the incline?

Why isn't the centripetal force parallel to the inclined surface? It seems like if you were to draw a circle with one point being at the center of the curve and one point at the car the circle would be inclined and the radial vector would be too. See image below for basic diagram of the problem.

Answer

I'll give you a couple of ways to think about this.

First, geometrically, the circle you are thinking about drawing should contain the entire circular path of the car. If we're assuming that the car is remaining at a constant "elevation" on the banked surface, then the center of that circle has to be at the same elevation: otherwise you'd be drawing a cone or bowl of some sort. So the center of the circle in question is raised off the ground in the middle, so that it's level with the car's elevation, and the centripetal force vector points horizontally toward the center.

Second, and more in physics language, let's assume again that we want the car to remain at a constant elevation. Then the vertical components of both velocity and acceleration must be zero throughout the whole path. (If they weren't, it would have to move up or down, either sliding sideways along the track, taking flight and lifting off of it, or trying to dig down into it.) But then by Newton's second law, $\vec{F} = m\vec{a}$, the vertical component of the force must also be zero. So all that leaves is a horizontal centripetal force, which is what sustains the constant circular motion.

Hope that helps!

newtonian mechanics - Rod Falling on Frictionless Surface

A rigid rod with mass $m$ is initially held at an angle with respect to a frictionless plane by a string. Then the string is cut. What happens to the rod? My intuition suggests that the rod should slide to the left but I cannot figure out how this is supposed to happen when the only forced involved are vertical (reaction force and weight). Moreover, while initially the reaction force should balance the weight, the reaction force causes the rod to rotate about its centre of mass. This would cause the end of the rod in contact with the surface to leave it. But then the reaction force disappears and the weight of the rod places it back on the surface again. I am not sure how to deal with this "jerky motion" either.

Answer

I'm guessing from your description that the left end of the rod is the end on contact with the ground, and the right end is the in the air at the point that the string is cut.

I think that what will happen is this: When the string is cut, gravity will cause the rod to fall, such that its centre of gravity falls vertically. As the rod falls, it will rotate to become more horizontal (it must be horizontal when it lands on the ground). That will of course cause the left end of the rod, which is already on the ground, to move to the left (your intuition is correct in that regard), but the right end will move to the right, ensuring no horizontal movement of the centre of gravity.

What are the forces that cause this movement of the ends? They will be internal forces within the rod (caused because the external forces are acting to compress the rod, and the rod reacts to that). Those internal forces will act along the rod and so have a horizontal component. If you sum/integrate these horizontal components along the whole rod, they'll cancel out and so add up to zero, but at any particular point along the rod, the horizontal force acting on that bit of rod will be nonzero.

The left end of the rod will not leave the ground while the rod is falling: The rod is simultaneously falling and rotating, and the vertical component of the two motions will exactly cancel at the left end, so that the left end will only move horizontally along the ground. (The entire rod may of course bounce off the ground immediately after falling, as a result of the collision with the ground, but I don't think that's what you were asking about).

newtonian mechanics - Jumping into water

Two questions:

1. 
   

   Assuming you dive head first or fall straight with your legs first, what is the maximal height you can jump into water from and not get hurt?
   In other words, an H meter fall into water is equivalent to how many meters concrete-pavement fall, force wise? (I'm assuming the damage caused will be mainly due to amount of force and not the duration)

   
   


2. 
   

   Assume you jump head first and hold a sharp and strong long object that cuts the water before you arrive, will that make the entrance to the water more smooth and protect you?

   
   

Answer

Answering your questions in reverse order:

Yes, a long pointy object (like your arms over your head, in a dive, or your pointed toes in a feet-first entry) will make a big difference. Remember the tongue-in-cheek adage, "it's not the fall that kills you; it's the sudden stop?" That is exactly what differentiates a fall onto concrete from a fall into water: how sudden is the stop. And making that stop LESS sudden (decreasing the magnitude of deceleration during the stop) is exactly how airbags save your life in a car crash. One can decrease the magnitude of deceleration by reducing the ratio $(\Delta V / \Delta t)$. Since there is roughly a linear relationship between time and distance traveled during the instant of impact, you can achieve the same effect by reducing the ratio $(\Delta V / \Delta s)$ where $s$ = distance traveled during the deceleration event. The easiest way to do this is to lengthen $s$.

One thing to remember about the water fall statistics is that a large number of them are likely "unpracticed". These are not olympic divers working up to 250 feet. A large proportion of them are unconditioned people forced into a water "escape"; or, worse, are people TRYING to die.

Assuming you are doing the right thing, and optimizing your form for water entry, you will simultaneously be minimizing your wind resistance during the fall:

1.) A fall from 30 feet will result in a velocity of roughly 44 ft/s = 30 mph.

2.) A fall from 100 feet will result in a velocity of roughly 80 ft/s = 54 mph.

3.) A fall from 150 feet will result in a velocity of roughly 97 ft/s = 66 mph.

4.) A fall from 250 feet will result in a velocity of roughly 125 ft/s = 85 mph.

The first case is a tower jump I did for the Navy, and is trival for anyone who is HWP and doesn't belly flop. The second is an approximation of a leap from a carrier deck, which the tower jump was supposed to teach you how to survive (be able to swim after the fall). The third is only 20% faster entry speed (and force) and should be survivable by anyone in good shape and able to execute good form (pointed toe entry, knees locked, head up, arms straight up). The La Quebrada cliff divers routinely dive from 125 feet as a tourist attraction. If forced to choose, I'd pick a feet-first entry at 150 feet over a dive at 125.

So the interesting part is the stretch from 150 to 250 feet. My guess is that the limit for someone voluntarily performing repeated water dives/jumps from a height of $x$ will show $x$ to be somewhere around $225 \text{ feet} \pm 25 \text{ feet}$.

EDIT: There are documented cases of people surviving falls from thousands of feet (failed parachute) onto LAND. These freaky cases of surviving terminal velocity falls do not answer the question practically; but they are there. For example, Vesna Vulović is the world record holder for the biggest surviving fall without a parachute.

quantum mechanics - Matrix elements of momentum operator in position representation

I have two related questions on the representation of the momentum operator in the position basis.

The action of the momentum operator on a wave function is to derive it:

$$\hat{p} \psi(x)=-i\hbar\frac{\partial\psi(x)}{\partial x}$$

(1) Is it ok to conclude from this that:

$$\langle x | \hat{p} | x' \rangle = -i \hbar \frac{\partial \delta(x-x')}{\partial x}?$$

And what does this expression mean?

(2) Using the equations:

$$ \frac{\langle x | \hat{x}\hat{p} | x' \rangle}{x} = \frac{\langle x | \hat{p}\hat{x} | x' \rangle}{x'} = \langle x | \hat{p} | x' \rangle $$

and

$$\langle x | [\hat{x},\hat{p}]|x'\rangle=i\hbar \delta(x-x')$$

one can deduce that

$$\langle x | \hat{p} | x' \rangle = i \hbar \frac{\delta(x-x')}{x-x'}$$

Is this equation ok? Does it follow that

$$\frac{\partial \delta(x-x')}{\partial x} = - \frac{\delta(x-x')}{x-x'}?$$

Answer

1) Notice that by inserting a complete set of position states we can write $$ \hat p \psi(x) = \langle x|\hat p|\psi\rangle = \int dx'\langle x|\hat p|x'\rangle\langle x'|\psi\rangle =\int dx'\langle x|\hat p|x'\rangle \psi(x') $$ so if we set $$ \langle x|\hat p|x'\rangle = -i\hbar \frac{\partial}{\partial x}\delta(x-x') =i\hbar \frac{\partial}{\partial x'}\delta(x-x') $$ then we can use integration by parts to obtain $$ \hat p \psi(x) =i\hbar \int dx'\frac{\partial}{\partial x'}\delta(x-x') \psi(x') = -i\hbar \int dx'\delta(x-x') \frac{d \psi}{dx'}(x') = -i\hbar \frac{d\psi}{dx}(x) $$ So your expression is correct. The derivative of a delta function is essentially defined by the integration by parts manipulation that I just performed; in fact derivatives of distributions in general are defined in an analogous way. See this lecture for example.

Hope that helps; let me know of any typos!

Cheers!

electromagnetism - Why does electromagnetic induction actually occur?

In my book, it is written that "An emf is induced in a loop when the number of magnetic field lines that pass through the loop is changing" (Faraday's law)

I understand that whenever there is a change in magnetic flux, there will an emf induced in a loop and this in turn will induce a current.

However, I really want to understand what exactly goes on to cause this. I just want to check if I am right:

1. 
   

   There is a relative movement between the loop and the magnet (e.g. loop moving towards magnet)

   
   


2. 
   
   

   Thus, electrons in the loop are "moving" relative to the field

   
   


3. 
   

   A moving charge experiences a force perpendicular to the direction of the field

   
   


4. 
   

   As a force is exerted on the electrons, they move in a particular direction

   
   


5. 
   

   Current is induced

   
   
   

I do not know much calculus and studying high school physics but could someone please help me get an intuition behind Faraday's law or tell me if my way of thinking about it is right?

Answer

You are correct. Essentially it all boils down to the Lorentz force $\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})$. If you move into a frame of reference where the section of the loop is momentarily stationary, then you can do a Lorentz transformation to find out what $\vec{E}$ is in that frame. In that frame, $\vec{v}=0$ of course, so the second term is zero, but $\vec{E}$ will now be non-zero, and, in fact, equal to $\vec{v}\times\vec{B}$. In this frame, the force on the charges is purely electrical in nature, i.e., due to an electric field, and this is why you can define an EMF analogously to electric potential. But because wires are usually loops and each segment of the loop has it's own reference frame, there's no way to do this everywhere (globally).

astronomy - How to explain the Moon halo phenomenon?

Today, here in Brazil, I have observed (and is still observing) an interesting phenomenon.

The Moon is near to a big star in the sky, but this is normal. The interesting part is what's around them.

A huge circle in which the moon and the star is inside. Why does this circle appear? What is it?

(This isn't a cloud. It looks like more something like a light reflection of a lamp.)

Unfortunately, I hadn't a good camera to catch this. I would appreciate having an image posted for this.

Answer

You might be describing a moon aura.

It's caused by diffraction on tiny ice crystals in the atmosphere: When the moonlight hits the ice, you get interference effects that depend on the angle of the incident light. The idea is similar to powder diffraction in crystallography.

Relation between Black-Scholes equation and quantum mechanics

I am interested in the link between the Black & Scholes equation and quantum mechanics.

I start from the Black & Scholes PDE $$ \frac{\partial C}{\partial t} = -\frac{1}{2}\sigma^2 S^2 \frac{\partial^2C}{\partial S^2} -rS\frac{\partial C}{\partial S}+rC $$ with the boundary condition $$C(T,S(T)) = \left(S(T)-K\right)_+.$$ Performing the change of variables $q=\ln(S)$ this equation rewrites $$ \frac{\partial C}{\partial t} = H_{BS}C $$ with the Black & Scholes Hamiltonian given by $$H_{BS} = -\frac{\sigma^2}{2} \frac{\partial^2}{\partial q^2}+\left(\frac{1}{2}\sigma^2 - r\right)\frac{\partial}{\partial q} +r.$$

Now I compare this equation with the Schrödinger equation for the free particle of mass $m$ : $$i\hbar \frac{d\psi(t)}{dt} = H_0\psi(t),\quad \psi(0)=\psi$$ with the Hamiltonian (in the coordinate representation) given by $$H_0 = -\frac{\hbar^2}{2m} \frac{d^2}{dq^2}.$$

My problem comes from the fact that the various references I am reading for the moment explain that the two models are equivalent up to some changes of variables (namely $\hbar=1$, $m=1/\sigma^2$ and the physical time $t$ replaced by the Euclidean time $-it$). However, their justifications for the presence of the terms $$\left(\frac{1}{2}\sigma^2 - r\right)\frac{\partial}{\partial q} +r$$ in the Hamiltonian seem very suspicious to me. One of them tells that these terms are "a (velocity-dependent) potential". Another one tells this term is not a problem since it can be easily removed is we choose a frame moving with the particle.

I have actually some difficulties to justify why, even with this term, we can say that the Black & Scholes system is equivalent to the one coming from the quantum free particle. I don't like this potential argument, since (for me) a potential should be a function of $q$ (so it would be ok for example for the $+r$ term) but not depending on a derivative.

Could you give me your thoughts on this problem?

photon electron interaction such that the photon energy does not correspond to any state change

Although I've read answers to similar questions, they did not answer my doubt:

If a photon has an energy less than that required to move an electron from one energy level to another, then what happens when the electron and photon collide? Is the photon absorbed, or does it by some mysterious method, just pass through the electron?

Can a photon transmit a momentum to a neutron?

Is a photon able to transfer an impulse to a neutron or, and this is the same, can light accelerate a neutron?

Answer

Yes, a sufficiently energetic photon can accelerate a lone neutron. The kinetic energy imparted to the neutron reduces the photon's wavelength (redshifts it) by the same amount, so the total energy of the system remains the same. Outside the nucleus, the neutron has a half-life of about 10.5 minutes.

field theory - Differentiating Propagator, Green's function, Correlation function, etc

For the following quantities respectively, could someone write down the common definitions, their meaning, the field of study in which one would typically find these under their actual name, and most foremost the associated abuse of language as well as difference and correlation (no pun intended):

Maybe including side notes regarding the distinction between Covariance, Covariance function and Cross-Covariance, the pair correlation function for different observables, relations to the autocorrelation function, the $n$-point function, the Schwinger function, the relation to transition amplitudes, retardation and related adjectives for Greens functions and/or propagators, the Heat-Kernel and its seemingly privileged position, the spectral density, spectra and the resolvent.

---

Edit: I'd still like to hear about the "Correlation fuction interpretation" of the quantum field theoretical framework. Can transition amplitudes be seen as a sort of auto-correlation? Like... such that the QFT dynamics at hand just determine the structure of the temporal and spatial overlaps?

Answer

The main distinction you want to make is between the Green function and the kernel. (I prefer the terminology "Green function" without the 's. Imagine a different name, say, Feynman. People would definitely say the Feynman function, not the Feynman's function. But I digress...)

Start with a differential operator, call it $L$. E.g., in the case of Laplace's equation, then $L$ is the Laplacian $L = \nabla^2$. Then, the Green function of $L$ is the solution of the inhomogenous differential equation $$ L_x G(x, x^\prime) = \delta(x - x^\prime)\,. $$ We'll talk about its boundary conditions later on. The kernel is a solution of the homogeneous equation $$ L_x K(x, x^\prime) = 0\,, $$ subject to a Dirichlet boundary condition $\lim_{x \rightarrow x^\prime}K(x,x^\prime) = \delta (x-x^\prime)$, or Neumann boundary condition $\lim_{x \rightarrow x^\prime} \partial K(x,x^\prime) = \delta(x-x^\prime)$.

So, how do we use them? The Green function solves linear differential equations with driving terms. $L_x u(x) = \rho(x)$ is solved by $$ u(x) = \int G(x,x^\prime)\rho(x^\prime)dx^\prime\,. $$ Whichever boundary conditions we what to impose on the solution $u$ specify the boundary conditions we impose on $G$. For example, a retarded Green function propagates influence strictly forward in time, so that $G(x,x^\prime) = 0$ whenever $x^0 < x^{\prime\,0}$. (The 0 here denotes the time coordinate.) One would use this if the boundary condition on $u$ was that $u(x) = 0$ far in the past, before the source term $\rho$ "turns on."

The kernel solves boundary value problems. Say we're solving the equation $L_x u(x) = 0$ on a manifold $M$, and specify $u$ on the boundary $\partial M$ to be $v$. Then, $$ u(x) = \int_{\partial M} K(x,x^\prime)v(x^\prime)dx^\prime\,. $$ In this case, we're using the kernel with Dirichlet boundary conditions.

For example, the heat kernel is the kernel of the heat equation, in which $$ L = \frac{\partial}{\partial t} - \nabla_{R^d}^2\,. $$ We can see that $$ K(x,t; x^\prime, t^\prime) = \frac{1}{[4\pi (t-t^\prime)]^{d/2}}\,e^{-|x-x^\prime|^2/4(t-t^\prime)}, $$ solves $L_{x,t} K(x,t;x^\prime,t^\prime) = 0$ and moreover satisfies $$ \lim_{t \rightarrow t^\prime} \, K(x,t;x^\prime,t^\prime) = \delta^{(d)}(x-x^\prime)\,. $$ (We must be careful to consider only $t > t^\prime$ and hence also take a directional limit.) Say you're given some shape $v(x)$ at time $0$ and want to "melt" is according to the heat equation. Then later on, this shape has become $$ u(x,t) = \int_{R^d} K(x,t;x^\prime,0)v(x^\prime)d^dx^\prime\,. $$ So in this case, the boundary was the time-slice at $t^\prime = 0$.

Now for the rest of them. Propagator is sometimes used to mean Green function, sometimes used to mean kernel. The Klein-Gordon propagator is a Green function, because it satisfies $L_x D(x,x^\prime) = \delta(x-x^\prime)$ for $L_x = \partial_x^2 + m^2$. The boundary conditions specify the difference between the retarded, advanced and Feynman propagators. (See? Not Feynman's propagator) In the case of a Klein-Gordon field, the retarded propagator is defined as $$ D_R(x,x^\prime) = \Theta(x^0 - x^{\prime\,0})\,\langle0| \varphi(x) \varphi(x^\prime) |0\rangle\, $$ where $\Theta(x) = 1$ for $x > 0$ and $= 0$ otherwise. The Wightman function is defined as $$ W(x,x^\prime) = \langle0| \varphi(x) \varphi(x^\prime) |0\rangle\,, $$ i.e. without the time ordering constraint. But guess what? It solves $L_x W(x,x^\prime) = 0$. It's a kernel. The difference is that $\Theta$ out front, which becomes a Dirac $\delta$ upon taking one time derivative. If one uses the kernel with Neumann boundary conditions on a time-slice boundary, the relationship $$ G_R(x,x^\prime) = \Theta(x^0 - x^{\prime\,0}) K(x,x^\prime) $$ is general.

In quantum mechanics, the evolution operator $$ U(x,t; x^\prime, t^\prime) = \langle x | e^{-i (t-t^\prime) \hat{H}} | x^\prime \rangle $$ is a kernel. It solves the Schroedinger equation and equals $\delta(x - x^\prime)$ for $t = t^\prime$. People sometimes call it the propagator. It can also be written in path integral form.

Linear response and impulse response functions are Green functions.

These are all two-point correlation functions. "Two-point" because they're all functions of two points in space(time). In quantum field theory, statistical field theory, etc. one can also consider correlation functions with more field insertions/random variables. That's where the real work begins!

planets - Why does Jupiter emit more energy than it receives?

I hear that Jupiter and Saturn emit more energy than they receive from the Sun. This excess energy is supposedly due to contraction.

• Is this accepted as fact (or is it controversial)?


• Does this mean that Jupiter is shrinking a little bit (its diameter decreases), or are there just changes in the internal density distribution?

Answer

The book Jupiter: The Planet, Satellites and Magnetosphere, edited by Fran Bagenal, Timothy Dowling, and William McKinnon has, in its third chapter, the text: "[Jupiter] is still contracting at a rate of ~3 cm per year while its interior cools by ~1 K per million year."

The chapter does not give a specific source for that, but the chapter has an extensive list of references. I believe that the subject first came up in 1966 in a paper by Low who measured the infrared heat flux to be 1.9 times the incident solar. I was under the impression that this has been measured, but I was unable to find a direct reference to this; at this point it is likely to be a model and that's the amount required for the observed level of heating. One early paper that modeled this is by W.B. Hubbard in the journal Icarus, published in 1977 under the title "The Jovian surface condition and cooling rate."

electromagnetism - Magnetic Force on a Ferromagenetic Material

I am currently working on a project involving solenoids, and I needed a force(Newtons, not a measure of magnetic field strength) equation. What I came up with after some digging around on the internet, is the equation:

$$F = (NI)\mu_0\frac{\text{Area}}{2g^2}$$

Where $F$ is force (in Newtons), $N$ is the number of turns in the coil, $I$ is the current being passed through the coil, $μ_0$ is the magnetic permeability of vacuum, and $g$ is the gap between the coil and the ferromagnetic material. (Area $A$ and $g$ can be any units, as long as you're consistent with the usage)

I don't know in which plane exactly the area $A$ is taken.

Assuming I have a rod, moving lengthwise into a solenoid, which plane would $A$ represent?

Plane a, plane b, or another plane that I did not consider relevant to this problem?

Rod:

Edit: I was looking for the force an electromagnet would exert on a ferromagnetic material moving into the coil. something like this.

Edit: If the equation I was using before does not work, I don't suppose anyone has the correct one?

Edit: After looking at the equation some more, I realized I had written it wrong. It should be:

$$F = (NI)^2\mu_0\frac{\text{Area}}{2g^2}$$

astrophysics - Why are there spectral lines at all?

My somewhat basic understanding of the concept comes from lectures I've attended about the Bohr-model, which explains the phenomenon as arising from the fact that certain configurations of an atom can only absorb certain wavelengths of light and other configurations can emit the same wavelength and change into the first configuration.

Now what I cannot understand is why these effects do not cancel out, and why in some instances absorption wins out and we observe absorption lines, and in other instances it is the other way around. Also, if I haven't misunderstood there are also some instances where we observe both absorption and emission at once.

So my question boils down to this: Why?

Answer

At your level of understanding, the Bohr atom will do.

Atoms are neutral, composed of orbiting electrons ( negatively charged) around a nucleus (positively charged).

The very basic question coming out of this fact, that an atom is composed of orbiting electrons around a positive nucleus is: how can it be possible when we know that accelerating charges makes them radiate electromagnetic waves away and lose momentum. The electrons should fall into the nucleus since a circulating charge has a continuous acceleration, and matter as we know it could not exist.

Enter the Bohr model: It postulated that there were some orbits where the electrons could run around without losing any energy , quantized orbits.

Enter the absorption lines: The electrons could only change orbits if kicked up by an electromagnetic wave of an energy specific to that particular orbit and discrete. There fore if one shone that specific frequency of light ( E=h*nu) on a specific atom there was a probability to kick an electron up to a higher, called excited, orbit.

Enter the emission lines: Once excited there was a probability for the electron to fall back emitting the specific energy it had absorbed before. This could be observed.

Different experiments will show the different behaviors, even though absorption and emission will be happening continuously in the material.

An experiment shining light on the material and looking at the reflected spectrum will see absorption lines at those frequencies, because the relaxation of the excited electrons will emit back radiation all around randomly, whereas the reflected spectrum is at a specific angle.

An experiment out of the line of the exciting photons will see the emission spectrum . Absorption spectra are useful for identifying elements in stars, the absorption happening in the star's atmosphere and appearing as dark lines in the black body spectrum.

It goes without saying that physics has moved to new horizons from the time that the Bohr atom was news. It has been supereceded by Quantum mechanics which gives tools to accurately predict and classify all spectra as the result of a coherent theory of the way the universe behaves ( i.e. quantum mechanically).

hilbert space - What is quantum entanglement?

What is quantum entanglement?

Please be pedagogical.

Edit: I have updated my background under my profile.

Answer

Entanglement is a quantum correlation between two (or many) objects - a correlation means that these two objects' properties are not independent of each other - which was created in the objects' common past when they were close to one another i.e. when they were two parts of the same physical system.

Quantum mechanics changes the character of possible "properties" that objects may have (the quantities describing objects's properties are usually called "observables" and they are represented by Hermitian operators on the Hilbert space), as well as the way how these properties are measured and predicted (just probabilistically), so it also changes the character and magnitude of correlations that the objects may exhibit.

In particular, quantum correlations may often be stronger - and affecting a large fraction of measurable properties of the objects - than what would be possible according to classical (i.e. non-quantum) physics. In classical physics, correlations have to satisfy e.g. the so-called Bell's inequalities in various situations but quantum mechanics - and the real world - can easily surpass these bounds.

Technically, objects and their properties in quantum mechanics are described by wave functions. To describe the state of two mostly independent objects, one has to take a wave function from the tensor product $H_1\otimes H_2$ of the Hilbert spaces describing the individual objects. The wave function in the tensor product implies probabilistic predictions for any pairs of properties of the first and second object; in general, they're not independent, and for each combination of the objects' properties, quantum mechanics (and the wave function) may remember an independent probability.

Any vector in the tensor product that can't be written as a tensor product of vectors from $H_1$ and $H_2$ (instead, it can only be written as a linear combination of such tensor products of vectors) is called entangled. In other words, it is non-entangled if it is a simple tensor product of two simpler vectors. If it is a simple tensor product, all probabilities of "coupled properties" of the pair of objects simply factorize to the probability of the first object, and probability of the second object, as you know from probabilities of independent phenomena.

The simplest pedagogical example of an entangled state (as well as the entangled state that is most often found in literature) is $$\frac{X_1\otimes Y_1 + X_2\otimes Y_2}{\sqrt{2}}$$ Because there are two terms with four different factors, you can't use the distributive law in any way that would allow you to rewrite it as a simple product. The letters $X,Y$ refer to the two objects and the labels $1,2$ refer to two different states of each of the two objects.

In this state, if the property "1 or 2" is measured on $X$, one obtains the answers 1 or 2 with 50% probability for each: the coefficient of the wave function is $1/\sqrt{2}$ because these complex coefficients have to be squared to obtain the probability. However, because $X_1$ is "coupled" to $Y_1$ and $X_2$ is coupled to $Y_2$, the state and the machinery of quantum mechanics predict that the object $Y$ will be measured to have the same property: if $X$ is in 1, $Y$ is in 1, and the same for the state 2.

Linear algebra - which is crucially important for quantum mechanics - allows one to reinterpret the state above as an "identity operator" so the correlation will exist regardless of the type of measurement that we perform both on $X$ and $Y$. For example, if the two states represent spins, the two particles will be correlated so that you will find out that they're polarized with respect to the same axis, if you measure both particles' polarizations with respect to the same, particular, but arbitrary axis.

This would be kind of impossible for two separated particles in classical physics that could only be perfectly correlated for one choice of the axis - but not another axis rotated by 45 degrees, for example - without a communication in between them. However, quantum mechanics predicts that such a 100% correlation "regardless of the axis" is not only possible but guaranteed by the state above and it requires no communication. Indeed, one can prove that relativistic theories in quantum mechanics - especially quantum field theory - don't allow one to transmit a single bit of information faster than light even though this would be needed in classical physics to guarantee the perfect correlation that quantum mechanics predicts for these experiments (and that the experiments confirm).

quantum mechanics - Why doesn't the momentum exchange (or lack thereof) between photon and beam-splitter destroy the interference?

I have a question (my very first here) related to 50/50 beam splitters as used in the Mach-Zehnder interferometers (see for example the Wikipedia page).

Let's concentrate on the input beam splitter: A continuous light beam (the input) is split 50/50, one in the 90° direction and one in the forward direction of the input beam. Momentum/energy conservation teaches us that deflected part of the beam must exert a (tiny) force on the beam splitter. The forward part of the beam will not.

In the single photon (quantum) case, the beam-splitter should get a slight "kick" if the photon is deflected, while the beam splitter is left in its original state if the photon goes straight through.

If I got this right so far, we can now increase the energy (and momentum) of the photon and decrease the mass of the beam splitter (now coupled to a sensitive piezo transducer or something) so that a photon "kick" can be recorded if deflected. Now "which way" information is available and interference should be destroyed as I understand this.

Would interference be restored if the transducer is disconnected? If yes, how do the photon "know" whether the transducer is connected or not? Maybe this is related to how much the photon/beam splitter system can become entangled to the environment?

If the answer is "No", how weak must the photon kick be in order for the interferometer to work? We already know that it works for ordinary light (very tiny photon kicks).

In other words: Why doesn't the momentum exchange (or lack thereof) between the photon and the beam-splitter (and the trace this leaves in the environment) destroy the interference?

Alternatively, couls someone just point me to relevant literature which discusses the transfer of momentum between between photons and optical elements?

quantum mechanics - Rigged Hilbert space and QM

Are there any comprehensive texts that discuss QM using the notion of rigged Hilbert spaces? It would be nice if there were a text that went through the standard QM examples using this structure.

Answer

I don't know of any books which use this language exclusively, but the basic idea is pretty straightforward:

All Hilbert spaces are isomorphic (if their dimensions match). This would present conceptual problems in quantum mechanics if we ever talked about the Hilbert space alone; how could we distinguish them? But it's OK because we are actually interested in a Hilbert space $\mathcal{H}$ equipped with an algebra of operators $\mathcal{A}$.

For example, the real difference between $\mathcal{H} = L^2(\mathbb{R})$ and $\mathcal{H} = L^2(\mathbb{R}^3)$: When we talk about the former, we're talking about $L^2(\mathbb{R})$ with the natural action of the 1d Heisenberg algebra $\mathcal{A}_1$ (generated by $P$ and $Q$ such that $[Q,P] = i\hbar$). When we talk about the latter, we're talking about the Hilbert space with the natural action of the 3d Heisenberg algebra $\mathcal{A}_3$.

Neither algebra actually acts on the entirety of $\mathcal{H}$. $(Q\psi)(x)= x\psi(x)$ doesn't necessarily lie in $L^2$. Likewise, the action of the differentiation operator $P = -i\hbar \frac{\partial}{\partial x}$ on a vector $v \in \mathcal{H}$ isn't defined if $v$ is not a differentiable function. And $P^2$ is only defined on twice-differentiable functions. However, there are some functions on which the action of any power $P^nQ^m$ is defined: If $v$ and all of its derivatives vanish faster at infinity than any polynomial, the action of any element of $\mathcal{A}_1$ is defined. Likewise, $\mathcal{A}_3$ really acts on the set $\mathcal{S}$ of functions in $L^2(\mathbb{R}^3)$ whose partial derivatives all vanish fast enough at infinity.

In general, if you have a Hilbert space and an algebra $\mathcal{A}$ of operators with continuous spectrum, there's a maximal subspace $\mathcal{S} \subset \mathcal{H}$ on which $\mathcal{A}$ acts. This is the subspace of $v \in \mathcal{H}$ for which $av$ is defined and $||a v|| < \infty$ for any $a \in \mathcal{A}$. It is called the space of smooth vectors for $\mathcal{A}$. (Exercise: $\mathcal{S}$ is dense in $\mathcal{H}$.)

$\mathcal{S}$ gets a topology from being a subspace of $\mathcal{H}$, but it actually has a much stronger topology from the family of seminorms $v \mapsto ||a v||$ (for $a \in \mathcal{A}$). This topology makes it a nuclear vector space.

Given $\mathcal{S} \subset \mathcal{H}$, you can construct the space $\mathcal{S}^* \supset \mathcal{H}$ of continuous (wrt the nuclear topology) complex-linear linear functionals on $\mathcal{S}$. (Here we are using the Riesz representation theorem to identify $\mathcal{H}$ with its dual $\mathcal{H}^*$.) This space should be thought of as the space of bras, in the Dirac bra-ket sense. The bra $\langle x |$ is the linear function which maps $\psi \in \mathcal{S}$ to $\psi(x) =\langle x | \psi \rangle $, aka, the Dirac delta function $\delta_x$ with support at $x$. (The space of kets is the conjugate space, consisting of conjugate-linear functionals on $\mathcal{S}$. The ket $|x \rangle$ maps a state $\psi \in \mathcal{S}$ to $\psi^*(x) = \langle \psi| x\rangle$.)

This space $\mathcal{S}$ is worth considering because it gives rigorous meaning to the idea that elements of $\mathcal{A}$ with continuous spectrum have eigenvectors, and that you can expand some states in these eigenbases. The elements of the algebra $\mathcal{A}$ can't have eigenvectors in $\mathcal{H}$ if they have continuous spectrum. But they do have eigenvectors in the space of bras. The definition is a standard extension-by-duality trick: $v \in \mathcal{S}^*$ is an eigenvector of $a \in \mathcal{A}$ with eigenvalue $\lambda$ if $(av)(\psi) = \lambda v(a^*\psi)$ for all $\psi \in \mathcal{S}$. (Exercise: $\langle x|$ is the eigenbra with eigenvalues $x$ of the position operator $Q$.)

The triplet $(\mathcal{S}, \mathcal{H}, \mathcal{S}^*)$ is a rigged Hilbert space. The language of rigged Hilbert spaces was invented to capture the ideas I've outlined above: the smooth vectors of an algebra of operators with continuous spectrum, and the dual vector space where the eigenbases of these operators live. The language actually matches the physics very nicely -- especially the bra-ket formalism -- but it provides a level of precision that's not really necessary for most calculations (e.g., with floating point arithmetic).

homework and exercises - Time taken for collision

We have three particles at the vertices of equilateral triangle of side $d$. At $t=0$ they start moving in such away that at all instant of time each of them has a speed $v$ towards adjacent one. We have to find the time after which they will collide. I know that you can solve this by calculating from reference frame of any one particle, but I want a way to solve this by calculating the total distance traveled first in ground frame and then dividing by speed.

Answer

Let me first describe the usual way one solves the problem as mentioned by the OP. I will then work the problem out in the ground frame.

Nice Answer: Let us work in the frame of particle 2. In this frame, particle 3 is spiralling in towards 2, while particle 1 has a complicated motion that is always point towards 3. It is easy to show that the component of the velocity of 3 along the line joining 2 and 3 is $\frac{3v}{2}$. The net time taken is $t_0 = \frac{2d}{3v}$.

Nice Answer in ground frame: In the ground frame, there is a clear symmetry. The particles slowly move in towards the center. Due to the symmetry, the component of the velocity of each particle towards the center of the triangle is always equal and is a constant. At $t=0$, the component in this direction is $v \cos \frac{\pi}{6} = \frac{\sqrt{3}v}{2}$. The total distance every particle travels in that direction is $\frac{d}{\sqrt{3}}$. The total time is then $\frac{2d}{3v}$.

Full answer in ground frame: Choose the origin to be the center of the equilateral triangle. At any time $t$, let the polar coordinates of the $i$th particle be $(r_i(t),\theta_i(t))$ with $0 < \theta_i(t) \leq 2\pi$ and $i=1,2,3$. The position vector is then given by $$ \vec{r}_i = r_i \left( \cos\theta_i {\hat i} + \sin\theta_i {\hat j} \right) $$ Now, before we proceed, we will use symmetry. Note that, no matter how the particles travel they will always form an equilateral triangle. Mathematically this implies the following equations $$ r_1(t) = r_2(t) = r_3(t) = r(t) $$ $$ \theta_1(t) - \theta_2(t) = \theta_2(t) - \theta_3(t) = \frac{2\pi}{3} $$ $$ \implies \theta_1(t) = \theta(t) + \frac{2\pi}{3} ,~\theta_2(t) = \theta(t),~\theta_3(t) = \theta(t) - \frac{2\pi}{3} $$ Thus, the original 6 variables have been reduced to two variables $r(t)$ and $\theta(t)$. Let us now write down the equations of motion. These are $$ \frac{d}{dt}\vec{r}_2(t) = \frac{v \left( \vec{r}_1 - \vec{r}_2\right) }{|\vec{r}_1 - \vec{r}_2|} $$ and similarly for $ \vec{r}_1(t) $ and $ \vec{r}_3(t) $, but given symmetry, these equations can be derived from the equation above. Let us now explicitly write out the equation above. $$ \vec{r}_1 - \vec{r}_2 = r(t) \left[ \left( \cos\theta_1(t) - \cos\theta_2(t) \right) {\hat i} + \left( \sin\theta_1(t) - \sin\theta_2(t) \right) {\hat j} \right] $$ $$ = \frac{\sqrt{3}}{2} r(t) \left[ \left( - \sqrt{3} \cos\theta - \sin\theta \right) {\hat i} + \left( \cos\theta - \sqrt{3} \sin\theta \right) {\hat j} \right] $$ Also $$ |\vec{r}_1 - \vec{r}_2 | = \sqrt{3} r(t) $$ The full equation is then $$ \frac{d}{dt} \left[ r(t) \left( \cos\theta {\hat i} + \sin\theta{\hat j} \right) \right]= \frac{v}{2} \left[ \left( - \sqrt{3} \cos\theta - \sin\theta \right) {\hat i} + \left( \cos\theta - \sqrt{3} \sin\theta \right) {\hat j} \right] $$ This splits into two equations $$ {\dot r} \cos\theta - r {\dot \theta} \sin\theta = - \frac{v}{2} \left( \sqrt{3} \cos\theta + \sin\theta \right) $$ $$ {\dot r} \sin\theta + r {\dot \theta} \cos\theta = \frac{v}{2} \left( \cos\theta - \sqrt{3} \sin\theta \right) $$ The boundary conditions are $r(0) = \frac{d}{\sqrt{3} },~\theta(0) = \frac{7\pi}{6} $. We now wish to decouple the two first-order differential equations into two decoupled second order differential equations. This is quite easy to do. Multiply the first equation by $\cos\theta$, the second by $\sin\theta$ and add the two. We get $$ {\dot r} = - \frac{\sqrt{3}}{2} v $$ We must now integrate this. At time $t=0$, $r(0) = \frac{d}{\sqrt{3}}$. The radial coordinate at arbitrary time $t$ is then $$ r(t) = - \frac{\sqrt{3}}{2} v t + \frac{d}{\sqrt{3} } $$ Though we are done here, for completeness, we can also solve for $\theta(t)$. The solution is $$ \theta(t) = \frac{7}{6} \pi - \frac{1}{\sqrt{3} } \log \left( 1 - \frac{3 v t }{2 d } \right) $$ The total time of travel is the time when $r(t_0) = 0 \implies t_0 = \frac{2d}{3v}$. Let us also make sure that we understand what the total distance is. Squaring and adding the two equations of motion we wrote above, we find $$ {\dot r}^2 + r^2 {\dot \theta}^2 = v^2 $$ The total distance travelled is then $$ s = \int_0^{t_0} \left[ {\dot r}^2 + r^2 {\dot \theta}^2 \right]^{1/2} = v t_0 = \frac{2d}{3} $$ Note that as $t \to \frac{2d}{3v}$, $\theta(t) \to \infty$, i.e. the particles spiral an infinite number of times before reaching the center of the equilateral triangle.

electromagnetism - Why are some materials diamagnetic, others paramagnetic, and others ferromagnetic?

1. 
   

   Why are some materials diamagnetic, others paramagnetic, and others ferromagnetic?

   
   
   


2. 
   

   Or, put another way, which of their atomic properties determines which of the three forms of magnetism (if at all) the compound will take?

   
   


3. 
   

   Is paramagnetism to ferromagnetism a continuous spectrum, or is there no grey zone in between?

   
   

Answer

There are a few decent rules of thumb for para- and diamagnetism.

A system is paramagnetic if it has a net magnetic moment because it has electrons of like (parallel) spins. These are often called triplet (or higher) states. In atoms and molecules, they occur when the highest occupied atomic/molecular orbital is not full (degeneracy > 2 * # of valence electrons). In this case, Hund's rules suggest that the electrons lower their energy by aligning their spins.

In contrast, a diamagnet has no magnetic moment because all electrons are paired.

Nearly all free atoms are paramagnetic because nearly all atoms have unpaired spins. The exceptions are the the last column of the s, p, d, and f block (2, 12, and 18). (Any that I'm missing?) For instance, that's an important property for the Stern-Gerlach experiments and magnetic trapping.

Most molecules, however, have fully paired spins. First off, most molecules have an even number of spins, except for free radicals, which are relatively unstable. To figure out if the molecule has a net magnetic moment (paramagnetic) or not (diamagnetic), you need to look at its molecular orbitals. The classical example is oxygen, which has a half-full (or half-empty) $\pi_{2p}^\ast$ orbitals, and nitrogen, which has a full $\pi_{2p}^\ast$ orbital. See: http://www.mpcfaculty.net/mark_bishop/molecular_orbital_theory.htm.

For crystals and solid-state materials, the question is more challenging, but it ends up coming down to the same question: is there a net magnetic moment because of unpaired spins, in which case it's a paramagnetic? or is there no net magnetic moment because all spins are paired, in which case it's a diamagnet?

Of course, in solid-state, there is a third situation, a ferromagnet. This is rather difficult to predict in real systems and is a major field of research. Some model systems (model system: a much simplefr mathematical model of a system) are solvable and give hints of what to look for. For instance, free spins in a lattice create a paramagnet by the argument above: the crystal has a net magnetic moment. You expect, in a magnetic field, the spin of one electron creates a magnetic field that can effect its neighbors. Since the system is paramagnetic, you might expect that the neighbors align with their local magnetic field, which is induced by their neighbors, and the whole crystal polarizes itself, creating a ferromagnet. This explanation is a mean-field Ising model. It gives a good intuition even though it's too simple to describe any real system.

cosmology - How can something finite become infinite?

How can the universe become infinite in spatial extent if it started as a singularity, wouldn't it take infinite time to expand into an infinite universe?

Answer

If the Universe is spatially infinite, it always had to be spatially infinite, even though the distances were shortened by an arbitrary factor right after the Big Bang.

In the case of a spatially infinite Universe, one has to be careful that the singularity doesn't necessarily mean a single point in space. It is a place - the whole Universe - where quantities such as the density of matter diverge.

In general relativity, people use the so-called Penrose (causal) diagrams of spacetime in which the light rays always propagate along diagonal lines tilted by 45 degrees. If you draw the Penrose diagram for an old-fashioned Big Bang cosmology, the Big Bang itself is a horizontal line - suggesting that the Big Bang was a "whole space worth of points" and not just a point. This is true whether or not the space is spatially infinite.

At the popular level - and slightly beyond - these issues are nicely explained in Brian Greene's new book, The Hidden Reality.

newtonian mechanics - Why do things accelerate?

Why do things accelerate? Let's say you are pushing an object in space for example, why would the object accelerate? I know that when a net force acts on an object, it accelerates but that is my question, why does this happen? Why wouldn't the velocity stay the same with a net force acting on an object?Because isn't it the same force? I'm trying to figure this out and it's making me crazy! EDIT: I changed the scenario because what i'm asking has nothing to do with gravity

Answer

Many sciences are a posteriori, and physics, the study of nature, attempts to form theories and models through induction. We observe stuff, we take data, and we try to come up with generalizations. The fact is, we aren't really sure if such generalizations exist or not. We simply belief so, because it makes life easier, and so far this belief has not been shown wrong.

Back to your question: why do things accelerate? This is a question that might never be answered. What physics is doing is trying to answer how things accelerate, or become hot or cold, or increase or decrease in mass. The reason why "truths" from the past are no longer true today, and "truths" today may very likely be no longer true some day, is that we are going bottom up, instead of top down: we can only see the manifestations of those generalizations, if they do exist, and we are doing nothing more than guessing what they are.

Imagine an alien seeing an Earth person pressing gas pedal, then the car moved. The alien never saw the interior of a car, and when she went back, she theorized that some mechanism makes the action of pressing gas pedal move the car. After 1 million observations of this correlation and none of them showing anything that contradicts her theory, she was so sure that pressing gas pedal will lead to the movement of the car.

But, she does not really know what kind of mechanism it is, that is hidden inside the car, that makes this theory valid. Although her theory is sound, as pressing gas pedal will 100% lead to the car moving (given the right conditions), she has no idea why, and she can never find out why if she is unable to disassemble a car and see for herself what is within.

We are aliens, and nature is the car. The sad thing is, we might never have a chance to see what's within the car. All we can do is observe, theorize, and use these theories to make our lives easier.

nuclear physics - Use of fission products for electricity generation

Why can't we use fissions products for electricity production ?

As far has I know fissions products from current nuclear power plants create enough 'waste' heat to boil water; and temperature decreases too slowly for an human life. So why can't we design a reactor to use this energy.

Answer

Here are some "order-of-magnitude" arguments:

Quoting https://en.wikipedia.org/wiki/Decay_heat#Spent_fuel :

After one year, typical spent nuclear fuel generates about 10 kW of decay heat per tonne, decreasing to about 1 kW/t after ten years

Now since this is heat, you can't convert it to electricity with 100% efficiency, the maximum possible efficiency is given by the Carnot efficiency $\eta$:

$$ \eta \le 1 - \dfrac{T_\mathrm{cold}}{T_\mathrm{hot}} $$

where $T_\mathrm{hot}$ would be the temperature of the spent fuel rods (in Kelvin) and $T_\mathrm{cold}$ would be the temperature of a cold reservoir against which a generator would work. One would have to do another calculation what a reasonable temperature of the fuel rods would be (in practice they currently seem to be kept at 50 degrees C).

With 'primary' fuel typically 55 Gigawatt days per tonne can be produced, i.e. a 1 Gigawatt powerplant would use 365.25 / 55 = 6.6 tonnes per year.

Even assuming you would be able to convert this to electricity with 100% efficiency and assuming an average 5 kilowatts per tonne over 10 years, this would yield about 18'000 Kilowatt days or 0.018 Gigawatt days, about 0.03% of the primary energy production.

You'll also see from the Carnot efficiency above that higher temperatures imply a higher possible efficiency, i.e. if one can spend some energy to extract the still fissionable material to be used in a reactor again, that is likely to be more efficient in terms of electricity generation.

It's true on the other hand that radioisotope thermoelectric generators (radioactive sources combined with thermocouples) have been used on satellite missions.

quantum mechanics - Pauli principle for particles very far apart from each other

Can two electrons be in the same state, when they belong to two different atoms, which are "far enough" (whatever that means) apart from each other? With "same state" I mean that (as far as specifiable) the states are really identical, except for the position of the electrons.

More specifically:
I am still not clear of how the separation of particles is taken into account for the Pauli exclusion principle. E.g. in a crystal the electrons seem close enough for the exclusion principle to become meaningful, for particles "a universe apart" from each other, it seems that it is kind of redundant (is there specific maths to that?), but what about everything in between?

(This is closely related to the talk Brian Cox gave, see this question: physics.stackexchange.com/q/18527/16689 )

newtonian mechanics - Will the momentum be conserved in this scenario?

I faced a problem:

A truck filled with sand has a mass of $6*10^3$ kg. It's moving at a constant speed of 20 $kms^-1$. If at one moment, sand starts to fall through a hole normal to the motion of the truck from the truck with rate 5 kg/s(due to a downward gravitational force); what would be the force needed to keep the truck at same constant initial speed? Neglect air drag, friction etc.

I am only concerned with if the momentum of the truck in the direction of it's motionn would conserve.

My attempt: No, it won't. Because conservation of momentum only works for closed systems. The sand pouring out and the object creating the gravitational field are in a closed system (not in terms of invaraint mass, but in terms of no external force). But, all other sand particle and the truck is out of this closed system. So, according to inertia of motion, all other particles will keep up the same velocity. So, necessary force $0N$

Feel free to point out my mistakes.

Answer

If you just mean the momentum of the truck, then yes, that's conserved. There's no net forces acting on it so whatever. This seems closest to the question being asked, which is a question about what force is needed on the truck.

If you mean the momentum of the truck plus the sand it happens to be carrying, then no, that's not conserved, because the amount of sand it's carrying happens to be decreasing.

If you mean the momentum of the truck plus the sand it originally carried, then the answer is "probably not, but this technically depends on things that you have not yet told us." So for example the truck could be moving alongside a conveyor belt that moves at exactly the same speed, and the sand could be passed out from a chute onto that conveyor belt. If we assume that these sorts of shenanigans haven't happened yet then yes, when the sand meets the road there will be a net force on it which will change its momentum.

If you mean the momentum of the truck plus the sand it originally carried plus Earth and everything on it, then we again come back to "yes, that's conserved."

The key take-away from all of this is that when you want to ask about "is the momentum conserved?" you have to be really, really clear about what collection of objects you are evaluating the momentum of.

The easy way to see the very first result is just to think about a grain of sand the very instant after it has fallen. It was moving forward with velocity vector $\vec v = [v_0, 0]$ when it was stored in the truck, and the instant after it falls, now it has picked up some small downward speed, $\vec v = [v_0, -\epsilon].$ Because it has not changed in its horizontal speed, no force was needed to speed it up or slow it down, so it cannot put a third-law reaction force on the car. After this instant, it is totally disconnected from the car and cannot put a third-law force on the car. So this force is 0.

But you can also imagine, I'm sure, a little nozzle where we let a tank of fluid spray out of it, and we could aim this little nozzle "backwards" or "forwards" from the truck, having some velocity $[v_0 \pm u, -\epsilon]$. If we aim it straight down we get the same result of course; but we could use it to accelerate the truck (pointing it backwards) or decelerate the truck (pointing it forwards). Sand will do the same thing if it falls out of a chute pointing one way or the other. So that phrasing "a hole normal to the motion of the truck" is very important, too.

pressure - What is the explanation of Pascal's law at molecular level?

What is the molecular explanation of Pascal's law?

Let us now consider what happens when we change the pressure on a fluid contained in a vessel. Consider a horizontal cylinder with a piston and three vertical tubes at different points. The pressure in the horizontal cylinder is indicated by the height of liquid column in the vertical tubes.It is necessarily the same in all. If we push the piston, the fluid level rises in all the tubes, again reaching the same level in each one of them.

I know this follows from Pascal's law but what is happening in the liquid when piston applies force is it being compressed or sth. How does piston change the magnitudes of pressure at all points inside the liquid. What happens if the fluid is confined and not confined when pressure is applied? I only need a qualitative intuitive explanation.

I can think that the transmission of fluid is by one molecule pushing the other but after it comes back to normal so how is pressure increased overall?

rotational kinematics - Why is the derivative of the quaternion equal to the Kronecker product of the quaternion itself and the angular velocity?

the dimension of $ \dot{\mathrm{Q}}(t)$ is 4x1 but the following product is 4x4 $$ \dot{\mathrm{Q}}(t)=\frac{1}{2}\mathrm{Q}(t)\otimes\overline{\Omega}(t)\ , (1) $$ equation (1) is part of the dynamic attitude system of rigid body

syms q0 q1 q2 q3 w1 w2 w3 real
Q=[q0 q1 q2 q3]' % quaternion
W=[w1 w2 w3]     % angular velocity
W_bar=[0 W]
Q_dot=0.5*kron(Q,W_bar)  % derivative of the quaternion

the result is :

Q =

 q0
 q1
 q2
 q3


W =


[ w1, w2, w3]


W_bar =

[ 0, w1, w2, w3]


Q_dot =


[ 0, (q0*w1)/2, (q0*w2)/2, (q0*w3)/2]
[ 0, (q1*w1)/2, (q1*w2)/2, (q1*w3)/2]
[ 0, (q2*w1)/2, (q2*w2)/2, (q2*w3)/2]
[ 0, (q3*w1)/2, (q3*w2)/2, (q3*w3)/2]

Answer

Your expression for quaternion multiplication is wrong. It is not the Kronecker (or outer) product, although some people use the same symbol $\otimes$ to represent it. Multiplying two 4-element quaternions together yields another 4-element quaternion. We need to take care with the sign convention, and the convention of body-fixed or space-fixed angular velocity, but I believe the formula you are looking for is (in matrix multiplication form) $$ \begin{pmatrix} \dot{Q}_0 \\ \dot{Q}_1 \\ \dot{Q}_2 \\ \dot{Q}_3 \end{pmatrix} = \frac{1}{2} \begin{pmatrix} Q_0 & -Q_1 & -Q_2 & -Q_3 \\ Q_1 & Q_0 & -Q_3 & Q_2 \\ Q_2 & Q_3 & Q_0 & -Q_1 \\ Q_3 & -Q_2 & Q_1 & Q_0 \end{pmatrix} \begin{pmatrix} 0 \\ \bar{\Omega}_1 \\ \bar{\Omega}_2 \\ \bar{\Omega}_3 \end{pmatrix} $$

Generally, if we write a quaternion as a combination of a scalar and a vector $\mathbf{A}= \bigl(a_0,\mathbf{a}\bigr)=\bigl(a_0,a_1,a_2,a_3\bigr)$, and similarly for $\mathbf{B}$ and $\mathbf{C}$, then the quaternion product may be expressed $$ \mathbf{C} = (c_0, \mathbf{c}) = \mathbf{A}\otimes\mathbf{B} = \bigl( a_0b_0-\mathbf{a}\cdot\mathbf{b},a_0\mathbf{b}+b_0\mathbf{a}+\mathbf{a}\times\mathbf{b} \bigr) $$ where $\cdot$ is the usual scalar product of two vectors, and $\times$ is the vector cross product. If we expand this into individual terms, we can see that it may be written in a similar form to my answer above $$ \begin{pmatrix} c_0 \\ c_1 \\ c_2 \\ c_3 \end{pmatrix} = \begin{pmatrix} a_0 & -a_1 & -a_2 & -a_3 \\ a_1 & a_0 & -a_3 & a_2 \\ a_2 & a_3 & a_0 & -a_1 \\ a_3 & -a_2 & a_1 & a_0 \end{pmatrix} \begin{pmatrix} b_0 \\ b_1 \\ b_2 \\ b_3 \end{pmatrix} $$

NB in the dynamics community these parameters can be called either "quaternions" or "Euler parameters". If you look on Wikipedia, you will see that the term "quaternion" refers to Hamilton's extension of complex numbers, and that usage is probably mathematically correct. There is also a close connection with (complex) Cayley-Klein parameters. However, in the current context, the parameters are all real, and we commonly refer to them as quaternions.

quantum field theory - The concept of particle in QFT

I never learnt QFT and I apologize for my (probably) elementary question. Somebody told me that in QFT a particle is viewed as an irregularity in the field.

On the other hand, in an article in Wikipedia I see the sentence "A QFT treats particles as excited states of an underlying physical field, so these are called field quanta."

Which one of the true is a better description? The 1st description hints that the particle is a localized phenomenon inside a field that maybe occupies a big region in space. The 2nd description speaks of an "underlying" field. So, is there a field and in addition there is a particle? If it is, then what is the occupation number of that "underlying" field?

None of these approaches is clear to me, I know the approach in QM, and none of them resembles the QM.

The motivation behind my question is a certain similarity that I find between the above descriptions and the Bohm interpretation of QM, (i.e. the background field - in Bohm's interpretation there is a background quantum potential - and a particle floating in it.)

In all, is a particle treated in QFT as a localized phenomenon inside a field occupying a wider volume? I would appreciate a simple and direct answer.

Answer

Somewhat surprisingly, the "generic" particle of QFT is in fact totally delocalized.

More precisely, particles are thought to come from the mode expansion of free fields. Since every free relativistic field $\phi$ fulfills the Klein-Gordon equation $(\partial^\mu\partial_\mu - m^2)\phi = 0$, a Fourier transform shows that it can be expanded as

$$ \phi(x) = \int \frac{\mathrm{d}^3p}{(2\pi)^3}\frac{1}{\sqrt{2p^0}}(a(\vec p)\mathrm{e}^{\mathrm{i}px} + a^\dagger(\vec p)\mathrm{e}^{-\mathrm{i}px})$$

where Lorentz invariance is not manifest, but can nevertheless be shown. A quantum field is operator-valued, and the operator valued objects $a(\vec p),a^\dagger(\vec p)$ fulfill exactly the correct commutation relations to be interpreted as creation and annihilation operators. The $n$-particle state of particles that are associated with the field $\phi$ is now defined as

$$ \lvert n;p_1,\dots,p_n \rangle := a^\dagger(p_1)\dots a^\dagger(p_n)\lvert \Omega \rangle$$

where $\lvert \Omega \rangle$ is the (mostly) unique vacuum state. In this way, you first create all particle states that are sharply localized in momentum space (and hence completely delocalized in position space) and you can build localized particle states by the usual building of "wavepackets" with fuzzy momentum out of the sharp momentum states:

A QM wavepacket of width $\sigma_x$ localized at $x_0$ is constructed out of the pure momentum states $\lvert \vec p \rangle$ as something like $$\lvert x_0,\sigma_x\rangle = \int \frac{\mathrm{d}^3 p}{(2\pi)^3}\mathrm{e}^{\mathrm{2i\sigma_x^2(x - x_0)^2}}\lvert p \rangle$$It works exactly the same for localized QFT particles, except that one should multiply the measure with $\frac{1}{\sqrt{2p^0}}$ to have a Lorentz invariant integration, and, of course, $\lvert p \rangle = a^\dagger(p)\lvert \Omega \rangle$.

The idea that "particles are local excitations of the fields" comes from the observation that this mode expansion is almost completely analogous to a classical field fulfilling a wave equation like the Klein-Gordon equation, where the $a(\vec p),a^\dagger(\vec p)$ would directly represent an excitation of the field of wavenumber $\vec p$. It cannot be made precise in the context of QFT because the quantum field is operator-valued and has no definite values, so it is wholly unclear what rigorous sense could be given to it being "excited". It is a nice picture, but nothing you should take too literally.

Also, take note that this is for the free field. The true interacting field of a QFT cannot be mode expanded in this way, and particle states are (through the LSZ formalism) only obtained in the asymptotic past and future (when they were far enough apart for interactions to be effectively non-existent) of the theory - the Hilbert space (and hence any states you could or could not identify as particles) of interacting QFTs is essentially unkown.

Furthermore, more mathematical methods of constructing QFTs often first construct the $a,a^\dagger$ and the Fock space of particle states, and then define the field out of it - then, the roles of particle and field as "fundamental" and "derived" are somewhat reversed.

waves - Since cables carry electricity moving at the speed of light, why aren't computer networks much faster?

Why can't cables used for computer networking transfer data really fast, say at the speed of light?

I ask this because electricity travels at the speed of light. Take Ethernet cables for example, I looked them up on wikipedia.

Propagation speed   0.64        c

Why only 64% What does propagation speed mean? I know there are other variables affecting the latency and perceived speed of computer network connections, but surely this is a bottle neck.

In other words, I'm asking, what is it about a fiber-optics cable that makes it faster than an Ethernet cable?

Answer

As you've probably guessed the speed of light isn't the limitation. Photons in a vacuum travel at the speed of light ($c_o$). Photons in anything else travel slower, like in your cable ($0.64c_o$). The amount the speed is reduced by depends on the material by the permittivity.

Information itself is slower still. One photon doesn't carry much information. Information is typically encoded in the change of states of the energy. And these changes of states can only be propagated at lower rates than the fundamental transmission speed.

Detecting both the energy and the rates of change require physical materials to convert the photons into something more usable. This is because the channel used for transmission usually conducts energy at a maximum rate called bandwidth. The bandwidth of the channel is the first limit in network speeds. Fiber optics can transmit signals with high bandwidths with less loss than copper wires.

Secondly the encoded signals have a lot of overhead. There is a lot of extra data transmitted with error correction, routing information, encryption and other protocol data in addition to the raw data. This overhead also slows down data throughput.

Lastly the amount of traffic on a network can slow down the overall system speed as data gets dropped, collisions occur and data has to be resent.

---

EDIT: I see you've changed your question some....

In other words I'm asking, what is it about a fiber-optics cable that makes it faster than an Ethernet cable?

Fiber optics has the ability to conduct higher energy charges. Photons with higher energies, by definition are at higher frequencies.

$E_{photon}=hf$ where $h$ is the plank constant (h=6.63*10^-34 J.s) and $f$ is the frequency of the photon.

Why does frequency matter? Because of how communication systems work. Typically we setup a strong signal oscillating at the most efficient frequency for the transmission channel to conduct it. If the frequency is too low and we lose our signal's power and likewise too high and we lose power. This is due to how the medium responds to different levels of charge energy. So there's a $F_{max}$ and a $F_{min}$.

Then we add information to the oscillation by changing it at some rate. There are a many ways to add information but in general the amount of information you can add is proportional to the rate the channel can respond to or bandwidth of the system. Basically you have to stay in between $F_{max}$ and $F_{min}$.

It just so happens that the higher the operating frequency the easier it is to get wider and wider bandwidths. For example a radio at 1GHz with 10% channel width only allows for 100MHz max switching rates. But a fiber optic signal at 500THz a 10% channel width means a 50THz max switching rate. Big difference!

You might be wondering why channels have frequency limits and why 10%. I just picked 10% as a typical example. But transmission channels of all types have limits to what kind of energy levels they absorb, reflect, and conduct. For a rough example x-rays which are high frequency or high energy charges, they go right though a lot of materials, whereas heat which is a frequency lower than optical light doesn't transmit well through paper but it can through glass. So there are frequencies where photons can be used to carry energy and frequencies where they can't.

Yes they do all travel at $c_o$ in free space and slower in other media, but they can't carry information at that same rate or higher. You might be interested to read Shannon-Hartley's Theorem.

electromagnetism - Poynting vector in a simple DC circuit

This image I took from Wikipedia shows the directions of E(red) and B(green) and poynting vectors(blue) at different locations on the circuit. I don't seem to understand why the E fields(red) are pointing from up to down in that manner in between the circuit at plane P. Is it because the wire on the upper part is connected to the positive terminal and hence must E field lines must emanate from every part of the wire?

density operator - Is it possible to go from the Master Equation formalism to Heisenberg-Langevin equations

If I have derived a master equation (e.g. in the Lindblad form) and solved for the density matrix, $\rho(t)$ I can get the mean value of an operator, A as:

$ = \mathrm{Tr}A\rho $.

But this is just the mean value. Suppose I want an equation for the operator in the Heisenberg picture

$ A_H (t) $

For this I can derive quantum-Langevin equations (aka. Heisenberg-Langevin equations see e.g. API p.340ff) which gives this in terms of the Hamiltonian for the full system including the reservoir OR in terms of quantum Langevin forces.

Is there a way to go from the density matrix in the Schrödinger picture to the operator in the Heisenberg picture (in the case of an open system like described above)?

API: Cohen-Tannoudji et al., Atom Photon Interactions, 1998 & 2004.

Answer

For a linear master equation as the one described by the Lindbladian there is no problem in passing from the Schrödinger to the Heisenberg picture, apart from an eventual loss of regularity wrt time.

This is because there is a duality chain between the "set of normal states" $V_*$ (to be precise the normal states are positive elements with norm one of this predual space), the von Neumann algebra of observables $V$ and the states $V^*$ (again the positive elements with norm one): $$V^*=(V)^*=(V_*)^{**}\;,$$ where the $*$ operation stands for the topological dual.

So, suppose you are given a (strongly-continuous) semigroup (Lindblad semigroup) $(e^{\,\cdot\,\mathscr{L}})_*: \mathbb{R}_+\times V_*\to V_*$, that to a time $t\geq 0$ and a normal state (density matrix) $\rho\in V_*$ associates another normal state $(e^{\,t\,\mathscr{L}})_*\,\rho$.

By the aforementioned map we can define the evolution $e^{\,\cdot\,\mathscr{L}}: \mathbb{R}_+\times V\to V$ on observables $V$ easily by duality: $$\forall A\in V,\rho\in V_*, t\geq 0,\quad \rho(e^{\,t\,\mathscr{L}}A)=(e^{\,t\,\mathscr{L}})_*\,\rho(A)\; .$$

The semigroup $e^{\,\cdot\,\mathscr{L}}$ is called the dual semigroup of $(e^{\,\cdot\,\mathscr{L}})_*$. The only care that has to be taken is that this dual semigroup is, in general, less regular than the original one: if the original one is strongly continuous, the dual is a priori only weakly continuous. There is however a subspace $V^{\dagger}$ of the observables where the restriction of the dual semigroup is again strongly continuous. Nevertheless you have defined your Heisenberg evolution, starting from the Schrödinger one (and you can do the same thing to obtain the evolution for general states as well).

Remark. Here I have considered the Lindblad generator $(\mathscr{L})_*$ as a linear operator on the states (density matrices); its explicit form is the usual $$(\mathscr{L})_*\rho= -i[H,\rho]+\frac{1}{2}\sum_j\Bigl([U_j\rho,U_j^{\dagger}]+[U_j,\rho U_j^{\dagger}]\Bigr)\; .$$ The generator for the observables semigroup $\mathscr{L}$ (acting in $V^\dagger$ where it can be defined) is simply $$\mathscr{L}X= i[H,X]+\sum_j\Bigl(U_j^\dagger X U_j -\tfrac{1}{2}\{U^\dagger_jU_j,X\}\Bigr)\; .$$