knowledge - Gladys unchained

_{This puzzle is part 2 of Gladys' journey across the globe. If you're new to the series, feel free to start at the beginning: Introducing Gladys.}

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Dear Puzzling,

Today I spent the entire day outdoors. There are so many enjoyable parks and gardens in this area. Who says you can't enjoy nature in a big city? The puzzle I'm sending you is a word chain. The name of my destination consists of two words. Have fun!

Wish you were here!
Love, Gladys.

1. A pope essentially munched alongside me during a period of fasting (7)
2. American car's air conditioning is beyond breaking point (7)

3. Black Prince, extremely exhausted, abruptly draws back (6)
4. Fantastic Potter has Elizabeth I's crown (7)
5. Football club's anthem botched by small child (9)
6. For every leading actor group, a prediction (8)
7. Half of meal prepared before specified time (7)
8. Make up a guess about jolly couple's birds (8)
9. Minister's corrupt circle (6)
10. Permanent building (6)
11. Prominent Republican host with a murderous brother (6)
12. Real corruption in Karnataka's bordering state (6)

13. Saloon outside Kazan's central marketplace (6)
14. Senseless gun mob regularly after central banks (4)
15. Short person essential to hybrid warfare (5)
16. Superman, finally social, upset Kal-El's dad (7)
17. Untalented writer has trouble finishing diacritical mark (5)
18. Vegetable food course by sun god (6)

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_{Gladys will return in "T-Rex goes cooking".}

Answer

Gladys is in the

https://en.wikipedia.org/wiki/Emerald_Necklace.

Clues and solutions:

CLEMENT    1. A pope essentially munched alongside me during a period of fasting (7)
PONTIAC    2. American car's air conditioning is beyond breaking point (7)
EDWARD     3. Black Prince, extremely exhausted, abruptly draws back (6)
TREETOP    4. Fantastic Potter has Elizabeth I's crown (7)

TOTTENHAM  5. Football club's anthem botched by small child (9)
FORECAST   6. For every leading actor group, a prediction (8)
ALREADY    7. Half of meal prepared before specified time (7)
SEAGULLS   8. Make up a guess about jolly couple's birds (8)
CLERIC     9. Minister's corrupt circle (6)
STABLE    10. Permanent building (6)
MCCAIN    11. Prominent Republican host with a murderous brother (6)
KERALA    12. Real corruption in Karnataka's bordering state (6)
BAZAAR    13. Saloon outside Kazan's central marketplace (6)
NUMB      14. Senseless gun mob regularly after central banks (4)

DWARF     15. Short person essential to hybrid warfare (5)
NICOLAS   16. Superman, finally social, upset Kal-El's dad (7)
HACEK     17. Untalented writer has trouble finishing diacritical mark (5)
RADISH    18. Vegetable food course by sun god (6)

Many of these were also solved by @hexomino, but I haven't looked at their answers ... except that I wasn't able to figure out #10 and peeked at that one, which apparently was actually solved by Chris Cudmore in comments. Kudos to Chris!

Piecing these together we get:

  A C E R O F R A W D
  S N E T T O T N E R
  T H A H S I D A M A

  R A C Y D A E R E W
  E M E . . . R A L D <--
N E C K . . . L A C E <--
I T C E R A L A Z I L
C O A I N U M B A R B
O P O N T I A C L E A
L A S E A G U L L S T

There is one thing I don't understand:

the definition part of #16. The wordplay is clear enough, but the definition has to be "Kal-El's dad" and I am failing to see why that should be NICOLAS rather than either JOR-EL or JONATHAN. [EDITED to add:] Thanks to El-Guest for setting me straight on this one. It turns out that Nicolas Cage actually named his son Kal-El. Gosh.

newtonian mechanics - Centrifuge speed of an object higher than a stationary orbit

In the question At what altitude above equator do gravitational and centrifugal forces cancel each other?, I asked how high a tower on the equator has to be such that at its top, gravitational and centrifugal forces are the same magnitude (and opposite sign).

Now imagine that the tower is a little bit higher, and an object is released from the top of the tower. The object will float away and "climb" higher. Does the object move slower or faster around the world than the tower?

It would be nice if the answer was explained in terms of the conservation of energy.

Answer

Now imagine that the tower is a little bit higher, and an object is released from the top of the tower. The object will float away and "climb" higher. Does the object move slower or faster around the world than the tower?

Slower. Provided that the space elevator is geo-stationary, the higher you go above geosynchronous orbit, the greater your velocity exceeds what's needed for a circular orbit. This means that the point at which you let go of the elevator, you're at perigee (the point of closest approach). Yes, you "climb" after that point.

Think of it in terms of rate of change of angle. All points on the tower move a full circle in exactly 1 day. They all have the same angular velocity. As you begin to climb, your angular velocity decreases. This is because:

• The circumference at a higher altitude is greater


• The spacecraft's speed decreases

The 2nd point is due to the energy balance within Earth's gravity well. As you drift further from the Earth, more of your kinetic energy is converted to gravitational potential energy. Both factors push in the same direction, so this is what happens. The orbital period will be greater than 24 hours for things released from the tower beyond the balanced geosynchronous orbit.

mass - Precision of Coulomb's law

Up to which precision has the coulomb law proven to be true? I.e. if you have two electrons in a vacuum chamber, 5 meters appart, have the third order terms been ruled out? Are there any theoretical limits to measure the precision ( Planck's constant?). Obviously there are practical limitations ( imperfect vacuum, cosmic rays, vacuum fluctuation). Still, does anyone know what was the smallest amount ever correctly predicted by that law?

---

Edit : Summary

On the high end of the energy spectrum a precision of 10^-16 has been shown ( 42 years ago )

For electron point charges at large distances the law might brake down due to practical reasons.

For moving particles QED gives a correction to the law: http://arxiv.org/abs/1111.2303

Answer

Quoting from my copy of the 2nd edition of Jackson's book on Classical Electrodynamics, section 1.2:

Assume that the force varies as $1/r^{2+\epsilon}$ and quote a value or limit for $\epsilon$. [...] The original experiment with concentric spheres by Cavendish in 1772 gave an upper limit on $\epsilon$ of $\left| \epsilon \right| \le 0.02$.

followed a bit later by

Williams, Fakker, and Hill [... gave] a limit of $\epsilon \le (2.7 \pm 3.1) \times 10^{-16}$.

That book was first published in 1975, so presumably there has been some progress in the mean time.

homework and exercises - Torque on electric dipole placed in non uniform electric field

When electric dipole placed in non uniform electric field, what is the approach to calculate torque acting on it? Can it be zero?

Answer

The torque $ \tau $ on an electric dipole with dipole moment p in a uniform electric field E is given by $$ \tau = p \times E $$ where the "X" refers to the vector cross product.

Ref: Wikipedia article on electric dipole moment.

I will demonstrate that the torque on an ideal (point) dipole on a non-uniform field is given by the same expression.

I use bold to denote vectors.

Let us begin with an electric dipole of finite dimension, calculate the torque and then finally let the charge separation d go to zero with the product of charge q and d being constant.

We take the origin of the coordinate system to be the midpoint of the dipole, equidistant from each charge. The position of the positive charge is denoted by $\mathbf r_+ $ and the associated electric field and force by $\mathbf E_+$ and $ \mathbf F_+$, respectively. The notation for these same quantities for the negative charge are similarly denoted with a - sign replacing the + sign.

The torque about the midpoint of the dipole from the positive charge is given by

$$ \mathbf \tau_+ = \mathbf r_+ \times \mathbf F_+ $$

where

$$ \mathbf F_+ = q\mathbf r_+ \times \mathbf E_+(\mathbf r+) $$

Similarly for the negative charge contribution

$$ \mathbf \tau_- = \mathbf r_- \times \mathbf F_- $$

where

$$ \mathbf F_- = -q\mathbf r_- \times \mathbf E_-(\mathbf r-) $$

Note that

$$ \mathbf r_- = -\mathbf r_+ $$

We can now write the total torque as

$$ \mathbf \tau_{tot} = \mathbf \tau_- + \mathbf \tau_+ =q\mathbf r_+ \times (\mathbf E(\mathbf r_+)+\mathbf E(\mathbf r_-))$$

It is clear that in taking the limit as the charge separation d goes to zero, the sum of electric fields will only contain terms of even order in d.

Noting that $$ \mathbf |r_+| = \frac{d}{2} $$

and defining in the usual way $$ \mathbf p = q\mathbf d = q(\mathbf r_+ - \mathbf r_- ) $$

We can write that $$ \tau_{tot} = \mathbf p \times \mathbf E(0) + \ second \ order \ in \ d $$

As we take the limit in which d goes to zero and the product qd is constant, the second order term vanishes.

Thus, for an ideal (point) dipole in a non-uniform electric field, the torque is given by the same formula as that of a uniform field.

Note that it is not correct to start with the expression for a force on an ideal/point dipole in a non-uniform field and then calculate torque from this force. To derive this expression one ends up first taking the limit of a point dipole (on which there is zero force in a uniform field) and then one finds a torque of zero, which is incorrect. One must start with the case of a finite dipole, calculate torque and only then pass to the limit.

When p and E are parallel and anti-parallel, the torque is zero, so yes zero is possible. But the case in which p and E are anti-parallel is one of an unstable equilibrium, and a small angular perturbation will cause the dipole to experience a torque which attempts to align the dipole with the electric field.

gauge theory - Cutkosky rule for the triangle diagram

Outline - the anomalous vacuum polarization correction

Suppose the abelian anomalous gauge theory (with axial gauge field $A$, vector gauge field $V$ and single massless fermion $\psi$): $$ \tag 1 L = -\frac{1}{4}V_{\mu\nu}V^{\mu\nu} - \frac{1}{4}A_{\mu\nu}A^{\mu\nu} + \bar{\psi}\gamma^{\mu}(i\partial_{\mu} + V_{\mu} + \gamma_{5}gA_{\mu})\psi, $$ and assume the 6th order of perturbation theory vacuum polarization diagram

The corresponding amplitude $M_{A\to A}$ has the form $$ \tag 2 M_{A \to A} = \int \frac{d^{4}q}{(2 \pi)^{4}}\epsilon_{\mu}(p)\Gamma^{\mu\nu\lambda}(p,q,p+q)\Gamma_{\lambda \nu \mu'}(p+q,q,p)\epsilon^{*\mu'}(p) $$ The effective vertex $\Gamma^{\mu\nu\lambda}$ corresponding to the triangle sub-diagram is anomalous and has the pole structure; we have $$ p^{\mu}\Gamma_{\mu\nu\lambda}(p,k_{1},k_{2}) \sim \epsilon_{\nu \lambda \rho \sigma}k_{1}^{\rho}k_{2}^{\sigma}, \quad k_{1}^{\nu}\Gamma_{\mu\nu\lambda}(p,k_{1},k_{2}) \sim \epsilon_{\mu \lambda \rho \sigma}k_{1}^{\rho}k_{2}^{\sigma}, \quad k_{2}^{\lambda}\Gamma_{\mu\nu\lambda}(p,k_{1},k_{2}) \sim \epsilon_{\mu \nu \rho \sigma}k_{1}^{\rho}k_{2}^{\sigma} \Rightarrow $$ $$ \tag 3 \Gamma^{\mu\nu\lambda}(p,k_{1},k_{2}) \sim \frac{p^{\mu}}{p^{2}}\epsilon^{\nu\lambda\rho\sigma}k_{1\rho}k_{2\sigma} + \frac{k_{1}^{\nu}}{k_{1}^{2}}\epsilon^{\mu\lambda \rho \sigma}k_{1\rho}k_{2\sigma} - \frac{k_{2}^{\lambda}}{k_{2}^{2}}\epsilon^{\mu\nu\rho \sigma}k_{1\rho}k_{2\sigma} $$

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The specific problem

In order to check the unitarity of the theory (i.e., to check whether the optical theorem holds) I need to calculate the imaginary part of $(1)$. There is the complication because of the pole structure of the vertex $(3)$. Note that typically vertices are free from poles, so usually for checking the unitarity of the gauge theory we don't need to worry about the imaginary part of vertices. An example is anomaly free non-abelian gauge theory, where the proof of the unitarity is simple (see, for example, section 6.5.4 on p. 237 here); if we have the abelian gauge theory, the checking of the unitarity is even elementary.

My book ("Advanced gauge quantum field theory" by P. van Nieuwenhuizen) "avoids" this problem by introducing the Higgs mechanism for the axial $U(1)$ sector: $$ \Delta L = \frac{1}{2}|\partial_{\mu} - 2igA_{\mu})\varphi|^{2} +\frac{\mu^{2}}{2}|\varphi|^{2} - \lambda |\varphi|^{4} - G (\bar{\psi}_{L}\varphi \psi_{R}+\bar{\psi}_{R}\varphi^{*} \psi_{L}) $$ It generates the mass for the axial gauge field and for the fermion, and then (by manipulating with the $G$ parameter) we can choose the fermion mass to be very large relative to the $A_{\mu}$ mass. Therefore, if the ingoing energy is smaller than the fermion mass, the imaginary part of the triangle diagram vanishes (loop fermions are always off-shell). However, the theory with the lagrangian $L + \Delta L$ is complicated theory whose quantization involves ghosts, and I'm not interested in it. Moreover, it turns out that the massless limit of the modification doesn't exist because of the anomaly, and the theory isn't equivalent to $(1)$.

But specifically in the case of the above diagram (surprisingly) the book states:

In this case these axial vector bosons could even be massless. The cutting relations for such graphs are again based on Ward identities, and if there are anomalies, there are extra terms in these Ward identities which lead to a break down of unitarity.

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My question

It seems that the statement above means that the imaginary part of the triangle diagram vanishes is not relevant independently on the fermion mass (since the masslessness of the axial gauge field leads to the masslessness of the fermion field). But I don't understand why.

Could You help?

collision - What would the effect be of a small black hole colliding with the earth?

If a small black hole (say about .1 mm radius or 1% of Earth's mass) came flying along at the speed of a comet or higher and impacted the earth, what would happen? Would it pass through the earth (and atmosphere) with minimal damage? Would it leave a crater (entry or exit?) Would it create a big enough shock wave to significantly damage the earth? Would it be captured and wander around inside the earth?

I read a book a few years ago with this as the primary part of the plot, and I have wondered about the physics ever since.

Answer

There would, of course, be no "impact" since the black hole won't interact with the earth the same way that a solid object would.

However, the gravitational effects from the black hole would be catastrophic. Since the black hole is about as massive as the Moon, it would significantly deform the earth as it passes through, causing mega-earthquakes, mega-tsunamis, and mega-volcanoes.

(the actual accretion rate of matter into the black hole would be negligible, as mentioned in the other question)

Per Alan's comment: some of the matter that gets pulled near the black hole will be turned into extremely powerful radiation (in a relativistic jet along its axis of rotation) which will surely do additional damage.

mathematical physics - What has "quantisation" to do with associated graded algebras?

I was currently reading an introduction into spin geometry by José Figueroa-O’Farrill. The first chapter handles Clifford algebras. When discussing the connection of the Clifford algebra to the exterior algebra, the author states:

Filtered algebras whose associated graded algebras are commutative (or supercommutative) can be interpreted as quantisations of their associated graded algebra, which inherits a Poisson bracket from the (super)commutator in the filtered algebra. This is precisely what happens for the Clifford algebra as we will now see.

Later on, this is made more precise: (the "$\mod F^{p+q-4}C$" just means "forget all terms of degree less than $p+q-4$)

We define a bracket $$[\cdot, \cdot]: \Lambda^pV \times \Lambda^qV\rightarrow \Lambda^{p+q-2}V$$ by $$[\alpha, \beta] := \alpha\beta - (-1)^{\vert\alpha\vert\vert\beta\vert}\beta\alpha \mod F^{p+q-4}C$$ It is an exercise to show that this is a Poisson bracket making $\Lambda V$ into a Poisson superalgebra. It is in this sense that $Cl(V,Q)$ is a quantisation of $\Lambda V$. We can think of $\Lambda V$ as the functions on the “phase space” for a finite number of fermionic degrees of freedom and $Cl(V,Q)$ as the corresponding quantum operator algebra. The Hilbert space of the quantum theory is then an irreducible representation of $Cl(V,Q)$.

(for more in-depth information see pages 8-10 of the above linked document.)

My question is therefore as follows: How is a "quantisation" of an algebra to be understood from a physical point of view? Can the usual way of quantization (i.e. replacing the position and momenta scalars by their corresponding operators) be understood in this picture?

Answer

The mathematical term of "quantization" of algebras is much broader than the physical notion of quantization of a classical theory, however, a very prominent quantization in the mathematical sense occurs during the physical process of quantization, which possibly is the origin of the mathematical terminology:

During (canonical) quantization, we turn the commutative algebra of classical observables (functions on phase space with the Poisson/Moyal bracket) into an anti-commutative algebra of operators on phase space. It is known (by the Groenewold-van Howe theorem) that's there's no proper quantization map with all desired properties if we insist on mapping the Poisson bracket to the quantum commutator, but there is if we use the Moyal bracket, which crucially has a parameter suggestively called $\hbar$ in it that controls how much it deviates from the classical Poisson bracket. Therefore, the algebras of quantum observables are quantizations in the mathematical sense of the algebra of classical observables, and the latter is obtained from the former by letting the quantization/deformation parameter go to zero, i.e. $\hbar\to 0$.

So this is the archetypical "quantization". In the case you're asking about, the two algebras in play are the "classical" exterior algebra with $v\wedge v = 0$ and the "quantum" Clifford algebra with $v\wedge v = Q(v)$ where $Q$ is a quadratic form, usually the Minkowski or Euclidean metric. Just like the quantum algebras of observables give the classical algebra of observables in the limit $\hbar\to 0$, the Clifford algebra gives the exterior algebra in the limit $Q\to 0$.

However, the text you quote suggest the similarity is more than just formal. If we allow for classical fermionic degrees of freedom (as they are needed, for example, for the BRST formalism), then the classical algebra of observables on a purely fermionic phase space of dimension $n$ is just the exterior algebra in $n$ dimensions. Introducing an $\hbar$ by which the quantum (anti-)commutator may deviate from the classical exterior algebra is akin to introducing the $Q$ from earlier, so for such a purely fermionic system, the Clifford algebra is indeed the quantum algebra to a system which has the exterior algebra as its classical algebra of observables.

However, it must be noted that this is not the main physical importance of the Clifford algebra. To physicists, the Clifford algebra is most useful because it defines the Dirac spinor representation by its irreducible representation.

notation - Uncertainty in parenthesis

In a physics text book I read the following:

$$e/m=1.758820150(44) ×10^{11} \mathrm{C/kg} $$ In this expression, $(44)$ indicates the likely uncertainty in the last two digits, $50$.

How should I understand this uncertainty? Does it mean $\pm 44$ on the last two digits?

Answer

$$e/m=\color{red}{1.758\,820\,1}\color{blue}{50}\,\color{magenta}{(44)}\color{green}{×10^{11}} \ \mathrm{C/kg}=\left(\color{red}{1.758\,820\,1}\color{blue}{50}\color{green}{×10^{11}} \pm \color{red}{0.000\,000\,0}\color{blue}{44}\color{green}{×10^{11}}\right) \ \mathrm{C/kg}$$

orbital motion - Calculating true anomaly of a hyperbolic trajectory from time

I'm creating a simulation for orbiting bodies with my primary challenge being calculating the true anomaly of a trajectory with respect to time. I have figured out circular, elliptical, and parabolic orbits, but i'm struggling with hyperbolic.

Everything I have found in my searches provides equations relating true anomaly v, to hyperbolic anomaly F, but nothing to determine F on its own.

The closest i've found to a solution is from this site. With the equation But deriving a solution was a little beyond me.

Can anyone help me find or derive the equations to calculate the true anomaly of a hyperbolic trajectory with respect to time? Thanks!

probability - Socks which may or may not match

You just won 2016 socks. Some of them are white, some are blue. The color of each sock was randomly chosen, with a 50/50 probability. Is it more probable that the socks can be paired, or that you will remain with two unmatching socks?

Answer

The probability of "pairability" is...

exactly 50%.
Consider what happens when all but one of the socks has been chosen. You'll have one color that has an even number of socks, and one that has an odd number. The last sock is equally likely to be either color: if it's the even color, then you'll have leftovers, and if it's the odd color, you'll be able to match every sock up.

gravity - What kind of energy gravitates, and why?

When listing energies for the purposes of keeping track of conservation, or when writing down a Laplacian for a given system, we blithely intermix mass-energy, kinetic energy and potential energy; they are all forms of energy, they all have the same units, and so this looks OK. For example, in the LHC, turning kinetic energy into new particles of mass-energy is routine. We just converted "energy which does not gravitate" (kinetic energy) into "energy which does gravitate". Isn't it a bit peculiar that this same thing called energy can manifest into two different kinds of forms - those forms which gravitate, and those which do not?

How about potential energy? It would be of course ridiculous to calculate your potential in relation to the galactic centre and expect that huge (negative, by convention) quantity of energy to gravitate; and yet if we allow its conversion into kinetic energy, and thence into particle creation, lo and behold we end up with something that does gravitate.

We know that the massless photon gravitates, because it can be "bent" around a star, per GR. A photon also expresses energy in the form E = p c. So clearly finite rest mass is not a requirement for certain forms of energy to gravitate.

So what's the rule here? When does energy gravitate, and why? Isn't it all supposed to be "just energy"?

Then there's the flip side of the equivalence principle - inertia. Do fields have inertia? - they do gravitate, so if they possess no inertia, doesn't that break EEP?

optics - Total Internal Reflection

Which laws are obeyed during TIR, laws of reflection or refraction? I have worked on it and I found that both can't be obeyed during this. Please explain.

Answer

Which laws are obeyed during TIR, laws of reflection or refraction?

Both.

Before the total internal reflection (TIR) threshold, the ray can be both reflected and refracted at the same time: some of it stays in the material, the rest leaves it, so you have two rays leaving the interface. That the TIR takes place is precisely a consequence of the "refraction law".

both can't be obeyed during this [TIR].

It doesn't make sense to state that: because these laws don't apply to the same ray, as the names in the Original Post makes clear. Beyond the critical angle, there is no refracted ray, and the reflected ray keeps obeying $\theta_r=\theta_i$, as it does for any angle.

thermodynamics - Is there really such a thing as an irreversible process?

If an isolated system goes from a state A to B, will it always eventually fluctuate back to state A? If not, give an simple example. Is it right to say that entropy only says that the probability for reversal is very low, but not actually 0, ie that it will take a very long time for some processes to reverse themselves, but they will infact eventually happen with probability 100%?

So will the universe eventually fluctuate back into yesterdays state? Or do we add some other physical law to make it irreversible?

electromagnetism - Dirac's quantization rule

I first recall Dirac's quantization rule, derived under the hypothesis that there would exit somewhere a magnetic charge: $\frac{gq}{4\pi} = \frac{n\hbar}{2} $ with $n$ a natural number.

I am wondering how the quantization of electric charge can be deduced from it. The quantization of the product $gq$ is certainly not enough; what else is demanded?

Photon emission and absorption by atomic electrons

Assume a photon is produced by an atomic electron making a transition down from a certain energy level to another.

Can that photon only be absorbed by another atomic electron making exactly the opposite transition?

Is there any chance that the photon could be absorbed by an atomic electron undergoing a transition with a slightly different energy difference?

electromagnetism - Why is the electromagnetic four-potential $A_{mu}$ not an observable?

Why within classical field-theory the electromagnetic four-potential (usually $A_{\mu}$) not an observable?

In classical mechanics we don't have problems with energy measurements and in quantum mechanics we talk all the time about the Hamiltonian which "is the observable of energy".

So why is $A_{\mu}$ not also considered as an observable? If we don't have a smart way to measure it how can we be 100% certain that tomorrow some dude won't figure out how to do it?

This confuses me greatly, especially with respect to the Aharonov–Bohm effect which is within quantum mechanics "a way to measure $A$".

The non-observable nature of $A$ seems to be the reason why we "gauge" it in order to do all kinds of stuff. That is why I'm interested in it.

Answer

The four-potential is not an observable because it is not invariant under a change of gauge. And no predictions of any physical theory are dependent on the choice of gauge, so the four-potential is not observable.

What is gauge invariant and observable is the integral of the four potential around a loop, and that is what is observed in the AB effect. However, it should be noted that the AB effect can be explained entirely in terms of local action by fields. As such, there is no reason to invoke the four potential to explain what's going on. The potential may sometimes be useful for doing calculations, but that is a different issue from whether it is an observable.

Alice and the Fractal Hedge Maze

This is an entry to the 12^th fortnightly challenge.

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Alice: Would you tell me, please, which way I ought to go from here?
Cheshire Cat: That depends a good deal on where you want to get to.
Alice: I don't much care where...
Cheshire Cat: Then it doesn't much matter which way you go!
Alice: ... so long as I get somewhere.
Cheshire Cat: Oh, you're sure to do that, if only you walk long enough.

Alice is in the most puzzling part of Wonderland yet. Following the white rabbit, she emerged found herself in the middle of a hedge maze. The rabbit provided her with a map before scurrying off, but it only seemed to make her more confused. She needs your help to figure out how to escape the maze.

The maze has 12 potential exits, numbered on the map. Each of the squares labeled A, B, C and D represent smaller copies of the entire maze. These submazes each have their own submazes, like infinitely many nested Matryoshka dolls, except that every doll has four dolls nested inside it.

Below the map is a bird's eye view of the actual maze, where you can see how the passages become smaller and smaller in a fractal fashion (only three levels of recursion are actually pictured). Fortunately, Alice has an ample supply of cakes and elixirs to change her size as necessary.

One last note: the little orange curve between B and D is a bridge which can be crossed over and walked under, but jumping from the bridge to the path below is not allowed.

Though I created this particular puzzle, the concept of a fractal maze is nothing new. Here are some other notable examples of cool fractal mazes, which served as inspiration for this one.

• 
  

  As far as I can tell, the concept of a fractal maze was created my Mark J. P. Wolf. He has made at least two mazes, taken from mathpuzzle.com.

  
  
  


• 
  

  These are from the blog Skeptic's Play:

  
  
  



• 
  

  Two devious looking mazes which I found referenced in this forum, but couldn't find the original sources for.

  
  
  

Answer

Here is a link to a prezi. If there is anyway to export it that would be great but I'm not exactly sure how it works.

Just keep clicking next through it.

https://prezi.com/oh2efo-ejbv9/untitled-prezi/?utm_campaign=share&utm_medium=copy

GIF

And the path in the format from-to (level). If the number includes a letter it comes from/goes to an internal maze box, if it does not it goes to the outside of the current level.

IN-A3 (1)
3-B1 (2)
1-12 (3)
B12-B11 (2)
11-A10 (3)
10-C4 (4)
4-B6 (5)

6-D8 (6)
8-10 (7)
D10-7 (6)
B7-7 (5)
D7-9 (4)
A9-8 (3)
B8-D3 (2)
3-B4 (3)
4-B6 (4)
6-D8 (5)

8-10 (6)
D10-7 (5)
B7-D10 (4)
10-8 (5)
D8-6 (4)
B6-4 (3)
D4-5 (2)
A5-1 (1)
OUT

logical deduction - Soccer balls in the stadium

The coach asks you take as many soccer balls as possible and put those balls onto the field with the condition that

For any arbitrary set of three balls, at least two of those balls are exactly 10 metres apart.

With this condition,

What is the maximum number of balls can you put into the field?

Answer

You can get at least:

7

Take:

Two 60-120 rhombi with 10 metre sides, join them at a 60 degree tip and rotate one so that the opposite tips are 10 metres apart.

Like so (not to scale, lines indicate distances of 10m):

Then place a soccer ball at each vertex.

Why this works:

By the pigeonhole principle, two balls must be in the same rhombus. Now, if they were in the middle two vertices or adjacent to each other around the edge of that rhombus, they would be 10m apart. Otherwise, they must be at opposite tips, but then there is no place for the third ball.

EDIT: Nicer diagram from here:

EDIT 2: Pjotr5 has a near proof of optimality - kudos to them. Let's finish it off.

We will:

Try to construct a unit graph by going around the 8 vertices clockwise from the top-left vertex

Process:

But then:

When we try to construct F, it must be one unit from D and one unit from E. However, they are two units apart and A is already halfway between them, so F and A would have to coincide. So we cannot construct a unit-distance graph with this graph either.

general relativity - How exactly and WHY does matter affect space-time?

According to general relativity, inertial mass and gravitational mass are the same, and all accelerated reference frames (such as a uniformly rotating reference frame with its proper time dilation) are physically equivalent to a gravitational field of the same strength.

Refer: Einstein's thought experiment of a Physicist in an Accelerated Box.

I find that the 'How' part of this question is somewhat easy to understand Mathematically but hard to visualize physically and on the 'Why' part I am totally blank.

Simply Cryptic #5

No back story, no setting, no theme, simply a cryptic clue to solve:

Starts to taste rather Mediterranean, some lemon too, regularly mixed with another's top condiment.

Since cryptic clues traditionally indicate the number of letters in the solution, I'll provide that as a hint:

(12)

_{After some feedback on my previous questions, hopefully this is a little better. However, criticism is still welcome. It's been a while since I've written one of these so I may well have forgotten some of the rules.}

---

Questions: Previous

Answer

It looks like

TARAMASALATA

as this is

A condiment and the initial letters of (Starts to)

Taste Rather Mediterranean, Some Lemon Too
mixed (anagrammed) with Anothers top - i.e. the letter A's

classical mechanics - How do you derive Lagrange's equation of motion from a Routhian?

1. 
   
   

   Given a Routhian $R(r,\dot{r},\phi,p_{\phi})$, how do you derive Lagrange's equation for $r$? Do you just solve the following for $r$? $$\frac{d}{dt}\frac{∂R}{∂\dot{r}}-\frac{∂R}{∂r}=0.$$

   
   


2. 
   

   And as a related question, what is the motivation for using a Routhian?

   
   

Answer

No. The coordinate $r$ stil follows the Euler-Lagrange equation, but $\phi$ and $p_\phi$ follow the Hamilton equations. But these are trivial, which is the whole point of the Routhian. The motivation is that the Routhian isn't really $R(r, \dot r, \phi, p_\phi)$ but just $R(r, \dot r)$ with a constant parameter $p_\phi$. $\phi$ isn't a coordinate because by definition, it was a cyclic coordinate in the Lagrangian and so it doesn't appear in the Routhian either. The momentum $p_\phi$, meanwhile, is conserved so it's really just some constant. With both of these conditions, we can simply take them both out, call $p_\phi$ a constant, and we wind up with a problem with 1 fewer effective dimensions. The same would be true using the complete Hamiltonian, but sometimes Lagrangians are easier to work with for the "hard" part of the problem (the non-cyclic coordinates), and the Routhian lets you stay "Lagrangian" for the hard part.

Here is a concrete example. Take the Lagrangian describing a harmonic potential in polar coordinates:

$$ \mathcal L = \frac{1}{2} m \left( \dot r^2 + r^2 \dot \phi^2 \right) - \frac{1}{2} k r^2 $$

Since $\phi$ does not appear in the Lagrangian, it is a cyclic coordinate. From the Euler-Lagrange equation

$$ \frac{d}{dt} \frac{\partial \mathcal L}{\partial \dot \phi} = \frac{\partial \mathcal L}{\partial \phi} = 0 $$

it follows that

$$ p_{\varphi} =\frac{\partial \mathcal L}{\partial \dot \phi} = m r^2 \, \dot \phi $$

is a conserved quantity. It would be nice to take $\dot\phi$ out of the set of coordinates and just replace it with the constant $p_\phi$. Then we'd essentially have a Lagrangian whose only coordinates are $r$ and $\dot r$, just with a constant parameter $p_\phi$. The way to do this correctly is to make the Routhian

$$ \mathcal R = \mathcal L - p_\phi \dot \phi = \frac{1}{2} m \dot r^2 - \frac{1}{2} k r^2- \frac{p_\phi^2}{2 mr^2} $$

(Aside: Try solving $\dot \phi$ in terms of $p_\phi$ and substitute into the Lagrangian. You'll get something very different that will lead to very wrong results. If you want to remove the cyclic coordinate, you must do a Legendre transform w.r.t. that coordinate.).

Since the Legendre transform has been done w.r.t. $\theta$ and $\phi$, these coordinates follow Hamilton's equations (with an appropriate change of sign). But these is trivial:

$$ \dot \phi = - \frac{\partial \mathcal R}{\partial p_\phi} = \frac{p_\phi}{mr^2}, \quad \dot p_\phi = \frac{\partial \mathcal R}{\partial \phi} = 0 $$

What we've achieved here is essentially separation of variables. $\phi$ has its own equation of motion, which involves $r$, but the equation of motion of $r$ is independent of $\phi$. We can solve the equation of motion of $r$, which is an effectively one-dimensional problem, and then go back to find out how $\phi$ evolves.

The effective one-dimensional problem has an effective potential

$$ V(r) = \frac{1}{2} k r^2 + \frac{p_\phi^2}{2mr^2} $$

This extra term, which blows up at small $r$, is called the centrifugal barrier, and accounts for the fact that angular momentum conservation makes it harder / impossible for the particle in question to reach the origin. The coordinates $r$ and $\dot r$ didn't have a Legendre transform applied, so they still follow the Euler-Lagrange equation

$$ \frac{d}{dt} \frac{\partial \mathcal R}{\partial \dot r} = \frac{\partial \mathcal R}{\partial r} \Rightarrow m \ddot r = -kr + \frac{p_\phi^2}{3mr^3} $$

A follow up question might be "Yes, its convenient to not stay with a full Lagrangian, but why not just go full Hamiltonian?" The answer here probably depends on the context of the specific problem. With the Routhian, we have a 2nd order differential equation to solve, and then an integral to find the e.o.m. of $\phi$. With a full Hamiltonian, we'd have a system of two 1st order differential equations, followed by the same integral. One might be easier to solve, or better computationally, etc.

mathematical physics - Can binary sequences generated from ergodic maps be chaotic?

Briefly, the way symbols are generated is:

Consider a one-dimensional chaotic map $T: [0,1]→[0,1]$ and a time series $\{x_n\}_{n=1}^N$ generated with this map. Define a threshold $A$ and a thresholding function $π$:

$$\pi(x) := \begin{cases} 1 &\text{if} \quad x>A \\ 0 & \text{else}\end{cases}$$

Consider a symbolic sequence $\{s_n\}_{n=1}^N$ obtained by applying $π$ to the elements of $\{x_n\}_{n=1}^N$, i.e., $s_i := π(x_i)$

For example, let $A =0.5$ be the threshold and the time series $x = [0.1,\, 0.56,\, 0.6,\, ...]$. Then

$\pi(x_1) = 0$;

$\pi(x_1) = 1$;

$\pi(x_2) = 1$.

Thus, the symbolic representation is $s = [0,1,1]$

I am facing technical difficulties in following the paper. My question is: Such binary sequences are just i.i.d. random variables of some distribution. How can we say that these sequences $s$ are chaotic?

Answer

When such 0/1 sequences are generated, they are just i.i.d random variables of some distribution.

No, they aren’t.

As for identically distributed, consider any sequence when changing $A$. The higher $A$, the higher the probability that the symbol is $1$.

As for independently, consider the sequence generated by a tent map and then transform all values via $$f(x) := \left(x-\tfrac{1}{5}\right) \bmod 1.$$

The corresponding sequence could as well have been generated by a somewhat shifted tent map, which looks like this:

If you select $A=\tfrac{7}{10}$ (see plot), it should become obvious that the probability that a $1$ is followed by another $1$ is higher than the probability that that a $0$ is followed by a $1$, so the symbol sequence has some memory.

---

How can we say that these sequences S are chaotic?

It all boils down to your definition of chaoticity at the end, but consider the following: Let $T$ be the classical tent map (forget the above shift) and let $A:=\tfrac{1}{2}$. Now, if we represent numbers in binary, the effect of the map on a number can be understood as removing its first digit after the decimal point, e.g., $$T(0.10100101011101) = 0.0100101011101,$$ $$T(0.0100101011101) = 0.100101011101,$$ and so on. Due to our threshold selection, we also have that $s_i$ is the digit removed from $x_i$ by $T$. Thus your sequence $s$ is the sequence of digits of $x_0$.

Thus, the more precisely, we know $x_0$, the longer can we accurately predict the sequence $s$, but any finite precision will make it impossible for us to predict $s$ forever. This is very reminiscent of the Butterfly Effect.

knowledge - A Sea O' Letters

ACROSS
1    Printer type
5    Stepping sound on muddy ground
12   Lord's labourer
16   ___-I-Am from "Green Eggs and Ham"

19   Naked
20   Holy radiance in art
21   Plant used in skin care
22   Greek cross?
23   Italian currency before the Euro
24   Magic the Gathering's Oona and Rubinia Soulsinger, for two
25   Scored less than the opposing team perhaps
26   2013 "Best Original Screenplay" sci-fi film
27   Portuguese greeting
28   "Jet Set Radio" gaming publisher

29   It might be flipped when flipped?
31   Take the world by _____
33   Restraining cord
35   Raison _____ (reason for being)
36   Lav
37   "Love is an Open Door", say
38   Large pitcher
40   Meditative syllables
42   What a good song might be left on
45   Mystery Hunt locale

46   ___ de Graaff generator
47   Decalogue receiver
49   No and Evil, say
50   Illuminated by flames
53   Ground
54   Stink strongly (of)
55   Tranquil
58   Lachrymose
60   Bullets with visible trajectories
62   Became apparent

64   Desiccates
66   Bakery item
67   Reddit Q&A sess.
68   Grassy meadow
70   The "D" of CAD, in engineering
73   Quebec theme park "La _____"
77   Jeanne d'___
79   Spread out
81   Online trouble-maker
83   Historical period

84   Relating to wedlock
86   Largest non-polar desert in the world
88   Catch forty _____ (take a nap)
89   Star Wars character Kylo
90   The ____ of Orléans
92   Off-road sport for two-wheelers (abbr.)
93   2016 animated film "Your ____"
94   Fastening device
97   Adulates
100  Red Monopoly playing piece

101  Feel unwell
102  Warren G. Harding's birth state
104  The "T" in TV
105  Use needle and thread
106  Complaint, informally
108  Frequently
110  Cryptocurr. with "smart contracts"
112  Proper's partner
114  This crossword's hidden theme, hinted by the circled cells 
116  Set free

119  "Fire in the ____!"
120  Hazy weather
122  Underwater locator
123  Simplicity
125  Cleopatra's alleged killer
126  Whip
128  Capital of Italia
130  Habitual twitch
131  It takes _____ tango
134  Cocktails Bond likes shaken

136  Lariat
138  Makes a mistake
139  Lyrical poem
141  With dash, it's a form of theft
142  "Punch-Out!!" protagonist Little ___
143  Ice skating venues
144  See 151-across
146  Confronts aggressively
150  ___-mo
151  With 144-across, Bellini ingredient

152  "Monty Python's Life of _____"
153  Day light
155  _____-weensy (tiny)
157  Motion sickness symptom
160  In the past
161  It follows 153-across
163  Anger
165  Like Bert during much of "Mary Poppins"
167  Foodie for fine dining
170  Halloween month

174  Theatre attendant
176  Giraffes' relatives
177  Lingerie items
178  They might be crushed to make Cookies 'n Cream
180  Spoken defamation
182  Clumsy lout
183  Quick cuts
184  Cleansing hydroxide
185  Sound intensity unit
187  Horn of ______ (cornucopia)

189  Letters on a tombstone
191  Price
192  At the peak of
194  "Dark web" network
195  Tennis follower?
197  Pick up (the phone)
199  Bubbly Nestlé bars
201  Requests help from the divine
202  PNG colour space
204  Driving need?

205  Edmund Hillary or Don Bradman, say
206  Meaningless chatter, when said thrice
208  Free from bacteria
210  Bawled
211  Tertiary education loc.
212  Roof overhang
213  Apportion
214  Indonesian city Banda ____
215  Greek vowel
216  Granger's activist organisation, in the books

217  Aids
218  Psychiatrist, informally

DOWN
1    Rorschach test component
2    Locked up
3    Printing mistake
4    Darjeeling, e.g.
5    Less risky
6    Marshy area, for short

7    Fertilising amide
8    Always, poetically
9    Hang around
10   Nautical rope securer
11   Monopoly publisher
12   What a spruiker might advertise
13   Rating system typically used in chess
14   Largest Antarctic ice shelf
15   Removes ceramic marks
16   Peanuts pianist

17   Interrupting throat clearing
18   Former Soviet space station
28   Preparing, for a 29-across
30   Subsidiary theorem
32   Alley-___, a basketball play
34   Young cow
35   Not given access
38   Anticipatory night
39   Entrepeneur Disney
41   "Bubble" or "merge", perhaps

42   Prog. a user may need
43   Son of 100-down
44   Non-verbal disapproval, when doubled
48   Python func. that converts 0 to '0'
51   Lands (a fish)
52   Late
53   Snake ____ (dice roll)
54   Casino city that rhymes with "casino"
55   Large body of water
56   "Beauty and the Beast" actress Watson

57   Gear up for battle again
59   "Toradora!" voice actress Kugimiya
61   wget alternative
63   USA's Homeland Sec., e.g.
65   "Star Wars: Episode III - Revenge of the ____"
69   In the style of
71   Up for _____ (available)
72   It can be p- or p-adic in mathematics
74   Inexperienced with
75   Wonderland bottle label

76   Canvas support
78   2006 Pixar film
80   ____ mater (school formerly attended)
82   Not strict
85   Loc. for version controlled files
86   Puzzling Stack Exchange, for example
87   "Rolling in the Deep" singer
91   Bus. letter directive
93   Not at all, archaically
94   Uber alternative

95   Falsehoods
96   _____-null (natural numbers cardinality)
97   Processing ord. for a queue, as opposed to a stack
98   "We have a _____ ground to cover"
99   Like 162-down games
100  Zeus' wife and sister
103  "To be at peace" in Egyptian, as in Nyarlat_____
105  Whale type
107  IEEE 754 numbers in programming
109  World War II bomber "_____ Gay"

111  Mathematician and physicist Poincaré
113  Ham and lamb, say
115  As well
117  Vientiane's locale
118  Nintendo handhelds with cameras
121  Italy's largest lake
122  Lower leg part
124  ___-friendly
127  Mix with an implement
129  "____, poor Yorick!" - Hamlet

131  To the point
132  Target for an iron
133  Director Welles
134  Islamic holy city
135  Close to
137  Plays a role
139  cos(0)
140  Faculty head
142  Selfie stick, e.g.
145  Consequently

147  "Vault 7" leaks' alleged source org.
148  4'33" composer
149  Recapitulate
151  Fiery Team Fortress 2 class
152  Soaks
154  Wrapper in sushi
156  Joins derrière with chair
158  ____-vide, a cooking technique
159  Dawn goddess in Greek mythology
162  JP : Famicom :: EN : ___

164  Ivory's counterpart
166  Pikachu's colour
168  Art of flower arrangement
169  Chevron-like symbols
170  Face-to-face exam
171  Eatery with trays
172  "Toradora!" voice actress Kitamura
173  Python func. that converts 0 to '0'
175  Catch some ____
177  With "It!", a toy by 11-down

179  Willow type
181  Court divider
183  Flashes intermittently
186  Like a slide rule, as opposed to a calculator
188  Universal logic gate
190  It's considered to be the fourth state of matter
191  In group theory, they can be left or right
193  Voyeur
196  They're twice the size of nibbles
197  217-across, for a crime

198  Consider again
199  YAML ____ Markup Language
200  Possible physical response to a rude remark
201  Word of relief
202  Spanish rivers, e.g. Grande and Tinto
203  Oversupply
205  Take to court
207  ___ Maria ("Hail Mary")
209  "Neon Genesis Evangelion" pilot Ayanami
210  Used to be

Answer

The fully solved grid looks like this:

The circled squares contain the word INK (so, for instance, 1-across is an INKJET printer). And the central row tells you the theme - SPLATOON RELEASED. (Splatoon 2 is a video game for the Nintendo Switch, a sequel to a Wii U game. It's a family-friendly first-person shooter game in which the goal is to cover as much territory with your color of ink as possible before time runs out.)

That explains the title: "Sea o' colors" is the Japanese name of an idol duo from the first game. But why is the grid so big?

If you look carefully at the finished grid (and know the theme of this puzzle) you might find that

there are some weapons hidden in the grid. For instance, the top row contains "Jet Squelcher", one of the weapons from the first game.

...Actually, it contains "jet squelchser". And other rows contain weapons with an extra letter too!

Those extra letters spell STAY FRESH, the catchphrase of the pop duo mentioned earlier.

The grid is available in this Google sheet. Thanks to feelinferrety, Sconibulus, Forklift, and Mike Q for the help solving.

radioactivity - How can quantum tunnelling lead to spontaneous decay?

I have never understood what measuring process (if any) is supposed to be continuously polling the quantum state of an unstable bound system subjected to decay via quantum tunnelling. The reason I reckon some kind of polling process should exist in the first place is the following:

According to the QM postulates, the unitary evolution of such a system should by definition keep it reversible, so it is only when measured that a decay can be observed or not. But this would make the decay rate dependent on the measurement rate, while we well know that the decay probability is constant, and the decay deemed "spontaneaous".

What am I missing?

Edit:

From the discussion in the comment section I gather I have not been clear about what I am asking here exactly. Let me try to reformulate the question.

It is about how quantum tunnelling is supposed to explain the exponential dynamics of a decay process. I am not asking about the Zeno effect; quite the opposite, actually: why, in the absence of any measurement, do we have an exponential decay at all? I just do not understand at what point in the unitary evolution of the unstable system the tunnelling effect is spontaneously happening.

The polling process I imagined is just a way to ask "why does the tunnelling effect manifest iself ?" because I cannot see how it can manifest without a measurement. Please do not infer that I am making up my pet theory here. I am only looking for a way to picture the situation, which at this time I don't get at all.

Answer

I am adding a few points that Lawrence's answer didn't yet address, but should help clarify the nature of the quantum tunneling process:

According to the QM postulates, the unitary evolution of such a system should by definition keep it reversible, so it is only when measured that a decay can be observed or not. But this would make the decay rate dependent on the measurement rate, while we well know that the decay probability is constant, and the decay deemed "spontaneaous". What am I missing?

The reversibility of the unitary evolution does not preclude the irreversibility of single tunneling events. This is because "reversibility" means different things in the two contexts:

1) A unitary evolution $\hat U$ is said to be time-reversible if whenever $|\psi(0)\rangle \rightarrow |\psi(t)\rangle = {\hat U}(t)|\psi(0)\rangle$ is a valid dynamics, then for $|{\bar \psi}(0)\rangle = {\hat T} |\psi(t_0)\rangle$ at some time $t_0$ and with ${\hat T}$ the antilinear time-reversal operator, $ |{\bar \psi}(0)\rangle \rightarrow |{\bar \psi}(t)\rangle = {\hat U}(t)|{\bar \psi}(0)\rangle \equiv {\hat T} |\psi(t_0-t)\rangle$ gives the exact time-reversed dynamics. This implies that $[{\hat T}, {\hat U}] = 0$, and similarly for the generator of ${\hat U}$, the Hamiltonian ${\hat H}$. For a spin-0 particle the time-reversal operator ${\hat T}$ amounts to complex conjugation in the position representation, but for spin-1/2 and higher it also includes a unitary component acting on the spin degrees of freedom.

2) A tunneling event is irreversible simply because once the particle tunnels through the barrier, the probability that it will tunnel back at some later time is practically null. The actual tunneling corresponds to the $|\psi(0)\rangle \rightarrow |\psi(t)\rangle = {\hat U}(t)|\psi(0)\rangle$ evolution. But let $D$ denote the spatial domain "inside" the barrier, and let $$ P_D(t) = \int_D{dV\;|\langle {\bf x} |\psi(t)\rangle|^2} $$ be the probability to localize the particle within $D$. In general it is assumed that the particle is initially localized entirely within D, such that $P_D(0) = \int_D{dV\;|\langle {\bf x} |\psi(0)\rangle|^2} = 1$. The statement that the probability of reverse tunneling is null then amounts to saying that $P_D(t)$ vanishes asymptotically, $$ \lim_{t \rightarrow \infty}{P_D(t)} = \lim_{t \rightarrow \infty}{\int_D{dV\;|\langle {\bf x} |\psi(t)\rangle|^2}} \;\rightarrow \;0 $$ The fact that the tunneling evolution ${\hat U}(t)$ is still time reversible means that if we now take as initial state the time reversed of the state at some asymptotically large time $t_0 \rightarrow \infty$, $|{\bar \psi}(0)\rangle = {\hat T} |\psi(t_0)\rangle$, when the particle is definitely localized outside of $D$, $\int_D{dV\;|\langle {\bf x} |{\bar \psi}(0)\rangle|^2} \rightarrow 0$, then the evolution of $|{\bar \psi}\rangle$ will be the exact time-reversed of the original one for $|\psi\rangle$, and will asymptotically drive the particle back into domain $D$ through reverse tunneling, $$ \lim_{t \rightarrow \infty}{\int_D{dV\;|\langle {\bf x} |{\bar \psi}(t)\rangle|^2}} \;\rightarrow \;1 $$ In the sense of an individual event, this reverse tunneling into $D$ is just as irreversible as the original tunneling out of $D$, despite the fact that ${\hat U}(t)$ is a time-reversible evolution.

If you prefer, think of an analogy with a free particle wave-packet of well-defined average momentum ${\bf p}$: it always travels in the direction of $\bf p$ and statistically speaking never turns around, despite inherent zitterbewegung and dispersion. But the free particle evolution is time-reversible, and the time-reversed of this wave packet will travel in the $-\bf p$ direction, etc.

[…] why, in the absence of any measurement, do we have an exponential decay at all? I just do not understand at what point in the unitary evolution of the unstable system the tunneling effect is spontaneously happening.

It is not so much at what point in time tunneling takes place, since this is essentially a statistical process, but what is the probability that at time $t$ the particle is located outside the inner domain $D$. To determine this probability we do not really need to follow the dynamics of the same system through multiple queries. In general we cannot even do this without altering the entire dynamics. What we must do is to query ensembles of identically prepared systems at different times $t$, ideally using a different ensemble for each such query. In principle, this would ensure that the measurement procedure does not interfere with the dynamics up to that point in time, and that the measured probability is indeed reliable.

Exponential decay then means that the fraction of particles detected outside domain $D$ in each such query diminishes exponentially with the time elapsed since the preparation of the respective ensemble. Note that the average tunneling time is statistically well defined, although we do not need to consider an average over exact individual tunneling times for each of the particles in an ensemble.

(From a comment to Lawrence's answer) to get a neat exponential law, it seems to me that some quasi-continuous interaction is needed. I have not seen that point discussed.

Actually it has been extensively researched in relation to a multitude of distinct fields, from nuclear physics to chemical kinetics and biophysics. See these "Lectures on Dissipative Tunneling", and google the Leggett-Caldeira model.

thermodynamics - How do we know that heat is a differential form?

In thermodynamics, the first law can be written in differential form as $$dU = \delta Q - \delta W$$ Here, $dU$ is the differential $1$-form of the internal energy but $\delta Q$ and $\delta W$ are inexact differentials, which is emphasized with the replacement of $d$ with $\delta $.

My question is why we regard heat (or work) as differential forms? Suppose our system can be described with state variables $(x_1,\ \cdots ,\ x_n)$. To my understanding, a general $1$-form is written as $$df = f_1\ dx_1 + f_2\ dx_2 + \cdots + f_n\ dx_n$$ In particular, differential forms are linear functionals of our state variables. Is there any good reason to presuppose that $\delta Q$ is linear in the state variables?

In other words, if the infinitesimal heat transferred when moving from state $\mathbf{x}$ to $\mathbf{x} + d\mathbf{x}_1$ is $\delta Q_1$ and from $\mathbf{x}$ to $\mathbf{x} + d\mathbf{x}_2$ is $\delta Q_2$, is there any physical reason why the heat transferred from $\mathbf{x}$ to $\mathbf{x} + d\mathbf{x}_1 + d\mathbf{x}_2$ should be $\delta Q_1 + \delta Q_2$?

I apologize if the question is a bit unclear, I am still trying to formulate some of these ideas in my head.

P.S. I know that for quasi-static processes, we have $\delta Q = T\ dS$ (and $\delta W = p\ dV$) so have shown $\delta Q$ is a differential form in this case. I guess my question is about non-quasi-static processes in general.

Answer

We want to show that "infinitesimal" changes in heat along a given path in thermodynamic state space can be modeled via a differential 1-form conventionally called $\delta Q$.

The strategy.

1. 
   

   We introduce a certain kind mathematical object called a cochain.

   
   


2. 
   

   We argue that in thermodynamics, heat can naturally be modeled by a cochain.

   
   


3. 
   
   

   We note a mathematical theorem which says that to every sufficiently well-behaved cochain, there corresponds exactly one differential form, and in fact that the cochain is given by integration of that differential form.

   
   


4. 
   

   We argue that the differential form from step 3 is precisely what we usually call $\delta Q$ and has the interpretation of modeling "infinitesimal" changes in heat.

   
   

Some math.

In order to introduce cochains which we will argue should model heat, we need to introduce some other objects, namely singular cubes and chains. I know there is a lot of formalism in what follows, but bear with me because I think that understanding this stuff pays off in the end.

Cubes and chains.

Let the state space of the thermodynamic system be $\mathbb R^n$ for some positive integer $n$. A singular $k$-cube in $\mathbb R^n$ is a continuous function $c:[0,1]^k\to \mathbb R^n$. In particular, a singular 1-cube is simply a continuous curve segment in $\mathbb R^n$. Let $S_k$ be the set of all singular $k$-cubes in $\mathbb R^n$, and let $C_k$ denote the set of all functions $f:S_k\to\mathbb Z$ such that $f(c) = 0$ for all but finitely many $c\in S_k$. Each such function is called a $k$-chain.

The set of chains is a module.

It turns ut that the set of $k$-chains can be made into a vector space in the following simple way. For each $f,g\in C_k$, we define their sum $f+g$ as $(f+g)(c) = f(c) + g(c)$, and for each $a\in \mathbb Z$, we define the scalar multiple $af$ as $(af)(c) = af(c)$. I'll leave it to you to show that $f+g$ and $af$ are $k$-chains if $f$ and $g$ are. These operations turn the set $C_k$ into a module over the ring of integers $\mathbb Z$, the module of $k$-chains!

Ok so what the heck is the meaning of these chains? Well, if for each singular $k$-cube $c\in S_k$ we abuse notation a bit and let it also denote a corresponding $k$-chain $f$ defined by $f(c) = 1$ and $f(c') = 0$ for all $c'\neq c$, then one can show that every singular $k$-chain can be written as a finite linear combination of singular $k$-cubes: \begin{align} a_1c_1 + a_2c_2 + \cdots + a_Nc_N \end{align} For $k=1$, namely if we consider 1-chains, then it is relatively easy to visualize what these guys are. Recall that each singular 1-cube $c_i$ in the chain is just a curve segment. We can think of each scalar multiple $a_i$ of a given cube $c_i$ in the chain as an assignment of some number, a sort of signed magnitude, to that cube in the chain. We then think of adding the different cubes in the chain as gluing the different cubes (segments) of the chain together. We are left with an object that is just a piecewise-continuous curve in $\mathbb R^n$ such that each curve segment that makes up the curve is assigned a signed magnitude.

Drumroll please: introducing cochains!

Now here's where we get to the cool stuff. Recall that the set $C_k$ of all $k$-chains is a module. It follows that we can consider the set of all linear functionals $F:C_k\to \mathbb R$, namely the dual module of $C_k$. This dual module is often denoted $C^k$. Every element of $C^k$ is then called a $k$-cochain (the "co" here being reminiscent of "covector" which is usually used synonymously with the term "dual vector). In summary, cochains are linear functionals on the module of chains.

Heat as a $1$-cochain.

I'd now like to argue that heat can naturally be thought of as a $1$-cochain, namely a linear functional on $1$-chains? We do so in steps.

1. 
   

   For each piecewise-continuous path $c$ (aka a $1$-chain) in thermodynamic state space, there is a certain amount of heat that is transferred to a system when it undergoes a quasistatic process along that path. Mathematically, then, it makes sense to model heat as a functional $Q:C_k\to\mathbb R$ that associates a real number to each path that physically represents how much heat is transferred to the system when it moves along the path.

   
   
   


2. 
   

   If $c_1+c_2$ is a $1$-chain with two segments, then the heat transferred to the system as it travels along this chain should be the sum of the heat transfers as it travels along $c_1$ and $c_2$ individually; \begin{align} Q[c_1+c_2] = Q[c_1] + Q[c_2] \end{align} In other words, the heat functional $Q$ should be additive.

   
   


3. 
   

   If we reverse the orientation of a chain, which physically corresponds to traveling along a path in state space in the reverse direction, then the heat transferred to the system along this reversed path should have the opposite sign; \begin{align} Q[-c] = -Q[c] \end{align}

   
   


4. 
   

   If we combine steps 2 and 3, we find that $Q$ is a linear functional on chains; it is a cochain! To see why this is so, let a chain $a_1c_1 + a_2c_2$ be given. Since $a_1$ and $a_2$ are integers, we can rewrite this chain as \begin{align} a_1c_1 + a_2c_2 = \mathrm{sgn}(a_1) \underbrace{(c_1 + \cdots + c_1)}_{\text{$|a_1|$ terms}} +\mathrm{sgn}(a_2) \underbrace{(c_2 + \cdots + c_2)}_{\text{$|a_2|$ terms}} \end{align} and we can therefore compute: \begin{align} Q[a_1c_1 + a_2c_2] &= Q[\mathrm{sgn}(a_1) (c_1 + \cdots + c_1) +\mathrm{sgn}(a_2) (c_2 + \cdots + c_2)] \\ &= \mathrm{sgn}(a_1)|a_1|Q[c_1] + \mathrm{sgn}(a_2)|a_2|Q[c_2] \\ &= a_1Q[c_1] + a_2 Q[c_2] \end{align} In summary, by thinking about heat as a functional on paths, and by imposing physically reasonable constraints on that functional, we have argued that heat is a $1$-cochain.

   
   

From cochains to differential forms.

Now that we have argued that heat can be thought of as a $1$-cochain, let's show how this leads to modeling "infinitesimal" changes in heat with a differential $1$-form.

This is where things get really mathematically interesting. We first recall the definition of a differential $k$-form over a $k$-chain. If $\omega$ is a $k$-form, and $c = a_1c_1 + \cdots a_Nc_N$ is a $k$-chain, then we define the integral of $\omega$ over $c$ as follows: \begin{align} \int_c\omega = a_1\int_{c_1} \omega + \cdots + a_N\int_{c_N}\omega. \end{align} In other words, we integrate $\omega$ over each $k$-cube $c_i$ in the chain multiplied by the appropriate signed magnitude $a_i$ associated with that cube, and then we add up all of the results to get the integral over the chain as a whole. For example, if $k=1$ then we have an integral of a $1$-form over a $1$-chain which is usually just called a line integral.

Now notice that given any $k$-form $\omega$, there exists a corresponding cochain, which we'll call $F_\omega$, defined by \begin{align} F_\omega[c] = \int_c\omega \end{align} for any $k$-chain $c$. In other words, integration of a form over a chain can simply be thought of as applying a particular linear functional to that chain.

But here's the really cool thing. The construction we just exhibited shows that to every differential form, there corresponds a cochain $F_\omega$ given by integration of $\omega$. A natural question then arises: is there a mapping that goes the other way? Namely, if $F$ is a given $k$-cochain, is there a corresponding $k$-form $\omega_F$ such that $F$ can simply be written as integration over $\omega$? The answer is yes! (provided we make suitable technical assumptions). In fact, there is a mathematical theorem which basically says that

Given a sufficiently smooth $k$-cochain $F$, there is a unique differential form $\omega_F$ such that \begin{align} F[c] = \int_c\omega_F \end{align} for all suitably non-pathological chains $c$.

If we apply this result to the heat $1$-cochain $Q$, then we find that there exists a unique corresponding differential $1$-form $\omega_Q$ such that for any reasonable chain $c$, we have \begin{align} Q[c] = \int_c \omega_Q \end{align} This is precisely what we want. If we identify $\omega_Q$ as $\delta Q$, then we have shown that

The heat transferred to a system that moves along a given path ($1$-chain) in thermodynamic state space is given by the integral of a differential $1$-form $\delta Q$ along the path.

This is a precise formulation of the statement that $\delta Q$ is a one-form that represents "infinitesimal" heat transfers.

Note. This is a new, totally revamped version of the answer that actually answers the OP's question instead of just reformulating it mathematically. Most of the earlier comments pertain to older versions.

Acknowledgement.

I did not figure this all out on my own. In the original form of the answer, I reformulated the question in a mathematical form, and I essentially posted this mathematical question on math.SE:

https://math.stackexchange.com/questions/658214/when-can-a-functional-be-written-as-the-integral-of-a-1-form

That question was answered by user studiosus who found that the theorem on cochains and forms to which I refer was proven by Hassler Whitney roughly 60 years ago in his good Geometric Integration Theory. In attempting to understand the theorem, and especially the concept of cochains, I found the paper "Isomorphisms of Differential Forms and Cochains" written by Jenny Harrison to be very illuminating. In particular, her discussion of theorem on forms and chains to which I refer above is nice.

electromagnetic radiation - How come Wifi signals can go through walls, and bodies, but kitchen-microwaves only penetrate a few centimeters through absorbing surfaces?

The only difference that I know of between kitchen microwaves and WiFi signals is how much power is pumped through them.

Why is it that WiFi signals, being 1000 times weaker can travel so much further, while something like a concrete block can absorb $1000 \, \mathrm{W}$ of microwaves from a kitchen magnetron being blasted at it? Or am I mistaken about the materials that WiFi signals can pass through?

calculation puzzle - Number Sequence Series-Question 7

This is a series of questions that are created by me.

Here is question 7:

(23)，(55)，(155)，(1255)，？，？，(56139611)

I will give a hint to you guys in one month if nobody can solve it.

physical chemistry - Is there anyway to use a scientific instrument to measure the density of electron around the atomic orbital?

Is there anyway to use a scientific instrument to measure the density of electron around the atomic orbital? Please list both old way and more modern ways.

Answer

Old ways used Schrodinger's equation's solutions for the atoms and mapped the square of the wave function.. Since the solution fitted the spectrum of the atom it was accepted that the orbital was also correct.

Recently there has been an experiment that measured the orbitals of the hydrogen atom

The abstract from the link:

To describe the microscopic properties of matter, quantum mechanics uses wave functions, whose structure and time dependence is governed by the Schrödinger equation. In atoms the charge distributions described by the wave function are rarely observed. The hydrogen atom is unique, since it only has one electron and, in a dc electric field, the Stark Hamiltonian is exactly separable in terms of parabolic coordinates (η, ξ, φ). As a result, the microscopic wave function along the ξ coordinate that exists in the vicinity of the atom, and the projection of the continuum wave function measured at a macroscopic distance, share the same nodal structure. In this Letter, we report photoionization microscopy experiments where this nodal structure is directly observed. The experiments provide a validation of theoretical predictions that have been made over the last three decades.

. A popularization is here.

After zapping the atom with laser pulses, ionized electrons escaped and followed a particular trajectory to a 2D detector (a dual microchannel plate [MCP] detector placed perpendicular to the field itself). There are many trajectories that can be taken by the electrons to reach the same point on the detector, thus providing the researchers with a set of interference patterns — patterns that reflected the nodal structure of the wave function.

And the researchers managed to do so by using an electrostatic lens that magnified the outgoing electron wave more than 20,000 times.

Please note that the orbitals are a probability distribution for finding an electron in a specific (x,y,z) around the nucleus, not a matter density in the classical sense. This experiment is for one electron and from the description it does not seem it would work for higher atomic numbers, at least not as simply.

word - Riley Riddles in Reverse, batch 2

_{"post some more sometime :)" -- user @Dragonrage in the comments of Riley Riddles in Reverse}

---

Here's another batch of "inverted" Riley riddles. These still work exactly like the ones before:

You get three words. You must find one solution word that you can attach..

• to the beginning of one of the three words, (you have to figure out which one)


• somewhere in the middle of another, and


• to the end of the remaining word,

so that in each case, a new, single word is formed.

All the words here are very random, and some of them are even very rare, but every clue, result, and solution word is going to be strictly valid in Scrabble.

1. sailing - cement - bi


2. rate - onions - scam



3. tan - caste - ka


4. one - regal - stat

If you get stuck, you can write the four solution words together, and the result will be yet another random Scrabble-valid word.

---

_{PS. This puzzle format is still very experimental, so please leave feedback and improvement suggestions in the comments. Thanks to those, especially @Riley, who have already done so.}

Answer

sailing - cement - bi

ASsailing, cASement, biAS

rate - onions - scam

PIrate, oPInions, scamPI

tan - caste - ka

RATtan, castRATe, kaRAT

one - regal - stat

IONone, regIONal, statION

Leading to:

ASPIRATION

homework and exercises - Time to fall in a non-uniform gravitational field

In an uniform gravitational field the loss in potential energy (the work done) by gravity is given by $\text{-}mg\Delta h$ and the time to do so (assuming no friction and no initial velocity) is given my $\sqrt{\frac{2 \Delta h}{a}}$, where $a$ is the strength of the field as commonly referred to the local gravity.

Now, you can do the same for a non-uniform field. To get the work done you simply take the difference in gravitational potential energy: $\int_\infty^{h_1} G\frac{Mm}{{h_1}^2} - \int_\infty^{h_2} G\frac{Mm}{{h_2}^2}$.

And here is my problem. I'm stuck. How does one calculate the time it takes to fall from $h_1$ to $h_2$ with respect to the changing field strength?

quantum field theory - Why is there a flux of radiation in the Hawking effect but not in the Unruh effect? (and other questions)

This question is slightly related to this one Do all massive bodies emit Hawking radiation?, which I think was poorly posed and so didn't get very useful answers. There are several questions in this post so I hope people who answer will follow the question marks and give their opinion on each one.

Let's look at flat spacetime first. Inertial observers detect no particles in empty space. Non-inertial observers do detect particles because there is a non-trivial Bogolyubov transformation that mixes positive and negative frequency modes of the matter field. And finally, if the non-inertial observers see a horizon as in the Unruh case, the spectrum of detected particles is precisely thermal.

Now look at a black hole spacetime. Free falling observers will not detect any particles but distant/stationary observers will detect a flux of radiation from the black hole outward to infinity. Q1: Why is there a flux in this case and none in the Unruh case? Shouldn't the two situations be nearly identical for the case of a very large black hole or very near to the horizon? How would you physically distinguish between the two anyway? (Maybe this is equivalent to asking why the black hole should lose mass and evaporate)

Q2: Is it natural to expect that an observer who is not in free fall but is not stationary either (and is on some weird trajectory) will detect particles but that they will not be in a thermal state?

For massive compact objects without a horizon, free fallers should see no particles but other non-inertial observers should detect particles because of the Bogolyubov transformation as in the Unruh case. This probably depends a lot on the answer to Q1 but Q3: suppose you bring such an object arbitrarily close to the point of forming a black hole with a horizon (by adding mass) -- what would a stationary observer see? Would radiation and evaporation be observed only once the horizon is formed?

Q4: In the Unruh case, the energy from the radiation is accounted for by the agency that accelerates the detector. In the black hole case, it is accounted for by the evaporation of the black hole. How is it accounted for in the case of a stationary detector outside a massive compact object without a horizon (which doesn't evaporate)?

quantum gravity - Why one-dimensional strings, but not higher-dimensional shells/membranes?

One way that I've seen to sort-of motivate string theory is to 'generalize' the relativistic point particle action, resulting in the Nambu-Goto action. However, once you see how to make this 'generalization', it becomes obvious how to write down the action, not just for a string, but for manifolds of higher dimension as well. In fact, Becker-Becker-Schwarz (the main source I happen to be learning from) actually do this. But (as far as I have read), they merely write down the action and do nothing further with it.

My question is: what happens when we proceed along the same lines as string theory, but when replacing a string with a 2-manifold, the simplest example of which would be the 2-sphere, a "shell/membrane"? Assuming 3 spatial dimensions, this is the highest dimensional manifold we can consider (because we don't allow for non-compact manifolds). Furthermore, there is only one compact manifold of dimension 1; however, there are infinitely many compact manifolds of dimension 2, which could potentially make the theory much richer (and probably much more difficult). For example, in string theory, we are stuck with $S^1$ (if you insist upon having no boundaries), but if you allow for 2 dimensions, we could consider the sphere, the torus, etc.

Because it seems that this path is not ever presented (in fact, I've never even heard of it), I would presume that something goes wrong. So then, what exactly does go wrong?

general relativity - Why do objects follow geodesics in spacetime?

Trying to teach myself general relativity. I sort of understand the derivation of the geodesic equation $$\frac{d^{2}x^{\alpha}}{d\tau^{2}}+\Gamma_{\gamma\beta}^{\alpha}\frac{dx^{\beta}}{d\tau}\frac{dx^{\gamma}}{d\tau}=0.$$ which describes "how" objects move through spacetime. But I've no idea "why" they move along geodesics.

Is this similar to asking why Newton's first law works? I seem to remember reading Richard Feynman saying no one knows why this is, so maybe that's the answer to my geodesic question?

What is the strategy to solve Simon Tatham's Twiddle?

Consider the goal 3x3 position:

You must reach this position by rotating 2x2 blocks by some multiple of 90 degrees like so:

Orientation of each individual number is preserved.

What is the strategy to solve this puzzle? How about for larger boards, such as this 6x6 board with 4x4 rotating blocks?

I'd like to observe that a "row by row" approach does not seem guaranteed to work, likely due to the dark forces of parity.

As an extension, what is the strategy if orientation of each number is NOT preserved with block rotation?

Answer

Update: Simon Tatham's Twiddle has been solved!

Almost complete answer, which solves for ALL board sizes, accounts for the variation in which orientation is not preserved with rotation, but cannot show how puzzles with larger block rotation can be solved.

Consider a puzzle of any size of side length $\geq 3$, with $2 \times 2$ rotating blocks. We can ignore orientation for now.

We prove that with orientation of numbers ignored, the puzzle is always solvable.

Firstly, in any $2 \times 2$ grid contained in the puzzle, there exists an algorithm to preform a 3-cycle on 3 of the numbers.

1 2 3
4 5 6
7 8 9

Converts to:

1 2 3

4 8 5
7 6 9

The 5, 6, and 8 got cycled.

The algorithm to preform this is as follows: Pick a $3 \times 3$ grid containing the 3 numbers in question, and orient the contained $2 \times 2$ grid so the 3 numbers, A, B, and C, are in these relative positions:

1 2 3
4 A B
5 C 6

Action X rotates the bottom-left four numbers counter clockwise, and action Y rotates the top-right four numbers counter clockwise. The algorithm is:

$XY'X'Y$

Which is demonstrated as follows:

1 2 3
4 A B
5 C 6

1 2 3
A C B
4 5 6

1 C 2
A B 3
4 5 6

1 C 2
4 A 3
5 B 6

1 2 3
4 C A
5 B 6

While this seems clever, I don't really have a completely logical explanation other than just proof by demonstration.

Now fun fact: The given 3 - cycle is equivalent to a swap of two adjacent numbers. Consider the $2 \times 2$ grid:

1 2
3 4

Now preform a 3 - cycle to obtain:

3 1
2 4

A single counter clockwise rotation reveals:

1 4
3 2

The 2 and the 4 are swapped. Thus, we can swap any 2 edge adjacent numbers, and now the proof that any board size can be solved is trivial...

...until we introduce non-preservation of orientation. Then it gets hard.

But, maybe not so hard, after a simple observation: If the puzzle is solvable, then once the numbers are sorted using the method previously described, the orientation of all numbers are either correct or at 180 degrees! Oh, and this time, I have proof!

Proof: Consider a number in the puzzle with correct orientation and position. Color the puzzle like a checkerboard, such that the square in which this number lies is black. If this number is moved by a rotation, it moves exactly one square, to a white square, and its orientation is changed by 90 degrees, to either 90 or 270. Another rotation brings this number back to black, and orientation is brought to either 0 or 180. Clearly, if this number is on its proper parity, then it will always have orientation of 0 or 180. In the state in which all numbers are sorted, all numbers are on their proper parity. Therefore, they must all have either correct orientation, or are oriented at 180 degrees.

We have a second observation: If the puzzle is solvable, then when the numbers are sorted, the number of numbers with 180 degree orientation is even.

Proof: Consider the sum of the degrees of the orientations of all the numbers in the puzzle. Each rotation changes the orientations of 4 numbers by 90, in one direction. This either adds or subtracts 90 * 4 = 360 degrees to the total sum. The solved state has a total orientation sum of 0, so any set of rotations must bring the total orientation sum to a multiple of 360. If the puzzle is sorted but there is an odd number of numbers with orientations of 180, then the orientation sum is 180 (mod 360), contradiction.

I now present an algorithm that orients 2 pairs of 2 numbers by 180 degrees, preserving position, like this:

1↑ 2↑ 3↑
4↑ 5↑ 6↑

To this:

1↓ 2↑ 3↓

4↓ 5↑ 6↓

Let action X rotate the left 4 numbers 90 degrees counter clockwise, and let action Y rotate the right 4 numbers 90 degrees counter clockwise. The algorithm is:

$XX, Y, XX, YY, X, YY$

Demonstration:

1↑ 2↑ 3↑
4↑ 5↑ 6↑

5↓ 4↓ 3↑
2↓ 1↓ 6↑

5↓ 3← 6←
2↓ 4→ 1→

4← 2↑ 6←
3→ 5↑ 1→

4← 1← 5↓
3→ 6→ 2↓

4← 1← 5↓
3→ 6→ 2↓

1↑ 6↑ 5↓
4↓ 3↑ 2↓

1↓ 2↑ 3↓
4↓ 5↑ 6↓

Let's call this procedure A. Careful observation yields a logical explanation to explain the effects of this algorithm, but I... uh... will leave it as an exercise to the reader.

I can now demonstrate that using procedure A, we can orient 2 corner adjacent numbers by 180 degrees, like so:

1↑ 2↑ 3↑
4↑ 5↑ 6↑

To,

1↑ 2↑ 3↓
4↑ 5↓ 6↑

X and Y have the same definition from the last algorithm, and we also have procedure A. The algorithm is:

$AYAY'$

...which doesn't warrant much explanation at all, especially once I demonstrate:

1↑ 2↑ 3↑
4↑ 5↑ 6↑

1↓ 2↑ 3↓
4↓ 5↑ 6↓

1↓ 3→ 6→
4↓ 2← 5←

1↑ 3→ 6←
4↑ 2← 5→

1↑ 2↑ 3↓
4↑ 5↓ 6↑

This doesn't yet solve the puzzle, But let's call this procedure B.

I think we can all agree that if we can invent an algorithm to orient two adjacent numbers 180 degrees, we basically win the game for all board sizes. Right? Do I have to prove it?? Please don't make me, I think we can all imagine a dozen informal proofs for why it would definitely solve the puzzle... ;-;

Such an algorithm exists though! We can convert,

1↑ 2↑ 3↑
4↑ 5↑ 6↑

to,

1↑ 2↓ 3↓
4↑ 5↑ 6↑

Using this algorithm!

$X'BX$

Demonstration:

1↑ 2↑ 3↑
4↑ 5↑ 6↑

4→ 1→ 3↑
5→ 2→ 6↑

4→ 1→ 3↓

5→ 2← 6↑

1↑ 2↓ 3↓
4↑ 5↑ 6↑

And that's it! We've proven that with $2 \times 2$ rotating blocks, we can logically solve all solvable boards of whatever size.

As it stands, the procedure for solving these puzzles is inefficient. To optimize, start by manually solving with a row-by-row approach from top to bottom until the last 2 rows. Then, start using a column-by-column approach from left to right until you are left with the unsolved $2 \times 2$ grid. Additionally, one can orient two side-adjacent numbers 180 degrees more quickly, by applying algorithm A, repositioning a wrongly oriented number, then reapplying A.

I absolutely do not want to think about larger rotating block sizes. That sounds horrifying and looks like puzzling hell. I'm proud of my contribution to the puzzling society as it stands (I couldn't find a strategy online), and you can try out the strategy here:

http://www.chiark.greenend.org.uk/~sgtatham/puzzles/js/twiddle.html (Run by javascript)

A solution to larger block sizes, a generalization for any rotating block size, or a procedure more optimal than mine is welcome.

Sample Solve:

---

I have made extremely significant progress on the $N \times N$ board with $3 \times 3$ rotating blocks. There is only one special case that I cannot resolve.

Special case: $N=3$

Trivial.

Special case: $N=4$

See the bottom of this answer.

General case: $N \geq 5$

Our plan is as follows:

• Reduce the $N \times N$ board to a $5 \times 5$ board.


• Solve the case $N=5$.

Part I: Reduction

The board reduction process is rather simple. We simply need to reduce an $N \times N$ board to a $(N-1) \times (N-1)$ board, for some $N \geq 6$, and repeat this process. Really, we just need to solve the following squares:

For $N \geq 6$, the board is large enough to provide sufficient space and mobility to maneuver each individual "edge" number to its correct square, until the last 3 numbers along each edge. As you might expect, this is easily resolved with a simple intermediate construction. The details are left to the reader; We have bigger fish to fry!

Part II: $N=5$

Divide the squares of odd parity into the following classes:

Our scheme is as follows:

1. Solve the four Class A numbers.


2. Solve the eight Class B numbers.


3. Maneuver the rest of the numbers into place with iterated commutators.

Step 1: Solve Class A

We need to solve the numbers $7$, $9$, $17$, and $19$. It is not hard to maneuver the first two one at a time. I recommend an intermediate construction to solve the next two numbers at the same time, in one move.

Step 2: Solve Class B

Trivial. Why?

If we restrict our allowed moves to only the rotations of the blocks centered at Class A squares (so they stay fixed), observe that the Class B squares form a $3 \times 3$ board, and our moves are just $2 \times 2$ rotating blocks. But we solved this already, remember?

Step 3: Solve the rest of the squares

I now present four algorithms in one:

The moves $X$ and $Y$ are described in the graphic above. If $A$ is some move, it is implied to be counterclockwise, and $A'$ is the clockwise variant.

Consider the following algorithm:

$$XYX'Y'\ XYX'Y'$$

This is reminiscent of Rubik's Cube commutators. It has the following effect:

Essentially, we have created a three-cycle.

This algorithm is one of a set of four algorithms. The other three come from reversing the direction of each move, or switching the side of each move. Essentially, our four algorithms are:

• $XYX'Y'\ XYX'Y'$


• $X'Y'XY\ X'Y'XY$


• $YXY'X'\ YXY'X'$


• $Y'X'YX\ Y'X'YX$

As for some guidelines as to what each one does:

• If your first move is counterclockwise, then the "slant line" of the three-cycle is from bottom-left to top-right. If your first move is instead clockwise, this diagonal slant line is in the other direction, from top-left to bottom-right.



• The side that your first move is on is the side that the central number moves to under the three-cycle.

Arguably, any of these four algorithms are already sufficient. The full set is just for your convenience. Really, any three numbers located on the even parity can be three-cycled by moving them into place using some intermediate moves, preforming our algorithm, then reversing our intermediate concessions. Though, I'd say it is quite likely that the game is solvable using these four algorithms alone.

Regardless, we must prove that three-cycles of even parity numbers will suffice to win the game for completeness.

Assume there exists a board, with odd parity solved, that is unsolvable with three cycles of even numbers, but solvable with other means. First, we follow our scheme, solving the numbers of odd parity, then using three-cycles to solve all numbers of even parity, except three of them. Then, preform three-cycles on the three unsolved numbers until one of them is solved. Since the game was unsolvable with three-cycles, we must have that the last two numbers remaining are unsolved, and are therefore switched. Hence, we have achieved a position in which two numbers of even parity are switched, and everything else is in the correct position. We must now prove that such a position cannot be reached from the solved state.

Every move, a block rotation, we make is virtually two simultaneous 4-cycles of numbers. When composed, we see that each move is itself a permutation of even parity. In the aforementioned board state, two and only two numbers were switched, indicating that there is a composition of moves resulting in an odd permutation. But our moves are even permutations, and no composition of these permutations can result in an odd permutation, hence the proof.

In layman's terms, we can imagine each $3 \times 3$ block rotation as a cycling of the numbers at the corners, and then a cycling of the numbers along the edges. The number in the center stays put. Each cycle itself can be "decomposed" into three switches of numbers. So, to get from 1234 to 2341, our switches look like 1234 -> 2134 -> 2314 -> 2341. So this cycle is an odd permutation. Since there are two of these cycles, we have an odd + odd = even permutation. In other words, each move we make contains an even number of switches. Now, if we start from the solved board, and we make moves to convert this to the board with only two numbers switched, then the net result is that we made one switch. That means that in total, using our moves, we must have made an odd number of switches. But each move made an even number of switches, contradiction.

I glossed over the fact that an even permutation cannot be odd; Any position reached with an even number of switches cannot also be reached with an odd number of switched. Trustworthy Wikipedia has some stuff on that: https://en.wikipedia.org/wiki/Parity_of_a_permutation

Sample Solve:

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$3 \times 3$ rotating blocks: The final nail in the coffin

We still haven't solved the $4 \times 4$ board. Understandable, since each move permutes more than half the board. Nevertheless, it is possible!

For the same reasoning concerning the parity of permutations, it is impossible to switch two numbers while leaving everything else untouched. Hence, it suffices to find a family of three-cycles, and abuse them to solve everything.

Ignoring their clockwise counterparts, there are four legal moves. They are situated in the top-left, top-right, bottom-left, and bottom-right. Denote these counterclockwise rotations as $A$, $B$, $C$, and $D$ respectively.

I claim that the following 10-move algorithm is a three-cycle:

$$ADA'D'C'DA'D'AC$$

Obviously, this three-cycle is confined to one parity. It is rather trivial to rotate or mirror this algorithm to uncover a family of 8 different three-cycles, that encompasses both parities.

The above algorithm seems to fly out of nowhere. I'll show how one may motivate this algorithm. Go to https://www.chiark.greenend.org.uk/~sgtatham/puzzles/js/twiddle.html, set the game type to "4x4', rotating 3x3 blocks", then click "solve game" so you can follow along.

We first experiment with the natural commutator:

$$ADA'D'$$

Observe that this actually results in two swaps of numbers. Namely, $5$ and $10$ are swapped, and $7$ and $12$ are swapped.

We now would like to use this commutator, except to undo the swap of $5$ and $10$, and swap either the $7$ or $12$ with something else. This creates a three-cycle. The easiest way to do this is $C'A'$. This retains the $5$ and $10$ in the "hot zone", so their swap will be undone. Now, notice that the $7$ is also still in the hot zone, though $12$ is not. Hence, this will create the three cycle after we apply the commutator again with $ADA'D'$ and undo our intermediate steps with $AC$. Composing all these moves reveals the algorithm $ADA'D'C'A'ADA'D'AC$, which reduces to $ADA'D'C'DA'D'AC$.

As said before, this three-cycle kills the game. A quick solving scheme goes like this:

1. Solve $1$, $3$, $8$, and $16$ with some clever maneuvering. After you solve the $1$ and $3$, it is easiest to separate $8$ and $16$, and then "prep" them by placing $8$ at the bottom-right corner, moving $16$ two squares to the left of it, and then $D$.


2. Iterate $C$ until it can be observed that the $6$, $9$, $11$, and $14$ can be solved with a three-cycle. Of course, you will have to maneuver the three offending numbers carefully into place before you do this. Do not panic if you encounter a state where seemingly only two of these numbers are switched. The solution is to either $C$ or $C'$, one of them will reveal which three-cycle we need (Do you recall in the beginning of this answer the observation that a 3-cycle can be converted to a 2-cycle? This is a similar problem). Remember that while a single rotation is an even permutation acting on the board, it is actually an odd permutation when acting on a single parity. Hence, it is very possible for there to be two numbers of the same parity swapped during this step. Since we have not solved the other parity, we would conclude that in such a state, the parity of the permutation of both parities is odd. They will be both even once we solve the first parity.


3. We're out of tricks. Monotonously abuse three-cycles on the other parity until the game is solved. Make sure you frequently use intermediate manipulations to speed up this process.

Sample solve:

And with that, I'm proud to say that $7/8$ of Simon Tatham's Twiddle variants have been annihilated! Now about those $4 \times 4$ rotating blocks... Eek.

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The final level: $6 \times 6$ with $4 \times 4$ rotating blocks

Jaap Scherphuis discovered the following algorithm:

Consider a $ 5 \times 4$ board, long side horizontal. Let the two possible moves be $L$ and $R$. Then the following is a three-cycle:

$$RLRL'RLRL'RLRLLR'L'R'LR'L'R'LR'L'R'LL$$

In short:

$$X = RLRL'$$ $$Y = R'L'R'L$$ $$X^3L'Y^3L$$

This results in an extremely annoying three-cycle. Nevertheless, there is sufficient maneuvering space to get any three desired numbers into position (which can be the pattern of the three-cycle flipped, reversed, translated, etc.) and then perform our algorithm. This is extremely tedious. Regardless, we now have a winning strategy.

Sample solve:

https://www.youtube.com/watch?v=zhT3kYMQfLU

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It's been a fun run. It surprises me how little research has been done on this game and its strategies, given the relatively intuitive mechanics. Granted, it is rather difficult to emulate the game mechanically, but I do think this nice combination game is pure enough to deserve an official name. I do hope that this game can be popularized a bit.