thermodynamics - Can Kinetic Theory obtain fluids behavior?

My understanding is that it can't. But when presenting it here I received comments that contradict this. So, as suggested, I am asking the question to be able to receive concrete approaches on this point from other users.

While I have put thought into it, and researched to some extent, I recognize I, by no means, am expert on this. So I will present my line of reasoning and arguments, and the question here is the one on the title.

My view of Kinetic Theory of Matter is that it visualizes macroscopic behavior of systems with sufficiently large number $N$ of constituents, as a result of the microscopic dynamics. That is, knowing the interactions between the $N$ particles to some extent, you can obtain macroscopic characteristics of the big system.

This theory is able to explain completely the macroscopic behavior of the ideal gas, which is to say real gases in a large portion of their phase space. It does an excellent job for the case of thermodynamic equilibrium, but is also used in describing macroscopic diffusion of some magnitude in gases, and is the choice when describing diffusion at molecular level.

But when we talk about liquids, there are some problems and I am only address here a system in thermodynamic equilibrium.

We can imagine a liquid made of the same molecules of the corresponding gas, interacting with the same potentials, but they just are closer and hence more restrained in movement because now they spend more time close to each other than in the gas.

Here is the first problem with that picture: if it was correct, we would never observe a gaseous and a liquid phase as different as they are. The gas would slowly turn into a liquid as energy was removed from it: the compressibility, the density, etc. would change continuously. But we know this is not true, because we have discontinuities in these magnitudes in phase transitions.

Another problem with this picture is the large incompressibility of liquids, while at the same time being so malleable. To give an idea, water at room temperature has a compressibility of about $450 MPa^{-1}$ (link)! This means that you need a pressure more than 4 times the atmospheric in order to compress it by $1\%$! And a simple calculation with the Lennard-Jones potential (used often in liquid water simulations) tells you that a decreasing the intermolecular distance in $1\%$ from the equillibrium distance ($r_m$) leads to an increase in the potential are less than $0.4\%$.

So this means that particles in the liquid overreact in the macro scale with respect to what we would expect if the average distance would scale linearly with macroscopic scale compression. But there seems to be non-linearity in the scaling to micro world, and this is not accounted for in KTM.

Furthermore, a liquid keeps his volume as opposed to how gasses fill their containers. This picture is more consistent with a microscopic system where the $N$ particles move freely, like in the ideal gas, but contained by some potential well of the size of the macroscopic system. However the KTM would describe it as particles interacting with those at some range, and is not been proven how these kinds of interactions could generate this type of big potential well, although admittedly it seems to be the case.

Also when a gas and a liquid in thermodynamic equilibrium with a clear boundary between them, how can this be explained? If molecules in the liquid are just closer to each other, since both are at the same temperature, the average kinetic energies should be the same in both phases, so what makes some of them remain at interacting distances while those in the gas, with no larger kinetic energies, move freely? Why do we see phenomena like vapor saturation, where inside the same container, the liquid and the gas are completely equilibrated, while the microscopic picture there is nothing preventing the liquid molecules from expanding, and the whole system becoming one with an average interparticle larger than what was in the liquid, and shorter than what was in the gas?

So these are my main concerns regarding KTM as applied to liquids, and I think they are not resolved. I saw other questions in this site like these (q1, q2, q3) and they seem to have not received attention from the KTM connoisseurs, or there is no valid answer to provide.

I will much appreciate to be proven wrong, but if not then it just means there is room for theory :). Also arguments on my case which I missed are also very welcome.

Big Bang Nucleosynthesis

In the very early universe, the hot plasma consisted of fixed amount of radiation (photons and neutrinos) and matter (electrons, protons, neutrons, etc). There were many competing reaction taking place and using statistical methods I understand that you can deduce the particle content of the universe when radiation and particles began to condense into nucleons (at roughly $100\times 10^9 K$).

According to ΛCDM, there is 5 times more dark matter than normal matter. These particles, ostensibly, had to form from the same budget of radiation yet I don’t see the reactions in any of the literature. How is it possible to accurately calculate the particle content of the early universe, specifically the proton to neutron ratio, when the reaction governing 80% of the matter creation isn’t known?

Answer

Standard big bang nucleosynthesis only involves particles which are part of the standard model - ie. excludes dark matter.

Is this justified? Well, dark matter is non-interacting (or so weakly interacting that we can't detect it), therefore it does not interact strongly with baryons and photons during the epoch of nucleon formation.

However there certainly is theoretical work that investigates non-standard big bang models that do include additional degrees of freedom due to new neutrino species or that include residual annihilation of weak scale massive dark matter particles. Apparently, these pieces of new physics have a significant effect on the nucleosynthesis of the light elements - deuterium, tritium, lithium. A reasonable review (I have not read it) appears to be given by Jedamzik & Pospelov (2009).

Some of these ideas have been used to suggest a solution to the problem that standard big bang nucleosynthesis appears to give too much Li (e.g. Bailly 2011).

Diffraction by small holes

This is a follow-up to this question: What happens to waves when they hit smaller apertures than their wavelenghts?

Hans Bethe wrote a paper in 1944, "Theory of Diffraction by Small Holes," Phys. Rev. 66, 163. I don't have access to the paper, but from descriptions online it sounds like he proved the following. Suppose a plane wave impinges on an absorbing sheet, and there is a hole in the sheet of diameter $d$, with $d<\lambda$. Let $P_0$ be the power incident on the hole, and $P$ the power diffracted through the hole. Then the transmission is $T=P/P_0=(d/\lambda)^4$.

As a practical application, I think this explains why microwaves don't leak strongly through the metal grille in the front of a microwave oven.

Questions:

1. 
   

   Does the thickness of the sheet matter? From references online, it sounds like the hole is treated as a waveguide...? This seems to relate to cutoff frequencies of waveguides, etc.?

   
   


2. 
   
   

   It seems to me that Huygens' principle would give $T=1$ for $d\ll\lambda$, since in this limit the wavelets are in phase. Why is Huygens' principle invalid here? Is this related to question #1?

   
   


3. 
   

   Is there a simple argument for the proportionality to $(d/\lambda)^4$? Or if not, how does one prove this using the gory details of Bessel functions, etc.?

   
   

electromagnetism - How to derive the expression for the electric field in terms of the potential?

How can I derive that $$\vec{E}=-\vec{\nabla}\phi-\frac{\partial \vec{A}}{\partial t}$$ where $\phi$ is the scalar potential and $\vec{A}$ the vector potential?

Answer

$\def\vA{{\vec{A}}}$ $\def\vB{{\vec{B}}}$ $\def\vD{{\vec{D}}}$ $\def\vE{{\vec{E}}}$ $\def\vH{{\vec{H}}}$ $\def\vS{{\vec{S}}}$ $\def\eps{\varepsilon}$ $\def\rot{\operatorname{rot}}$ $\def\div{\operatorname{div}}$ $\def\grad{\operatorname{grad}}$

Faraday's law: $$\rot(\vE)+\dot\vB=0$$ Source-less B-field: $$\div(\vB)=0$$ From this in a simply connected domain there follows the existence of a vector potential $\vA$ with $$\vB = \rot\vA$$ Therewith, Faraday's law reads $$\rot(\vE+\dot{\vA})=0$$ The curl-freeness of the vector field $\vE+\dot{\vA}$ (in a simply connected domain) implies the existence of a scalar potential with $$\vE+\dot{\vA} = -\grad\varphi$$ and that is your formula. You see $\varphi$ is just defined in the way you wrote it down.

The purpose of introducing $\vA$ is to solve $\div\vB=0$ and the purpose of introducing $\varphi$ is to solve Faraday's law. In most cases the only equation which remains to be solved is Ampre's law $$ \rot\vH = \vS + \dot\vD $$ which reads with our new independent variables $\varphi$ and $\vA$ as $$ \rot(\mu\rot \vA) = (\kappa + \eps\partial_t)(-\grad\varphi - \dot\vA). $$ Maybe, one has also a pre-defined space-charge density $\rho$. That would imply the equation $$ \rho = \div \vD = \div (\eps(-\grad\varphi - \dot\vA)) $$ For constant $\eps$ you have $$ -\frac{\rho}{\eps} = \Delta\varphi + \div \dot\vA $$ Now, you have some degrees of freedom in the choice of $\vA$. If $\vA$ is a vector potential for $\vB$ then for any smooth scalar function $\varphi'$ also $\vA':=\vA+\grad\varphi'$ is a vector potential for $\vB$ since $\rot\grad=0$. One possible choice is $\div\vA = 0$. Therewith, the equation for the space charge reads just $$ -\frac{\rho}{\eps} = \Delta\varphi $$ which we know from electro-statics. The nice thing of $\div\vA=0$ is that the above equation decouples from the magnetics. (But only if $\rho$ assumed to be predefined.) So one can solve the problem staggered.

The condition $\div\vA=0$ can always be satisfied. If we initially have a vector potential $\vA'$ with $\div\vA'\neq 0$ then we define $\vA := \vA'+\grad\varphi'$ such that $$ 0 = \div \vA = \div(\vA' + \grad\varphi')= \div\vA' + \Delta \varphi' $$ To find the function $\varphi'$ for the modification of $\vA'$ we just have to solve the poisson equation $$ \Delta\varphi' = -\div\vA' $$ for $\varphi'$. The coice $\div A=0$ is the so called Coulomb gauging.

A field is determined by its curl and sources. Under the assumption that we have already the required boundary conditions for $\varphi$ and $\vA$ then the fixation of the divergence for the vector potential (i.e., the gauging) determines the potentials uniquely. Beside the Coulomb gauging there are other forms of useful gauging (e.g., Lorentz-gauging).

lagrangian formalism - Euler's equations of rigid body motion from least action principle

I would like to derive Euler's equations of rigid body motion from least action principle.

Suppose we are in free space so we have no gravity so Lagrangian is equal to kinetic energy.

$$ L = T = \int_M \rho(x)(x\cdot \Omega_B)^2 d^3x $$

where $\rho$ is density, $\Omega_B$ is angular velocity bivector in reference body frame and $M$ is the rigid body.

The action principle says(if I haven't messed up something) that

$$ 0 = \frac{d}{ds}\bigg|_{s=0}\int_0^T \int_M \rho(x)[x\cdot (\Omega_B(t)+s \delta \Omega(t))]^2 d^3x $$

for every $\delta \Omega$ with zero ends i.e. $\delta \Omega(0)=\delta \Omega(T) = 0$. I should get these equations:

$$ I(\Omega_B) - \Omega \times I(\Omega_B)=0 $$

where

$$ I(\Omega) = \int_M \rho(x) (x\wedge (x\cdot \Omega)) d^3x. $$

But I can't see where I would get the time derivative of $\Omega_B$ from the action principle. Can someone tell me what I'm doing wrong? I guess there is some trouble with that $\delta \Omega$ can't be completely arbitrary.

Answer

I) A Lagrangian variational principle for Euler's equations for a rigid body

$$ \tag{1} (DL)_i ~=~M_i, \qquad i\in\{1,2,3\}, $$

is e.g. explained in Ref. 1. Here the angular momentum $L_i$, $i\in\{1,2,3\}$, along the three principal axes of inertia is tied to the angular velocity $\omega_i$, $i\in\{1,2,3\}$, by the formula

$$\tag{2} L_i~:=~I_i \omega_i, \qquad i\in\{1,2,3\}, \qquad (\text{no sum over }i).$$

The covariant time-derivative $D$ of a vector $\eta_i$, $i\in\{1,2,3\}$, is defined as

$$\tag{3} (D\eta)_i ~:=~ \dot{\eta}_i+(\omega\times\eta)_i, \qquad i\in\{1,2,3\}. $$

The angular velocity vector $\omega$ plays the role of a non-Abelian gauge connection/potential.

II) To see the $so(3)$ Lie algebra, we map an infinitesimal rotation vector $\alpha$ into an antisymmetric real $3\times3$ matrix $r(\alpha)\in so(3)$ as

$$\tag{4} \alpha_i \quad\longrightarrow\quad r(\alpha)_{jk}~:=~\sum_{i=1}^3\alpha_i \varepsilon_{ijk}. $$

The $so(3)$ Lie-bracket is given by (minus) the vector cross product

$$\tag{5} [r(\alpha),r(\beta)]~=~r(\beta\times \alpha). $$

Similarly, for the corresponding $SO(3)$ Lie group, a finite rotation vector $\alpha$ maps into an orthogonal $3\times3$ rotation matrix $R(\alpha)\in SO(3)$ as explained in this Phys.SE post. Infinitesimally, for an infinitesimal rotation $|\delta\alpha| \ll 1$, the correspondence is

$$\tag{6} R(\delta\alpha)_{jk} ~=~\delta_{jk} + r(\delta\alpha)_{jk} + {\cal O}(\delta\alpha^2). $$

III) A finite non-Abelian gauge transformation $\omega\longrightarrow\omega^{\alpha}$ takes the form

$$\tag{7} r(\omega^{\alpha})~=~R(-\alpha) \left(\frac{d}{dt}-r(\omega)\right)R(\alpha), \qquad \alpha\in \mathbb{R}^3.$$

An infinitesimal non-Abelian gauge transformation $\delta$ takes the form

$$\tag{8} r(\delta\omega)~=~\frac{d}{dt}r(\delta\alpha)-[r(\omega),r(\delta\alpha)],$$

or equivalently

$$\tag{9} \delta\omega_i~=~(D\delta\alpha)_i, \qquad i\in\{1,2,3\}, $$

where $\delta\alpha$ denotes an infinitesimal rotation vector corresponding to an $so(3)$ Lie algebra element $r(\delta\alpha)$.

We call (7)-(9) gauge transformations for semantic reasons, because of their familiar form, but note that (most of) them are not unphysical/spurious transformations. We stress that the angular velocity $\omega$ is a physical variable.

IV) Finally we are ready to discuss the action principle. The finite rotation vector $\alpha(t)\in \mathbb{R}^3$ plays the role of independent dynamical variables for the action principle. One may think of virtual rotation paths $\alpha:[t_i,t_f]\to \mathbb{R}^3$ as a reparametrization of virtual angular velocity paths $\omega:[t_i,t_f]\to \mathbb{R}^3$. The action reads

$$\tag{10} S[\alpha,\omega]~=~\int_{t_i}^{t_f} \! dt ~L $$

with Lagrangian

$$\tag{11} L~=~\frac{1}{2} L^{\alpha}\cdot \omega^{\alpha} + M\cdot \alpha , $$

where

$$\tag{12} L^{\alpha}_i~:=~I_i \omega^{\alpha}_i, \qquad i\in\{1,2,3\}, \qquad (\text{no sum over }i).$$

The Lagrangian (11) consists of rotational kinetic energy plus a source term from the torque $M$. Here $\omega^{\alpha}$ is the actual angular velocity vector, while $\omega$ here (in contrast to above) is a fixed non-dynamical reference vector, which is not varied. It is sort of a gauge-fixing choice. Infinitesimal variation yields

$$ \tag{13}\delta L ~\stackrel{(11)}{=}~ L^{\alpha}\cdot \delta\omega^{\alpha} + M\cdot \delta\alpha ~\stackrel{(9)}{=}~ L^{\alpha}\cdot \left(\frac{d}{dt}\delta\alpha + (\omega^{\alpha}\times\delta\alpha)\right) + M\cdot \delta\alpha ,$$

which (after integration by parts and appropriate boundary conditions) leads to Euler's equations (1) for the angular velocity vector $\omega^{\alpha}$.

References:

1. J.E. Marsden and T.S. Ratiu, Introduction to Mechanics and Symmetry, 1998.

quantum mechanics - What do atoms really look like?

When considering the orbital model of the atom it seems like the shape of each orbital corresponds the shape that contains a volume such that there is a 90% chance of an electron being there. I also assume that in the majority of the literature, the equation that defines the shape of the different orbitals corresponds to the solution of some kind of 3-dimensional version of the Schrödinger Equation the configuration of which corresponds to only one electron. Are we then Supposed to add up all the orbitals? Don't all those orbitals with different shapes intersect? (if they do, i would doubt its a good solution) Would the orbitals of atoms with multiple electrons get distorted instead of cutting trough each other?

Taking all of this into consideration, is there any way to represent the three dimensional probability density of the electrons in a carbon atom for example?

Answer

The classical pretty shapes of orbitals like this

are "hydrogen-like" orbitals. Essentially, they depict exact solutions of the Schrodinger equation for atoms with a point-nucleus and no other electrons.

Atoms with multiple electrons have orbitals that are similar to those hydrogen-like orbitals, but not quite the same because of electron-electron interactions. The wikipedia article puts it well:

For atoms with two or more electrons, the governing equations can only be solved with the use of methods of iterative approximation. Orbitals of multi-electron atoms are qualitatively similar to those of hydrogen, and in the simplest models, they are taken to have the same form. For more rigorous and precise analysis, the numerical approximations must be used.

So, in answer to your last question, you could represent the 3D probability density of electrons in a Carbon atom, but it would have to be solved numerically; there isn't a closed-form solution, and in the end the picture would look pretty similar to a bunch of hydrogen-like orbitals superimposed.

special relativity - Energy Conservation Dilemma

Assume that a man is travelling in a space ship at a certain relativistic speed with respect to a man at rest at some point in space, such that 3 minutes in the ship is equal to 5 minutes for the person at rest.

Also assume that the man in the ship has a lighter which contains gas of a certain amount such that the lighter can be lit for 5 minutes .

Now if the man in the space ship lights the lighter for 3 minutes, then he would have 2 minutes' worth of gas left, but the stationary observer would have seen light emitted for about 5 minutes (since 3 in that space ship = 5 minutes for the stationary observer)

How is it possible for the stationary observer to see light for 5 minutes? And in this case, how is energy conserved?

homework and exercises - Energy of a system of conductors

Electrostatic potential and charges on conductors that are closed to each other can be put in relation with the capacitance matrix .

Can the energy of the system of two (or more) conductors be rewritten as the sum of a part due to each conductor and another one that is due to a "shared" energy of the two conductors?

---

Consider two conductors of capacitance ,charges and potentials $q_1$, $C_1$, $V_1$, $q_2$, $C_2$, $V_2$.

The energy of the system is by definition $$U=q_1 V_1+q_2 V_2$$

The matrix of capacitance is the 2x2 symmetric matrix such that

$$\begin{pmatrix} q_1 \\ q_2 \end{pmatrix}=\begin{pmatrix}c_{11} & c_{12}\\ c_{12}& c_{21} \end{pmatrix}\begin{pmatrix}V_1 \\ V_2\end{pmatrix}$$

Can I express $U$ as something like the following?

$$U=\frac{1}{2}\frac{q_1^2}{2C_1}+\frac{1}{2}\frac{q_2^2}{2C_2}+...$$

Where $...$ stays for an expression that includes the charges, the potentials and the coefficients $c_{11},c_{12},c_{23}$. This expression should represent the "shared" energy of the two conductors.

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Example (which I wonder how to generalize)

Two conductiong spheres have the parameters indicated above and are at a big distance $x$ (induction influence is neglected). The energy of the system can be written as

$$U=\frac{q_1^2}{2C_1}+\frac{q_2^2}{2C_2}+\frac{q_1 q_2}{4 \pi \epsilon_0 x}$$

In this case the expression I'm looking for is $\frac{q_1 q_2}{4 \pi \epsilon_0 x}$, but how can one in general write this term (if it is possible to do it)?

Answer

If you assume that $$U=\frac {1}{2}q_1V_1+\frac {1}{2}q_2V_2$$ is correct for the total energy, you simply solve the matrix capacitance equation for $q_1$ and $q_2$, insert them into the system energy equation, and order the terms according to the products of $V_1, V_2$ Thus you'll obtain for the energy $$U=\frac {1}{2}( c_{11}V_1^2+2c_{12}V_1·V_2+C_{22}V_2^2)$$ where symmetry of the matrix $c_{12}=c_{21}$ is used.

In the general case of n conductors, you can always diagonalize the symmetric capacitance matrix by a suitable orthogonal transformation of the "coordinates" $V_1,V_2,...$ to coordinates $V_1^*, V_2^*,...$ to obtain a quadratic form for the energy $U=\frac {1}{2}\sum_{i=1}^n c_iV_i^*{^2}$.

Capacitance of a Grounded Capacitor

Suppose one plate of the capacitor is grounded which means there is charge present at only one plate. We know that the potential across the capacitor will be 0, i.e., V=0.

And capacitance of the Capacitor will be C=Q/V

C=Q/0 implying C=∞

So it means that the capacitance of a grounded capacitor is Infinite. I know this is not true as a conductor cannot store infinite electrical energy.

So where am I going wrong?

Please tell me. Thanks

Answer

Suppose one plate of the capacitor is grounded which means there is charge present at only one plate.

The electric potential of an ideal ground does not change no matter how much charged is added or removed. From the Wikipedia article Ground (electricity)

In electronic circuit theory, a "ground" is usually idealized as an infinite source or sink for charge, which can absorb an unlimited amount of current without changing its potential.

So, attaching one capacitor plate to ground simply fixes the electric potential of that plate; if the ungrounded plate has charge $Q$, the grounded plate will have charge $-Q$.

how could the grounded plate gain -Q charge.

The ideal ground supplies the $-Q$ charge to the plate without changing potential.

If somehow it gains -Q charge it will flow to the earth.

No, that's not correct. A common problem in electrostatics is calculating the surface charge on a grounded plane when a charge $Q$ is placed some height above the plane. Essentially, the charge flows until the electrostatic energy of the configuration is minimum which, in the case of the grounded plate capacitor, is when there is charge $-Q$ on the grounded plate which is as close to the $Q$ charge as is possible.

OK I got u but why the potential across a grounded capacitor is taken 0

It isn't taken to be zero unless $Q=0$.

pattern - Wordplay addition paradox

There are words from which you can remove a "chunk", leaving a new word. Like this:

    WISHBONE


   WI SHBO NE

 WI   SHBO   NE 

WI     NE

WI            NE

   WI     NE


     WINE

There are also words that work the other way, for which inserting a "chunk" produces a new word. For example, you can insert the chunk AUTIFI into the word BEER to make BEAUTIFIER.

A "chunk" is a string of consecutive letters. It must consist of at least two letters (no single-letter chunks). It does not need to be a valid English word.

Now, what if I told you there are words from which you can remove a chunk, then insert a different chunk with different letters, and get the original word again?

What the heck am I talking about?!

There are actually thousands of such examples. I'm just looking for a general description of the pattern that creates this strange phenomenon.

(Too easy? Too hard? Try the counterpart subtraction paradox.)

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Afterword:

Here is the specific example which motivated this post:

Start with the word QUARTERBACK and remove the chunk RTERBA to obtain QUACK. Now, take the new chunk ARTERB (which is obviously different from the chunk that was removed) and insert it into the word QUACK to obtain the original word QUARTERBACK again.

Perfectly identical, but not exactly the same! It's a little bit of a shell game, and if you followed it with sharp eyes, you might have noticed that it was the first "A" in QUARTERBACK which was retained in QUACK, but then it became the second "A" when QUARTERBACK was restored.

The two "A" are like sentinels which stand on either side of a middle string of letters. The chunk which is removed must contain the middle string of letters as well as one of the sentinels. The chunk which is inserted must contain the middle string of letters as well as the other sentinel.

@GarethMcCaughan did a good job below of exploring whether the sentinels can be more than a single letter. They can be!

Answer

I assume that

"with different letters" means only that the sequence of letters isn't the same, rather than that the (multi)set of letters isn't, because otherwise the thing seems to be genuinely impossible unless there's some sort of lateral-thinking nonsense going on.

In that case

D(ESP)ERATE can lose ESP to make DERATE and then gain SPE to make DE(SPE)RATE again. Or R(ESIGN)ED can lose ESIGN to make RED and then gain SIGNE to make RE(SIGNE)D again.

The general picture here is

that you have words ABCBD and ABD where A,B,C,D are arbitrary strings of letters. The easiest cases (as above) have B a single letter, but I bet there are some where B is longer. At any rate, you're then removing BC and inserting CB or vice versa.

[EDITED to add:]

Yes, B can certainly be longer. For instance, BANYANS can lose ANY and gain YAN or vice versa. Or consider HONEYMOONED; you can lose ONEYMO making HONED and then gain YMOONE to get HONEYMOONED again.

riddle - This day in history I

THIS DAY IN HISTORY is a series of puzzles I intend to create in which I choose an event a certain amount of years ago. I need you to figure the event and tell me how many years ago it happened

From a great maker of bikes
Comes something the government likes
Took on this day
And now there's a lot in the place
Of which it originates

I will release the answer when the next one is released

Answer

Is it

The Wright Brothers making the first successful plane?

From a great maker of bikes

The Wright Brothers built the first working plane, but before that they made bikes

Comes something the government likes

The government uses planes a lot

Took on this day

Planes are used very often these days

And now there's a lot in the place Of which it originates

The first plane was made in North Carolina, and there are a lot of planes in North Carolina now

If so, it happened in

1903

So it happened around

116 years ago

spacetime - Space-time in String Theory

I would like to understand how Physicists think of space-time in the context of String Theory. I understand that there are $3$ large space dimensions, a time dimension, and $6$ or $7$ (or $22$) extra dimensions, and all these dimensions need to fit together in a way such that the extra dimensions are compactified (with a Calabi-Yau or $G_2$ structure).

My question, however is not about the possible $10$, $11$ or $26$ dimensional manifolds that may be possible, but about whether string theorists consider space-time as somehow quantized (or discrete), or rather as a continuous manifold, or are both options possible? In other words, can strings move continuously through space, or are there a discrete set of locations where strings can be, and does string theory rule out one of the options?

How about the same question in loop quantum gravity (LQG)? Should I think of the spin networks in LQG as describing a discrete space-time?

Thanks for your insight, or any references you may be able to provide!

Answer

I think Anna s comment is correct, in LQG spacetime consists of discrete atoms and in ST it is continuous.

In addition, This article contains an interesting and quite accessible Nima talk related to the topic. Therein Nima explains why the present notions of spacetime are doomed and introduces the recent cutting edge ideas about how spacetime could emerge from a newly discovered and not yet fully explored structure called T-theory.

homework and exercises - Adiabatic expansion in van der Waals gas

Given a Van der Waals gas with state equation: $$\left( P+\frac{N^2 a}{V^2}\right)\left( V-Nb \right)=NkT,$$ show that the equation of an adiabatic process is: $$\left( V-Nb\right)T^{C_V}=\text{constant}.$$

I began by setting $đQ=0$ in $$\mathrm dU=đQ+đW,$$ one then gets $$0=\mathrm dU+P~\mathrm dV.$$

Now given $U=\frac{3}{2}NkT-\frac{N^2 a}{V},$ I plugged it's derivatives into $$\mathrm dU=\left( \frac{\partial U}{\partial T}\right)_V~\mathrm dT+\left( \frac{\partial U}{\partial V} \right)_T~\mathrm dV,$$ from which I obtained $$0=C_V~\mathrm dt+\left(P+\frac{N^2 a}{V^2} \right)_T~\mathrm dV=C_V~\mathrm dT + \frac{NkT}{V-Nb}~\mathrm dV,$$ using $V~ đW$'s equation.

Dividing by $T$ and integrating gives $$C=\log{T^{C_V}}+\log{(V-Nb)^{Nk}},$$ which is equivalent to $$C'=(V-Nb)^{Nk}T^{C_V},$$ for $C$ and $C'$ constants.

Now the expression so obtained seems very similar to what I was looking for, but I can't seem to get rid of the $Nk$ exponent. Anyone got a different approach to this problem, or a way to get the desired formula?

Answer

The correct answer is $(V-Nb)T^{C_V/Nk}=\text{const}$, the problem statement is just wrong.

homework and exercises - Killing vector and one-form

p. 21 in this paper (http://arxiv.org/abs/0704.0247)

$V$ is Killing vector, where $V^2 = −4b\bar{b}$, which means it is timelike Killing vector.

The authors say:

From $V^2 = −4|b|^2$ and $V = ∂_t$ as a vector we get $V_t = −4|b|^2$,

My question here is how did the authors set $V_t$ equal to this value?

They add,

From $V^2 = −4|b|^2$ and $V = ∂_t$ as a vector we get $V_t = −4|b|^2$, so that $V = −4|b|^2(dt+σ)$ as a one-form, with $σ_t = 0$.

Why did they assume that?

Answer

My question here is how did the authors set $V_t$ equal to this value?

On page 21 the authors say: "Let us choose coordinates $(t, z, x_i)$ such that $V = \partial_t$ and $i = 1, 2$."

So they chose the coordinates such that $V=\partial_t$, which means $V^t=1$. Note that the other components of $V^\mu$ are zeros. Next we have $$V^2=V_\mu V^\mu=-4|b|^2=V_t V^t+V_{x_1}V^{x_1}+V_{x_2}V^{x_2}+V_{z}V^{z}=V_t*1,$$ from which you find $V_t=-4|b|^2$. Here we used $V^{x_i}=V^z=0$.

Though what I don't get is their requirement that σt be equal to zero and why did they place a dt next to it. Why did they assume that?

$\sigma$ is a general one form on coordinates $x^i, z$, which means $\sigma=\sigma_{1}dx^1+\sigma_2 dx^2+\sigma_3 dz$, note that later they use the gauge freedom to set $\sigma_z=0$. They chose the coordinates to fix $V_t$ and the rest it is the most general one form on $x^i$. For example, the most general one form on coordinates $t,x^1, x^2,z$ is $\alpha=\alpha_0 dt+\alpha_1 dx^1+\alpha_2 dx^2+\alpha_3 dz$. Compare it to their expression for $V$ (after (4.7)) and you will see that they chose only the first components, the rest is arbitrary.

optics - Can very high power laser beams self-focus in vacuum?

I first recall reading about such an effect in a SF story entitled "Rails Across the Galaxy" which involved self focusing laser beams. And in a science paper here

soft question - Church-Turing hypothesis as a fundamental law of physics

The Church-Turing hypothesis says one can not build a computing device which has more computing power (in terms of computability) than the abstract model of Turing machine. So, there is something in our laws of physics which prevent us from making devices which are more powerful than Turing machine, so in this respect it can be viewed as a law of physics.

What is the physicists' view of Church-Turing hypothesis?

Can the Church-Turing hypothesis be deduced from other fundamental law of physics?

homework and exercises - Equivalent resistance in ladder circuit

I have stumbled upon a given question I really have a hard time to solve. Basically I need to find an equivalent resistance in some form of "ladder" configuration. Where the chain is an infinite sequence of resistors.

I have really no good idea how to find this equivalent resistance. Trying the old fashioned rule of parallel and resistors in series I came to a very messy formula: $$R + \frac{1}{\frac{1}{R} + \frac{1}{R+\frac{1}{\frac{1}{R}+....}}}$$

I know however that the solution should be much more simple.

Now I tried using Kirchhoff's loop rule. Which states that the power difference in a closed loop must be 0. Naming the "potential currents" between AB $I_1$, between BC (through the single resistance) $I_2$ and the current "from B to the right" $I_3$ Considering the loop containing BC & the rest of the structure this rewrites to: $$I_1 = I_2 + I_3$$ $$I_3 \cdot R - I_2 \cdot R_{eq} = 0$$

The problem is, 2 variables, 2 functions doesn't really bring me closer to an answer :(. What am I missing?

Answer

HINT:

Notice that $$R_{eq}=R+\frac1{\frac1R+\frac1{R_{eq}}}$$

electromagnetism - Does light interact with electric fields?

We know that light is an electromagnetic wave and it does interact with charges.

It contains magnetic field and electric field oscillating perpendicularly but when we apply an electric or magnetic field in any direction to the wave the applied electric field or magnetic field vector doesn't alter the magnetic or electric field in the electro magnetic wave (according to vector addition rule)....why?

thermodynamics - Ideal gas temperature and pressure gradients?

Consider an ideal gas in a $d\times d\times L$ box with the $L$ dimension in the $x$-direction. Suppose that the opposite $d\times d$ sides of the box are held at temperatures $T_1$ and $T_2$ with $T_2>T_2$ and that the system reaches a steady state. According to these notes, the thermal conductivity of an idea gas scales as the square root of temperature; $k=\alpha\sqrt{T}$ in which case by Fourier's Law one gets that the temperature gradient in the $x$-direction is $$ T(x) = \left[T_1^{3/2}+(T_2^{3/2}-T_1^{3/2})\frac{x}{L}\right]^{2/3} $$ What is the corresponding pressure gradient $P(x)$ in the steady state?

Answer

It's a steady state. If there were a pressure gradient, there would be net force on the gas (ignoring gravity). There's no net force here because the air isn't accelerating. Thus the pressure is constant.

The number density varies across the box inversely to the temperature so the ideal gas law holds.

general relativity - What does it mean to go from a co-variant vector to a contravariant vector?

In most presentations of general-relativity I see the following statement,

We can change from a covariant vector to a contravariant vector by using the metric as follows, ${ A }^{ \mu }={ g }^{ \mu \nu }{ A }_{ \nu }$

My questions are,

1. What is the need to do this particular change in relativity?


2. The covariant components represent the components of a vector the contravariant components represent the components of a dual-vector, for finite dimensional vector spaces the two spaces are isomorphic. What is the significance of representing a quantity in contravariant or convariant forms? Is the need purely mathematical?

electromagnetism - Transverse polarizations of a massless spin 1 particle

Physical polarization vectors are transverse, $p\cdot{\epsilon}=0$, where $p$ is the momentum of a photon and $\epsilon$ is a polarization vector.

---

Physical polarization vectors are unchanged under a gauge transformation $\epsilon + a\cdot{p}=\epsilon$, where $a$ is some arbitrary constant.

---

For a massless spin $1$ particle, in the Coulomb gauge,

one common basis for the transverse polarizations of light are

$$\epsilon_{\mu}^{1}=\frac{1}{\sqrt{2}}(0,1,i,0)\qquad \epsilon_{\mu}^{L}=\frac{1}{\sqrt{2}}(0,1,-i,0).$$

This describes circularly polarized light and are called helicity states.

---

In the centre of mass frame, in the positive $z$-direction, the polarization vectors of a photon are

$$(\epsilon_{\mu}^{\pm})^{1}=\frac{1}{\sqrt{2}}(0,1,\pm i,0)\qquad (\epsilon_{\mu}^{\pm})^{L}=\frac{1}{\sqrt{2}}(0,1,\mp i,0).$$

---

1. Why are physical polarization vectors transverse?



2. How is $\epsilon + a\cdot{p}=\epsilon$ a gauge transformation? The gauge transformations I know are of the form $A_{\mu}\rightarrow A_{\mu}+\partial_{\mu}\Lambda$.


3. How do you define transverse polarization of light? For example, why are $\epsilon_{\mu}^{1}=\frac{1}{\sqrt{2}}(0,1,i,0)$ and $\epsilon_{\mu}^{L}=\frac{1}{\sqrt{2}}(0,1,-i,0)$ the transverse polarizations of light?


4. Why do the polarization vectors $\epsilon_{\mu}^{1}=\frac{1}{\sqrt{2}}(0,1,i,0)$ and $\epsilon_{\mu}^{L}=\frac{1}{\sqrt{2}}(0,1,-i,0)$ correspond to transverse polarizations of light?


5. Why are these polarization vectors $\epsilon_{\mu}^{1}=\frac{1}{\sqrt{2}}(0,1,i,0)$ and $\epsilon_{\mu}^{L}=\frac{1}{\sqrt{2}}(0,1,-i,0)$ called helicity states?


6. In the centre of mass frame, in the positive $z$-direction, why are the polarization vectors of a photon given by $(\epsilon_{\mu}^{\pm})^{1}=\frac{1}{\sqrt{2}}(0,1,\pm i,0)$ and $(\epsilon_{\mu}^{\pm})^{L}=\frac{1}{\sqrt{2}}(0,1,\mp i,0)?$

Answer

I will answer the first two questions and leave the rest to others.

First question - We describe the photon with $A_{\mu}$ but this has 4 degrees of freedom.But we ultimately need just 2. To reduce the number we use the transverse condition and say the momentum is perpendicular to the polarization vector. So any polarization vector will be transverse to the direction of motion. This gives us 3 degrees of freedom. To get to 2, we set up an equivalence class of polarization vectors. This leads to second question

Second question - $ A_{\mu} \rightarrow A_{\mu }+\partial_{\mu} \lambda $ in momentum space becomes $ \epsilon \rightarrow \epsilon + \alpha p $ . This is gauge transformation because notice $ \epsilon \cdot p \rightarrow e \cdot p + \alpha p \cdot p $ but $ p \cdot p = 0 $ so in fact $ \epsilon\text{ and } \epsilon + \alpha p $ represent the same physical state in the same way that $ A_{\mu} $ and $A_{\mu }+\partial_{\mu} \lambda $ represent the same physical state.

story - Who stole my laptop?

It’s obvious that one of my “friends” was lying to me back in March this year. Someone who knew I’d be gone stole my $1,500 laptop and my crappy old broken microwave (I probably won’t ask for that one back).

I dug up this old chat log, hopefully you can tell me who’s full of it and who isn’t. You can assume only one person is/is aiding the thief, and everyone has perfect memory. Multiple backstabbers might just be too much for my fragile heart to handle.

4:11 PM  opened chat.
4:11 PM  joined chat.
4:11 PM  joined chat.
4:12 PM : hey alice
4:12 PM : hey, just back from your month in europe?
4:12 PM : a week ago, yeah, glad to be back in 'murica

4:12 PM  joined chat.
4:13 PM : sam, st patricks is tomorrow
4:13 PM : no **** alice im getting stoked early
4:14 PM : dude why we here anyway
4:15 PM : Wait until everyone gets here.
4:16 PM  joined chat.
4:16 PM : ssss
4:17 PM : hey scott how was europe with curmudgeon
4:18 PM : o didnt you hear? i had to stay home sick
4:18 PM : got ******* food poisoning like 2 days before the flight

4:18 PM : sucks to suck
4:19 PM : wheres john?
4:19 PM : said he’d be late, had a meeting or some ****
4:24 PM  joined chat.
4:25 PM : there he is, hurry up so my gf can paint me green
4:26 PM : okay guys, this might be a little uncomfortable, but I need you to answer honestly. I’m not going to sue or anything
4:26 PM : **** whats going on?
4:26 PM : hush up and listen john
4:27 PM : exactly one month ago, someone who knew my garage code got into my house and stole my laptop and microwave.
4:27 PM : you seriously think any of us would do that? wtf

4:28 PM : I wouldn’t be doing this if there weren’t a simple solution. since the culprit conveniently knocked over and broke my clock, which still displayed the date and time when I got home. I want all of you to tell me where you were at noon a month ago.
4:29 PM : were supposed to remember? i dont remember where i was 5 hours ago man
4:29 PM  Well thankfully, that was the day I called you all to see how you were doing, so you should have no trouble remembering.
4:30 PM : oh right, thats easy. i was still weak from the food poisoning and you called when i was mustering up the energy to go visit my man johnny
4:30 PM : i was probably at his house by noon
4:31 PM : you were, cause dude called me right before you got there and i told him to shove off
4:31 PM : can confirm, was told to shove off
4:31 PM : you called me while i was driving.
4:31 PM : yeah, i remember the roaring of the car when i called actually, it was hard to hear you
4:32 PM : was definitely still out around noon, since i grabbed a bite to eat

4:32 PM : i woke up, got ready, visited my grandma on one side of town, then back across to the post office to get stamps and mail a birthday card
4:32 PM : thanks. what about you sam?
4:33 PM : your call actually woke me up, after noon. i was asleep, man
4:33 PM : oh yeah, i remember that actually. sorry about that
4:34 PM : hey man, i saw sam that day around, walking down our street with his gf! dont bs me
4:34 PM : even if thats true, we dont live anywhere near curmudgeon, which means it couldnt have been me
4:35 PM : and while were on the topic, i know you werent with scott
4:35 PM : shut up sam
4:36 PM : he was with my gfs friend becca, probably stealing your microwave
4:37 PM : she doesnt live anywhere near him either and you know it

4:37 PM : fine sam, i made it up since my real alibi is so stupid. i was sound asleep at noon, i conked out right after i hung up with curmudgeon
4:38 PM : ya sorry, wouldnt have lied but i had to cover for this bumbling idiot
4:39 PM : idiots
4:39 PM : no, no guys, it’s okay, i believe all of you. was probably just somebody who watched me punch in the code sometime, it’s no big deal
4:40 PM : see you guys
4:40 PM  closed chat.

So maybe I’m a bit of an overbearing friend, but can you tell me if it paid off? Someone must have a contradiction in their story. Who’s lying? (There is a definitive answer, so don’t be too quick to flag this as "too broad" ye-who-likes-to-flag!)

Answer

The liar is

Alice

Because

She said she went to the post office. However, we know the chat conversation takes place the day before St Patrick's Day 2015, which means the 16th of March. The theft (and phone calls) occurred exactly 1 month previously, on Monday, February 16th. That was President's Day, which is a national holiday, so the post office would have been closed. Alice couldn't have bought stamps, so she must be lying.

mathematics - Paint numbers from 1 to 8 with two colours

Can you paint every number from 1 to 8 with two colours, such that there are no distinct numbers $a, b, c$ of the same colour with $a+b=c$? For example, you cannot have 2, 3 and 5 of the same colour since $2+3=5$.

Good luck!

Answer

How about this

Red - $\{1,2,4,8\}$
Blue - $\{3,5,6,7\}$

electromagnetism - Equation describing magnetic hysteresis

So when you're looking at B-H curves for ferromagnetic substances, you often see these magnetic hysteresis curves, which occur, I gather, largely because of domain formation which has some reversible and some irreversible components:

I've been reading through many papers, web sites and this book (Hysteresis in Magnetism), but I haven't seen any equations for how to generate these curves. I recognize that there may be no easy way to express the entire curve in a single equation, but clearly people are generating these plots, and I think they're doing it from some equations, presumably based on some characteristic parameters like the saturation remanance, the coercive force, etc. Is there anyone out there who can help me with this?

Personally, my first choice would be an equation that actually describes the system given some set of material conditions (subject to some constraints is fine, too) but if that's quite complicated and for the most part, people just stitch together two sigmoid functions until it looks about right, then I'm fine with that answer as well, so long as there's some justification for why this is done in there somewhere.¹

¹_{Full disclosure: I am in thesis deadline mode and I'm currently probably a 5 on the Stanford Sleepiness Scale, so I apologize if some of the words in this post don't make sense - I'll try and edit it if people find that I haven't adequately conveyed my question.}

Answer

The Jiles-Atherton model of ferromagnetism is used in some circuit analysis programs. It may be overkill for this question, but it does give pretty pictures.

I'm going to work in MKS units exclusively.

The model equations are: $$ B= \mu_0 M \quad , \quad M = M_{rev} + M_{irr} $$ where $B$ is the magnetic flux density, and the magnetization $M$ is composed of reversible ($M_{rev}$) and irreversible ($M_{irr}$) components.

Physically, during the magnetization process:

• $M_{rev}$ corresponds to (reversible) magnetic domain wall bending (the S-shaped magnetization curve, but without hysteresis)


• $M_{irr}$ corresponds to (irreversible) magnetic domain wall displacement against the pinning effect (the hysteresis).

These components are calculated according to:

$$ M_{rev} = c (M_{an} - M_{irr}) $$ $$ M_{an} = M_s \left[ \coth \left(\frac{H + \alpha M}{A}\right)-\frac{A}{H+ \alpha M}\right] $$ $$ \frac{dM_{irr}}{dH} = \frac{M_{an}-M_{irr}}{k \delta - \alpha (M_{an}-M_{irr})} $$

Here the anhysteretic magnetization $M_{an}$ represents the magnetization for the case where the pinning effect is disregarded. (This case corresponds to $c=1$, where $M=M_{an}$ and $M_{irr}$ therefore does not contribute to $M$.)

The quantity in the expression for $M_{an}$ in square brackets is the Langevin function $\mathcal{L}$: $$ \mathcal{L}(x) = \coth(x) - \frac{1}{x} \quad , \quad \mathcal{L}(x) \approx \left\{\begin{array}{ccc} \frac{x}{3} & , & |x|<<1 \\ 1 & , & x >> 1 \\ -1 & , & x<<-1 \end{array} \right.$$

and $\delta$ is the sign of the time rate of change of the applied magnetic field $H$: $$ \delta = \left\{ \begin{array}{ccc} +1 & , & \frac{dH}{dt}>0 \\ -1 & , & \frac{dH}{dt}<0 \end{array} \right. $$

$M_{rev}$ can be eliminated from this system of equations to reduce their number by 1: $$ M = c M_{an} + (1-c)M_{irr} $$

The equations for $M, M_{an},$ and $M_{irr}$ are inter-dependent and so are to be solved simultaneously.

There are 5 parameters (listed here together with sample values):

• $M_s$, the saturation magnetization [1.48 MA/m]


• $c$, the weighting of anhysteretic vs. irreversible components [0.0889]


• $\alpha$, the mean field parameter (representing interdomain coupling) [0.000938]


• $A$ sets the scale for the magnetic field strength [470 A/m]


• $k$ sizes the hysteresis [483 A/m]

For the values listed, a crude spreadsheet produced this plot:

The horizontal axis is the applied magnetic field H, in A/m, sweeping from 0 up to 2500, then down to -2500, and then up again to 2500. The vertical axis is the flux density B in T.

This example comes from a 1999 IEEE paper by Lederer et al, "On the Parameter Identification and Application of the Jiles-Atherton Hysteresis Model for Numerical Modelling of Measured Characteristics".

It appears that choosing the parameters to match a given material is a chore, but that's another story...

riddle - Room 11: Bits and pieces of Chinese history

Credit: Brandon X.

This is part of a larger riddle room puzzle. Other rooms:

You have just woken up. You recall the night before, being driven from your home to a luxurious hotel by some people of a company or organization or something-The Ofgelenkt Institute something or other. They requested you to be part of their testing program-with a reward of up to 500,000 dollars! After signing a few papers, you were off and away! But now you look up. You are in a barren room, with just a few pieces of furniture and pictures and stuff. You are confused, until you hear the intercom from a hidden speaker:

“You have been selected for your special skills to go through a testing program. This program will challenge your intelligence in thinking in a different way. As you can see, you are locked in a room with four doors out, labeled over the entrances. There are clues around the room to help you select the right door. If you get knocked out(choose the wrong door), you get $200 for your participation."

“We wish you luck, and may you be guided in the right direction.” The intercom shuts off suddenly.

You begin looking around the room. On the opposite end of the room, you see the four doors, with 1, 2, 3, and 4 above them, and you notice the table with the buttons in front of them, labeled with the logo of a fish on a sea of red, with the words Ofgelenkt Institute printed in pretty cursive on it.

(the logo of Ofgelenkt Institute)

You continue looking around the room. On one side, you see a table with a trifold on it. There are a lot of photos with captions on it.

• 
  

  One is a picture of the Terracotta Warriors. Its label: The terracotta warriors were constructed to escort the Chinese Emperor Qin Shihuandi(秦始皇) to the afterlife, fighting battles for him

  
  


• 
  

  Another picture is of a man you do not recognize. Its caption: John Marzaroli, founder and former CEO if Ofgelenkt Institute, had died of a heart attack when starting a branch of his institute in China. (this caption had a logo on it)

  
  
  


• 
  

  Another picture is of a tomb. Its caption: This is a picture of the tomb of Yue Fei(岳飛) and his son. Kneeling before him are statues of the people who got him killed, which were once spit on by visitors until it was recently outlawed.

  
  


• 
  

  Another picture is of a battlefield covered in shells. Its caption: This picture is from the Battle of Triangle Hill during the Korean War. A Chinese soldier of the name of Huang Jiguang( 黄继光) sacrificed himself by blocking American machine gun fire(hurling himself at a machine gun), allowing his comrades to win the battle.

  
  


• 
  

  The next image was a picture of some Chinese characters put in a certain way. The Caption:This is a picture of the only existing calligraphy made by famous poet Li Bai(李白). He is possibly one of the most famous poets in China and lived to the age of 61, supposedly dying of drowning in the Yangtze River.

  
  

To your right, you see a bookshelf with 3 shelves. It is labeled with the logo of the Ofgelenkt Institute. It has a sticker on it, saying Item A1Z26. You walk over to it and look at the shelves:

• The first shelf contains The Titan’s Curse, The Battle of The Labyrinth, and The Last Olympian by Rick Riordan


• The second shelf contains Harry Potter and the Sorcerer’s Stone, Harry Potter and the Goblet of Fire, and Harry Potter and the Half-Blood Prince by JK Rowling


• The third shelf contains Flatland by Edwin Abott Abott, Lord of the Flies by William Golding, and Of Mice and Men by John Steinbeck

The books all have the logo of the Ofgelenkt Institute, as well as a number, found in this order(left to right, top to bottom):

3, 21, 18, 22, 5, 2, 1, 12, and 12

You look behind you. There is a map of the world in it, labeled Product Shipping Code:

3-15-13-13-15-14-19-5-12-12-5-18.

There are several pins in it. You look at the countries that have been pinned:

United States

Japan

Mexico

South Korea

Netherlands

Germany

India

Underneath the map, it says:

Item A1B2

You unpin the map, and turn it over. It says:

Made in the United States

Product of Ofgelenkt Institute

The logo is there as well.

As you keep walking around, you notice books strewn across the floor.

• 
  

  The first book: Aliens: Har vi møtt dem? (Aliens: Have We Met Them?) by Davis Erino, Translated by Ikke Bruk

  
  


• 
  

  The second book:死亡和失踪(Death and Disappearance), by Seth Oshico, Translated by 选择这.

  
  


• 
  
  

  The third book: Spanish For Beginners, by Juana Estratagema

  
  


• 
  

  The fourth book: À cause du renne (Because of the Reindeer) by Henry Borichos, Translated by Francis Jamais

  
  


• 
  

  The fifth book: Neden Savaşıyoruz (Dünya Tarihi Üzerine Bir Araştırma) (Why We Fight (A Research on World History)) by Vera Cabull, translated by John Dokunmayin

  
  

Of course, you would love to obtain the 500,000 dollars. So, which door do you go through?

Hint #1:

Lucky numbers

Hint #2:

What do the stories on the trifold all have in common?

P.S. This is supposed to be part of a much larger puzzle, which have different rooms that are linked to one another. This is the first room. Other rooms might be posted later.

interactions - Reason behind fundamental forces

Can anyone please explain the basic most fundamental reasons behind fundamental forces, i.e. what causes electromagnetic, nuclear and gravitational forces.

homework and exercises - Lie derivative of Riemann tensor along killing vector ( = 0 )

I'm currently learning the mathematical framework for General Relativity, and I'm trying to prove that the Lie derivative of the Riemann curvature tensor is zero along a killing vector.

With the following notation for covariant differentiation, $A_{a||b} $ (instead of $\nabla_b A_a$ ), I have the following:

$\it\unicode{xA3}_\xi R_{amsq} = R_{amsq||x} \xi ^x + R_{xmsq} \xi^x{}_{||a} + R_{axsq} \xi^x{}_{||m} + R_{amxq} \xi^x{}_{||s} + R_{amsx} \xi^x{}_{||q}$.

I suspect that I need to invoke the second Bianchi identity. However, before I can do this, I need to somehow get this into a different form. There has to be some property of either killing vectors or maybe covariant derivatives that I'm forgetting/failed to learn. Any help would be appreciated.

Answer

However, I had difficulty understanding that answer and would like to understand how to do it this way. That is to say, I'd really like to know what property or identity that I'm missing before I can use use the Bianchi identities to show that it is manifestly zero.

The other proof uses the first Bianchi identity. That's where the starting assumption $R^a{}_{bcd}\xi^d = \xi^a{}_{;bc}$ comes from. If you want to use the second Bianchi identity, it is $$(\nabla_\xi R)(X,Y) + (\nabla_X R)(Y,\xi) + (\nabla_Y R)(\xi,X) = 0\text{,}$$ and therefore applying it and the Leibniz rule produces: $$\begin{align} \underbrace{\nabla_\xi[R(X,Y)]+\nabla_X[R(Y,\xi)]+\nabla_Y[R(\xi,X)]}_\mathrm{foo} = \underbrace{R(\mathcal{L}_\xi X,Y) + R(\mathcal{L}_XY,\xi) + R(\mathcal{L}_Y\xi,X)}_\mathrm{bar}\text{,} \end{align}$$ where it was assumed that the torsion vanishes, so that $\mathcal{L}_AB = \nabla_AB-\nabla_BA$. Additionally, $$\begin{eqnarray*} (\mathcal{L}_\xi R)(X,Y) &=& \mathcal{L}_\xi[R(X,Y)] - R(\mathcal{L}_\xi X,Y) - R(X,\mathcal{L}_\xi Y)\\ &=&\mathcal{L}_\xi[R(X,Y)] - R(\mathcal{L}_\xi X,Y) - R(\mathcal{L}_Y\xi,X)\\ &=&\underbrace{\mathcal{L}_\xi[R(X,Y)] - [\mathrm{foo}] + R(\mathcal{L}_XY,\xi)}_\mathrm{qux}\text{.} \end{eqnarray*}$$ So the objective is to show that the right-hand side, $\mathrm{qux}$, is identically zero whenever $\xi$ is a Killing vector field.

Let's write $S^a{}_b = [R(X,Y)]^a{}_b = R^a{}_{bcd}X^cY^d$, and just crank it out: $$\begin{eqnarray*} \mathcal{L}_\xi S^a{}_b &=& \nabla_\xi S^a{}_b - S^e{}_b\xi^a{}_{;e} + S^a{}_e\xi^e{}_{;b}\\ &=& \nabla_\xi S^a{}_b + X^cY^d(R^a{}_{ecd}\xi^e{}_{;b} - R^e{}_{bcd}\xi^a{}_{;e})\\ &=& \nabla_\xi S^a{}_b + X^cY^d(\nabla_c\nabla_d-\nabla_d\nabla_c)\xi^a{}_{;b}\text{,} \end{eqnarray*}$$ where the last step is actually valid for arbitrary $Z^a{}_b$, not just $\xi^a{}_{;b}$. The first term of this cancels with the first term of $\mathrm{foo}$. So far we have not used the fact that $\xi$ is a Killing vector field. Let's do so now by considering the other two terms of $\mathrm{foo}$: $$\nabla_X[R(Y,\xi)]^a{}_b - \nabla_Y[R(X,\xi)]^a{}_b = \nabla_X\nabla_Y\xi^a{}_{;b} - \nabla_Y\nabla_X\xi^a{}_{;b}\text{,}$$ where the starting identity $R^a{}_{bcd}\xi^d = \xi^a{}_{;bc}$ was used. The same identity also gives: $$R(\mathcal{L}_XY,\xi)^a{}_{b} = \nabla_{[X,Y]}\xi^a{}_{;b}\text{.}$$ Therefore, we have shown that for any vector fields $X,Y$, $$\begin{eqnarray*} X^cY^d(\mathcal{L}_\xi R^a{}_{bcd}) &=& \left[X^cY^d(\nabla_c\nabla_d-\nabla_d\nabla_c)-(\nabla_X\nabla_Y-\nabla_Y\nabla_X) + \nabla_{[X,Y]}\right]\xi^a{}_{;b}\\ &=& 0\text{.}\end{eqnarray*}$$ (If you have trouble with the last step, check Christoph's answer to the other question and modify appropriately.) Thus $\mathcal{L}_\xi R^a{}_{bcd} = 0$, QED.

general relativity - String theory and background independence

I have read that string theory assumes strings live in spacetime defined by general relativity which make the theory background dependent (although general relativity is a background independent theory). Background independence dictates that spacetime emerge from more fundamental ingredient than spacetime. Quoting Brian Greene, “Then, the theories ingredients - be they strings, branes, loops, or something else discovered in the course of further research - coalesced to produce a familiar, large-scale spacetime” (Greene, The Fabric of the Cosmos, 2004: 491). My question: why couldn’t spacetime just be spacetime, a fundamental entity? If so, string theory, based on general relativity, would not be a “great unsolved problem” facing string theory.

Answer

I think you might be slightly misinterpreting what string theorists believe their theory says about quantum spacetime.

Start with a classical spacetime with a certain pseudo-Riemannian metric, and consider the theory of strings propagating on this spacetime. There are two very important results:

1. Unless $R_{\mu \nu} = 0$, the theory acquires a non-zero beta function under conformal transformations. Afaik this is widely believed to be unphysical, because strings theorists don't know how to make sense of the theory when it isn't invariant under conformal transformations. Therefore we conclude that string theories exist only on Ricci-flat spacetimes, that is, on spacetimes solving the vacuum EFE.


2. In the spectrum of the string, you will find spin-2 states which have been conjectured to model physical gravitons. They have all properties that one would expect gravitons of General Relativity to have, except that their ultraviolet completion is finite.

The question then becomes – why did we have to choose the classical background in the first place? After all, isn't quantum gravity supposed to model classical spacetime as a certain limit of the full theory?

An interesting observation is that if you try to build a "coherent" state from string theory gravitons, the setup can be re-interpreted exactly as if the string was in the vacuum state, but propagating in a different spacetime.

So it can be conjectured that there's a concealed duality between the excited states of the string/superstring propagating in one classical spacetime, and an unexcited string propagating in another spacetime. Thus, maybe string theory is background-independent after all, even though the perturbative formulation that we've found isn't manifestly background independent?

Though afaik there's no background independent nonperturbative formulations of string theory known (I am not considering any of the AdS/CFT stuff here because it probably isn't directly related to the perturbative string/superstring theories and still remains just a conjecture).

From my personal correspondence with people working in the field, I conclude that there's no consensus on this subject, despite some individuals' belief that there is consensus :) I personally know people working on topics related to string theory, who are absolutely convinced that background independence is a must-have property of the non-perturbative definition of string theory, whatever it is. But I also know at least one senior lecturer who is perfectly satisfied with a special-relativistic flat spacetime entering the definition of the theory.

newtonian mechanics - What causes a force field to be "non-conservative?"

A conservative force field is one in which all that matters is that a particle goes from point A to point B. The time (or otherwise) path involved makes no difference.

Most force fields in physics are conservative (conservation laws of mass, energy, etc.). But in many other applications, the time paths DO matter, meaning that the force field is not "conservative."

What causes a force field to be "non-conservative?" Could you give some examples (probably outside of physics)?

Answer

A force field $F_i(x)$ is conservative if for every curve $C$ from a point $y_1$ to a point $y_2$, we have $\int\limits_C F_i(x)\mathrm{d}x^i$, so that the energy difference between $y_1$ and $y_2$ is independent of the curve taken from one to the other. Equivalently, the integral around a closed curve must be zero, $\oint\limits_C F_i(x)\mathrm{d}x^i=0$ for every closed curve $C$. Alternatively, we require $\nabla\times F=0$, so that we can write $F=\nabla V$; that is, the curl of the force field is zero so that the force field can be expressed as a divergence. Generalizations of this elementary account to higher dimensions in terms of differential forms are possible.

Although Shuhao Cao's comment that whether a physical theory is macroscopic or microscopic will determine whether the theory is conservative is very often correct, nonetheless phenomenological microscopic theories may find it convenient to include nonconservative force fields. For example, the effect of an externally imposed magnetic field on an otherwise microscopic model may be nonconservative. (see Ron's comment below, which points out that variation of an externally imposed magnetic field over time may be used to give an example of a nonconservative field in 4D. The implication of restriction to 3D that is established by my first paragraph has to be removed.)

electrostatics - How to avoid getting shocked by static electricity?

sometimes I get "charged" and the next thing I touch something that conducts electricity such as a person, a car, a motal door, etc I get shocked by static electricity.

I'm trying to avoid this so if I suspect being "charged" I try to touch something that does not conduct electricity (such as a wooden table) as soon as possible, in the belief that this will "uncharge me".

• Is it true that touching wood will uncharge you?


• How and when do I get charged? I noticed that it happens only in parts of the years, and after I get out of the car...

logical deduction - Hats and Aliens

Here's a neat little puzzle taken from a Google interview question:

10 humans are abducted by aliens; each represents 10% of the entire human population. The aliens give each abductee either a purple hat or a green hat. The 10 are lined up in a single file line, each facing forward, such that the last person can see the remaining 9's hats, the second to last person can see the remaining 8's hats and so on. No one can see his or her own hat.

The aliens then proceed, starting from the last person, to ask each of the abductees what the color of their hat is. If they guess correctly, they and the 10% of the human population they represent survives; if not, the opposite happens.

Assuming the abductees are given a chance to develop a strategy before they are lined up and questioned: what is the optimal strategy they can utilize (i.e. the one with the highest expected number of survivals)?

During the questioning, the abductees are not allowed to say anything besides their guess for the color of their hat when it is their turn.

---

Hint:

The optimal strategy will always ensure 9 survivals, and will have a 50% chance of the 10th survival.

Answer

I think this would work? Sorry if it's confusing. I sort of confused myself when I came to this conclusion.

The first guy would count the number of hats before him of a particular color... like, I dunno, I guess purple? And if the count of purple hats is odd, then he would say that his hat is purple. But if the count of purple hats is even, he'd say his hat was green.

So the next guy should then be able to determine the color of his hat by:

The remaining hats. So if the questioning went:
Alien: What color is your hat?

10th person: Purple.

Then the 9th person would look ahead and:

Count the number of purple hats. If it's an odd number, he has a green hat. If it's an even number, his hat is purple. So on and so forth. They just need to remember the "number" of purple hats.

But the first guy has no way of knowing his own hat color, so he's got a 50/50 chance of dying either way.

If anyone has trouble understanding this, a visual explanation can be found here:

https://www.youtube.com/watch?v=N5vJSNXPEwA

biology - Why are there limits on physics' applicability?

(Not sure if this is the right SE for this question).

Physics is great at predicting what inanimate stuff do. Why can't it also predict what living things do? For example, if I throw a ball off a cliff then I can predict that it'll move in a parabola. If I throw a person off a cliff I can also predict that he'll move in a parabola, but I can't predict if he'll land with his left hand pointing up or not, since he can choose to make my prediction wrong.

In principle, the motion of all particles that make up the person is deterministic (up to quantum uncertainty), which should imply that the motion of the person is also deterministic. But it isn't. The only thing I can think of that can create this uncertainty is quantum mechanics, but by Bell's theorem QM is fundamentally unpredictable. If QM is the root reason we can't predict living behavior, that should imply everything we do is random, e.g. "I choose to drive to work today, but halfway there I might choose to drive to a cinema instead", which is obviously not the case.

It seems clear to me that physics fundamentally fails to predict living beings. Hence the question: why?

EDIT: I don't consider this a duplicate of Are living organisms deterministic? because it assumes that living organisms are not deterministic, and asks why. The answers in that question also deal with "total" determinism, not determinism in the quantum sense, which is also addressed in this question.

quantum mechanics - What is the real interpretation of Planck's constant and what are its origins?

In the physics texts I have read and from other online information, I gather that Planck's constant is the quantum of action or that it is a constant specifying the ratio of the energy of a particle to its frequency. However, I'm still not understanding exactly what it is?

From other things I have read, I understand that Planck did a "fit" of data concerning others' experiments and came up with this value; exactly what other data exactly did he fit to arrive at this really small value? Or maybe he did it some other way? Perhaps an answer concerning its origins will help me understand my first question better?

Answer

In point particle classical mechanics, the action $S$ is the time integral of the Lagrangian $L$

$$S=\int Ldt$$

You can check its dimensions are of $[ML^2T^{-2}][T]=[ML^2T^{-1}]$ this is, energy times time. The constant ratio is due to the energy $E$ and frequency $\nu$ relation for photons:

$$E=h\nu \Rightarrow h=\frac{E}{\nu}$$

The "fit" that you are talking about comes of the blackbody radiation spectrum. If we use as variables temperature $T$ and frequency $\nu$ in classical physics we have two laws:

High frequency law: Wien's law $$I(\nu,T)=\frac{2h\nu^3}{c^2}e^{-\frac{h\nu}{kT}}$$

Low frequency law: Rayleigh-Jeans law $$I(\nu,T)=\frac{2h kT\nu^2}{c^2} $$

There is no intermediate frequency law. Planck assumed that radiative energy is quantized via $E=h\nu$ and interpolated the energy fitting for an expression of the type

$$I(\nu,T)=F(\nu,T)e^{g(\nu,T)}$$

that should satisfy both limits ($\nu \approx 0, h\nu >> kT$). Finally he obtained

$$I(\nu,T)=\frac{2h\nu^3}{c^2}\frac{1}{1-e^{\frac{h\nu}{kT}}} $$

However there is a much more nicer and physical derivation of Planck's law due to Einstein that you can find in Walter Greiner Quantum Mechanics an Introduction chapter 2

mathematics - Touching Matchsticks

You are asked to place matchsticks on a flat surface such that each matchstick end meets three others, and no matches cross. It is easy to achieve this for patterns that extend indefintely:

The challenge is to truncate such patterns to finite 2D networks. How small a matchstick network can you create?

Further clarifications: the matchsticks all have equal length and can be thought of as mathematical line segments. At each point of contact, exactly four ends meet. All matches lay flat on the surface, no gluing allowed!

Answer

I must admit that I found It with google but this is the solution:

homework and exercises - Inconsistency in Lagrangian vs Hamiltonian formalism?

Can both Lagrangian and Hamiltonian formalisms lead to different solutions?

I have a simple system described by the Lagrangian \begin{equation} L(\eta,\dot{\eta},\theta,\dot{\theta})=\eta\dot{\theta}+2\theta^2. \end{equation} The equations of motion are obtained from Euler-Lagrange eq.: \begin{eqnarray} 4\theta-\dot{\eta}=0\; \mathrm{and}\; \dot{\theta}=0, \end{eqnarray} yielding the solution $\eta(t)=4\theta_0t+\eta_0$ where $\eta_0$ and $\theta_0$ are constants.

But when I obtain one of the equations of motion from the Hamiltonian (via Legendre transformation), \begin{equation} H=\left(\frac{\partial L}{\partial\dot\eta}\right)\dot\eta+\left(\frac{\partial L}{\partial\dot\theta}\right)\dot\theta - L =-2\theta^2, \end{equation} \begin{equation} \dot\eta=\frac{\partial H}{\partial p_\eta}=0, \end{equation} the situation is surprisingly different from the Lagrangian approach because $\eta$ is now a constant!

Can someone give a proper explanation for this inconsistency? Am I doing something wrong here?

Answer

The problem here is that, because there exist constraints of the form $f(q,\,p)=0$, the phase space coordinates of the usual Hamiltonian formulation aren't independent. I'm not sure how you encountered this Lagrangian, but this issue is a common hiccup in electromagnetism and (if you'll pardon a more obscure example) BRST quantisation. The good news is you can still form a Hamiltonian description equivalent to the Lagrangian one. The trick is to append suitable terms to the "naïve" Hamiltonian, as explained here, and as a result the Poisson brackets are upgraded to what are called Dirac brackets.

For your problem the full Hamiltonian is $H=-2\theta^2+c_1 p_\eta+c_2( p_\theta-\eta)$, where the $c_i$ remain to be computed as functions of undifferentiated phase space coordinates. In fact $c_1=\frac{\partial H}{\partial p_\eta}=\dot{\eta}=4\theta$ while $c_2=\frac{\partial H}{\partial p_\theta}=\dot{\theta}=0$, so $H=-2\theta^2+4\theta p_\eta$. You can verify this gives you the right equations of motion.

newtonian mechanics - What is the relationship between force and momentum in collisions?

I know that $ \Sigma F = \Delta mv/\Delta t$. But if we had a marble that moves in a straight line at a constant velocity and colloids with another marble. Because of the law of conservation of momentum, the second marble now had the velocity of the first marble. But what is the force that the first marble applied one the second marble? The collision is almost instantaneous. Wouldn't that make the force in $ \Sigma F = \Delta mv/\Delta t $ insanely large because $ \Delta t $ is so small?

Answer

The force can be surprisingly large, but $\Delta t$ is not zero, and the force is not infinite.

Make some estimates: the duration of the collision is so short that our eyes and brain cannot perceive it. Make an estimate for an upper limit for the duration. (There's no right answer, but a lot of wrong answers. For example, I would think that a duration of 0.1 s would be perceivable, and "wrong". My upper limit should be smaller.)

From this you can get a lower limit on the force.

You can improve your estimate for $\Delta t$. You know the speed of the marble. You can make a guess at the size of the deformation of the ball that occurs during the collision. It's certainly less than one tenth of the radius. Probably less that 1/100 ... (Make your own estimate.) From there calculate a $\Delta t$.

Try this: make a crude model of the force generated when two satellites collided in 2009. The relative speed of the collision is known. The masses and sizes can be found or estimated. Calculate the duration of the collision and the force generated. (They did not collide head-on, so divide by 2 to crudely account for this. :) They also disintegrated. So there the analysis will have flaws. But it's an interesting exercise as long as you keep in mind that it is unrealistic. )

Is pair production only with $gamma$ photons?

in my revision guide the section on pair production only mentions it happening with gamma photons, so the question arose whether this is the only way it can happen?

This is what the book says: "Pair production is when a particle-antiparticle pair is produced from a single gamma photon. The gamma photon must have enough energy to produce that much mass. Pair-production usually happens near a nucleus, which helps to conserve momentum."

So does pair production only happen with gamma photons?

Thanks

Answer

Pair producing an electron and a positron requires an energy at least equal to their masses, $2m_ec^2$. This would just create them stationary, and you'd need an even greater energy to give them some momentum.

Since $m_e = 0.511$ MeV/c$^2$, the minimum energy required for pair production to occur is $\sim 1$ MeV.

What photons have an energy close to that value? Gamma rays.

Now, if you tried to do the maths of energy & momentum conservation, a single photon pair converting to an electron and a positron would never conserve both.

You can see it by noting that, for the outgoing electron+positron system, there exists a centre of mass frame where the total momentum is 0. But for the photon, there is no such thing since zero momentum would mean 0 energy ($E =pc$). The presence of a nucleus makes you have more 'elements' in the energy-momentum conservation calculations, which allows it to give/take some energy/momentum thus enabling the pair production process to occur.

quantum field theory - Why do negative norm states break unitarity?

I often hear my teachers say that the negative norm states break unitarity. And I can also read this elsewhere, such as at this place

In this gauge the relation between unitarity and gauge invariance is thus obvious: by breaking the gauge invariance we thus break the unitarity, since the negative norm states enter the game.

Whose unitarity is broken? How to break it?

Answer

I asked Mark Srednicki about this, and he told me that it's not really correct to say that negative-norm states break unitarity, because negative-norm states don't exist by the definition of the inner product. It's often a convenient calculational trick to formally expand your state space so that it's no longer a Hilbert space by adding in negative-norm ghosts, and the presence of physical states formally appearing to couple to ghosts indicates the presence of a quantum anomaly that prevents you from consistently quantizing your theory. But this is just a calculational trick to see the anomaly - the anomaly is real, the ghosts aren't.

In particular, you can always in principle see the existence of the anomaly without introducing ghosts. For example, the usual explanation for the fact that bosonic string theory can only be formulated in 26 dimensions is that that's the only number of dimensions in which the ghosts decouple. But we can alternatively work in light-cone gauge with only positive-norm states, and we find that only in 26 dimensions do the Lorentz generators close. This is another way to see the anomaly that doesn't require any mention of ghosts.

Mark also said that another reason it's incorrect to say that ghosts "break unitarity" is that they really just prevent you from consistently quantizing your theory at all - there's no reason to specifically single out unitarity as being broken.

electrostatics - How to determine whether a statically charged object is positively or negatively charged?

for example, I rub a ballon on carpet. What is an experiment I can do to tell which object is net positive charged and which is negative?

Answer

Find some material which you know takes up or gives away electrons. Bring it close to the object in question. If they are of like charge, they will come together, otherwise they will repel.

Take advantage of the Lorentz Force and move the object in question near a strong magnet. Use the Lorentz to figure out which way the particle ought to move, and observe which direction the object actually does move. This would be easiest, according to my opinion, if you ran it next to a wire which produces a magnetic field.

quantum mechanics - Fermi-Dirac distribution derivation?

I am trying to derive the Fermi-Dirac statistics using density matrix formalism. I know that

$$= Tr \rho A.$$

So I started from

$$= Tr \rho n(\epsilon_i)=\frac {1}{Z} \sum e^{-\beta \epsilon_i n_i}n_i=\frac {1}{Z} e^{-\beta \epsilon_i}. $$

In the last passage I used the pauli principle ($n_i=0,1$). Now to derive the correct Fermi-Dirac distribution I have to use for $Z=1 +e^{-\beta \epsilon_i}$. Why I have not to use the general form of

$$Z=\prod_i (1 +e^{-\beta \epsilon_i})~?$$

Can anybody give me a good explanation?

Answer

The derivation of the Fermi-Dirac distribution using the density matrix formalism proceeds as follows:

The setup.

We assume that the single-particle hamiltonian has a discrete spectrum, so the single-particle energy eigenstates are labeled by an index $i$ which runs over some finite or countably infinite index set $I$. A basis for the Hilbert space of the system is the occupation number basis \begin{align} |\mathbf n\rangle = |n_0, n_1, \dots\rangle \end{align} where $n_i$ denotes the number of particles occupying the single-particle energy eigenstate $i$. For a system of non-interacting identical fermions, the set $\mathscr N_-$ of admissible occupation sequences $\mathbf n$ consists of those sequences with each $n_i$ equal to either $0$ or $1$. Let $H$ be the hamiltonian for such a system, and let $N$ be the number operator, then we have \begin{align} H|\mathbf n\rangle = \left(\sum_{i\in I}n_i\epsilon_i\right)|\mathbf n\rangle, \qquad N|\mathbf n\rangle = \left(\sum_{i\in I} n_i\right) |\mathbf n\rangle \end{align} where $\epsilon_i$ is the energy of eigenstate $i$. We can also define an observable $N_i$ which tells us the occupation number of the $i^\mathrm{th}$ single-particle energy state; \begin{align} N_i|\mathbf n\rangle = n_i|\mathbf n\rangle \end{align}

Note that we are attempting to determine the ensemble average occupation number of the $j^\mathrm{th}$ energy eigenstate. In the density matrix formalism, this is given by \begin{align} \langle n_j\rangle =\mathrm{tr}(\rho N_i) \end{align} where \begin{align} \rho = \frac{e^{-\beta(H-\mu N)}}{Z}, \qquad Z = \mathrm {tr}\big(e^{-\beta(H-\mu N)}\big) \end{align}

The proof.

1. Show that \begin{align} Z = \sum_{\mathbf n\in \mathscr N_-}\prod_{i\in I}x_i^{n_i} \end{align} where $x_j = e^{-\beta(\epsilon_j-\mu)}$, the sum is over admissible sequences $\mathbf n$ of occupation numbers of single-particle energy states, and the product is over indices $i$ labeling an orthonormal basis of single particle energy eigenstates.


2. Show that the ensemble average occupation number of the $j^\mathrm{th}$ state can be computed as follows: \begin{align} \langle n_j\rangle = x_j\frac{\partial}{\partial x_j}\ln Z \end{align}


3. Show that the product and the sum in the partition function can be "exchanged" to give \begin{align} Z = \prod_{i\in I}\sum_{n=0}^1 x_i^n \end{align} where the product is now over single-particle energy eigenstates, and the sum is over admissible occupation numbers of a single-particle state.



4. Combine the results of steps 2 and 3 to show that \begin{align} \langle n_j\rangle = \frac{1}{e^{\beta(\epsilon_j-\mu)}+1} \end{align} which is the desired result.

electrostatics - Electric field lines?

Obviously, electric field lines for 2 charges, one stronger than the other, behave like this: https://www4.uwsp.edu/physastr/kmenning/images/pop4.19.f.19.gif. Is there a way to measure angles for particular field lines? For example, the field line at theta=0 is just a straight line going from the positive charge to the negative one. Similarly, theta= 180 goes from the positive charge to negative infinity. Are there similar rules for some arbitrary value of theta? Do any theta values do something interesting besides theta=0 and 180?

lateral thinking - Sum other numbers

Begin with a flagrantly erroneous summation and a woefully vacant substitution table.

           234

         +   5                   Digit        2    3    4    5    6    7    8
        -------           Substitute digit    _    _    _    _    _    _    _
          5678

How can the substitution table be filled out to correct this summation?

This is almost too easy if you just follow these guidelines.

• 
  
  

  Assign 7 unique substitute digits from 0 through 9 for digits 2 through 8 in the table (one digit per digit)

  
  


• 
  

  Replace digits in the summation by their substitutes in the table (no other kinds of edits, as the summation and table should be taken at face value)

  
  


• 
  

  All numbers and digits are decimal (no notation tricks are involved)

  
  


• 
  

  No leading zeros in the total or either summand

  
  
  


• 
  

  The summation has a unique solution

  
  

Added:   Regular arithmetic pretty much forces the resultant summation. lateral-thinking allows the guidelines to attain it.

Answer

Making an assumption:-

That if a substitute digit is itself in the lookup table, it will be replaced again.

 Digit               2    3    4    5    6    7    8
 Substitute digit    3    4    9    1    7    8    0 

The Summation becomes:

999 + 1 = 1000 because:
2->3->4->9,
3->4->9,
4->9,
5->1,
5->1,

6->7->8->0,
7->8->0,
8->0

Process:

As the question states, if you follow the guidelines, it should lead you towards the answer

First, as mentioned in the question, there is one possible summation. It must be 999 + 1 = 1000 as a 3 digit number plus a 1 digit number must equal a 4 digit number, and the first digit of the 4 digit number has to be the same as the 1 digit number.

Then, knowing that 6,7,8 must equal 0 we can first assign any one of those digits the substitute digit of zero, lets choose 8.

Since 0 is now used (and the question states the substitute digits must be unique) in order for 6 or 7 to equal 0, the only substitute digit we can assign is 8 (since 8 = 0).

This same logic is then applied to 2,3,4 since they all need to equal 9

visible light - Why do diamonds shine?

I have always wondered why diamonds shine. Can anyone tell me why?

Answer

Diamond is one of the hardest material. We know that it's an allotrope of carbon. A diamond (crystalline in nature) has a three dimensional arrangement of carbon atoms linked to each other by strong covalent bonds. What you've shown a round brilliant cut diamond.

Actually, the secret that's rattling inside a diamond is refraction, total internal reflection (not to be confused with ordinary reflection) & dispersion. The refractive index of diamond is pretty high (2.417) and is also dispersive (coefficient is 0.044). Due to this fact, diamond is an important application in optics.

Consider an ideal cut diamond. I explain according to the figure below. When the light is incident at an angle $1$, it refracts inside and travels through the lattice. At the surface which separates air & diamond media, the incident angle $2$ is very well above the critical angle ($c_a$) and simultaneously ($3$ & $4$) the reflection takes place at different surfaces of the diamond. Finally, the light refracts out.

The first one shows the mechanism of internal dispersive reflection. The second figure shows the reflections inside ideal cut, deep and shallow cut diamonds.

                                      

            

Note: For total internal reflection to take place, light must travel from an optically denser medium to a relatively rarer medium. Also, the incident angle should be far high above thee critical angle.

There are youtube goodies regarding the topic...