How to determine the direction of a wave propagation?

In the textbook, it said a wave in the form $y(x, t) = A\cos(\omega t + \beta x + \varphi)$ propagates along negative $x$ direction and $y(x, t) = A\cos(\omega t - \beta x + \varphi)$ propagates along positive $x$ direction. This statement looks really confusing because when it says the wave is propagating along $\pm$ x direction, to my understanding, we can drop the time term and ignore the initial phase $\varphi$ while analyzing the direction, i.e. $y(x, 0) = A\cos(\pm\beta x)$, however, because of the symmetry of the cosine function, $\cos(\beta x)\equiv \cos(-\beta x)$, so how can we determine the direction of propagation from that?

I know my reasoning must be incorrect but I don't know how to determine the direction. So if we don't go over the math, how to figure out the direction of propagation from the physical point of view? Why $-\beta x$ corresponding to the propagation on positive x direction but not the opposite?

Answer

For a particular section of the wave which is moving in any direction, the phase must be constant. So, if the equation says $y(x,t) = A\cos(\omega t + \beta x + \phi)$, the term inside the cosine must be constant. Hence, if time increases, $x$ must decrease to make that happen. That makes the location of the section of wave in consideration and the wave move in negative direction.

Opposite of above happens when the equation says $y(x,t) = A\cos(\omega t - \beta x + \phi)$. If t increase, $x$ must increase to make up for it. That makes a wave moving in positive direction.

The basic idea:For a moving wave, you consider a particular part of it, it moves. This means that the same $y$ would be found at other $x$ for other $t$, and if you change $t$, you need to change $x$ accordingly.

Hope that helps!

electromagnetism - Why does a dielectric have a frequency dependent resistivity?

This question has come about because of my discussion with Steve B in the link below.

Related: Why is glass much more transparent than water?

For conductors, I can clearly see how resistivity $\rho\,\,(=1/\sigma)$ can depend on frequency from Ohm’s law, $\mathbf{J}=\sigma\mathbf{E}$. So if the E-field is an electromagnetic wave impinging on a conductor, clearly the resistivity is frequency dependent. In a similar fashion, the frequency dependence of the electric permittivity $\epsilon=\epsilon_0n^2(\omega)$ can be derived through the frequency dependence of the electric polarization and impinging electromagnetic wave (see How Does $\epsilon$ Relate to the Dampened Harmonic Motion of Electrons?).

1. 
   

   What does it mean physically for a dielectric to have a frequency dependent resistivity from (i) classical and (ii) quantum viewpoints? I am especially interested in the optical frequency range.

   
   


2. 
   

   Can a simple mathematical relationship be derived similar to the frequency dependent resistivity (for conductors) and electric permittivity (for dielectrics)?

   
   

Thank you in advance for any help on this question

Answer

A simple model that explains the frequency dependency of the resistivity of metals reasonably well is the Drude model (http://en.wikipedia.org/wiki/Drude_model). There we have frequency dependency because the electrons in a plasma are not moving arbitrarily fast, which is consistent with Xurtio's explanation. The cutoff frequencies are usually in the optical domain. For dielectrics similar models exist, which are often a sum of Lorentzian resonances. These have their origin in resonant absorption which is a quantum physical effect.

The imaginary part of the permittivity is related to the conductivity. This can be seen as follows: Amperes law is

$\nabla \times \mathbf{H} = \mathbf{J} +i \omega \epsilon_r \epsilon_0 \mathbf E$

and insert Ohms law in differential form

$\mathbf{J} = \sigma \mathbf{E}$

then you get

$\nabla \times \mathbf{H} = i \omega (\epsilon_r \epsilon_0 -i \sigma/\omega) \mathbf E$

which is just of the same form of as original form of amperes law but without the explicit $\mathbf{J}$ term. In conclusion Ohms law can be integrated in free space Maxwells equations (without the source terms) when the relative permittivity $\epsilon_r$ is taken as a complex value ($\widetilde\epsilon_r = \epsilon_r - i \sigma/(\omega \epsilon_0)$), where an imaginary part is added related to the conductivity. This essentially models the effect of moving charges under the influence of an oscillating field (light).

So the relation between polarization ($\mathbf D = \widetilde{\epsilon}_r \epsilon_0 \mathbf E = \mathbf P + \epsilon_0 \mathbf E$) and conductivity $\sigma$ is given as

$\mathbf{P} = \epsilon_0 (\epsilon_r - i \sigma/\omega - 1) \mathbf E$.

Since the real part of the permittivity is frequency dependent, so is the conductivity. This is because of the Kramers-Kronig relations which follow from a causality relation.

angular momentum - How much effort would be required to fix the Earth's rotation?

Given that the earth's rotation has been slowing down by very slight amounts over time, forcing us to introduce leap seconds and so forth into our clocks and calendars, I would like to ask if this could be fixed by "generating" more spin via some sort of power plant like structure(s) with a massive spinning object. (would it be ideal for these to be on the equator?)

How much energy would be needed in order to change the length of a year by 1 second?

what about in order to eliminate February 29th such that we don't need a "leap day" anymore? (24 hours over 4 years) - note that for this, the spin would need to be altered such that the day is removed, but then returned to near its original state such that we don't keep losing days

related Q/A: the earth IS slowing down (rotationally)

Answer

A day is currently about 86400.002 seconds long. If we could just increase the Earth's rotation rate by a mere 2 milliseconds per day we would get rid of the need for those pesky leap seconds. No problem! We only need something that rotates with an angular momentum of 1.4×10²⁶ joule-seconds about an axis pointing due south.

One way to do this would be to build a train track around the Earth at the equator. I'll assume a 20 meter long train car with a gross mass of 150,000 kg moving at bullet train speeds, 320 km/hour. There's room for about two million cars on this track. That gets us to 1.7×10²⁰ joule-seconds. We would only need 800,000 such circum-equatorial trains. Alternatively, we would only need one such train if we could make the train move at 0.23 c.

Another approach would be to place a large rotating disc at the South Pole. For example, a uranium disc with a radius of 20 kilometers and a height of 28 meters rotating at 10,000 RPM will just about do it.

In other words, it can't be done.

strategy - Five Card Magic Trick with $N$ card Deck

Last week, I saw a magic show, and was the volunteer for climactic act. The magician introduced this act by talking about the Fitch-Cheney card trick, but then explained that she was going to do an even more impressive trick. Namely, instead of using a $52$ card deck, she would use a larger deck, with $N$ cards, though she didn't say exactly how big $N$ was.

Here's what happened:

The magician left the room. An assistant asked me to look through the $N$ card deck, and pick out any $5$ cards. The assistant inspected the cards I chose, then gave one back to me, which I put in my pocket. He arranged the remaining four cards in a neat stack, and placed this stack face down on a table. The magician returned, inspected the stack, and successfully guessed the card in my pocket.

What is the largest value of $N$ for which the trick can be performed? For that $N$, how is it done?

Note that there is no slight of hand or secret communication. Each card is distinct, the faces are rotationally symmetric, and the stack is placed at the exact same spot on the table every time the trick is performed. The assistant is only allowed to choose which card the volunteer gets ($5$ choices), then permute the four other cards ($4!=24$ choices).

---

^{Aside: From this question, we know $N=53$ is possible.}

Answer

Stealing from 2012rcampion's answer, we need to create a one-to-one mapping from five-card hands to four-card messages (four cards plus permutation). By symmetry, each four-card message can be mapped to an equal number of five-card hands, so by the regular version of Hall's Theorem there must exist such a one-to-one mapping (this is laid out in more detail in GentlePurpleRain's comment link).

Thus, the trouble is actually constructing a matching, and ideally doing so in a way that would work well as a magic trick. The following works for the former, and you can decide if it works for the latter.

---

Suppose the cards are numbered 0 to 123. Call the five cards, from smallest to largest, $c_0$, $c_1$, $c_2$, $c_3$, and $c_4$. To find the pocketed card, the assistant adds up the numbers on all five cards, reduces the answer modulo $5$ to get $S$, and then hands back $c_S$.

Now the magician can see four cards. Suppose these four cards sum to $S'$ modulo $5$. Suppose also she mentally removes these four cards from the deck and then relabels the remaining cards in order from $0$ to $119$ (call these "reduced numbers"). Then we have

the missing card must have a reduced number equivalent to $(-S')$ modulo $5$*.

This means there are only $24$ possibilities, and hence with a code that takes permutations of four cards to the set $\{1, 2, ..., 24\}$, the assistant can supply enough information for the magician to identify the card.

---

*This is a bit hard to see, so let's break it down into cases. If $c_0$ was removed, that means the sum of the five cards is $0$ mod $5$, and hence the missing card must be $-S'$ mod $5$ to counter the $S'$ sum of the other four cards. If $c_1$ was removed, then the missing card must have had a number of $-S' + 1$, but since it appears after $c_0$ that we mentally removed from the deck, it's reduced number is $-S' + 1 - 1 = -S'$. Similar logic shows this works for all five cases.

schroedinger equation - Classical limit of a quantum system

If we have a one dimensional system where the potential

$$V~=~\begin{cases}\infty & |x|\geq d, \\ a\delta(x) &|x|

where $a,d >0$ are positive constants, what then is the corresponding classical case -- the approximate classical case when the quantum number is large/energy is high?

Answer

Here we derive the bound state spectrum from scratch. Not surprisingly, the conclusion is that the Dirac delta potential doesn't matter in the semi-classical continuum limit, in accordance with Spot's answer.

The time-independent Schrödinger equation reads for positive $E>0$,

$$ -\frac{\hbar^2}{2m}\psi^{\prime\prime}(x) ~=~ (E-V(x))\psi(x), \qquad V(x)~:=~V_0\delta(x)+\infty \theta(|x|-d), \qquad V_0~>~0, $$

with the convention that $0\cdot \infty=0$. Define

$$v(x) ~:=~ \frac{2mV(x)}{\hbar^2}, \qquad e~:=~\frac{2mE}{\hbar^2}~>~0 \qquad k~:=~\sqrt{e}~>~0\qquad v_0 ~:=~ \frac{2mV_0}{\hbar^2}. $$

Then

$$ \psi^{\prime\prime}(x) ~=~ (v(x)-e)\psi(x). $$

We know that the wave function $\psi$ is continuous with boundary conditions

$$\psi(x)~=0 \qquad {\rm for}\qquad |x|\geq d.$$

Also the derivative $\psi^{\prime}$ is continuous for $0<|x|

$${\lim}_{\epsilon\to 0^+}[\psi^{\prime}(x)]^{x=\epsilon}_{x=-\epsilon} ~=~v_0\psi(x=0). $$

We get $$\psi_{\pm}(x)~=~A_{\pm}\sin(k(x\mp d))\qquad {\rm for } \qquad 0 \leq \pm x \leq d.$$

1. 
   

   $\underline{\text{Case} ~\psi(x=0)=0}$. Then $$n~:=~\frac{kd}{\pi}~\in~ \mathbb{N}.$$ We get an odd wave function $$\psi_n(x)~\propto~\sin(kx).$$ In particularly, the odd wave functions do not feel the presence of the Dirac delta potential.

   
   


2. 
   

   $\underline{\text{Case} ~\psi(x=0)\neq 0}$. Then continuity at $x=0$ implies that the wave function is even $A_{+}+A_{-}=0$. Phrased equivalently, $$\psi(x)~=~A\sin(k(|x|-d)).$$ The kink condition at $x=0$ becomes $$ v_0A\sin(-kd)~=~2kA \cos(kd), $$ or equivalently, $$ v_0\tan(kd)~=~-2k.$$ In the semiclassical continuum limit $$k \gg \frac{1}{d}, \qquad k \gg v_0,$$ this becomes $$\frac{kd}{\pi}+\frac{1}{2}~\in ~\mathbb{Z}, $$ i.e., in the semiclassical continuum limit the even wave functions do not feel the presence of the Dirac delta potential as well.

   
   

homework and exercises - Why can velocity and acceleration be negative?

Why is speed and acceleration negative when $V_1$ of an object is say 150m/s, $V_2$ is 0 m/s and $\Delta d=0.50\,\rm m$? I found the time it takes which is 0.0033s and the acceleration to be 90909.09 m/s$^2$.

My questions:

1. What do negative velocity and negative acceleration mean?


2. Is it negative because it is slowing down?


3. Why is the acceleration so big? Is that even reasonable?

physics - Fisherman boat and sea level

A fisherman rowing his boat on a very small lake throws his anchor into the water. Does the water level of the lake rise, fall or stay the same?

riddle - Dungeon Crawler: Level 2.5

_{You are the protagonist in a game designed by NeedAName Inc. Proceed through the levels using your video game knowledge to solve the puzzles therein. Solve all levels and beat the game!}

---

Level 2.5: A Poetic Riddle

After a weird glitch in Level 2, you find yourself transported to Level 2.5 . You wonder what happened in that weird level, but find that you don't want to dwell on it; the game developers probably should've hired beta testers. Moving on, you find a riddle scrawled on the walls:

One kills with range or mythic pet
No passive vet this fighter yet.

Another kills by proxy host

A frightful host he proudly boasts.

One bashes skulls with just one fist
A brute, for this does he exist.

A shielded elemental mage
Dare not enrage this spellwise sage.

A righteous man, this holy knight
He conquers night with hallowed light.

To shadows this one is confined
Both flesh and mind be struck in kind.

The last a man of varied shapes

The beasts he tames, their forms he drapes.

Based off the riddle, you recognize the game and enter it's name into the keypad at the door. What game is it?

Previous Level Next Level

Answer

Sounds like Diablo 2

One kills with range or mythic pet
No passive vet this fighter yet.

Amazon

Another kills by proxy host
A frightful host he proudly boasts.

Necromancer - has undead minions

One bashes skulls with just one fist
A brute, for this does he exist.

Barbarian smash

A shielded elemental mage
Dare not enrage this spellwise sage.

Sorcerer

A righteous man, this holy knight
He conquers night with hallowed light.

Paladin

To shadows this one is confined
Both flesh and mind be struck in kind.

Assassin

The last a man of varied shapes
The beasts he tames, their forms he drapes.

Druid

Proofs that Earth is not flat?

Well, I do not believe that Earth is flat, but I met some conspirationistas believing that all physics we learn in school is twisted, all the Earth's pictures from space are photoshopped by NASA, etc, etc.

So those guys wanted to put a satellite into orbit, take some pics, and see for themselves.

Of course, a satellite won't stay above a flat Earth, but what other very simple means to prove/check that earth is round (well, geoid) do we have?

I am looking for simple means, available to anyone, that do not need any external tool or knowledge that a naysayer would reject.

Some things I've considered:

• 
  

  Meteorological balloon with camera and a phone with GPS, with a parachute to recover it. It has been done by amateurs, and in good conditions, you can see the Earth's curvature.

  
  


• 
  

  Measure the Sun's height above horizon at noon at two different latitudes. The difference, along with some simple geometry would give you Earth's diameter.

  
  

Answer

Watch a boat through binoculars as it sails away from you. Much easier.

optics - Why does electromagnetic refraction happen?

Why does refraction happen? In high school textbooks, it is stated that it happens because the speed that electromagnetic waves propagate in the media change. But why cant they continue propagating in the same direction with a different speed?

quantum mechanics - Is Schrödinger’s cat misleading? And what would happen if Planck constant is bigger?

Schrödinger’s cat, the thought experiment, makes it seem like as if measurement can cause a system to stop being in a superposition of states and become either one of the states (collapsed).

So does a system always exist in a superposition? In this sense, do things in the macroscopic world have a wavefunction? Is it because of the size of the everyday object, so it won't behave so much like an electron? Theoretically, if the Planck constant is to be bigger, everyday object would start behaving more like particles in a quantum scale?

Answer

In principle everything exists in a superposition of states. However everything interacts with it's environment, and this collapses the superposition through a mechanism called decoherence. In the particular case of Schrodinger's cat there is a brief discussion of this here. Also search this site for decoherence as there have been lots of questions about it.

For any system, like the cat, there is an associated decoherence time that tells you approximately how long it takes the superposition to collapse. For objects on the atomic scale this can be quite long, but for macroscopic objects the decoherence time is extremely short and we can never observe these objects to be in a superposition of states.

Someone has actually done an experiment to prepare a virus in a superposition of states . Because a virus is so much smaller than a cat the superposition can be observed. Luboš Motl has written a fascinating article about this on his blog.

logical deduction - Mate in one with NO PIECES?

The following pictures depict chess positions one move before white checkmated black:

The problem is, I seemed to have completely lost which pieces each letter/number represents. All I know is that the number substitution of the pieces is consistent across the board and that

1-7 include {2 pawns, 1 knight, 1 bishop, 1 rook, 1 queen, and 1 king}

A-F include {1 pawn, 1 knight, 1 bishop, 1 rook, 1 queen, and 1 king}.

Text boards:

-------A
E-------
-----4--
----CB3-
------26
------D-
--------

-------7

---B----
-16--F--
---3-E--
-A--4---
---7----
------D-
--------
--------


-3------
574-----
--------
--------
--------
--------
--------
---C-D--

So... what piece does each character represent?

---

Took inspiration from the snakes and ladders retrograde analysis puzzle and decided to make a non-retrograde analysis chess puzzle (which doesn't count in the fortnightly topic challenge :( )

Answer

The #3 looks like a

back rank mate, so 3 is the black king. C is missing from #2, so that can't be the king, which means that the white pieces in #3 are D=king and C=queen/rook. For the position to be a mate in 1, 4 needs to be a knight/pawn and 7 a rook/pawn.

The pieces in #1 and #2

can't be attacking the opposing king, so
- B can't be a queen or a rook (#1)
- E can't be a queen or a rook (#2)
- F can't be a knight (#1)
- 2 can't be a queen or a rook (#1)
- 4 can't be a bishop or a queen (#2)
- 6 can't be a bishop or a pawn (#1)
- 7 can't be a knight (#1)

Pawns can't be

on the 1st/8th rank, so A, B and 7 can't be pawns.

In #1 the mating move looks like

Qh6 by A=queen, protected by B=knight.

So the white pieces are

A queen, B knight, C rook, D king, with E and F being a bishop or a pawn.

Substituting the known pieces in #2,

Having F be a pawn would make it possible to mate with f8=Q (or f8=B for extra pizzazz), with all escape squares covered by white pieces.

Now we just give the black pieces some combination which makes the #2 position a checkmate after that move.

1 pawn, 2 bishop, 3 king, 4 pawn, 5 queen, 6 knight, 7 rook.
A queen, B knight, C rook, D king, E bishop, F pawn.

The boards with all pieces:

symmetry - Can Noether's theorem be understood intuitively?

Noether's theorem is one of those surprisingly clear results of mathematical calculations, for which I am inclined to think that some kind of intuitive understanding should or must be possible. However I don't know of any, do you?

Independence of time <=> energy conservation.
Independence of position <=> momentum conservation.
Independence of direction <=> angular momentum conservation.

I know that the mathematics leads in the direction of Lie-algebra and such but I would like to discuss whether this theorem can be understood from a non-mathematical point of view also.

Answer

It's intuitively clear that the energy most accurately describes how much the state of the system is changing with time. So if the laws of physics don't depend on time, then the amount how much the state of the system changes with time has to be conserved because it's still changing in the same way.

In the same way, and perhaps even more intuitively, if the laws don't depend on position, you may hit the objects, and hit them a little bit more, and so on. The momentum measures how much the objects depend on space, so if the laws themselves don't depend on the position on space, the momentum has to be conserved.

The angular momentum with respect to an axis is determining how much the state changes if you rotate it around the axis - how much it depends on the angle (therefore "angular" in the name). So the symmetry is linked to the conservation law once again.

If your intuition doesn't find the comments intuitive enough, maybe you should train your intuition because your current intuition apparently misses the most important properties of time, space, angles, energy, momentum, and angular momentum. ;-)

quantum gravity - Why does Stephen Hawking say black holes don't exist?

Recently, I read in the journal Nature that Stephen Hawking wrote a paper claiming that black holes do not exist. How is this possible? Please explain it to me because I didn't understand what he said.

References:

1. 
   

   Article in Nature News: Stephen Hawking: 'There are no black holes' (Zeeya Merali, January 24, 2014).

   
   


2. 
   
   

   S. Hawking, Information Preservation and Weather Forecasting for Black Holes, arXiv:1401.5761.

   
   

Answer

The paper by Dr. Stephen Hawking doesn't say that black holes don't exist. What he says is that black holes can exist without "event horizons". To understand what an event horizon is, we first have to understand what is meant by escape velocity. This last one is the speed you need to escape a body. Now, here is where the event horizon and the escape velocity comes in play: the event horizon is the boundary between where the speed needed to escape a black hole is less than that of light, and where the speed needed to escape a black hole is greater than the speed of light.

So Hawking says that instead of event horizon, there may be "apparent horizons" that would hold light and information only temporarily before releasing them back into space in a "garbled form".

classical mechanics - Why does the kinetic energy of a particle moving in circular motion increase when the turn radius decreases and no torque is acting?

Why does the kinetic energy of a particle moving in circular motion increase when the turn radius decreases and there is no torque acting? E.g. if a planet is rotating about its axis and it shrinks to half of its radius then according to conservation of angular momentum velocity will doubled, i.e. its kinetic energy will increase. Does this mean that without any external force the kinetic energy of any thing can be increased? Please explain!

Answer

Does this mean that without any external force the kinetic energy of any thing can be increased?

No, of course not. Just because there is no torque that doesn't mean that there isn't a force. In this particular case, you have an inwards radial force that is performing work by countering and exceeding the centrifugal force. This work is what causes the increase in kinetic energy.

statistical mechanics - How to obtain the Bose-Einstein distribution from the canonical ensemble?

In the wikipedia page

https://en.wikipedia.org/wiki/Fermi%E2%80%93Dirac_statistics#Canonical_ensemble

the Fermi-Dirac distribution is obtained from the canonical ensemble in the following manner:

The average occupation number is given by

\begin{equation} \bar{n}_{i}=\frac{\displaystyle \sum_{n_1,n_2,...}n_{i}\,e^{-\beta(n_1E_1+n_2E_2+...)}}{\displaystyle \sum_{n_1,n_2,...}e^{-\beta(n_1E_1+n_2E_2+...)}} \end{equation}

where the summation is over the sets of $n_r$ that satisfy $\sum\,n_r=N$, being $N$ the total number of particles in the system. The derivation followed in Wikipedia then separates the state of energy $E_i$ and calls

\begin{equation} \bar{n}_{i}=\frac{\displaystyle \sum_{n_i} n_{i}\,e^{-\beta\, n_iE_i} \, Z_i(N-n_i)}{\displaystyle \sum_{n_i}\,e^{-\beta\, n_iE_i} \, Z_i(N-n_i)} \end{equation}

where $Z_{i}(N-n_i) = \sum_{n_1,n_n,...} \,e^{-\beta\,(n_1E_1+n_2E_2+...)}$. The summation here does not consider the state $n_i$. Since for the Fermi-Dirac case $n_i=0,1$, it is possible to write the summation easily (it has only two terms) to obtain

\begin{equation} \bar{n}_{i}=\frac{\displaystyle 1}{\displaystyle \bigg[Z_i(N)/Z_i(N-1)\bigg]\,e^{-\beta\,E_i} +1} \end{equation}

which will, in turn, result in the Dirac-Fermi distribution once the ratio between the partition functions is related to the chemical potential.

Now, I'm trying to follow the same procedure to obtain the Bose-Einstein distribution but in this case $n_i$ can vary between $0$ and $\infty$ and the summation cannot be done in any direct way I could think of. How can I approach this?

Thank you very much.

Answer

The expression: $$\begin{equation} \bar{n}_{i}=\frac{\displaystyle \sum_{n_i} n_{i}\,e^{-\beta\, n_iE_i} \, Z_i(N-n_i)}{\displaystyle \sum_{n_i}\,e^{-\beta\, n_iE_i} \, Z_i(N-n_i)} \end{equation}$$ can be rewrite in a usual for statphysics manner: $$\bar{n}_{i}=-\frac{1}{\beta}\frac{\partial\ Ln {\ Z(N)}}{\partial E_i}$$ We denoted denominator as $Z(N)$ -canonical partition function of the system (do not confuse it with $Z_i(N)$ - partition function with "punctured" i-th one particle state. Then, denoting $z_i=e^{ \beta \mu}$ -"punctured" fugacity, we can write: $$Z(N)=\sum_{n_i=0}^ {n_i=N}\ e^{-\beta\, n_iE_i} \, Z_i(N-n_i) = \sum_{n_i=0}^ {n_i=N}\ (e^{-\beta\,E_i})^{n_i} (\frac{1}{z_i})^{N-n_i} =$$ $$=z_i^{-N}\sum_{n_i=0}^ {n_i=N}\ (z_ie^{-\beta\,E_i})^{n_i} $$ Then, sequentially we write down the sum formulae for a finite geometric series, take the logarithm, tend $N$ to infinity, and differentiate by $E_i$. Finally, acting in the wiki spirit, we replace $\mu_i$ =$\mu$

PS. After I wrote this answer, I looked at the link in Wikipedia, it turns out the derivation of the B-E distribution by a similar method is in the same link ((Reif 1965) on p. 342

Can diffraction bend light more than ninety degrees?

Is it possible for light to be diffracted more than ninety degrees?

When light diffracts through a narrow slit or pinhole, it travels in all forward directions. The width of the slit determines how much the light will cancel itself.

Light travels from each point in the slit to everywhere. Light from different points in the slit must travel different distances. So it will be in different phases. At some spots all the light cancels with light in different phase. At other spots only some of it cancels.

Obviously you can't detect light that is bent more than 90 degrees, because the flat wall the slit is in would absorb it if there was any such thing.

What if the wall wasn't flat? What if we had, say, two razor blades that make the slit between them, and they are turned at 90 degrees to each other.

Is there theoretical reason to say that light cannot diffract backward?

Has the experiment been done?

special relativity - Deriving the geodesic equation using a Lagrange multiplier to fix affine parametrisation

The geodesic equation can be derived using the action $$S_0 ~=~ \int d\tau \sqrt{-\dot{x}_\mu\cdot \dot{x}^\mu}\tag{1}$$ (I am using the (-+++) convention and $\dot{x} = \frac{dx}{d\tau}$). To simplify calculations one then chooses an explicit parametrization namely the arc length $\tau$ i.e. $$\dot{x}_\mu\dot{x}^\mu = -1.\tag{2}$$ From my point of view this means that I minimize the action with the constraint: $$\dot{x}_\mu\dot{x}^\mu + 1 = 0.\tag{3}$$ So the resulting equation should be the same if I use the following action instead $$S = \int d\tau \sqrt{-\dot{x}_\mu\cdot \dot{x}^\mu} + \lambda(\dot{x}_\mu\dot{x}^\mu + 1)\tag{4}$$ where $\lambda$ is a Lagrange multiplier.

Let's find the eom in Minkowski space: $$0 = \dot{p}_\mu = \frac{d}{d\tau}\left(\frac{-\dot{x}_\mu}{\sqrt{-\dot{x}_\mu\dot{x}^\mu}} + 2\lambda\dot{x}_\mu\right)\tag{5}$$ $$\dot{x}_\mu\dot{x}^\mu + 1 = 0.\tag{6}$$

The square root in the first equation equals 1. So $$p_\mu = (2\lambda - 1)\dot{x}_\mu.\tag{7}$$ From the second equation I find $$\ddot{x}^\mu \dot{x}_\mu = 0.$$ Using this $$\frac{d}{d\tau} \dot{x}^\mu p_\mu = \ddot{x}^\mu p_\mu + \dot{x}^\mu \dot{p}_\mu = 0.\tag{8}$$ So $$\mathrm{const.} = \dot{x}^\mu p_\mu = 1-2\lambda \Rightarrow \dot{\lambda} = 0\tag{9}$$ Putting all together yields: $$ \dot{p}_\mu = (2\lambda - 1) \ddot{x}_\mu = 0.\tag{10}$$

In the case $\lambda \neq \frac{1}{2}$ this simply gives the old eom $\ddot{x} = 0$. However in the case $\lambda = \frac{1}{2}$ there is no restriction to $\ddot{x}$.

I don't understand where this case $\lambda = \frac{1}{2}$ comes from. How do I deal with it? Can I simply neglect it? Or have I forgotten something?

Answer

1. 
   

   First of all, we should stress that what OP calls $\tau$ is not$^{\dagger}$ proper time off-shell but just some world-line (WL) parameterization. However, the constraint $$\dot{x}_\mu\dot{x}^\mu ~\approx ~-1\tag{A}$$ will imply that the WL parameter $\tau$ is the proper time on-shell.

   
   


2. 
   

   Since the EOM depends on the first derivative $\frac{d\lambda}{d\tau}$ of the Lagrange multiplier, we should specify a single condition, e.g. an inertial condition (IC) for $\lambda$. If we choose the IC different from $1/2$, we avoid the problem when $\lambda$ is $1/2$.

   
   
   


3. 
   

   The nature of the $\lambda=1/2$ pathology is a degeneracy of the constraint force/missing rank issue. To see this more clearly note that we can get an equivalent action $$ \tilde{S} ~=~ \int_{\tau_i}^{\tau_f}\!\mathrm{d}\tau \left(\sqrt{1} + \lambda(\dot{x}_\mu\dot{x}^\mu + 1)\right) \tag{B}$$ by inserting the constraint (A) into the first term in OP's action (4). If we repeat OP's calculation for the equivalent action $\tilde{S}$ we will see that the trouble has shifted to $\lambda=0$. Clearly, the case $\lambda=0$ corresponds to a degenerate case where the stationary action principle (B) is ill-defined.

   
   

--

$^{\dagger}$ If the WL parameter $\tau$ is proper time off-shell as well, it would mean that OP's action (4) is just $S=\tau_f-\tau_i$, which is fixed by boundary conditions (BC). In other words, the action would not depend on the WL, i.e. the variational problem would be ill-defined.

riddle - Discover the meaning of life!

You've trekked through jungle and tundra, across mountaintops and deserts, into giant caverns and beneath the ocean to find the person known as The Oracle. Finally, after all your hard work, you find him. He sits alone in an opulent palace, never moving, never blinking, allowing each visitor to ask one question of him. You take a moment to catch your breath after climbing the steps to his throne, gather yourself, and ask:

"What is the meaning of life?"

You were told that he would answer one question for each person, and this question has been nagging on your mind.

What you were not told is that his answers were not quite straightforward.

As he begins to speak, you scramble to grab your notepad and begin writing. At the end of his message to you, you read back what he's said. "What?", you ask out loud, audibly confused.

Of course, The Oracle does not answer.

---

To start, stay within the pathway of general glee,
though stay clear of the law faery, who flits about wildly.

The road third-least taken leads to you and I in flight.
Head north 'til you see the reverse, then once more north to the burial site.

When five becomes square, it may lead to confusion;
behead the basilisk, an ironic inclusion.

Before two sharp curves and the ring of sorcery,
stand still, close your eyes, and count aloud past three.

In the air - do you hear it? The water baby screams.
Let our own misgivings start and end in our dreams.

A quixotic elf and I stand at the gates -
succumb to the virus, and the city awaits.

As you approach the arbor, I wait in between.
Clutch your keepsake as the first enemy's heard and seen.

Will being stuck in all this goo truly help you see?
Primarily, primarily; whip around, forcefully.

Between the lines you'll find your answer, definitively twisted;
yet in the shadows aligned so clearly you'll be surprised you missed it.

---

Can you discover the meaning of life from The Oracle's message?

---

HINT 1:

The Oracle speaks quite cryptically, don't you think? (I've added a new tag to the puzzle that should add some clarification to it. Hopefully that should give you some idea as to how the puzzle should be solved - and more importantly, some knowledge on how it shouldn't.)

HINT 2:

The two most important words to figuring out this puzzle's big gimmick are "definitively twisted".

Stanza-Specific Hints:

Stanza 1: This couplet gives you a key cryptic crossword indicator, but don't get caught up on the result making any sense. In fact, the couplet tells you exactly which words to use and which to avoid. Keep in mind somewhere in there is a definition!
Stanza 2: Pretty much everyone has identified 'u' and 'i' from this couplet, and you're all on the right track. Most of the words in this couplet aren't indicative of words, per se.
Stanza 3: If five becoming square actually meant the number 25, that wouldn't lead to confusion at all, would it? Harry Potter's version of the basilisk is not the only one out there.
Stanza 4: The last number reference leads to confusion. This one does not. Make sure you're counting aloud. To figure out the rest of the couplet, getting out a pen and paper and illustrating might help.
Stanza 5: I already mentioned that 'in our dreams' is just in this couplet as a rhyming mechanism. There's only one other place where I do that, and I'll identify it. The cryptic here has already been solved. Which words weren't part of it? Those may help you with another stanza...
Stanza 6: This one is a fairly straightforward clue. Open your cryptic-solving mind, remember the gimmick of the puzzle, and look for a keyword or two to get things going.
Stanza 7: You already know at least one word here that doesn't belong with this couplet. The rest of the couplet should be easily solvable, especially if you don't worry about the definition until you solve the cryptic.

Stanza 8: Will 'truly help you see' truly help you see? Not really.
Stanza 9: The first line means two things. You already know what 'definitively twisted' means, which is a good start. 'In the shadows' is a reference to the cryptic clues, which are apparently aligned so clearly! I'm surprised you missed it.

All stanzas: You will find pretty quickly that the words all share a specific property. It's not an obscure property, either (I promise).
All stanzas: The best way to approach this puzzle is to solve a stanza's cryptic clue, find its definition, then solve that stanza's cryptic clue with the words in the previous stanza's definition excluded (obviously); lather, rinse, repeat. Luckily, you've already been given a head-start with both a cryptic clue solved and its definition identified.

---

6/11/2015 Update: Hey, guys! I've opened a bounty on this puzzle as well as my other unanswered puzzle as a way of saying thanks for the bounty you helped me earn on this excellent puzzle. As such, I want this puzzle to be solved - my bounty shan't go to waste! If there are any questions I can answer to confirm whether or not you're going in the right direction, please don't hesitate to ask them in the comments. If I feel they don't give too much of the puzzle away, I'll be happy to answer them.

---

6/15/2015 Update: 2 days left on the bounty - someone better solve this! Pete and Fillet have given you all you need to work with. I can also conclusively tell you that the ninth stanza is specifically instructions, rather than another clue to the puzzle. Once again, I'm happy to answer any questions! (I've also added stanza-specific hints, in hopes that one or two might kickstart the solution to this puzzle.)

---

6/16/2015 Update: Hopefully this will be the last time I update this puzzle! I just want to note that partial answers are totally awesome for this puzzle. If the final answer is spread among multiple posts, I'll post a community wiki answer encompassing everything (in fact, I may do that anyways). When all's said and done (or, tomorrow before it expires), I'll give the bounty to whomever's solved the most clues! Good luck, happy puzzlers! You're doing awesome so far (just sometimes off in the wrong direction)!

Answer

So, standing on the shoulders of giants, and adding in the answers to two remaining stanzas, we get the final complete solution (as noted by Quassnoi), to discover that life is indeed:

PUZZLING

---

Looking at the riddle, it's fairly obvious that the clues are cryptic, but there's a little more to it than that. As hinted, the final stanza tells us how to approach the puzzle beyond these cryptic clues.

Between the lines you'll find your answer, definitively twisted;
yet in the shadows aligned so clearly you'll be surprised you missed it.

The definitively twisted part hints to us that the stanzas not only contain a word discovered through cryptic clues, but also the definition of the answer to one of the other stanzas. So as we solve cryptic clues, we can validate these answers by looking for words/phrases elsewhere to find synonyms/definitions thereof.

The in the shadows aligned part of the clue will become apparent at the very end...

---

Now, stanza by stanza we have:

Stanza 1:

To start, stay within the pathway of general glee,
though stay clear of the law faery, who flits about wildly.

Giving us:

PATHOGEN, by staying within PATHway Of GENeral, avoiding each of the letters from "law faery" (once each), as shown by Pete. This is of course synonymous with virus, from stanza 6.

Stanza 2:

The road third-least taken leads to you and I in flight.
Head north 'til you see the reverse, then once more north to the burial site.

Giving us:

QUINCUNX, by way of Q (the third least common letter for a word to start with), U & I (in flight), then N(orth) CU (U C reversed), N(orth), X (marks the spot). The definition of which ("an arrangement of five objects with four at the corners of a square or rectangle and the fifth at its center") can be found in the first line of stanza 3.

Stanza 3:

When five becomes square, it may lead to confusion;
behead the basilisk, an ironic inclusion.

Giving us:

WIZARDRY (again as from Pete), from a beheaded basilisk (aka lizard) to get IZARD, included in WRY (irony). The synonym, sorcery, is found in stanza 4.

Stanza 4:

Before two sharp curves and the ring of sorcery,
stand still, close your eyes, and count aloud past three.

Giving us:

FORZANDO, once more found by Pete, based on FOR (what you hear when you count past three), Z (two sharp curves) AND O (a ring). The definition of forzando in music is to play forcefully, as seen in stanza 8.

Stanza 5:

In the air - do you hear it? The water baby screams.
Let our own misgivings start and end in our dreams.

Giving us:

HEIRLOOM, again from Pete, because HEIR (sounds like air) plus the start letters of L(et) O(ur) O(wn) M(isgivings) at the end. With the synonym being keepsake from stanza 7.

Stanza 6:

A quixotic elf and I stand at the gates -
succumb to the virus, and the city awaits.

Giving us:

FELICITY, as solved by Quassnoi, derived from FEL (anagram of elf), I, and then CITY. Felicity means happiness, or glee, as seen in stanza 1.

Stanza 7:

As you approach the arbor, I wait in between.
Clutch your keepsake as the first enemy's heard and seen.

Giving us:

AIRBORNE, again from Quassnoi, since we put I into ARBOR (airbor), and add N (first letter "heard" in enemy), and E (first letter "seen" in enemy). The definition, "in flight" is found in stanza 2.

Stanza 8:

Will being stuck in all this goo truly help you see?
Primarily, primarily; whip around, forcefully.

Giving us:

POLLIWOG, by whipping around (reversing), the letters from WILL stuck in the letters of GOO (i.e. gowillo reversed), but starting with P (start of primarily). A polliwog is a tadpole, or a water baby from stanza 5.

---

Bringing all these words together, we end up with eight eight letter words:

Pathogen
qUincunx
wiZardry
forZando
heirLoom
felicIty
airborNe
polliwoG

And finally, coming back to the "in the shadows aligned" hint, we see aligned down the diagonal, the final solution.

newtonian mechanics - Are Newton's "laws" of motion laws or definitions of force and mass?

If you consider them as laws, then there must be independent definitions of force and mass but I don't think there's such definitions.

If you consider them as definitions, then why are they still called laws?

geometry - How many different pentagons in this grid?

The 3x3 grid is given as below;

How many pentagons could be drawn by connecting the dots by lines in the grid?

Rules:

• Pentagons could be convex or concave.


• The lines cannot intersect each other.


• If reflecting or rotating a pentagon forms the same pentagon you have already counted before, it should not be added.

Example:

Note: The other question referred as the duplicate question's answers are not right. and the OP is missing. That's why I believe this question should be solved here and accepted.

Answer

I get 23 pentagons. There are 126 ways of choosing 5 out of the 9 dots. After reducing for symmetry, only 23 are left, which you can check using Burnside's lemma. I drew these in the picture below. For any of these you can put a convex hull around the points. If that is a pentagon, then that is the unique one you can make from those points. If all the points are on the hull but it is a quadrilateral or triangle, then no pentagons are possible. If it has an interior point, then you have a choice of which side of the convex hull to replace by two sides connecting to the interior point.

Can Quantum Entanglement and Quantum Superposition be considered the same phenomenon?

Quantum entanglement is known to be the exchange of quantum information between two particles at a distance, while quantum superposition is known to be the uncertainty of a particle (or particles) being in several states at once (which could also involve the exchange of quantum information for a particle that is known to be in several locations simultaneously). I was wondering if all of this was nothing more but the exchange of quantum information between different masses, and if this could clear up all the confusion in terms of how quantum systems connect in this field of science. A clear explanation for how both of these quantum phenomena work, and if they really are connected (the exchange of quantum information?) would be much appreciated.

energy - Is it possible to create matter?

Is it possible to create matter? In a recent discussion I had, it was suggested that with enough energy in the future, "particles" could be created.

It seems like this shouldn't be possible due to conservation but perhaps I could be wrong. Would any of you Physics masters care to elaborate?

(Note... I will understand the basics, I am by no means an expert.)

connect wall - I make billions (#6)

I make billions (#6)

This is a two part connect wall puzzle. The sixteen words should be arranged into four groups of four words, with each group's words related to each other by a common theme. The four group names can then be used to find the final answer.

    box     farm    fire    flat

    games   giant   hill    monkey  
    pig     queen   roll    space
    tire    vein    wing    wolf

Hint for after you solve the groups:

Although the group names might relate to a man, the answer should be the name of a company.

_{Previous puzzles in this series: #1 #2 #3 #4 #5}

Answer

I reckon the wall can be resolved as follows:

X = Box, Games, Wing, Space (Xbox, the X Games, X-wing from Star Wars, and SpaceX);
ANT = Farm, Queen, Hill, Fire (ant farm, queen ant, anthill, fire ant);
IRON = Pig, Flat, Giant, Tire (pig iron, flatiron, The Iron Giant, tire iron);
SPIDER = Monkey, Wolf, Roll, Vein (spider monkey, wolf spider, spider roll (sushi), spider vein).

These 4 can of course be resolved with:

MAN (X-man, Ant-Man, Iron Man, Spider-Man)

Since these are all:

Characters in the Marvel franchise as created by Stan Lee et al.

Meaning the company in the final answer is:

MARVEL, a company which not only makes movies which rake in more than a billion dollars, but sometimes also feels like they make a billion movies!

quantum mechanics - Hydrogen atom, what's the wave equation for the atom's nucleus?

I learned from the class about the equation for hydrogen atom's electron where textbook assumed that the center/nuclei of hydrogen atom was fixed at origin.

However, since every particle was a wave, the nuclei of the hydrogen atom (say only contain one proton) could be seen a wave as well.

My question was that:

1. 
   
   

   What's the wave equation for the proton in hydrogen atom? Was it simply a traveling wave when the atom was moving, and a Dirac Delta function when it was "fixed"? (Further, what if there was a neutron?)

   
   


2. 
   

   In the case when hydrogen was traveling, say along $x$ axis, would there be an extra influence/interaction towards the electron's wave equation?

   
   

Answer

By conservation of momentum, the center of mass of the atom is what actually stays fixed. This implies that there is a perfect correlation between the wavefunctions $\Psi$ of the electron and $\Phi$ of the proton:

$$\Phi(x)=\Psi(-(M/m)x),$$

where $M$ is the mass of the proton and $m$ is the mass of the electron.

The effect on the energy levels is to replace the electron's mass with the reduced mass.

seasonal - Compound word cousins

Find a single word answer with the property that compound words formed from the answer have a word in common with compound words formed from each of the following :

Text version:

BERRY  BOTTLE  CRYSTAL   CUSHION   FLY      HIGH  
HIKE   INDEX   KETTLE    MARKET    MUSTARD  PLAYER  
POT    RED     RING      ROCKET    SALAD    SNOW    
SWEET  TIME    TROPICAL  VACUUM    VOICE    WINE    

For example: the words FAMILY, RADISH, SKI are connected in this way by CHESTNUT:

family tree - chestnut tree, horseradish - horse chestnut, waterski - water chestnut.

---

Updates and clarifications upgraded:

1. The answer is associated with the two types of compound words used: open and closed.


2. There is a different link word for each of the 24 grid words.


3. The grid shape was cracked by M Oehm in a comment:
   

   
   

   The 6x4 grid looks a bit like an advent calendar.

   
   

   
   


4. Inspiration: To Do, Or Do Without by Hugh Meyers.

Answer

I think the answer is:

window.

The associated compound words could be:

bayberry — bay window
glass bottle — windowglass
crystal lattice — lattice window

seat cushion — window seat
flyscreen — windowscreen
high time — time window
tax hike — window tax
index case — window case
electric kettle — electric window
transfer market — transfer window
french mustard — french window
player-manager — window manager
pot marigold — marigold window

rose-red — rose window
keyring — Windows key
rocket launch — launch window
salad dressing — window dressing
snowmobile — Windows Mobile
sweet shop — shop window
time frame — window frame
tropical storm — storm window
vacuum cleaner — window cleaner
voice box — window box

winemaker — windowmaker

Some of these work better than others, but a few, especially cleaner and dressing, fit remarkably well. I'm sure I haven't got all the intended answers and some are a bit of a stretch. Googling "window" yiedls a lot of stuff related to our favourite OS and consequently I've taken the liberty to include references. (I find Windows key acceptable, but Windows Mobile a bit of a stretch. Perhaps a windows mobile might be something hanging in the window of a nursery. Funny how many things you can combine with French and with bay.)

Possible links I didn't use: sash, front, shopping, light, pane, sill, picture.

Finally, the hints:

The answer is associated with the two types of compound words used: open and closed. — Like compound words, windows may also be open and closed.

The 6x4 grid looks a bit like an advent calendar. — A classical advent calendar has 24 small windows.

Inspiration: To Do, Or Do Without by Hugh Meyers. — That puzzle was about doors, which are related to windows in that they are openings in walls.

(My first thought was that the answer could be door here, too. There were some promising pairs, for example door / bottle opener and keyring / door key. But when I learned about the window tax and saw the window cleaner and widow dressing solutions, I changed my mind to window.)

forces - Homework about spinning top

I have a top of unknown mass that has a moment of inertia $I=4\times 10^{-7} kg \cdot m^2$. A string is wrapped around the top and pulls it so that its tension is kept at 5.57 N for a distance of .8 m.

Could somebody help me derive some equations to help with this? Or to get me in the right direction? I have been trying to derive some sort of equations from $E=\frac{I \cdot \omega ^{2}}{2}$ but I cant get anywhere without ending up at radius = radius or mass = mass.

I need the final angular velocity.

lie algebra - Lie group Homomorphism $SU(2) to SO(3)$

The Lie algebra of $ \mathfrak{so(3)} $ and $ \mathfrak{su(2)} $ are respectively

$$ [L_i,L_j] = i\epsilon_{ij}^{\;\;k}L_k $$ $$ [\frac{\sigma_i}{2},\frac{\sigma_j}{2}] = i\epsilon_{ij}^{\;\;k}\frac{\sigma_k}{2} $$

And of course, there is an isomorphism between these two algebras, $$ \Lambda : \mathfrak{su(2)} \rightarrow \mathfrak{so(3)} $$ such that $ \Lambda(\sigma_i/2) =L_i $

Now is it possible, using $\Lambda$, to construct a group homomorphism between $SU(2)$ and $SO(3)$?

I was checking up on Lie group homomorphism, and in Wikipedia, there is a beautiful image

In this image's language, how are $\phi$ and $\phi_*$ related to each other (just like the algebra and group elements are).

Note : I know there is a one-to-two homomorphism between these two groups which can be directly found using the group elements. I am not looking for this.

EDIT 1 : In $ SL(2,\mathbb{R}) $ the generators, say $X_1,X_2,X_3$, they obey the following commutation rules :

$$ [X_1,X_2] = 2X_2 $$ $$ [X_1,X_3] = -2X_3 $$ $$ [X_2,X_3] = X_1 $$

And in the case of $ SO(3) $ with a different basis, $ L_{\pm} = L_1 \pm i L_2 $ and $ L_z = L_3 $ with the commutators being,

$$ [L_z,L_{\pm}]= \pm L_{\pm} $$ $$ [L_+,L_-]= 2 L_z $$

This algebra is very similar to the algebra of the previous one, so why is that we can't define a map ?

EDIT 2:

Can the group homomorphism between these two groups be written like this (Something like what I expected) : $$ R = \exp(\sum_k i t_k L_k) = \exp\left(\sum_k i t_k \frac{\sigma_k}{2}\right) = \exp\left(\sum_k i t_k \frac{1}{2}ln(U_k)\right) $$

Now this seems like the map $\phi$,

$$ R = \phi(U) = \exp\bigg(\sum_k i t_k \frac{1}{2}ln(U_k)\bigg) $$

Answer

First notice that the generators are $-i\sigma_k/2$ and $-iL_k$, since the groups are real Lie groups and thus the structure tensor must be real.

The answer to your question is positive. In principle it is enough to take the exponential of the Lie algebra isomorphism and a surjective Lie group homomorphism arises this way $\phi : SU(2)\to SO(3)$: $$\phi\left(\exp\left\{-\sum_k t^k i\sigma_k/2\right\}\right) =\exp\left\{-\sum_k t^k iL_k\right\}\:.$$ The point is that one should be sure that the argument in the left-hand side covers the whole group. For the considered case, this is true because $SU(2)$ is compact.

If you instead consider no compact Lie groups, like $SL(2,\mathbb C)$, the exponential does not cover the group. However it is possible to prove that products of exponential do. In that case a product of two exponentials is sufficient, in practice decomposing an element of $SL(2,\mathbb C)$ by means of the polar decomposition, mathematically speaking, or as a (unique) product of a rotation and a boost physically speaking.

quantum mechanics - How does a photon drive out the electrons in a solar cell?

We know that solar cells work when a photon hits the n-type the photon's energy drives free the electrons in the n-type to generate a current. But we also know that when a photon hits the atoms it makes the electrons excited. So why doesn't the photon make the electron excited and makes the electron drive out?

OR is it like this that in solar cells the photon gives so much energy to the electron that when it goes to a higher energy state and changes shell it gets out of the atom shells and becomes a free electron carrier?

Thanks,
Bhavesh

Answer

I think a simple view is this:

The solar cell must have a PN junction, which is a junction between p-type (many holes, no electrons) and n-type (many electrons, no holes) materials. Right where they meet there is actually a "depletion width" within which there is hardly any of either. Within this region, as photons come in they generate electron-hole pairs, which really just means that an electron has been excited from the valence to conduction band, leaving a hole behind. The electron is then pushed back to the n-side by the "built-in" electric field, while the hole is pushed to the p-side. Think of this is terms of energy: both end up where their energy is lower, so electrons prefer the n-type material, while holes prefer the p-type. Some understanding of semiconductor doping and Fermi statistics helps greatly to understand this.

statistical mechanics - How and why can random matrices answer physical problems?

Random matrix theory pops up regularly in the context of dynamical systems.

I was, however, so far not able to grasp the basic idea of this formalism. Could someone please provide an instructive example or a basic introduction to the topic?

I would also appreciate hints on relevant literature.

Answer

The basic idea is that statistical properties of complex physical systems fall into a small number of universal classes. A very known example of this phenomenon is the universal law implied by the central limit theorem where the sum of a large number of random variables belonging to a large class of distrubutions converges to the normal distribution. Please see Percy Deift's article for a historical and motivational review of the subject. Of course, one of the most motivating examples (mentioned Percy Deift's article) in is the original Wigner's explanation of the heavy nuclei spectra. Wigner "conjectured" the universality and looked for a model which can explain the repulsion between the energy levels of the large nuclei (two close energy levels are unlikely) which led him to the Gaussian orthogonal ensemble having this property built in. Now, the heavy nuclei Hamiltonian matrix elements are not random, but since by universality , for large N, the distribution of the eigenvalues does not depend on the matrix elements distribution, then the random matrix eigenvalue distribution approximates that of the Hamiltonian.

logical deduction - Council of Magic - a palace of deceptive spellcasters

Council of Magic:

The King of Puzzlington has sent you to a distant palace to gain the aid of the Council of Magic. As your luck would have it though, spellcasters are a tricky sort.

There are 4 types of spellcasters in Puzzlington: Wizards, Witches, Priests, and Warlocks.

Wizards:
Wizards follow one of two paths: the path of Fire or the path of Water.
Wizards on the Path of Fire always tell the truth when asked a question.
Wizards on the Path of Water always lie when asked a question.

Witches:
Witches come in two styles: Light Witches or Dark Witches.
Light Witches tell the truth during the day and lie at night.
Dark Witches lie during the day and tell the truth at night.

Priests:
Priests worship one of two Gods: Yes, god of life or No, god of death.
When asked a question, instead of answering, priests just say their God's name.
That is, a Priest of Yes will always answer "Yes" to any question.

Warlocks:
Warlocks are unpredictable tricksters.
When asked a question, Warlocks will tell the truth or lie, as they wish.
They will do one or the other, however.

The Council:
The Council of Magic consists of 4 powerful spellcasters. The spellcaster's names are Alice, Bob, Claire and Dave. The Council has one Wizard, one Witch, one Priest and one Warlock; you do not know which Council member is which type of spellcaster. You do not know the Path of the Wizard, Style of the Witch or God of the Cleric.

The members of the Council have full knowledge of each other. That is, they know which member is which type of spellcasters and which sub-type each is. The sub-type of a spellcaster is their Path, Style or God, as appropriate.

You arrive at the palace at noon. You can, once every 12 hours, ask any one member of the Council any one question that can be answered Yes or No (that is, after each question, it switches from night to day or day to night). If a spellcaster is asked a question they can't answer (because they don't know the answer), they remain silent. Priests are an exception to this rule; to them, their god is always the answer, no matter what the question.

Further complicating matters is the Warlock, who hates you; when asked a question, they will choose to tell the truth or lie, whichever they think will hurt you the most.

Your task is to learn the type and sub-type of all Council members as fast as you can.

---

I know a way to solve this puzzle, but it takes 6 days (12 questions) in the worst case. I'm fairly sure there is a faster way to do it (there is just over 7 bits of information you need to find); but I don't know it. I'd sure like to know it if anyone can find one.

Answer

I think I can do it in:

at most 7 questions

First question:

This borrows some ideas from Gareth McCaughan's answer. He notes the following: "if I ask a witch or wizard "what would you say if I asked you ...?" I always get a truthful answer.". With this in mind, my first question's purpose is to tell whether the recipient is in the (wizard/witch) pair or the (warlock/priest) pair. I ask of A: " If I asked the warlock whether 2+2=4, what would he say? ". If A was in the (witch/wizard) group, he would be silent. If A was in the (warlock/priest) group, he would not be silent, as these 2 always give an answer.

Next 2 questions:

I now ask the same question to B. This tells me what group B is in. In the worst case, I now have A in one group and B in the other group, so I have to ask the same question to C. After this, I know everyone's group, but I don't know anything else (since the warlock always tries to screw up my strategy as much as possible, he will give a different answer to the priest).

Question 4:

Let's say that A,B are (warlock/priest) and C,D are (witch/wizard). There are 2 possibilities. (1st possibility): The only person who hasn't answered a question yet is in the (warlock/priest) group. Call that person X. I then ask of C: "what would you say if I asked you what X would say if I asked him if 2+2=4". If he gives an answer, I know that X is the priest and I know what god he follows (whatever the answer was), and I know who the warlock is. If no answer, then the warlock is X, the priest is the other member of that group and his god is whatever he answered before. (2nd possibility): The only person who hasn't answered a question yet is in the (witch/wizard) group. In this case I ask of C: "what would you say if I asked you what A would say if I asked him if 2+2=4" and again I know everything about the priest and warlock.

Question 5:

I now ask of C: "what would you say if I asked you if you are the wizard" This sorts out who is witch and who is wizard.

question 6 and 7:

These are to determine the types of the wizard and the witch.

simulations - Can we simulate the whole universe in computer?

Here is an idea: As human kind we discovered many laws, formulas, events, etc. in our surrounding environment.here is a way to discover the unknowns :

Now if we simulate a virtual world with our knowledge in computers we expect it to work, because we guess that we know all of it. If a bug occurs in the simulation or something acts differently from the expected in the real world, that may tell us that there is something new; a thing (law, formula, ...) that we didn't know, which causes that bug or that anomaly.

For examples imagine that we created a virtual simulation for motion dynamics and we didn't add the friction to our simulation. Everything in the simulation will work fine but in a different manner. So if we hadn't discovered friction yet, would it help us to think about if there is something we didn't add? Or if there is something we don't know?

But can we simulate the whole universe, everything we know, in one simulation? I know it will require a deadly high computational power. Something like what they did for DNA; two years of computation to analyse their pattern. can we simulate everything we know to create a virtual universe to compare it to our real universe?

general relativity - Closed timelike curves in the Kerr metric

I just read in Landau-Lifshitz that the Kerr metric admits closed timelike curves in the region $r \in (0, r_{hor})$ where $r_{hor}$ is the event-horizon ( I am talking about the case $|M|>|a|$ (subextremal case) here ). Now, unfortunately they don't give an example of such a curve. Could anybody of you write down explicitly such a CTC so that I could go through the computation once by myself. I would really like to see this once.

If anything is unclear, please let me know.

special relativity - Time dilation formula

In the time dilation equation $$t'=t/\sqrt{1-(v/c)^2}$$ where $ t'$ is the time measured by an observer in motion for the same event, where $t$ is time measured by the observer at rest.

Imagine a situation where both of the observes measure a time of 5 sec on their watches. These events (an observer seeing the fifth tick on his watch) are not simultaneous.

Now, my question: is the time taken by the observer in motion given by the equation $$t'=t/\sqrt{1-(v/c)^2}.$$ Does this signify that as time is travelling slower for the observer in motion it takes more time than the observer at rest to measure the time interval of 5 sec?

Why only light nuclei are able to undergo nuclear fusion not heavy nuclei?

Is it because of the binding energy or the binding energy per nuclei . I am having trouble with this whole binding energy and nuclear fusion concept.

Answer

Heavier nuclei can also undergo fusion, but that's not very useful for energy production. One of the reasons is, as you've mentioned, the binding energy per nucleon. Let's have a look at the binding energy curve (image taken from Wikipedia):

Iron-56 has the highest binding energy per nucleon, which means it is the most stable nucleus. Roughly speaking, elements left of iron-56 in this graph can release energy by fusion. You could also fuse elements heavier than iron-56, but this will only cost you energy.

Another restriction on which fusion reactions are useful for energy production is the Coulomb barrier. When you want to fuse two nuclei together, you have to bring them very close together, within the range of the nuclear force. Remember, however, that the nuclei consist of neutrons and protons, and so they are positively charged. That means to bring two nuclei very close together, you have to also do work against the coulombic force which is trying to push them apart.

When you use heavier nuclei, they will have more protons, and so the coulombic repulsion between the nuclei will increase. The means that you need to put in more energy in order to fuse the nuclei together. Because this only detracts from the efficiency of the process, it is favorable to use lighter nuclei for nuclear fusion reactors.

Resources for introductory quantum statistical mechanics

I am currently struggling to understand my basic introductory course on quantum statistical mechanics and I have done a basic course on single particle quantum mechanics.

I was wondering whether anyone knows of any good resources (online/books) that I could read to help my understanding of this subject.

I would like to read about the following topics : quantum microcanonical ensemble, quantum canonical ensemble, quantum grand canonical ensemble, quantum harmonic oscillators, ideal quantum gases, spin and statistics and many particle states.

Answer

There are of course many books out there, quantum statistics is a really well-established field, so regardless of suggestions here you should really look further on your own as well and find one that suits you best. But here are a few that I've used in the past that you may find useful:

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  Statistical mechanics: A survival guide by Mike Glazer and Justin Wark. It's a book based on the lectures notes given by the two authors, so the material does not go into the advanced topics too much and sticks with minimum necessary basics. After 3 nice chapters on classical stat. mech. mainly discussing distinguishable particles, it moves over to indistinguishable particles and first develops the classical version then introduces the concept of Gibbs paradox, with this transitions toward quantum statistics for the remaining part of the book. Einstein statistics, quantum dist. functions, photon gas etc. Bear in mind it may not contain all that you're looking for, but it is a good start and very easy to read.

  
  



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  Landau & Lifshitz Vol. 5 is of course always recommended, and you have it also online for free. Big part of the book is dedicated to classical stat. mech. but starting from chapter 5 it moves over to the quantum version, a lot of in-depth discussions, but does also cover some simple examples of photon gas or lin. harmonic oscillator.

  
  


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  One that comes closest to the specific topics you mentioned is Statistical mechanics A set of Lectures by Richard P. Feynman. Note it is not part of the 3-volume Feynman lectures, it's another book solely on stat. mech. based on Feynman's lecture notes. It's really a modern book, in the sense that it is mainly focused on quantum statistics and adopts the more common notations of Dirac's bra-ket, traces etc. Basically all the calculations are shown in this book (pages and pages of it), no real steps omitted. Chapter 1 and 2 are of main interest to you: Discussing the partition function for a system of discrete E states, LHO, Q-statistics for many particle systems, then chapter two is really a priceless in-depth discussion of density matrices and their use in stat. mech., after a long intro (covering some QM basics as well) it starts discussing specific examples such LHO, 1D free particle all using density matrices (more advanced topics come next).

  
  


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  Anyway again, there are many books, impossible to tell which one is best for you, but here's another one: Introduction to Modern Statistical Mechanics David Chandler. One of my favorites, it's one of those books that does tell all the tricks and intuitive ideas that other books skip out on. You can start from any chapter and still be able to read through while understanding most of it. The chapter 4 starts discussing non-interacting systems by re-discussing the occupation number in quantum stats, photon gases etc. Then moves over to ideal gases of real particles discussing the grand canonical potential for bosons and fermions.

  
  


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  Finally if you want to go for some of thick well known (covering almost everything, mainly for references (not many calcs shown)), then here are two that I recommend: Statistical Mechanics 3rd edition by Pathria (chapter 5 to 9 cover all the topics you mention), and Statistical Mechanics by McQuarrie with a nice first chapter reminding everything on thermodynamics, QM etc. This book first discusses quantum statistics and only few chapter later returns to classical stat mech (briefly) (chapter 4,6 and 10 discuss most elements of Q-statistics.)

  
  

Please take the time to check out the table of contents of all the books mentioned, read the reviews, see excerpts of the books on google books.

thermodynamics - Rubber band stretched produces heat and when released absorbs heat.. Why?

I always used to wonder why this happens.. when one stretches a rubberband to nearly it snapping point holding it close to your skin - preferably cheek(helps feel the heat), it emits heat. While releasing the stretched rubberband holding it close to the skin produces a cooling effect on the skin. Can some one explain the physics behind this pls?

Answer

This is a very interesting question with a very interesting answer. The key lies in the reason for the stretchiness of the rubber band.

Rubber is made of polymers (long chain molecules). When the elastic band is not stretched, these molecules are all tangled up with each other and have no particular direction to them, but when you stretch the elastic they all become lined up with one another, at least to some extent. The polymer molecules themselves are not stretched, they're just aligned differently. To a first approximation there's no difference in the energy of these two different ways of arranging the polymers, but there's a big difference in the entropy. This just means that there's a lot more different ways that the polymers can be arranged in a tangled up way than an aligned way. So when you release the elastic band, all the polymers are jiggling around at random due to thermal motion, and they tend to lose their alignment, so they go back towards the tangled state, and that's what makes the elastic contract. This is called an entropic force.

Now, I said earlier that there isn't any difference in energy between the stretched (aligned) and un-stretched (tangled) states. But it takes energy to stretch the elastic -- you're doing work to pull the ends apart, against the entropic force that's trying to pull them back together. That energy doesn't go into stretching the individual polymer molecules, but it has to go somewhere, so it ends up as heat. Some of this heat will stay in the elastic (making the polymer molecules jiggle around a bit faster) but some will be transferred to the surrounding air, or to your skin.

The reverse happens when you let the elastic contract. The molecules are jiggling around at random and becoming more and more tangled, which makes them contract. But to contract they have to do work on whatever's holding the ends of the elastic apart. That energy has to come from somewhere, so it comes from heat.

At first this might seem to run against thermodynamics - normally you can't just cool something down without heating something else up. But remember that the state with the tangled molecules has a higher entropy than when they're aligned. So you're taking heat out of the air, which reduces its entropy, but this reduction in entropy is countered by the increase in entropy of the elastic itself, so the second law is safe.

For further reading you can look into the ideal chain, which is an idealised mathematical model of this situation.

cosmology - How is pressure calculated in the early universe?

I'm trying to calculate the sound horizon from the start of time to decoupling. To do this I need to know the speed of sound and how it changes as the universe grows. The speed of sound in a fluid is:$$c_s^2=\frac{\partial p}{\partial \rho}$$Where $c_s^2$ is the speed of sound, $p$ is the pressure, $\rho$ is the density. I think I have a handle on calculating the density:$$\rho(z)=\Omega_b h^2 \rho_{crit} (z+1)^3 \space g\space m^{-3}$$But I have no idea how you calculate the pressure. I'm assuming that the pressure was mainly photonic up to the time of decoupling but I'm having trouble finding reference material.

Also, intuitively I would think that the pressure would fall in exact proportion to density since they're both related to the change in volume. So is it enough just to find the pressure at the time of decoupling and divide by the density and use it as a constant?

Answer

Before recombination baryons and photons are highly coupled and act like a single fluid, so the density is $\rho = \rho_{\rm bar} + \rho_\gamma$, with the usual scaling $\rho_{\rm bar} \sim (1 + z)^{3}$ and $\rho_{\gamma} \sim (1 + z)^{4}$. However, the pressure is the same for both components $P = P_{\rm bar} = P_{\gamma}$, this will lead to

$$ c_s^2 = \frac{c^2}{3}\left[\frac{3}{4}\frac{\rho_{\rm bar}}{\rho_\gamma} + 1\right]^{-1} \tag{1} $$

To derive this expression remember that in an uniform field $P_{\gamma} = \rho_\gamma c^2 / 3$ so that the adiabatic speed of sound is

$$ c_s^2 = \left( \frac{\partial P}{\partial \rho}\right)_S = \frac{c^2}{3} \left.\frac{\partial \rho_\gamma}{\partial(\rho_\gamma + \rho_{\rm bar})}\right|_S = \frac{c^2}{3}\left[1 + \left(\frac{\partial \rho_{\rm bar}}{\partial \rho_\gamma}\right)_S \right]^{-1}\tag{2} $$

We need to calculate $(\partial \rho_{\rm bar}/\partial \rho_\gamma)_S$. And to do that we can use the fact that $\rho_{\rm bar}\sim a^{-3}$ and $\rho_{\gamma}\sim a^{-4}$ so

$$ \frac{\partial \rho_\gamma}{\rho_\gamma} = -4 \frac{\partial a}{a} ~~~\mbox{and}~~~\frac{\partial \rho_{\rm bar}}{\rho_{\rm bar}} = -3 \frac{\partial a}{a}~~~\Rightarrow~~~ \frac{\partial \rho_{\rm bar}}{\partial \rho_\gamma} = \frac{3}{4}\frac{\rho_{\rm bar}}{\rho_\gamma} \tag{3} $$

If you replace (3) in (2) you will get (1)

electromagnetism - Temperature and resistance?

Why does resistivity increase with temperature?

The explanations I have heard so far are that increasing temperature increases vibrations in the lattice structure resulting in the number of collisions increasing.

But the vibrations from thermal energy are random, so around half of the vibrations would result in the atoms moving out of the electrons way and half would result in atoms moving into the way of the electrons.

And since half of the vibrations result in increasing the drift speed and half reduce it, shouldn't the average drift speed remain the same?

wordplay - Anime guess Riddle #2

Like in my first part, I'm searching for the name of an anime. Knowledge from the anime is required so maybe, if you havent seen it, you cant figure it out. I hope you have fun :)

My MC is the worlds best terrorist, his friends are his sidekicks and his worst nightmare.
His hobbies are ripping off royals and controlling people
But in the end he's a real good guy

Hint

Season 1 ended with one of the biggest cliffhangers I've ever seen, but Season 2 made Season 1 useless

Answer

This is definitely

Code Geass

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"My MC is the worlds best terrorist"

The main character is Lelouch Lamperouge, self-appointed leader of a terrorist movement opposing the Britannia Empire's occupation of Japan

"His friends are his sidekicks and his worst nightmare"

Lelouch's best friend, Suzaku Kururugi, works for the Britannian military, opposing the terrorists in his advanced Knightmare Frame mecha. Another of his schoolfriends, Kallen Stadtfeld, is a prominent member of the resistance.

"His hobbies are ripping off royals"

The members of the Britannian royalty are the main villains of the series.

"and controlling people"

Lelouch has the power of Geass, which allows him to give someone a single vocal command that they are compelled to obey.

"But in the end he's a real good guy"

He's the protagonist... though tbh I wouldn't say he's a "good guy". He does some pretty terrible things in the name of opposing the Britannian regime.

Hint (note, this actually does spoil the anime):

Season 1 ends with Lelouch and Suzaku holding each other at gunpoint. We hear a gunshot... cue end credits. I never actually watched Season 2, but from what I've heard it has a completely different storyline to Season 1.