# How Slow is “sufficiently slow” ?

A  quasi-static process that is reversible would be one which takes an infinitely long time to complete and during which the system and the surroundings are infinitessimally close to equilibrium. Sadly both are ideal situations and are not really achievable in practice

However, if I can execute the process sufficiently slow, I should be able to get close to reversibility. So the question is:

How slow do we have to go to get the reversible quasi-static assumption to be valid?

It depends…… Depends on what???

Let’s see…

Consider a system(piston) that is to be led along the quasi-static adiabatic process.

The constraint are to be moved slowly step by step, the system being permitted at each step to come to a new equilibrium state while fulfilling the condition $PV^\gamma=constant$. After a slight outward motion of the piston we must wait until the system fully achieves equilibrium, then we proceed with the next movement and we wait again, and so forth. Although this is the theoretical prescribed procedure, the practical realization of the process seldom follows this prescription. In practice the constraints usually are, relaxed continuously, at some “sufficiently slow” rate.

The characteristic time tau it takes for a system to reach equilibrium called the relaxation time. Although thermodynamics is a powerful theory that has been successfully predicts the fate of so many physical systems with some environment, strictly speaking thermodynamics does not deal with what is happening during relaxation. It describes the initial and final states only.

So, if the adiabatic expansion of the gas in the cylinder is performed in times much longer than this relaxation time the expansion occurs reversibly and isentropically. If the expansion is performed in times comparable to or shorter than the relaxation time there is an irreversible increase in entropy within the system and the expansion. The external pressure is decreased too rapidly that the resulting rapid motion of the piston is accompanied by turbulence and inhomogeneous flow within the cylinder. The process is then neither quasi-static nor reversible.

To estimate the relaxation time we first recognize that a slight outward motion of the piston reduces the density of the gas immediately adjacent to the piston. If the expansion is to be reversible this local “rarefaction” in the gas must be homogenized before the piston again moves appreciably. The rarefaction itself propagates through the gas with the velocity of sound, reflects from the walls of the cylinder and gradually dissipates. The mechanism of dissipation involves both diffusive reflection from the walls and viscous damping within the gas which is the damping due to imaginary part of the refractive index of sound wave in air.

Let’s assume that the second one is more dominant

Divide the air inside the piston into thin slices of thickness $dx$ that are perpendicular to the direction of sound propagation.The intensity of sound wave emerges from the layer immediately adjacent to the piston will decay as it travel through distance because some of the them have run into particles and did not make it to the next layer or end up having lower average velocity. The change in intensity due to a slice of air is

$dI=-kIdx$

To determine $k$, we need to know about the cross section of the particles . We assume that $dx$ is sufficiently small that one particle in the slab cannot obscure another particle in the slab when viewed along the $x$ direction. It follows that the fraction of particles collided when passing through this slab is equal to the total opaque area of the particles in the slab, $\sigma nAdx$, divided by the area of the slab $A$, which yields on $\sigma ndx$. Thus the fraction of particles collided when passing through the slab is given by:

$\frac{dI}{I}=-\sigma ndx$

Integrating both sides we get

$I=I_{0}e^{-\sigma nx}=I_{0}e^{-\sigma nv_{s}t}$

Where $v_{s}$  is the velocity of sound. Which is easily derived and given by

$v_{s}=\sqrt{\frac{\gamma kT}{\mu}}$

Thus the characteristic relaxation time or the time it takes for the the rarefaction pulse to be effectively dissipated is given by

$\tau=\frac{1}{\sigma nv_{s}}=\frac{1}{\sigma n}\sqrt{\frac{\mu}{\gamma kT}}$

So, if the adiabatic expansion of the gas in the cylinder is performed in times much longer than this relaxation time, the  quasi-static assumption is a good one.

Assume the following numerical values:

$v_s=340 m/s$

$\sigma=3 \times 10^{-19} m^2$ (hydrogen atom)

$n=10^{23} m^{-3}$

* the numerical value of the cross section of hydrogen is “stolen” from this paper http://arxiv.org/pdf/plasm-ph/9506003.pdf

Then we get

$\tau=9.8\times 10^{-8} s$

Which is surprisingly  small… Therefore we are safe to assume the adiabatic condition in real life

# Bernoulli Part I (gas)

Without doing any initial bullshitting, let’s go directly to the problem…

One of the most ubiquitously found explanation for the lift force on an airplane is quick, simple, and gives the correct answer. But yet, I think it introduces misconception and misleadingly invokes Bernoulli’s equation. Let’s take a look at the argument…

Using the argument, one should also expect a lift for a symmetric wing profile as shown above. The argument claims that the distance from this point to the trailing edge is greater along the upper surface than along the lower surface. Using the ‘equal time’ argument it is assumed that two neighbouring ﬂuid particles which ‘split’ at stagnation point should somehow meet again at the end point then this requires that the average flow velocity on the upper surface is greater than that on the lower surface.  Now Bernoulli’s equation is quoted, which states that larger velocities imply lower pressures and thus a net upwards pressure force is generated. The above explanation is extremely widespread, it can be found in many textbooks. The problem is why should two particles on either side of the wing take the same time to travel from A to B? I never saw any argument for that.

However, if we considers the problem from a microscopic point of view, we comes to a different conclusion: a symmetric wing profile won’t produce net lift.

Now let’s consider the case using ideal gas point of view. If the wing is stationary, the pressure on all parts of the wing is identical, i.e. there is no lift. If the wing is moving in the indicated direction, the front half of the upper wing surface experiences an increased pressure because of the increased speed  air molecules hitting it (this creates a downward force). On the other hand, the rear half experiences a reduced pressure because the of the reduced speed(creating a lift). Overall, there is consequently no lift. It is obvious that an overall lift is only achieved if the rear section of the wing has a larger area than the front section.

Bernoulli’s principle seemed to had successfully predicted or explained many strange behavior. But indeed in some of those cases Bernoulli’s principle does not give a correct explanation (e.g. blowing air between two pieces of paper, the paper appears to get closer to each other). it’s just misleadingly plausible, in many cases the “fast moving air area” has lower pressure than the “slow moving air area” has although the causes is not necessary due to the Bernoulli’s principle. In fact, for most cases the main effect of pressure decrease is due the to Coanda effect, which mainly reduce the air density in some area. The effect governed by the Bernoulli’s principle is often too small to be significant, It is beautifully shown in this demonstration:

http://www.physics.umd.edu/lecdem/services/demos/demosf5/f5-12.htm

# Small Bubble, Big Trouble

Simple yet interesting:

Consider a perfectly rigid box in a gravitational field. The box is completely filled with an incompressible ideal fluid(zero viscosity) except there is a small air bubble on the bottom. Assuming that the bubble is always spherical, what will happen next??

We know that the gravitational field will produce hydrostatic pressure gradient on the water so that the pressure decreases as the altitude increase. Because of this, the bubble will feel non-uniform pressure all over the surface, and as a result of it the bubble will be pushed upwards. As the bubble goes up, it will encounter a lower pressure region, thus it will expand….

Wait!! Expand??? Didn’t I just mentioned that the box is perfectly rigid and also the fluid is incompressible?? What’s possibly wrong here?

So… the bubble’s volume is conserved.. what should we do to account for this? hhmm….

(1). While maintaining the pressure gradient, the pressure of the fluid and the pressure on the walls increase as a whole so that the decrease in pressure with height is balanced with the increase of pressure overtime. Thus the bubble will always feels the same pressure.

(2). As the bubble rises, the temperature decreases precisely as to balance with the local pressure of the fluid.

Wait wait wait… the second one is suspicious! What if initially the bubble is colder than the fluid, the colder body will cools down while the hotter will gets even hotter. Thus it will violates the second law thermodynamics right?? NO! its not a problem it doesn’t violates the 2nd thermodynamics, since there is an external work done on the system. The graviational potential energy is decreased as the bubbles rises, (some fluid goes down).

And the first one… But how does the pressure of the fluid changes so easily? Does pressure suppose to be due to thermal collisions? Let’s imagine matter as consisting of jiggling molecules, each neighboring molecules are connected with a spring. There are two types of elastic energy stored in the material:

i. The one that is related to the average distance between neighbouring molecules, this is determined by the local density.
ii. Related with the “standard deviation” of the molecules movement, this one is determined by the local temperature.

If the the interaction between molecules is super strong, as in our case, the “standard deviation” of the molecules movement is unimportant. The “spring” only needs to change it’s length a little bit to account for the kinetic energy. Thus the temperature is unimportant, and we also know that if we change the temperature of water, it’s pressure won’t change significantly. In our case the pressure of fluid will acts like a normal force, it will respond to any pressure so that the volume of the fluid doesn’t change.

Eventually I think both (1) and (2) will happen at once, and the rate of heat transfer(compared to how fast the bubble moves) and the heat capacity of the bubble(number of air molecules) will determine which one is more significant.

# Why did he spin the egg before throwing it?

I’m bored cooking and eating an ordinary omelette so I try to search some videos about how to cook an omelette in youtube. Somehow I stumble on this video

Watch it! The rest of this post is pointless if you don’t watch it

I assume you already watch the video, so now you understand the title of this post. Okay let’s discuss about it..

First, we know that his goal was to cut the egg. In more detail, he wants the egg to be in horizontal orientation before hitting the rectangular knife(or rectangular what?). The orientation of the egg is described by the vector pointing from the “blunter” end to the “sharper” end. And let’s call everything inside the egg with the word “fluid”.

When the egg is spinning the centrifugal force pushes the fluid to the ends of the egg, blunt and sharp.  So Perhaps he spin the egg because he didn’t want to cut the yolk at the center of the egg? NO! I hope the purpose must be something much cooler than that, so let’s assume that this is not his goal.

Another possibility, he made the egg spinning about the vertical axis so that it becomes more stable against perturbation and more likely to remains in horizontal orientation. It’s much easier to throw and catch a stable egg. And he also needs to spin the egg for a long time because the fluid inside and the shell are not quite coupled, the friction between them is very small.

Yet another possible advantage… if we spin the egg around the vertical axis, it makes the egg harder to move in a straight line. It’s common sense! you can imagine it! Because when the egg is moving translationally it is also rolling, when the orientation change due to the rotation the egg still preserve it’s rolling angular velocity. Consequently the egg tends to move in a circle, the radius of the circle depends on the velocity of egg’s center of mass, in other words it makes the egg harder to move in a straight line. So what if the egg is harder to move in a straight line? Of course it’s very helpful! It’s very hard to throw an egg properly if it can rolls easily from the rectangular tool(or what?), it’s much easier to control when the egg is spinning.

But the pan and the rectangular tool is very oily, it hardly rolls at all, it might simply slide without rolling.. You can see that when the awesome guy was spinning the first egg, it’s clearly shown that the egg moved through a curve path. That means that the friction is significant! I also tried putting an egg on an oiled frying pan and then I tilted the pan. It turns out that although the pan is very slippery the egg was still rolling through it’s way down. But actually my egg is pretty rough, maybe his egg is not as rough as mine…

Maybe the last theory is right, but the guy’s hands are also very talented, I think he can balance the egg just by tilting the rectangular tool rapidly. I hope one day I can meet an omurice cook and ask him about the reason…

# +m.B or -m.B?

In The Feynman Lectures on Physics vol 2, RPF showed that the “true energy” of a magnetic dipole(current loop) in a constant external B field is actually $+\vec{m}.\vec{B}$ , not  $-\vec{m}.\vec{B}$ . Basically the reason as Feynman said is that we need to take into account the extra energy to keep the current in the loop constant. If we included the battery’s work to keep the current in the loop constant when performing the virtual work pulling a current loop from infinity to the final position, the total energy will be $+\vec{m}.\vec{B}$ .
But if we use  $+\vec{m}.\vec{B}$  instead of  $-\vec{m}.\vec{B}$ , we will get contraditory results in some cases. For example, we know that a magnetic  dipole tends to point along the field lines, but $+\vec{m}.\vec{B}$ predicted that it tends to point opposite to the field lines.

How should I explain this?

Let’s try to derive it by calculating the total magnetic field energy.

$U=\int \frac{1}{2\mu_0}B^2 dV$

$U=\int \frac{1}{2\mu_0}(\vec{B_{ext}}+\vec{B_{dipole}})^2 dV$

where

Consider a “real” dipole, a magnetized sphere of radius R, instead of an ideal one. A real dipole has a pure the magnetic field outside the sphere:

$B_{dipole}=\frac{\mu_0}{4\pi r^3}[3(\vec{m}.\hat{r})\hat{r}-\vec{m}]$  for r>0

and

inside the sphere, we assume the field is due to A spinning spherical shell of uniform charge with a magnetic dipole m. The magnetic field is:

$B_{dipole}=\frac{\mu_o\vec{m}}{2\pi R^3}$ for r->0

expanding the things inside the bracket

$U=\int \frac{1}{2\mu_0}(B_{ext}^2+B_{dipole}^2+\vec{2B_{ext}}.\vec{B_{dipole}}) dV$

The first and second term are independent of orientation, we are not interested in them, we can ignore them

$U=\int \frac{1}{2\mu_0}(2\vec{B_{ext}}.\vec{B_{dipole}}) dV$

$U=\int_{0}^{R} \frac{1}{2\mu_0}(\vec{2B_{ext}}.(\frac{\mu_o\vec{m}}{2\pi R^3})) dV+\int_{R}^{\infty} \frac{1}{2\mu_0}(2\vec{B_{ext}}.(\frac{\mu_0}{4\pi r^3}[3(\vec{m}.\hat{r})\hat{r}-\vec{m}])) dV$

The second integrand is an odd function, it involves

$\int_{0}^{\pi} sin\theta(3cos^2\theta-1)d\theta=0$

thus the only surviving term is from the first term

$U=\frac{2}{3}\vec{m}.\vec{B}$

Obviously I am just calculating something wrong,

But remember carefully…

that in the derivation of the energy of a magnetic field we get

$U=\frac{1}{2\mu_0}[\int B^2 dV-\int (\vec{A}\times\vec{B}).\vec{dA}]$

We can always extend this out to infinity, and if our currents are finite the last term vanishes. However if we have a constant magnetic field this is not the case, so we have to include the “surface” term

now we are left with

$U_{surface}=-\int (\vec{A_{ext}}\times\vec{B_{dipole}}).\vec{dA}$

$U_{surface}=-\int [\frac{\mu_0}{4\pi r^3}(3(\vec{m}.\vec{r})\hat{r}-\vec{m})\times(\frac{1}{2}\vec{B_{ext}}\times\vec{r})].\vec{dA}$

$U_{surface}=\int\frac{\mu_0}{4\pi r^3}(-\vec{m}\times ((\frac{1}{2}\vec{B_{ext}}\times\vec{r})).\vec{dA})$

$U_{surface}=\int\frac{\mu_0}{8\pi r^3}(\hat{r}(\vec{m}.\vec{B_{ext}})-\vec{B_{ext}}(\vec{m}.\hat{r})).\vec{dA})$

The calculations are pretty messy, there is nothing interesting to talk about, so I’ll just put the final result

$U_{surface}=\frac{1}{3}\vec{m}.\vec{B_{ext}}$

thus the total field energy is

$U=+\vec{m}.\vec{B_{ext}}$

as what Feynman said….. I think this is so cool.

on the other hand for electric dipole we will get

$U=-\frac{1}{3}\vec{p}.\vec{E_{ext}}$ for the energy density term

and

$U=-\frac{2}{3}\vec{p}.\vec{E_{ext}}$ for the surface term

so the total energy is $U=-\vec{p}.\vec{E_{ext}}$

I never thought that electricity and magnetism are very different. The differences between electric dipoles and magnetic dipoles are:

1. The “inner field” points from positive charge to negative charge in electric dipoles, while it points in the opposite direction for magnetic dipole.

2. Scalar potential $V$ behave in a different way with vector potential $\vec{A}$

But why does magnetic dipoles(elementary particles) doesn’t use the total field energy as the interaction energy?

Because only $-\vec{m}.\vec{B}$ can be converted into mechanical energy…

But the remaining $+2\vec{m}.\vec{B}$ also changes with the orientation! where does this energy goes?

It seems that the energy goes into some kind of mysterious energy that keeps elementary particles’ spin constant. If we leave it that way, the energy will not be conserved. To “protect” the interpretation of electromagnetic field energy and the tendency of potential energy to be minimal, we need to add a “mysterious energy” to the energy term.  But I think is not necessary, actually the field energy is derived by calculating the mechanical work to put the system together. In this case we don’t need to put an extra work to keep the magnitude of dipole moment constant but the field energy still takes that into account. So maybe the energy stored in the field is not always true in general. It does not work in this case. If we stick to the definition of potential energy $F=-grad(U)$, we wouldn’t have this trouble, we don’t need to care about the existance of the “mysterious energy”.

# Flying Relativistic Man

Several months ago, there was a week in which I spent all my time thinking about how to find the charge distribution of conducting disc(I end up with the distribution of cylindrical symmetric ellipsoid, that can’t be specialize into a disc, with sphere and needle as the two extreme case). Then I discuss with my friend hoping to find out another way to solve it, but we end up discussing about relativistic electrodynamics instead, which is triggered by the fact that if a sphere moving with relativistic velocity it will turns into an ellipsoid(without flight time effect) . During this discussion, somehow we found a  simple and very interesting paradoxical case of relativistic electrodynamics.

As shown below there are two point particles with charge q each, the distance between them for this instance is L . Then a relativistic human, for some reason, using his hands, starts moving particle b away from a with constant relativistic velocity $v$, while holding the particle a at rest .

At this instant, the force on b due to a is obviously

$F_{ba}=\frac{kq^2}{L^2}$

Now comes the problem, the force on a due to b is different

$F_{ab}=\frac{kq^2}{\gamma^2L^2}$

Which means that the relativisticman receives different force on his left and right hand. Which means…

The relativisticman can fly away like a superman just by moving his hands!!!

How can it be possible assuming that the relativisticman exists??

The explanation is as follows

The electric field due to a alone at a random point p is

$\vec{E_{a}}=\frac{kq\vec{r_{a}}}{|r_{a}|^3}$

The electric field due to b alone at a random point p is

$\vec{E_{b}}=\frac{kq\vec{r_{b}}}{\gamma^2|r_{b}|^3(1-(\frac{vsin\beta}{c})^2)^\frac{3}{2}}$

b will also produce magnetic field

$\vec{B_{b}}=\frac{\vec{v}\times\vec{E_{b}}}{c^2}$

$\vec{B_{b}}=\frac{kq(\vec{v}\times\vec{r_{b}})}{\gamma^2c^2|r_{b}|^3(1-(\frac{vsin\beta}{c})^2)^\frac{3}{2}}$

using geometry relation $r_{a}sin\alpha=r_{b}sin\beta$, $\vec{v}\times\vec{L}=0$ and $\vec{r_{a}}=\vec{r_{b}}+\vec{L}$, we can eliminate the dependence of $r_{a}$ in $B_{b}$

$\vec{B_{b}}=\frac{kq(\vec{v}\times\vec{r_{a}})}{\gamma^2c^2(r_{a}^2+L^2-2r_{a}Lcos\alpha-(\frac{r_{a}vsin\alpha}{c})^2)^\frac{3}{2}}$

The total poynting vector of 2 charges is

$\vec{S}=\frac{1}{\mu_{0}}[(\vec{E_{a}}+\vec{E_{b}})\times\vec{B_{b}}]$

the second term $\frac{\vec{E_{b}}\times\vec{B_{b}}}{\mu_{0}}$ is constant over time, so we are not interested in it. By neglecting the this term, the momentum hidden in the electromagnetic field becomes

$\vec{p}=\mu_{0}\epsilon_{0}\int \vec{S}dV$

$\vec{p}=\epsilon_{0}\int (\vec{E_{a}}\times\vec{B_{b}})dV$

substituting $\vec{E_{a}}$,$\vec{B_{b}}$, and $dV=r_{a}^2 sin\alpha d\phi dr_{a} d\alpha$ we end up with

$\vec{p}=\frac{\epsilon_{0}k^2q^2v}{\gamma^2c^2}\int \int \int \frac{r_{a}sin\alpha[\vec{r_{a}}\times(\vec{v}\times\vec{r_{a}})]}{(r_{a}^2+L^2-2r_{a}Lcos\alpha-(\frac{r_{a}vsin\alpha}{c})^2)^\frac{3}{2}}dr_{a} d\theta d\phi$

by symmetry we know that the momentum will be in the same direction with $\vec{L}$, so we can put $\vec{r_{a}}\times(\vec{v}\times\vec{r_{a}})=sin^2\alpha \hat{L}$ , substituting this and $\int d\phi=2\pi$ the integral becomes

$\vec{p}=\frac{2\pi\epsilon_{0}k^2q^2v}{\gamma^2c^2}\int_{0}^{\infty} \int_{0}^{\pi} \frac{r_{a}sin^3\alpha\hat{L}}{(r_{a}^2+L^2-2r_{a}Lcos\alpha-(\frac{r_{a}vsin\alpha}{c})^2)^\frac{3}{2}}dr_{a} d\alpha$

The integral of $r_{a}$ has the form of

$\int_{0}^{\infty} \frac{x}{(ax^2-bx+c)^\frac{3}{2}}dx$

Which can be solved without much effort by using wolfram alpha

http://www.wolframalpha.com/input/?i=%5Cint+x%2F%28ax^2-bx%2Bc%29^%283%2F2%29

After LOTS of mess, the integral becomes

$\vec{p}=\frac{2\pi\epsilon_{0}k^2q^2v}{Lc^2}(\int_{0}^{\pi} sin\alpha d\alpha+\int_{0}^{\pi}(\frac{sin\alpha cos\alpha}{\sqrt{1-(\frac{vsin\alpha}{c})^2}})d\alpha)\hat{L}$

We can see that the second integral is odd, then we get a surprisingly simple result

$\vec{p}=\frac{4\pi\epsilon_{0}k^2q^2v}{Lc^2}\hat{L}=\frac{kq^2v}{Lc^2}\hat{L}$

the rate of change of the momentum is

$\frac{d\vec{p}}{dt}=\frac{-kq^2v^2}{L^2c^2}\hat{L}$

which is equal to $F_{ab}-F_{ba}$. So it means that the force difference exerted by the Relativisticman is only used to change the momentum of electromagnetic field. Since the mechanical momentum is not changing at all, the Relativisticman must receives the same change in momentum but in the opposite direction. Thus the man can fly away like superman!

# Light in Matter

Recently I start to (re-)wondering about a question “What really cause light to appear slower in media?”

We know that if we solve the maxwell equation, we will end up with the phase velocity of light is related to the permeablity and the permitivity of the material. But this is not what I’m interested in, I want to go deeper than that. We know that the real speed of photons is actually not changing, the decrease in speed is just apparent. Material is mostly empty, the light will still travel with c in the spacing between atoms. The rare atoms will disturb the light in some way. So the problem is how exactly do the atoms affect the light.

Some textbooks that I read explain it in a way kind of like this:

In a material the photons is absorbed by an atom and then re-emitted a short time later, it then travels at $c$ a short distance to the next atom and get absorbed&emitted again and so on. How quickly the atoms in a material can absorb and re-emit the photon and how dense the atoms decides the speed of light in that material. So the light appears slower because it has a smaller “drift speed”.

It seems very convincing right? It makes the probability of the thought of questioning it comes to my mind pretty small. But an hour ago I luckily found out that this thingy is not as simple as that, it turns out to be pretty cool.

I realize an alternative “wave” explanation:

Atoms respond to the light by radiating electromagnetic wave. This “new light” interferes with the “old light” in some way that result in delayed light (advanced in phase),this can easily shown by using simple phasor diagram. Consequently effectively the light covers a smaller phase each second. Which gives the impression of a lower phase velocity $c/n$. However the group velocity is changing in a weird and difficult to imagine way.

I think that the first explanation does not explain the change in phase velocity of light. if we consider light travelling into a slab of non-dispersive negative refractive index material, let’s say the light is directed perpendicular to the slab. The phase velocity’s direction will be flipped, but group velocity’s direction in the material will not change. Only the second explanation can explain the flipped phase velocity direction. I guess that the velocity that we get in the first explanation is actually belongs to the group velocity. It makes sense to me that the front most of the photon stream determines the first information that the light delivers. Negative index of refraction means that the light get delayed in phase more that how much it gains when it travels between atoms.

And also, I still don’t really understand about the detail of the absorption-emission process for small light’s wavelength compare to the distance between neighboring atoms( for large lambda compare to the atoms spacing, the photons will be absorbed by atoms as a group( phonons)). The dispersion relation that we know is continous and also osme material is non-dispersive, therefore the absorption process must accurs in all frequency for a certain range. So definitely it doesn’t involve the atomic transition, otherwise it will be quantized. My guess is that or the relevant absorption process is the stimulated absorpsion/emission which makes a lot of sense to me, it agrees with my classical explanation.

But I’m not sure yet, I will correct everything as soon as I get some more convincing arguments.