enigmatic puzzle - Kitty's New Mission

A new year and a new mission for Kitty.

Having successfully located the bomb on the previous mission, Kitty is on the next level of training and is delighted to see conditions have improved considerably.

This time she's in a rather nice hotel room and now she knows where she is as there is a rather pleasant note from the management with her mission instructions.

Time to explore—the room is a big octagonal one with a double bed. It’s been done up very nicely for the time of year with tinsel decorating the walls, one of those dancing Santa Clauses jangling away on the windowsill and a rather jolly Christmas tree in the corner next to the door leading through to a separate bath and shower cubicle.

And there’s snacks too! Tea and coffee making facilities, a nice selection of biccies and funsize chocolate bars:a Mars, a Bounty and an Aero. Ooh a minibar too! On the top shelf is a selection of alcohol—gin, whiskey, brandy and on the bottom shelf a range of condiments—mint sauce, mayonnaise, cranberry sauce and mustard.

In the bookcase are a few novels, a couple of children’s books, ’The Mystery of the Locked Hotel Room’, a book of birthstones, a book of zodiac signs, a guide to alphabet substitution codes and ‘Fungal Fun—a study of amusing skin diseases.’

In the cupboard by the bath are some cleaning things—some rubber gloves, an apron, a small iron, some hand and face towels and a couple of small bottles of Febreze fabric conditioner and Surf washing powder. And a small dead spider… yuk…

So there you go Kitty. Lie back on the hotel bed, enjoy the contents of the minibar and try to deduce when and where this bomb will go off.

fermions - Why is supersymmetry a continuous symmetry?

Supersymmetry feels like a discrete symmetry to me, since the fermions are turning into bosons, and vice versa. I understand there is an infinitesimal parameter involved in the transformations, but I don't know what it actually determines physically.

experimental physics - Why did it take so long to find the Higgs?

The $W$ and $Z$ boson took a long time to be discovered because they were so heavy; we couldn't produce them in a particle collider until the 80's. But the Higgs boson isn't that much heavier than the $W$ and $Z$, so we presumably have been producing them for decades. The Higgs also couples to just about everything, so it's not hard to make.

Given the above, why did we have to build the LHC to find the Higgs? Was it just a matter of getting enough statistics, or was there another reason?

Answer

Electroweak theory told us where to look for the $W$ and $Z$ gauge bosons. For the Higgs, its mass is a free parameter, hence we didn't know where to look for it. Once you start to look in many places for a particle, you also have to factor in the look-elsewhere effect, which basically means that the more places you look for a particle, the higher the chances you only observe a statistical fluctuation.

LEP

Today we know the Higgs mass is roughly $125$ GeV. The LEP collider at CERN, reached a center of mass energy $\sqrt{s}$ of $209$ GeV. This would have been enough to produce the Higgs. But then, $e^+ e^-$ colliders do not have the highest Higgs production cross section. Your statement:

The Higgs also couples to just about everything, so it's not hard to make.

is not correct. The Higgs couples only very lightly to light particles. So you need to produce heavy particles first in order to have a good chance to also produce a Higgs.

At LEP, the Higgs must have been produced by vector boson fusion, just not enough for us to clearly distinguish it from background:

Associated production is also thinkable, but the cross section is very small, since $m_H + m_Z > \sqrt{s}_{LEP}$:

Even smaller would be $ttH$ production:

Long story short, at LEP, the center of mass energy was simply not high enough, since it's not enough to produce the Higgs only, but you have to produce heavy particles first, that then couple to the Higgs.

Tevatron

What about the Tevatron? Surely the Higgs must have been produced copiously at the Tevatron. Why didn't they see it? The answer is that they did see it. Both CDF and $D^0$ reported an excess. In the combined publication, they even reported an "evidence for the presence of a new particle consistent with the standard model Higgs boson". The combination observed a global significance of $3.1 \sigma$.

But the question remains, why "only" a $3.1 \sigma$ excess, although years of data taking? The answer is two-fold: First, the instantaneous luminosity of the Tevatron was relatively small, at least compared to the LHC. The problem is that while using a proton anti-proton collider helps you in raising the cross sections, the production of anti-protons is notoriously slow and you can't easily reach high instantaneous luminosities.

The second reason, and also the general answer to why finding the Higgs was not so easy, lies in the way the Higgs decays:

This image shows the decay channels of the Higgs boson for a given mass. At $125$ GeV, it mostly decays to a $b\bar{b}$ pair. $b$ quarks hadronize and what you see in the detector are two "jets". A two jets event is not a particularly clean signature. In a hadron collider machine, your background is overwhelming. As far as I know, even today, the observation of the $H \to b\bar{b}$ signal has not reached the $3 \sigma$ level in any LHC experiment.

There are two ways around this: Either look for a Higgs produced with another particle or rely on Higgs decaying to other particles. In a Higgs associated production, the $b\bar{b}$ pair is accompanied by a $W/Z$ boson and the sensitivity is much higher, since the background is much smaller. This is what the experiments at the Tevatron observed.

Higgs decaying to $ZZ$ or to $\gamma\gamma$ are particularly clean channels. But the branching fraction $H \to ZZ$ is $2.67$ %, and $H \to \gamma\gamma$ is even only $0.228$ %, so you need a large dataset to see these decays. Such a dataset was not available in time at the Tevatron.

It is interesting to note, that if the Higgs would have been lighter, it would have been produced copiously at LEP, and we would have found it a long time ago. If it would have been heavier, the branching fraction to $b\bar{b}$ would have been considerably smaller, and we would have observed it at the Tevatron. The mass of $125$ GeV simply turned out to be a signal which was hard to distinguish from background, for various, aforementioned reasons. That's why it took so long to find the Higgs.

general relativity - Why doesn't a global frame of reference exist for GR?

I only have at best a layperson's familiarity with GR, so forgive me if I am asking a basic question, but I have heard that in GR, we cannot have a global frame of reference, that is a frame of reference that is applicable to the entire universe. Can anybody explain why this is?

I have heard that it has something to do with the expansion of the universe, since we can still define it locally. At the same time I have also heard that it is due to our understanding of space-time as being curved. Can anybody expand on that?

newtonian mechanics - Heuristic equation for Friction force between materials

I'm programming a game where different types of objects will be sliding over different types of terrains (Top-down in two dimensions). At my current level of physics education we are given the coefficient of friction between any pairing of surfaces, but that's not practical for the amount of objects and terrains this game (will) have.

So my questions are,

• How do I find the force of friction between two objects, when I have their surface area and any other constants I need to add?


• How can I somewhat accurately simplify that to something on the order of $z=x*y+a*b$ ?


• I'd like to account for static vs kinetic friction, is there an easy way to do that?

Edit: failing any established approximations, if I make up some abitrary "grippiness" constant for all of the surfaces, what's the best way to combine them into a coefficient of friction?

mathematics - Guess five binary digits!

Person A thinks of a 5 digit binary number. Person B tries to guess the number. B can guess a 5 digit binary number and A will respond with the number of correct digits (digits in the right place).

What is the maximum number of guesses B will take to know the binary number (assuming he is playing optimally)?

resource recommendations - Where to find detailed measured emission spectra of all chemical elements?

I'd like to have something like this, but for single atoms and with more extended range of wavelengths.

All I could find e.g. for hydrogen was lots of talks about Rydberg formula etc. and plots of spectrum in visible region. But I need experimental data to look at, not something plotted from formulas like Rydberg's one.

Similarly for other elements, I mostly found only plots of spectral lines in visible range.

So, where can one find detailed measured spectra for all (or most) chemical elements, preferably with UV and IR parts? It'd be most useful to have it in numeric form (spreadsheet etc.) instead of pictures.

Answer

The NIST Atomic Spectra Database is a decent source for general-purpose lookup and identification of spectral transitions and levels. You will probably be more interested in their spectral line info. It also provides comprehensive references, both in a per-line basis (in the Line Ref column of the results to the Lines database) and via the bibliographic database query links at the top of the results pages.

The database also gives the relative intensities of the lines, as well as the transition lifetimes, in the form of the $A_{ki}$ of the lines results; for more information see here and here.

Regarding your comment:

Thanks for the reference. But while it's quite interesting, I'd still like to have raw spectroscopic measurement data, i.e. relative intensity as a function of wavelength, not result of processing it into a list of spectral lines.

The hard part of spectral measurements is the conversion of raw spectral data into a list of lines; the reverse direction (converting a line/strength list into a simulated spectrum) is easy, which is why the NIST database provides line lists, and not the raw experimental data. Also, the appearance of the raw data is highly dependent on the experimental conditions, whereas the lines/strengths is uniquely associated with each ion. You can simulate spectra from the line list if you know the ambient conditions (temperature, pressure).

Huygens Principle and Interference

If Huygens Principle states that every point on a wave acts as a source of secondary wavelets, then:

1. 
   

   Is the new wavefront formed by the interference of these secondary wavelets?

   
   


2. 
   

   If so, then how can a light source have uniform intensity if there is constructive and destructive interference happening during the light's propagation? (See image)

   
   

(Note: This image was downloaded from the internet. The yellow boxed portions, the black arrows that point to it and the write- up near the arrows illustrate my doubt. The rest of the write- up came along with the downloaded image.)

That is, in both cases, the waves interfere. But, in the propagation of a plane wavefront (according to Huygens Principle), the resultant wave (formed by the interference of secondary wavelets) has uniform intensity. Whereas, the intensity of the resultant wave in the image on the right, is not uniform. Why?

Answer

There is a key difference between the two cases, and you have to consider the phase of each wavelet. "Wavefronts" that create a double-slit interference pattern come only from the two slits. When the two spherical waves (wavelets) reach a point $P$ on the screen they have each picked up a phase proportional to the distance between $P$ and their slit, and the resultant amplitude at $P$ will depend on the difference between the two phases.

Now, for the propagating plane wave each point on the wavefront creates a wavelet, and each wavelet may contribute at a point $P$ (you may picture this as being equivalent to "placing a slit at each point on the wavefront"). Now you see that problem is harder, since we have to add up all the (infinite) contributions from the wavelets to find the amplitude at $P$! Fortunately, for points on the wavefront far from the projection of $P$ the phase picked up during propagation will be wildly different (since the distance traveled is considerably big), and because of this there's a tendency for mutual cancelation of these wavelets. Therefore only the wavelets created "near" the projection of $P$ contribute significantly, and since they were created near each other their phase will not change too much and they'll add up constructively at $P$.

To actually calculate and verify that the amplitude is the same is better to use directly the formal treatment of waves through a wave equation though.

special relativity - If traveling at the speed of lights stops time, why does it take light 8 minutes to reach Earth?

I just learned that, according to Einstein's relativity theory, time reaches zero for an observer (light) when traveling at the speed of light, so everything is supposed to be at the same place in the universe for light. But why does it take 8 minutes for light to travel from the Sun to Earth? Is it because we are observing it from earth? At 300.000 km/s distances in the universe are hardly zero? I can't seem to grasp this.

Answer

Light travels at the speed $c$ this speed is finite and with out using any relativity we can calculate the time it takes for something travelling at this speed to reach us: $\text{time} = \frac{\text{Distance}}{\text{speed}}$ or $ t= \frac{d}{c} = \text{8 minutes}$ in this case.

For a person travelling very close to the speed of light with velocity $v$ from the sun towards the earth time does slows down, and he goes past the earth in a matter of seconds. But for us time doesn't slow we see the person with almost the speed of light and the time it takes to reach us is again $ t= \frac{d}{v}$ which will be almost 8 minutes but slightly longer.

Now for light you say time freezes completely this is not really accurate, but for arguments sake I will accept it: Then the same logic applies as before. For light it seems that zero time has passed but for us it is still 8 minutes.

This might seem like a paradox, but time is relative in Einstein's theory of relativiy.

Note that your argumentation is backwards, "time reaches zero for light, because everything is at the same place". While the more "correct" way to say it would be that the photon does not experience time and therefore everything seems to be at the same place.

Hope this helps

chess - Checkmate all the kings #1

The purpose of this puzzle is to find the minimum number of moves to checkmate all the kings.

RULES

• You are playing as White and you can make as many moves as you want before Black's turn.


• During your moves you can take any black piece except kings.


• During your moves your king can not be in check position.


• At the end of your turn all the black kings must be check mate : if Black can make one move that ends with one king being safe, you don't win. Note that this one move can't be a king moving to a threatened position.


• One piece can be used in multiple checkmates (you don't have to take all the king, just to checkmate them)

Examples :

This is a valid ending position because both kings are check mate (A king can not move in a check position to defend another king)

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This is not a valid ending position because one of the king is not checkmate

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This is a valid ending : a single piece can checkmate 2 kings

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I have tried to make this first puzzle easy to introduce the concept. I will make harder similar puzzles later, feel free to create one yourself too...

Answer

I got a quick answer but not sure how good it is.

1 rook to g6(2 moves), the other rook at a8(1 move)
1 knight to d5(2 moves), the other knight to e5(3 moves)
total 8 moves

waves - Why is the angle of the wake of a duck constant?

Why is the angle of a wake of a duck constant? And why are some conditions on the water depth necessary?

I realize that this question turns up in google searches, but I did not see a good discussion. I will be quite happy with a link.

Edited to add:

Could anyone tell me how the two up-voted answers are related?

Answer

The ideal Kelvin boat wake ignores surface tension, and it assumes deep water waves with an (in general) broad spectrum of frequencies $\omega$ with dispersion relation $\omega^2=gk$, where $g\approx 9.8 \frac{m}{s^2}$. The ideal Kelvin wake furthermore assumes that the ship sails with a constant velocity, and that the wave amplitudes of the partial waves are so small that they obey a linear superposition principle. The Kelvin wake does not describe the narrow turbulent band behind a ship, nor shock waves. The Kelvin wake consists of two types of waves: transverse and divergent waves. There are two characteristic angles

$$\alpha\approx 19^{\circ} \qquad \mathrm{and} \qquad \beta\approx 35^{\circ},$$

corresponding to

$$\tan(\alpha)= \frac{1}{2\sqrt{2}} \qquad \mathrm{and} \qquad \tan(\beta) = \frac{1}{\sqrt{2}},$$

or equivalently,

$$\sin(\alpha)= \frac{1}{3} \qquad \mathrm{and} \qquad \sin(\beta) = \frac{1}{\sqrt{3}}.$$

In polar coordinates $(r,\theta)$ of a co-moving coordinate system, where the position of the boat is at the origin, the transverse waves are in the region $|\theta|\leq \beta$, and divergent waves are in the region $\alpha\leq |\theta|\leq \beta$.

The angles $\alpha$ and $\beta$ are constant in at least two ways: Firstly, they don't depend on the distance $r$ to the ship. This is because the speed of each partial wave (with frequency $\omega$) is independent of the position $(x,y)$. Secondly, $\alpha$ and $\beta$ are, evidently, universal angles, independent of, for instance, $g$. This is explained in the references below.

_{(source: wikiwaves.org)}

References:

1) Howard Georgi, "The Physics of Waves", Chapter 14. (Hat tip:user1631.)

2) MIT on-line open course ware, mechanical engineering, wave propagation, lecture notes, fall 2006, Chapter 4.7.

3) Wikiwaves.

Greatest volumetric heat capacity

Is there any substance with bigger volumetric heat capacity than water? According to this table water has the biggest known VHC. But I can't believe that in the 21. century we have no special material with larger VHC.

Answer

Here is a thesis that details construction and characterization of thin films with large volumetric heat capacities (some nearing $6\ MJ\cdot m^{-3}K^{-1}$):

Volumetric heat capacity enhancement in ultrathin fluorocarbon polymers for capacitive thermal management

homework and exercises - Derivation of Ohm's law using classical and quantum model

How can I derive Ohm's law using classical and quantum model? Please help me with this.

special relativity - Importance of Powers of Velocity in Classical Mechanics

Is there any general significance to calculated quantities that depend purely on general powers of the velocity of a particle/system/etc? The first power being momentum and the second being kinetic energy.

I know that in relativistic mechanics the momentum and energy become quantities that must actually be expressed in infinite orders of velocity since energy and momentum are functions that can be expressed as power series of velocity. So, if there is any significance to the momentum and energy being to first and second order in Newtonian mechanics respectively, why do they go to functions of infinite order when special relativistic effects become a concern.

Or am I making connections that lead nowhere?

electromagnetism - Where is the energy stored in an inductor?

In an inductor,

Most text books say that the $(1/2)Li^2$ is stored in the magnetic field.

But is there another way to explain this?

In a capacitor I understand that all the energy that the battery provides is used up to seperate the two oppositely charged plates.

Is there an analogy for Inductors? OR Could you explain how energy can be stored in a

magnetic field? Does that mean energy is stored in an electric field produced by a point charge?

Answer

Does that mean energy is stored in an electric field produced by a point charge?

The classical self energy of a point charge is formally infinite and, thus, somewhat of an embarrassment.

Could you explain how energy can be stored in a magnetic field?

It's clear that energy is stored in a magnetic field so I'm not sure what you're looking for here.

When work is done, energy is converted from one form to another. Work is being done by the battery when the current through an inductor is increasing.

This is simply due to the fact the product of voltage and current is power, the rate at which work is done.

And, when the magnetic field threading the inductor coils is changing, there is a voltage across the inductor.

Thus, when the current through the inductor is increasing, there is a voltage across, proportional to the rate of change of current, and thus, an associated power

$$p_L = i_L \cdot v_L = i_L \cdot L \frac{di_L}{dt} $$

Further, when the current is decreasing, work is being done by the inductor on the circuit. So, the work done increasing the current (and associated magnetic field) is stored as energy in the magnetic field.

kinematics - If I lift a box vertically, why is the work I do equal to the distance I lift it times the force of gravity on the box?

I have problems fully understanding the concept of work, so please forgive me if this is simple. If I take a box of mass $ m $, and lift it a distance $ d $ vertically, why is the work I have done equal to $ gmd $, where $ g $ is the force gravity exerts on the box? I understand that work is equal to force times distance--so I'm not asking about the definition of work--but if I exert an upward force equal in magnitude to gravity's, won't the box remain motionless, i.e., net zero force, in which case the velocity is constant, and displacement and work done will be equal to zero?

Edit: To be clear, what I'm asking is not a duplicate of "Why does holding something up cost energy while no work is being done?", because I'm not asking about work done on an object with zero displacement, nor is it a duplicate of "What exactly is F in W=∫baFdx?", because I'm not asking about the distinction between the work done by an individual force and net force.

Is there a "Size" Cutoff to Quantum Behaviour?

We all know that subatomic particles exhibit quantum behavior. I was wondering if there's a cutoff in size where we stop exhibiting such behavior.

From what I have read, it seems to me that we still see quantum effects up to the nanometer level.

Answer

The classic experiment demonstrating quantum effects, the 2 slit experiment, has been preformed with subsequently larger and larger particles as our technology available to do it advanced. Originally, it was performed with electrons, which are just as much matter as any other matter, but are extremely small. The largest particle it has been demonstrated with are Buckminsterfullerene, which contain 60 Carbon atoms. For size comparison:

$$m_e = 5.485 x 10^{-4}u$$ $$m_{buckyball} = 720.642 u$$

There is a good reason that the experiment gets more difficult with increasing mass, and to be sure, the buckball experiment was quite an accomplishment. To the basics of quantum mechanics:

$$\Delta x\, \Delta p \ge \frac{\hbar}{2}$$

Alternatively, the de Broglie wavelength is:

$$\lambda = \frac{h}{p} = \frac{h}{m v}$$

I believe that in order to obtain the same wavelength with the same mass you have to decrease the velocity. The reason this could be problematic for such experiments is that it is hard to successfully create the conditions needed with a large and slower moving particle, such as needing a better vacuum.

quantum mechanics - Calculating the Berry curvature in case of degenerate levels (Non abelian Berry curvature): issue

The Berry phase accumulated on a path can be described by a matrix when we look at adiabatic time evolution with a Hamiltonian with degenerate energy levels. The Berry phase matrix is given by $$ \gamma_{mn}= \int_\mathcal{C} \left\langle m(R) \right | \nabla_R \left| n(R) \right \rangle . d R. $$

here $R$ parametrizes the said path and $ A_{mn}= \left\langle m(R) \right | \nabla_R \left| n(R) \right \rangle$. Now what I want to do is calculate the Berry Curvature, something that, if I assume my path above is closed and has three determining coordinates$R_1$, $R_2$ , $R_3$ is $\vec{F}$ such that

$$ \mathbf{\gamma}=\int_\mathcal{S} \vec{F}.d\vec{s} $$ note that the $\gamma$ and $F$ here are matrices and we're integrating over the surface $\mathcal{S}$ enclosed by curve $\mathcal{C}$

What is stopping me from applying the stokes theorem to $\gamma_{mn}$ and getting $\vec{F}_{mn}=\nabla_R\times A_{mn} $?

It is said that the answer contains a matrix commutator $[A_i,A_j]_{mn}$ c because this berry phase is non abelian. but I seem to be missing something fundamental.

Edit: Note: this also corresponds to problem 2 Chapter 2 of Topological insulators and superconductors by Bernevig and Hughes

Answer

The short answer is that only if the Berry curvature is defined by: (in matrix notation): $$F_{\mu \nu} = \partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu} + [A_{\mu}, A_{\nu}]$$ it becomes gauge covariant, i.e., for a gauge transformation: $$A_{\mu} \rightarrow g^{-1}A_{\mu} g+g^{-1}\partial_{\mu}g$$ $g \in U(N)$ ($N$ is the degeneracy of the level), the curvature transforms as: $$F_{\mu \nu} \rightarrow g^{-1}F_{\mu \nu} g$$ The partial expression $F_{\mu \nu} = \partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}$ will be covariant (actually invariant) only with respect to Abelian gauge transformations and not the whole $U(N)$ group.

The reason that the Berry curvature cannot be obtained as in the Abelian case by a direct application of the Stokes theorem is that the Stokes theorem does not exist(in the usual sense) in the non Abelian case becuse the holonomy needs to be path ordered:

$$\mathcal{Hol}(A) = \mathrm{P} e^{\int_C A_{\mu}dx^{\mu}}$$

However, it can be applied sequentially for an infinitesimal path. let us choose a square path in the $(xy)$ plane centered at $(x,y)$ with infinitesimal sides of length $\Delta x$ and $\Delta y$ as depicted in the picture.

Therefore

$$ \begin{matrix} \mathcal{Hol}(A) \approx e^{\int_a^b A_{\mu}dx^{\mu}} e^{\int_b^c A_{\mu}dx^{\mu}} e^{\int_c^d A_{\mu}dx^{\mu}} e^{\int_d^a A_{\mu}dx^{\mu}} \\\approx (1+A_x\Delta x-\frac{1}{2}\partial_yA_x\Delta x \Delta y) (1+A_y\Delta y+\frac{1}{2}\partial_x A_y\Delta x \Delta y) (1-A_x\Delta x-\frac{1}{2}\partial_yA_x\Delta x \Delta y)(1-A_y\Delta y+\frac{1}{2}\partial_x A_y\Delta x \Delta y) \\ \approx 1+(\partial_x A_y - \partial_y A_x + [A_x, A_y])\Delta x \Delta y\approx (1+ F_{xy}\Delta x \Delta y) \end{matrix} $$

Please notice that the commutator term is created due to the need to exchange the order of the second and the third expressions in the product in order to cancel $A_x \Delta_x$.

general relativity - Intuitive meaning of Globally Hyperbolic

I am been studying differential geometry and spacetime and I keep coming across the term globally hyperbolic. I am having a hard time coming up with an intuitive understanding of this idea. What is an intuitive meaning of globally hyperbolic.

My background is mostly mathematics at a higher undergraduate level.

Answer

Globally hyperbolic refers to the fact that hyperbolic equations always have locally a well defined Cauchy problem, that is, a unique development given initial conditions. Which means that, given a matter field at a time $t_1$, there exists a unique solution of that field at a time $t_2$. It is boosted up to globally hyperbolic if that property holds globally.

Globally hyperbolic isn't directly related to this notion (you can still have hyperbolic equations for which this property does not hold up globally in a globally hyperbolic spacetime), but the two notions are close. A globally hyperbolic is, formally, a spacetime that is both causal and such that the intersection of the past and future of two different points is compact, which roughly translates to having no naked singularities (a point removed from spacetime might cause that set to fail to be compact).

Alternatively, global hyperbolicity can be defined by the existence of a Cauchy surface - an achronal (cannot be linked by any causal curve) spacelike hypersurface such that every causal curve crosses it exactly once. No more (no curves going to the same time more than once) and no less (no curve appearing out of nowhere).

Breaks in causality and naked singularities may cause a lack of existence or uniqueness of solutions for matter fields - corresponding either to matter going round in a circle, so to speak, or coming out of nowhere or disappearing. Globally hyperbolic spacetimes benefit from a theorem due to Hawking stating that, for such a spacetime with a reasonable enough matter field (basically if it already behaves nicely in flat space), any hyperbolic equation is globally hyperbolic.

homework and exercises - Why does the light bulb's brightness decrease?

The book said that the brightness of R1 would increase and R2 would decrease. I don't understand this at all.

Why does adding a wire from b to c change anything to R1?

Shouldn't charge still flow through R1 as they should and R2's brightness decrease because the charges at point b have an alternate route to travel to point 2?

Answer

The voltage drop across R₂ becomes 0, and the full voltage is applied across R₁ instead.

electrostatics - Help me understand static electricity

This is what I understand about electricity: (The following information is paraphrased from the book CODE by Charles Petzold.)

Atoms are made up of protons, electrons and neutrons. Protons and electrons tend to exist in equal numbers in the atoms to which they reside. Sometimes it is possible for an electron to jump from one atom to another. This is electricity.

The words electron and electricity come from the Greek word for amber. This is because the Greeks discovered static electricity when they experimented with rubbing amber and wool together. In these experiments, the electrons would jump from the amber over to the wool.

More modern day experiments can be conducted with shoes and carpet.

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Here is where I am confused:

Petzold says:

When the carpet picks up electrons from your shoes, eventually, everything gets evened out when you touch something and feel a spark. That spark of static electricity is the movement of electrons by a rather circuitous route from the carpet through your body back to your shoes.

After this explanation he moves on to a larger concept, but I am left slightly confused.

• 
  

  Exactly what is this "circuitous route"? Does the thing I touch also have to be touching the carpet?

  
  


• 
  

  Why did the electrons jump from my shoes to the carpet in the first place?

  
  


• 
  

  If the electrons were transferred to the carpet for a reason, what is it that makes them get transferred back to my shoes?

  
  

Answer

Exactly what is this "circuitous route"? Does the thing I touch also have to be touching the carpet?

Though I'm not a native English speaker I am pretty sure that a circuitous route is a path that combines you shoes and the carpet as were they a part of a circuit.

The thing you touch has to be connected to the carpet (by touching the carpet itself or by touching some other conductor touching the carpet. There must be a conducting path).

Why did the electrons jump from my shoes to the carpet in the first place?

Electrons are bound in atoms at different energy states. Certain amounts of energy could rip off an electron from an atom, and thereby off from your shoe. By brushing your shoe over the carpet, you are adding energy, and in some points that energy and the force done is intensive enough to rip loose an electron that is left on the carpet.

Moving a charge to a neutral location is not a place the charge would naturally go. At the same time the electron might be held tight to the original atom. Force must be done to add enough energy to overcome such energy barrier.

If the electrons were transferred to the carpet for a reason, what is it that makes them get transferred back to my shoes?

Since you moved an electron away from a neutral position, this original location is now missing one negative charge. There is now one too many positive charges at this location, and the net charge is positive.

The new location of the electron was neutral before the electron came. This location now has one extra unbalanced electron and has a negative net charge.

As mentioned before, too move this electron required energy, since you are moving it out of equilibrium. It wants to go back to equilibrium - that is, a negatively charged electron is attracted to a positively charge location. Whenever you create a path for this electron to move back to the positively (or any other positively) charged location, this electron takes the trip and moves there.

According to the amount of negative charge you have build up (which has got to do with how much energy you have put into your shoe-vs-carpet rubbing), the larger the electric "shock" will be, since more charge is moving.

visual - 3D nonogram, beginner's edition

Trying out the concept of 3D nonograms. This one should be pretty simple.

Name the depicted object.

Answer

nonogram:

this is depicting

a 6-sided dice!

classical mechanics - How does the Hamiltonian change if $Lto L + frac{dF}{dt}$?

The Hamiltonian is defined as the Legendre transform of the Lagrangian $$H = p\dot{q} -L .$$ In the Lagrangian formalism we are free to add the total derivative of an arbitrary function $F=F(q,t)$ to the Lagrangian $$ L \to L' = L + \frac{dF}{dt} $$ because such an additional term has no effect on the Euler-Lagrange equation.

For the momentum $ p = \frac{\partial L}{\partial \dot{q}} $ such a shift of the Lagrangian implies $$ p \to p'= p + \frac{\partial F}{\partial q} .$$

If we combine these two observations, we find the following transformation law for the Hamiltonian \begin{align} H(q,p) \to H' &= p' \dot q' - L' \\ &= \left( p + \frac{\partial F}{\partial q} \right) \dot q - \Big( L + \frac{dF}{dt} \Big) \\ &= H + \frac{\partial F}{\partial q} \dot q - \frac{dF}{dt} \end{align} Is this transformation law correct or am I missing something important? (In the literature I've only found several times the statement that the Hamiltonian like the Lagrangian is allowed at most to change be the total derivative of some function. But here it seem that there is an additional term...)

everyday life - Why is boiling water loud, then quiet?

Water in my electric kettle makes the most noise sixty to ninety seconds before the water comes to a full boil. I have been fooled many times by the noisy kettle, only to discover that the water was not yet hot enough for tea. The kettle is only at a full boil after the noise has subsided.

I have noticed the same phenomenon with many other kettles, including conventional kettles on kitchen ranges; it is not a peculiarity of this electric kettle.

Why does the boiling become quieter as the water reaches full boil?

condensed matter - Why does water ($mathrm{H_2O}$) only have two distinct fluid phases?

Water (and other substances) can exist in many distinct solid phases (with different crystallic micro-structure), but only in two fluid phases - liquid and gaseous, in which the molecules are oriented randomly (they is no long range order). Is there an explanation in the molecular theory, why there are just two "disordered" phases? Why isn't there just one? Or more than two?

lateral thinking - It's opposite day!

Explain to me that it is opposite day. On opposite day, everything you say is reversed. "No" means "yes", "is" means "isn't", and so on. Because you are trying to explain to me it is opposite day on opposite day, the task isn't simple. For example:

• 
  

  If you tell me it is opposite day, you will be saying that it isn't opposite day, and therefore won't convince me.

  
  


• 
  

  If you tell me it is not opposite day, I will simply believe and agree that it is not opposite day.

  
  

What can you tell me to convince me, without a doubt, that it is opposite day?

Edit: For those asking, the sentence's meaning is reversed. So "I am going to tell a lie" turns into "I am not going to tell a lie".

Edit #2: For those claiming the riddle is too broad, opposite day has been revised to changing meaning. There is a specific task and specific parameters.

Answer

I would say

I am going to tell a lie. Today is not opposite day.

Which means

I am not going to tell a lie. Today is opposite day.

Both will convey the same meaning that today is opposite day.

terminology - Meaning of dimension in dimensional analysis

I was wondering what dimension can mean in physics?

I know it can mean the dimension of the space and time.

But there is dimensional analysis. How is this dimension related to and different from the previous one? How is it related to and different from units (e.g. kilometer)?

Answer

A dimension (in dimensional analysis) is defined by the transformation law of an object under changes in scale. If I have an object which is twice as big, it has 4 times the surface area and 8 times the volume, so the surface area has dimension of length-squared, and the volume has dimension of length-cubed. Dimensional analysis is applied whenever you have a quantity where there is a scale that you can change.

There are two different notions of dimension of space which took a while to be disentangled. The topological dimension is defined inductively by the cutting properties of the space. If a space can be cut in two by a point, it's 1 dimensional. If it can be cut in two by a 1 dimensional shape, it's two dimensional. This type of definition requires care for wild shapes, but it produces an integer dimension of the space.

The scaling dimension, or fractal dimension, is defined differently, in terms of distances on the space. The scaling dimension counts the number of boxes of size A required to cover the space, and sees how this goes up as A gets small. The exponent is the scaling dimension.

quantum mechanics - How do experiments prove that fermion wavefunctions really pick up a minus sign when rotated by $2pi$?

Theoretically, after a rotation of $2\pi$, a fermion wavefunction picks up a minus sign, and it is after a rotation through $4\pi$ that it returns to its initial quantum state. Now, the wave-functions or the quantum state is not a directly measuable quantity. Then how will the fact that "the wavefunction picks up a minus sign" be reflected in measurements?

Answer

The main experiment to mention here is neutron interference. While you cannot detect the total phase of a state - $\lvert \psi \rangle$ and $-\lvert \psi\rangle$ represent the same quantum state and cannot be distinguished - you can detect relative phases, i.e. the $\phi$ in $$ \lvert \psi_1\rangle + \mathrm{e}^{\mathrm{i}\phi}\lvert \psi_2\rangle.$$ So the trick here is to "rotate" $\lvert \psi_2\rangle$ but not $\lvert \psi_1\rangle$. For this, we take advantage of the coupling of spin to magnetic fields, described for instance here. The reason we use neutrons is that they are uncharged and will therefore not be deflected by a magnetic field in the manner electrons would.

We split a neutron beam in two and subject one of the two resulting beams to a constant magnetic field $B_0$ in $z$-direction. The Hamiltonian (generically $\propto \vec S\cdot \vec B$) becomes $$ H = k B_0 S_z,$$ where $k = \frac{g_n e}{m_e c}$ is a bunch of constants determining the magnetic moment of the neutron. With $\omega := k B_0$ this now means that the phase $\phi$ is $\omega t/2$ (the half comes from the half-spin of the neutron and is precisely what we want to detect!), where $t$ is the time travelled, which you can relate to the distance travelled and thereby to the interference pattern on your screen.

That's basically it, once you see that your pattern is consistent with $\omega t/2$, you've shown that a $2\pi$ rotation, corresponding to $\omega t = 2\pi$ since the time evolution is then $\mathrm{e}^{\mathrm{i}2\pi S_z}$, aka a "full rotation", just acts as a half-rotation on the neutron state.

mathematics - Painting a 4x6 grid with 2 colours

Can you paint a 4x6 grid with 2 colours such that it doesn't contain any rectangles whose corners are all the same colour? Can you do it without a computer? Rectangles must be 2x2 or greater and parallel to the grid's sides.

Good luck!

Answer

Absent the no-computers tag, I just wrote a quick Python script:

The formulation of the problem was pretty easy too, just

Treat each grid cell as a position in a (in this case, 24-digit) binary number. Each rectangle can be represented as a bitmask with 1s at the positions of its corners. There are 90 such masks. Now just enumerate the first $2^{24}$ numbers, and for each mask you can check whether the number totally intersects it, or totally doesn't (i.e. the AND of the number and the mask is either equal to the mask, or zero). If none of the masks match this criterion, then the number is a solution. This is the first one I found, but it is one of 720 solutions (actually 360 solutions if you don't distinguish between a coloring and its inverse).

I did some more exploring, and

It seems that there aren't too many (types of) grids that can support this. All 1xN grids are trivially solutions. For 2xN grids, you can always make one row entirely one color, and the other the other color. For 3xN, you can do up to N=6, but 3x7 fails (and therefore anything bigger also fails). As shown here, 4x6 works. But 5x5 fails, meaning that this is an exhaustive list (1xN, 2xN, and anything that fits in a 4x6 grid).

magnetic fields - Physical interpretation of 2-forms dual to pseudovectors

Mathematically for every 3D pseudovector $x^i$ there is a 2-form $F_{ij}=\epsilon_{ijk}x^k$ such that the 2-form transforms properly under all orthogonal transformations. Therefore I would expect it would be more natural to write physical quantities such as angular momentum $\textbf{L}$ or magnetic field $\textbf{B}$ in terms of their corresponding 2-forms.

Is there any physical insight as to why these quantities behave the way they do apart from experimental verification. If it is simply the way they are, is there any insightful interpretation of their corresponding 2-forms? I seem to be able to get some intuition from looking at the vectors but none at all by analysing the 2-forms.

Is the way these vectors physically behave related to their pseudoness? For example the rather odd direction of magnetic force.

Answer

Angular momentum is a very instructive example to look at - in particular, to look at how the notion of angular momentum (or, in fact, rotation), changes when you consider more or fewer than the usual three spatial dimensions. The proper notion of angular momentum that generalizes to all dimensions is $L = \vec r\wedge \vec p$, i.e. a 2-form, the wedge product of position and momentum. In three dimensions, the Hodge dual of this form is the ordinary pseudovector of angular momentum, and in fact one might define the cross product in 3d as the Hodge dual of the wedge product. You should think of the 2-form $L$ as describing the plane in which the rotation happens, together with some numbers encoding its direction and speed.

Let's start in one dimension: There is no rotation for an object in one dimension - it can only move forward and backward, and nowhere else. This corresponds to the second exterior power of a one-dimensional vector space being zero - there are no 2-forms, hence no rotation.

In two dimensions, there is clearly rotation, imagine a one-dimensional "rod" spinning in a plane. You might be tempted to describe the rotation as the 3d vector of angular momentum perpendicular to the plane, but this is an extrinstic description. If the world were truly two-dimensional, this description would not be available - but the description by two-forms is available. 2-forms in 2 dimensions are dual to 0-forms, i.e. scalars, so rotation in a fixed plane is fully described not by a vector, but by a number - its magnitude tells you how fast the rotation is and the sign whether it is clockwise or counter-clockwise.

In three dimensions, we get the familiar duality between 2-forms and 1-forms/vectors. But note that there is really nothing about rotation that would force you to describe it as "rotation about an axis" rather than "rotation in a plane" - the two descriptions/interpretations are fully dual, and it is the latter that generalizes to all dimensions.

In four dimensions...well, I get that this is not visual anymore, but think about special relativity, and the Lorentz transformations, which are generalized rotations - they are generated by 2-tensors, not vectors, and their associated conserved quantity is a 2-tensor, the energy-momentum tensor, not a vector.

Note that I have nowhere relied on the "pseudoness" of the angular momentum vector in 3d. It's an artifact of the Hodge dual not commuting with reflections, but it's really not the defining property of a "pseudovector". A "pseudovector" is not a vector at all, it is intrinsically a 2-form, and especially when you generalize to other dimensions you must respect that, as I also pointed out here.

That the magnetic force is a pseudovector and not a vector is something you can only appreciate after switching to the covariant formulation of electromagnetism and recognizing the electric and magnetic fields as certain parts of the electromagnetic field strength tensor - and once again, you will find that going to other dimensions shows that the magnetic field is not fundamentally a "vector" at all - in particular, it is equivalent to a scalar in 2 spatial dimensions, and to a more complicated $d-2$-form in higher dimensions. You might think of the magnetic 2-form as a collection of planes the velocities of charged particles are "dragged along" in the sense that the more parallel their velocity is to these planes, the stronger is the magnetic force that tries to make them describe circles in those planes:

Consider that the vector $\vec B$ is perpendicular to the planes its Hodge dual 2-form describes and that the Lorentz force is $\vec v\times \vec B$. This is maximal when $\vec v$ and $\vec B$ are perpendicular, i.e. when $\vec v$ lies in the plane the dual encodes, and the Lorentz force is also also perpendicular to $\vec B$, so it always lies in that plane.

Text Rebus Extreme! 2!

The second rebus question in the series.

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here
____
bored

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bets
_

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all4114

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knightleg

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A
\
C

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<3 <-

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Requirements:

• You must answer each one


• You must explain how each one works


• Both of the above must be correct

Hint #1: They are all cliches.

newtonian mechanics - Deceleration rate of objects of different mass but the same otherwise

Using a tennis ball as an example object, if one ball weighs 1 ounce and the other is 2 ounces, and both are struck at 100 mph on the same trajectory, would there be any difference in the deceleration rate between the 2 different mass balls? (All other things about the two balls being equal). For instance, would the elapsed time for the ball to travel let's say 100 feet be different or the same?

FOLLOW UP QUESTION (based on answer #1): If it's true that the lighter ball would decelerate faster (and consequently take longer to travel a given distance), then what would be the difference in the initial velocity of the two balls (one 1 ounce, the other 2 ounces) if both are struck with the same implement with the same force (let's say a tennis racket travelling 100 mph).

I'm assuming the lighter ball would have a higher initial speed. If so, would the higher initial speed of the lighter ball offset the increased deceleration rate for the lighter ball. In practical terms: with the initial speed being different, which ball would arrive first in the above example of 100 feet, the lighter ball or the heavier ball?

If not too difficult, could you explain how this relationship (first the initial speed difference and then the total travel time difference for 100 feet) would be calculated?

gravity - Can a photon be made to orbit a known (or undiscovered theoretical) body?

Can a photon through some process be made to orbit a celestial or any other object?

Two follow-up questions.

1. 
   

   Can this orbit be described as the photon crossing its own path.

   
   


2. 
   

   Will this wave-function be effected by positive interference. To the effect of increasing frequency?

   
   

Answer

In short, yes. But there are 2 caveats for the orbit for a massless particle around a spherical body. Both can be seen from the following plot, borrowed form "Spacetime and Geometry" by Carroll pg.211:

i) There is only one possible orbit for $r = 3 GM = \frac{3 R_{s}}{2}$ where $R_s$ is the Schwarzschild radius. This orbital radius could very well be within the radius of the body being orbited, hence the particle may not even be able to get this trajectory in the first place.

ii) The orbit is unstable.

EDIT: As for the updated, additional questions:

First of all, the above analysis was done for a classical particle, so there is no notion of waves or interference there. In this sense, the photon will cross its own path the same way my dog crosses his own path when he runs in a circle.

If you really want to treat the photon quantum mechanically, I can only take a stab at it under some idealized circumstances. In one idealization we could take the photon state is very localized (high probability of detecting it in small range, virtually zero everywhere else). We can send this photon once around in the orbit - in this case it will act a lot like a classical particle in that it won't interfere at all with the 'tail' of the wavefunction from the prior pass. (That we can set up states like this is merely an intuitive notion on my part since I know we can set up single photon states and there is little chance of detecting them at say, Alpha Centauri, so localized states of a single photon seem to be possible). Now for the other idealization, we could take a state that isn't very localized at all, say a plane wave directed tangentially to the orbit. Now that the wave function isn't localized it's going to be able to 'feel' all around the space and I think its going to want to 'fall off' the trajectory where it can be at a state of lower energy. That is, over a long period of time its going to sense the instability and have a very small probability of being detected at along the radius at $ 3 G M$ rendering the question sort of moot. One thing you could do is stand on one side of the body being orbited and shoot a photon towards it and have your buddy on the other side detect it, which is pretty much the double slit experiment and will result in interference (the 2 paths are going left or right of the body being orbited). (I am more posting this part of my answer as a guess to see what other people think and less out of certainty).

I want to emphasize point (ii) above - something that is unstable is a bit like having a needle balanced on its tip - sure you can do it for a split second but the slightest movement will make it fall over so really no matter how many special circumstances you want to invent or questions you want to ask about the photon orbiting its just not going to stay there very long.

astrophysics - Storing a Planet-sized Chunk of Metal Inside a Star

Would it be physically possible to "store" a planet-size or larger sum of metal, say gold or platinum, inside a star by letting it fall to the core?

Would it be possible to detect which stars had these treasures inside them?

(This is for a Sci-Fi project, but I'd like to root it in reality).

Answer

In astronomy parlance, the Sun has a "metal"$^{1}$ mass fraction of about $0.02$. A solar mass is $\sim2\times10^{30}\;\rm{kg}$, so the sun contains about $4\times10^{28}\;{\rm kg}$ of "metals". That's about $20$ times the mass of Jupiter. A lot of that metal mass will be ${\rm C}$ and ${\rm O}$ and other elements a chemist would call non-metals, but I think there should be enough ${\rm Fe}$, ${\rm Na}$, ${\rm Mg}$, etc. to make at least a small planet or large moon.

The elements you drop into the star would be roughly sorted into concentric spherical shells, ordered with the heaviest elements in the middle, given enough time. There is a serious risk, depending on the masses and elements involved, that whatever you drop in starts fusing and making different elements, or if the right thresholds are exceeded, that the entire thing explodes in a supernova.

The Sun is a fairly typical star, not especially massive or puny, metallicity not remarkably high or low. I already showed that the Sun has a fair bit of metal without anyone dropping in any extra, and there are massive metal rich stars out there that have more than a solar mass worth of metals inside occurring naturally. In fact, very nearly every atom in the Universe that is not hydrogen, helium or lithium was made inside a star (and anything heavier than iron was most likely made when a star exploded). Some metals get ejected in supernovae and in stellar winds, but a large fraction of the metal budget of the Universe is already locked up in stars.

It would be possible to detect the contents of a star, with a carefully measured spectrum of the atmosphere and sophisticated stellar modelling (the spectrum serves as a boundary condition for the model). It would be more difficult than what astronomers do today since part of what goes into the models is guided by how metals are transported naturally in the Universe; artificially moving stuff around throws a wrench in the gears, but it's plausible that a concerted effort by an intelligent civilisation could develop the necessary science.

Olin Lathrop and John Rennie have raised some concerns about retrieval. I agree that a wormhole is probably a bad idea. Perhaps your best bet for retrieval is to set some carefully calculated extra mass on a collision course, stand (way, way, way) back, let the star go BOOM, wait a thousand years or so for things to cool off, then harvest the metals out of the gas of the nebula. A $1000$ year old supernova remnant is still pretty harsh conditions; the Crab nebula (exploded in the $11^{\rm th}$ century) has a temperature of about $10,000\;{\rm K}$, but rather low density. I'd call it plausibly survivable by a suitably advanced spacecraft.

$^{1}$Astronomers call everything that is not ${\rm H}$ or ${\rm He}$ a "metal".

mathematics - The town of Concentrië

Here you see a water map of the town of Concentrië. There are eight channels that run from the central boathouse of the city in a star shape.

There are three canals that lie concentric around the central boathouse. There is one-way traffic on the canals, indicated by the green arrows. On the canals you can sail both ways ; you cannot turn halfway ; at a crossing of a canal you may turn.

Every piece of a canal from boathouse to the first canal or from canal to canal has a length of 4.
Each piece of canal from canal to canal has a length of 3 (inner canal), 6 (middle canal) or 9 (outer canal).

From the central boathouse you have to deliver six packages, at the addresses indicated with the red triangles. Then you have to return to the central boathouse.

Specify the length of the shortest route you can take to return all packages as an answer.

knowledge - Guide to Codes and Ciphers

^{Note: This is a Guide not a Puzzle}

ciphers are quite common now on puzzling, and at first can seem quite confusing. But ciphers are bigger than this site, used worldwide by companies and secret services to encrypt data.

But what is a cipher, and what's the difference between a code and a cipher? What types of codes and ciphers are there and how can I make or solve them?

(The main purpose of this post is to help out newcomers to the site who may be a bit daunted or confused at the sight of ciphers, but who knows? An experienced user could learn something here too, I certainly did in my research :) )

Answer

This guide aims to explain various ciphers, help you understand how they work, and how to decode them with or without a key.

This answer is currently being split into multiple posts to improve scrollability and readability after some advice from other users. This may take a while, and apologies for the stop-start fashion of it.

Mission accomplished! This answer now contains links to separate posts of different types of ciphers, so there is no character limit allowing me to elaborate in more detail and to stop you having to scroll. Thanks a lot to @n_palum for helping!

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Index:

• What is a cipher?
  
  
  
  
  • Brief History
  
  
  • Definition
  
  
  • How to make a good one
  
  
  • Difference between Codes and Ciphers
  
  
  
  


• Types of cipher
  
  
  
  • Classes and definitions
  
  
  • Transposition ciphers
  
  
  
  • Monoalphabetic Substitution ciphers
  
  
  • Polygraphic Substitution ciphers
  
  
  • Polyalphabetic ciphers
  
  
  • Other ciphers
  
  
  • Mechanical Ciphers
  
  
  
  


• Cryptanalysis
  
  
  
  • Frequency Analysis
  
  
  
  • Index of Coincidence
  
  
  • Kasiski Examination
  
  
  
  


• Resources

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Brief History

Ciphers have played major parts in historical events dating back to around 1900 BCE where apparent nonsense hieroglyphics can be found. From there, ciphers have developed, a recipe found encrypted on a tablet from 1500 BCE, and Hebrew scholars using monoalphabetic ciphers in 600 BCE. Nowadays, ciphers are common, encryption used by companies, secret services and even everyday applications such as Whatsapp. They make the world a lot more secure, but what actually are these ciphers?

Definition

A cipher is, simply put, a way of hiding data using a disguised way of writing. It is usually an algorithm with the purpose of converting data to a code to stop outside parties from obtaining the data and allowing only the intended recipient access.

A cipher consists of at least two, often 3 pieces of data:

• The plaintext - the message or data which shall be encoded


• The key (Not used for all ciphers) - A piece of data which is required to decode the ciphertext to the plaintext


• The ciphertext - the encoded plaintext which is usually illegible

The process of encryption is

Plaintext -> Method of encryption (type of cipher) + Key (if required) -> Ciphertext

Decryption is the reverse.

How to make a good one

On puzzling, we don't want to just see a short string and be expected to solve it. For what to do and what not to do see this meta post

Difference between a Code and a Cipher

For everyone but cryptographers, the words code and cipher are synonymous. If you were to talk about codes and ciphers to someone you'd probably find they used the words interchangeably. But there is a difference.

Codes are everywhere, and you won't even notice the most of the time. A code replaces words or entire sentences or phrases with symbols or characters. The important thing here is that each set of symbols or characters have a meaning. These meanings are usually stored in a code book. For instance, telegraph communicators used code to convey messages quicker, here is an extract of one of their codebooks:

You can see different words on their own can mean whole sentences.

Codes are very common, and you use them without even thinking. A traffic light uses a colour code for the words 'stop', 'wait' and 'go'. Most people use code every day, probably including you, whilst talking in chat or texting things like 'brb', 'afaik' and 'idk'. The most common code, used for information interchange, is ASCII.

The point of codes isn't really to hide data, just converting it to an easier way to transmit.

A cipher, on the other hand, the ciphertext has no meaning whatsoever. Each character is replaced according to an algorithm. For instance, Morse code isn't a code, it's actually a cipher.

Most ciphers were invented to hide data.

The difference broken down:

Codes generally operate on semantics, meaning, while ciphers operate on syntax, symbols. A code is stored as a mapping in a codebook, while ciphers transform individual symbols according to an algorithm.

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Classes and definitions

There are two different categories of ciphers: Classical (pen and paper) and the more modern Mechanical (requires a machine).

There are several different classes of classical ciphers, as listed below:

• Transposition ciphers - Positions of the characters in the plaintext change, but the characters themselves remain the same


• Monoalphabetic substitution ciphers - Each character (not always true, but most) is replaced with a different character(s)


• Polygraphic substitution ciphers - Groups of characters are replaced


• Polyalphabetic ciphers - Characters are encoded using a different alphabet. Usually position dependent.


• Others - Completely different, or above classes are combined

There are a few mechanical ciphers, which I will write a brief note on after the classical ciphers below.

Transposition ciphers

Transposition ciphers involve moving the characters in the plaintext to different positions using an algorithm. The characters themselves remain unchanged, making this type of cipher insecure for short plaintexts.

See this separate answer for more details on different types of transposition ciphers.

Monoalphabetic substitution ciphers

Monoalphabetic substitution ciphers replace each letter in the plaintext with a different character/group of characters. If the plaintext is lengthy then these can be easily broken by frequency analysis.

See this separate answer for more details on different types of monoalphabetic substitution ciphers.

Polyalphabetic Substitution Ciphers

Polyalphabetic Substitution ciphers involve replacing characters in the plaintext with characters/groups of characters from an alternate alphabet.

See this separate answer for more details on different types of polyalphabetic substitution ciphers.

Polygraphic Ciphers

Polygraphic ciphers involve having groups of characters in the plaintext replaced.

See this separate answer for more details on different types of polygraphic ciphers.

Other ciphers

Other ciphers are out there and many don't fit into any of the above categories. They can be combination ciphers, combining elements above to make them stronger, or just be completely different.

See this separate answer for more details on different types of other ciphers.

Also, see this community wiki of other ciphers that have been missed out, and feel free to add to it!

Mechanical ciphers

Mechanical ciphers were invented in WWII. They rely on gearing mechanisms to shift letters through an alphabet to get the final message.

Most famous examples are the Enigma machine and the Lorenz machine. I won't be able to explain a machine very well, so I won't bother going into detail. See the links for more, or this list in Wikipedia.

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There are many ways to attempt to break a cipher without a key. Here are the best ways (taken from my answer here):

Cryptanalysis is defined as

'the art or process of deciphering coded messages without being told the key.'

If you have the key and know the encryption method, you can simply reverse the process to get to the plaintext.

If you have the key but not the encryption method, then this question covers how you can identify the cipher

However, if you have neither the key nor the encryption then you can use cryptanalysis.

This can be used to achieve a

• Total break — working out the key and the plaintext.


• Global deduction — discovering the method of encryption and finding the plaintext, but not the key.


• Distinguishing algorithm — identifying the cipher from a random permutation.

There are a couple of different ways to solve ciphers:

Frequency Analysis

Frequency analysis works best with substitutional or rotational ciphers, though both of those can have keys. Frequency analysis studies the frequency of letters in a ciphertext.

Computers have calculated that in the English language, the order of the most frequent letters from high to low is etaoinshrdlcumwfgypbvkjxqz.

Here is the stats for analysis on the English language, including unigram, bigrams, trigrams etc.

As you can see from this graph, 'e' is by far the most frequent letter. 't' - 'r' is a lot closer.

How to use

If the cipher is a substitution, and the ciphertext is quite large, then you can attempt to break the cipher.

Using an online tool such as this, you can find the most common letters and most frequent substrings.

The most frequent letter in the ciphertext is probably 'e', and so on.

Using this you can break a cipher, or get an almost correct plaintext which you can then deduce the correct plaintext.

Example

Example found online. This is a known rot cipher, but we don't know what number:

ymnxhtzwxjfnrxytuwtanijdtzbnymijyfnqjipstbqjiljtknrutwyfsyyjhmstqtlnjxfsifuuqnhfyntsymfyfwjzxjinsymjnsyjwsjy

Most common letters:

j = 13, y=13, n=11, t=10.

so we can assume either e = j or y. If e = j, then j is +5 from e so we can assume this is rot 5. Decoding using rot 21 (the reverse) gives:

thiscourseaimstoprovideyouwithdetailedknowledgeofimportanttechnologiesandapplicationthatareusedintheinternet

So we have solved it using just one substitution.

This method really works best with a quite lengthy ciphertext and is almost useless with short ciphertexts.

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Index of coincidence

The index of coincidence provides a measure of how likely it is to draw two matching letters by randomly selecting two letters from a given text, from the formula number of times that letter appears/length of the text

The calculation itself is complex. Here is the calculation, in its most basic form from Wikipedia.

How to use

The basis is that by splitting the ciphertext into groups of x, and stacking them, if the key length = x then the I.C. will be around 1.73 (index coincidence of English language). If it isn't the same as x it will be around 1.

Example

(From Wikipedia)

We have the following ciphertext:

QPWKA LVRXC QZIKG RBPFA EOMFL JMSDZ VDHXC XJYEB IMTRQ WNMEA IZRVK CVKVL XNEIC FZPZC ZZHKM LVZVZ IZRRQ WDKEC HOSNY XXLSP MYKVQ XJTDC IOMEE XDQVS RXLRL KZHOV

We can guess this is vigenere with a short key and its English. We can stack them in, say groups of 3 or any other number:

QPW
KAL
...

So if the key length is x, then the I.C should be around 1.73. Calculating all key lengths of 1-10:

1  1.12
2   1.19

3   1.05
4   1.17
5   1.82
6   0.99
7   1.00
8   1.05
9   1.16
10  2.07

We can see that 5 and 10 are the closest to 1.73, and as 10 is a factor of 5 then the key length will be 5.

Next stack the ciphertext in groups of 5, and using frequency analysis on each column we can find the key. When we try this, the best-fit key letters for each column are "EVERY". A vigenere decoder gives the message:

MUST CHANGE MEETING LOCATION FROM BRIDGE TO UNDERPASS SINCE ENEMY AGENTS ARE BELIEVED TO HAVE BEEN ASSIGNED TO WATCH BRIDGE STOP MEETING TIME UNCHANGED XX

Kasiski Examination

The Kasiski Examination is another way of deducing the key length. Works best with longer ciphertexts, though a computer is then usually required.

The Kasiski Examination finds the repeated strings in the ciphertext and the distance between them. The distances are likely to be multiples of the keyword length. Finding more repeated strings means it is easier to find the key length, as it is the highest common factor/greatest common divisor of the distances.

Example

(Courtesy of wikipedia, with some added elaboration.)

Take the plaintext

cryptoisshortforcryptography

'crypto' appears twice in the plaintext, the distance between is 16 characters. (Count from the first c to the r before the second)

If the key is 'abcdef' the length is 6, which doesn't go into 16 we don't get any repeats in the ciphertext:

abcdefabcdefabcdefabcdefabcdefab
cryptoisshortforcryptography
csasxtitukswtgqugwyqvrkwaqjb

'abcdef' matches 'crypto' the first time, but for the second crypto the key is 'efabcd' and as a result, the ciphertext doesn't match.

But if the key is 'abcd', the length is 4 which goes into 16. So the ciphertext repeats:

abcdabcdabcdabcdabcdabcdabcdabcd
cryptoisshortforcryptography
cqwmtngpsgmotemocqwmtneoaofv

You can see that 'abcdab' lines up with 'crypto' both times. And hey presto we get a repeat in the ciphertext: 'cqwmtn'.

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Resources

quantum mechanics - Generalized coherent states in arbitrary potential

It is well known that for Hamiltonian of harmonic oscillator there exist special states which saturate the Heisenberg uncertainity principle and under the time evolution follow closely classical trajectories. Of course they are wavepackets with finite volume in phase space, but that volume is not expanding with time evolution. These states are the famous coherent states: $$ | z \rangle = e^{za^{\dagger}-z^*a} |0 \rangle, $$ where $| 0 \rangle$ is the ground state of $H = a^{\dagger}a$ and $z \in \mathbb C$. My question is whether generalized coherent states with similar properties (holding possibly only approximately) exist for other Hamiltonians, either $$H= \frac{p^2}{2m}+V(x)$$ with arbitrary potential or even more generally in other quantum theories (also with infinte number of degrees of freedom). If they do, is there a general method for finding them? Do they share some nice analytic properties of coherent states (I suppose not, unless there is rich symmetry structure).

Remark: Title of my post was edited by moderation from "semiclassical" to "generalized coherent". I am not completely sure whether that is correct. I did a little bit of research and my impression is that what people usually mean by this second term are special states transforming between each other according to a represenation of some group and together forming a resolution of identity - essentially a topic of representation theory. This property is indeed the case in example I have shown, and the group under consideration is Heisenberg group. However I am interested if special states behaving approximately classicaly exist in general, when the underlying algebra of observables is not so simple.

Answer

To answer your question in its narrowest interpretation: yes but it depends on the potential (as you correctly guessed).

Perelomov generalized the notion of coherent states so it can be applied to a large number of Lie groups.

Your question actually hinges on the observation that different definitions of coherent states happen to coincide for HW.

Coherent states can be defined for HW as minimum uncertainty states, and then generalize to the notion of "intelligent" states. This definition has also been used when the Perelomov method (see below) does not work.

Coherent states can also be defined, in a spirit closer to the definition you gave, as transformed ground states; this is the generalization of Perelomov, and is applicable to a wide variety of groups. This is possibly the generalization you are looking for, and there is a lot of literature on spin coherent states for instance (as the next most common type of coherent states in applications). Perelomov coherent states also saturate the uncertainty relations suitably-defined observables.

Lastly, CS can be defined as eigenstates of the lowering operators; in this case they only exist for non-compact groups and were generalized to $SU(1,1)$ by Barut and Girardello.

Given the close relation between special functions and representations of Lie algebras, it is often possible to find an algebraic structure from which one can obtain coherent states when the potential yields analytical solutions to the Schrodinger equation in terms of special functions. Note that this is neither a necessary nor a sufficient condition.

Note also there are "coherent-like" states for various types of potentials: Morse, Poechl-Teller and the likes. There are also variations that will produce different types of distributions (this comes about by adjusting the energy levels of a 1d potential). Literature on all these can be found on arXiv.

knowledge - The Wall of Walls

I present to you... The Wall of Walls!

\begin{matrix} \text{sniper} & \text{bauxite} & \text{marshmallow} & \text{viper} & \\ & \\ \text{sparrow} & \text{absolution} & \text{modernism} & \text{pigeon} & \\ & \\ \text{joplin} & \text{golden} & \text{earth} & \text{smaug} & \\ & \\ \text{sergeant} & \text{gala} & \text{inspector} & \text{square} & \\ & \\ \text{circle} & \text{donut} & \text{berlin} & \text{springfield} & \\ & \\ \text{pentagon} & \text{granite} & \text{sandstone} & \text{falkor} & \\ & \\ \text{oreo} & \text{negro} & \text{predator} & \text{smith} & \\ & \\ \text{graduate} & \text{captain} & \text{independence} & \text{scholar} & \\ & \\ \text{money} & \text{office} & \text{triangle} & \text{constable} & \\ & \\ \text{lollipop} & \text{fuji} & \text{shockwave} & \text{albatross} & \\ & \\ \text{acrobat} & \text{hunger} & \text{madeira} & \text{wailing} & \\ & \\ \text{python} & \text{boa} & \text{anaconda} & \text{flash} & \\ & \\ \text{silent} & \text{calendar} & \text{marble} & \text{resurrection} & \\ & \\ \text{pelican} & \text{norbert} & \text{malone} & \text{covenant} & \\ & \\ \text{drive} & \text{great} & \text{columbia} & \text{prometheus} & \\ & \\ \text{tocantins} & \text{putumayo} & \text{illustrator} & \text{viserion} & \\ & \\ \end{matrix}

You must sort these 64 words into 16 groups of 4 words. However, once you're finished with that, your mission is to build another wall with 16 new words, those words being the key word in each of those category names, and then solve it too!

For instance, if you grouped the words galaxy, kart, party and odyssey in a "words in the title of Mario videogames" category, you should use Mario as a word for the second wall (and maybe that would be part of the group "Nintendo characters" or "Names beginning with an M" in that wall). Don't worry, I think the word to be used for each category should be quite obvious (if it's the right group).

The only thing left to say is that only one of the words on the second wall is a plural. I clarify this because since you will be working with groups, the category names will probably be plural, so you should take this into account and use the singular form of the key word in all but one case.

Good luck!

Edit: Since the first Wall was pretty much solved, here is the 2nd wall, to avoid discrepancies:

post snake dragon android
rock amazon alien missouri
adobe police apple bird

google shape wall hitman

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Although I've mentioned these in the comments, I'll add these here for clarification.

Hint 1:

The word that should be in plural in the second wall is very fitting.

Hint 2:

Stone instead of rock does work for the category I had in mind, but not as well.

Hint 3:

The first category for the second wall was already found. The tags for the remaining three are music, movies and entertainment.

Hint 4:

My other question on this site will pretty much tell you what one of the categories is (I made this before that one was solved, and I wanted someone to solve all of this first and then make the connection, but it didn't happen!)

Hint 5:

The words in another category are supposed to be preceded by "the" (unless you choose the alternative word in hint 2. That word is... cooler).

Answer

adobe: illustrator flash shockwave acrobat
google: calendar drive earth scholar
amazon: negro tocantins putumayo madeira (from @frogtown_j)
apple: golden gala fuji smith

These are all tech companies.

police: constable inspector sergeant captain

alien: covenant prometheus predator resurrection
android: marshmallow donut oreo lollipop (from @hexomino)
walls: berlin wailing great hunger (from @hexomino)

These are parts of song titles on Radiohead's OK Computer (thanks, hint 4): "Karma Police", "Subterranean Homesick Alien", "Paranoid Android", "Climbing Up the Walls".

shape: square circle triangle pentagon
bird: pelican albatross pigeon sparrow
post: malone office modernism graduate (from @FionaSapphire)
missouri: springfield joplin independence columbia (from @frogtown_j)

These are all parts of the titles of 2018 Best Picture Oscar candidates: "The Shape of Water", "Lady Bird", "The Post", "Three Billboards Outside Ebbing, Missouri".

snake: viper boa anaconda python
rock: granite sandstone marble bauxite
dragon: viserion smaug falkor norbert
hitman: silent money absolution sniper (from @frogtown_j)

These are all WrestleMania wrestlers: Jake "The Snake" Roberts, Dwayne "The Rock" Johnson, Ricky "The Dragon" Steamboat, Bret "Hitman" Hart.

general relativity - Extremal black hole with no angular momentum and no electric charge

A black hole will have a temperature that is a function of the mass, the angular momentum and the electric charge. For a fixed mass, Angular momentum and electric charge are bounded by the extremality condition

$$M^2 - a^2 - Q^2 \gt 0$$

Exactly at the extremality boundary, both entropy and temperature are zero.

Suppose i create a black hole with a spherically symmetric, incoming wavefront of electromagnetic radiation in a pure quantum state (that is, the density matrix satisfies the property $\rho^2 = \rho$). The wavefront is shaped in a way such that the whole energy of the packet will be inside the Schwarzschild radius, which will form an event horizon.

Since the wavepacket is as nearly as pure as it is physically possible to create, the quantum (Von Neumann) entropy is zero or nearly zero. But the formation of the black hole does not create nor destroy entropy, so the black hole must contain zero or nearly zero entropy as well. So the black hole seems to be extremal (it has zero temperature) but it nonetheless does not have any angular momentum (it is formed from a wavefront with zero net polarization over the whole sphere) and it does not have any charge (electromagnetic radiation is neutral).

Question: what "hair" does have a black hole formed from such a pure state, so that it can be extremal and still do not have angular momentum or electric charge (which are the classical hair that we come to expect from classical general relativity)

This question is a mutation of this question, but while that specific question tries to look what input black hole states create specific output (Hawking) radiation states that are far from thermal from a statistical point of view, this question is specific about extremality that is unrelated to angular momentum and charge

newtonian mechanics - Meaning of the word "Moment"?

This question is more of a question about the origin of a physical term moment used in many contexts. My question is about the linguistic or historical meaning of the word "moment".

Please don't provide the mathematical equations for the moment of inertia or the electric of magnetic dipole moment. My question is about the essence of the word.

I'm asking this question because as a non-native English speaker, I find it confusing as how did this word got to be used in the different contexts.

Answer

The origin of the word is the Latin verb movere which means "to move"; the noun is motus which means "motion". The root of the Latin word comes from the protolanguage Indo-European where meue/mew- means "to push away".

In other languages like Spanish we say momento or in French moment [like in moment cinétique=moment angulaire=angular momentum]. Amusingly, to refer to the vector $\mathbf{p}$ in English we say momentum, in Spanish momento lineal / cantidad de movimiento $\simeq$ "quantity of movement" and in French we just have the option quantité de mouvement.

For more info see here and here.