word - Twelve Labours - #03 Golden Hind

_{This puzzle is part of the ‘Twelve Labours’ series. Previous instalments can be found here: Prologue | 01 | 02}

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Hercules was grateful the morning sun wasn’t hotter as he pulled the heavy flatbed trolley laden with crates up Hippocrates Hill and along Aristotle Avenue to the Golden Hind. Arriving out of breath, he knocked at the deliveries entrance and the door was answered by the landlady, Artemis.

“Oh wow, Hercules – did you drag that all the way here?! You must be exhausted – come in, come in.”

Hercules picked up a crate and heaved it inside. Although he hadn’t recognised the name, the pub felt strangely familiar to him...

“Didn’t this use to be the White Swan?”

“Oh, a long time ago now!” chuckled Artemis. “It’s been the Golden Hind for the last month. Before that it was the Black Dog, the Golden Ram... You might say I’ve got a thing for coloured animals!”

Hercules lifted the lid of the crate and was met by the sight of 169 bottles of various colours, with letters on their caps, and a variety of odd symbols stencilled onto the wooden rim of the crate itself. A slip of paper on the top listed the names of the beverages to be found within. Hercules sighed.

“This is another puzzle, isn’t it?”

“Of course!” replied Artemis. “You see, I borrowed something from your mother and she would like it back. You just need to work out what that item was, given what you see before you...”

Hercules frowned. “Did you customise this crate yourself? Hang on – did you go down to the wholesalers already, open this up, do something weird to it, and then leave it there for me to bring all this way?!”

“May-be...” said Artemis, a little guiltily. “It is very heavy... Anyway, search around and see what you can find. Just beware – I’ve got a thing for coloured animals, remember...”

TASK: Solve this enigmatic puzzle to find the 13-letter object Artemis borrowed from Hercules’ mother. Watch out for those ‘coloured animals’...

_{Bottle-cap text reproduced below for copy-paste purposes:}

NOMTAWOSINODA  
RRIDRETEMEDSE  
ADLOGDETILFOX  

WSETUOBACKNYI  
AEMOHGRCHUMTS  
SDISEGEKSCANU  
YENNAIADSFLEE  
PMOOSNSERATWH  
PITGULURETDTT  
AHAREMUCHEAIE  
HCUOZSELSELEM  
NRRGWHERBREWU  
UAACHILLESELR

_{Grid of colours and list of unicode symbols available in source, if required.}

Answer

Take the descriptions of the bottles,

and index into the titles by each part of the percentages. (Indexing is a common puzzle hunt tactic: if you have a list of words/phrases and a number corresponding to it, take the (number)-th letter of the phrases.) For example, ALE ZEUS is rated 7.6%: take the 7th and 6th letters of ALE ZEUS and you get S and U.

Skipping all zeroes, the parts before the decimal spell START TOP LEFT. The parts after the decimal spell USE THE D-PAD.

So, to follow that new instruction:

Start in the top left corner, moving right. Whenever you hit a letter representing a direction, turn to move Up, Down, Left, or Right.

When doing this, the path takes you to nearly every bottle:

The path starts at the cyan here, then changes to yellow, then to orange, and then to red.

The unused letters spell out the item: a BACKSCRATCHER.

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A list of all of the other messages in the puzzle, useful or not:

The bottles:

Most of the bottle colors have a message in them.
- Dark green: NOT A CLUE
- Light green: ALSO NOT A CLUE
- Orange: FAKE TRACE
- Pink: WHITE LIES, reversed

- Light red: HERRING, reversed - Purple: RANDOM NOISE, anagrammed
- Dark red: RUBBISH MIXED UP, anagrammed
- Yellow: UNHELPFUL ANAGRAM, anagrammed

- White: WORD SEARCH
- Dark gray: DOT TO DOT AZUL
(The remaining three blue colors do not spell anything, and the light gray color that uses the majority of the bottles also does not spell anything.)

Following the first instruction from the bottles:

Solving the word search (using the beverage names as a word list), and looking at the remaining letters, you get: "NOT A WORD SEARCH, SEEK CLUES ELSEWHERE".

Following the second instruction from the bottles:

Drawing lines between bottles of each blue color, you get "ABV". This is a hint to turn to the ABV percentages given with the beverages, which leads you down the correct path.

The symbols:

The first letters of the symbols, reading around, spell a message: ONLY TWO CHARACTERS ARE RELEVANT. OTHERS ARE JUST DISTRACTING.

This refers to the greater than sign and the caret, which mark the start and end of the path in the true solution.

The beverage descriptions:

The first letters of the beverage descriptions tell you "ONLY FOR FLAVOUR", which says that the descriptions are otherwise irrelevant.

word - Riddles/puzzles having answer hidden in the question?

Can you give any example of riddles or puzzles in which the answers are someway "cleverly hidden" inside the question? I think that such examples are more common in word-related riddles, but maybe similar ones can be found also in numeric puzzles? Who knows?

I'm italian, and there's a very common riddle in italian language which says: "Nomina una parola qualunque contenente tre lettere U" (translated: "Name whatever word containing three 'U's "); the funny thing is that the italian word "QUALUNQUE" (a really really common word) is maybe one of the few actually containing three U letters...). That's what makes this riddle really clever and someway surprising when you reveal the answer...

So, do you know any other example in english language?

(Sorry if this is not a real riddle, but I think it can be interesting to discuss this topic)

Answer

I'll confess that I've always had a fondness for the Jester's riddle from Zork Zero:

"I once heard of a bookkeeper who, while working on the accounts of the Frobozz Magic Balloon Company, noted that the word 'balloon' has two double letters in a row! Stretching his limited imagination to the limit, this bookkeeper wondered if there were any words with THREE double letters in a row. He couldn't think of one -- but I'll bet that YOU can!"

material science - Is strain always relative to some initial state?

Let us say I am given a material with no knowledge about its history. Can I somehow calculate its strain ? Or a strain is always relative to some initial state (change in length/initial length) ?

particle physics - What exactly is a photon?

Consider the question, "What is a photon?". The answers say, "an elementary particle" and not much else. They don't actually answer the question. Moreover, the question is flagged as a duplicate of, "What exactly is a quantum of light?" – the answers there don't tell me what a photon is either. Nor do any of the answers to this question mentioned in the comments. When I search on "photon", I can't find anything useful. Questions such as, "Wave function of a photon" look promising, but bear no fruit. Others say things like, "the photon is an excitation of the photon field." That tells me nothing. Nor does the tag description, which says:

The photon is the quantum of the electromagnetic four-potential, and therefore the massless bosonic particle associated with the electromagnetic force, commonly also called the 'particle of light'...

I'd say that's less than helpful because it gives the impression that photons are forever popping into existence and flying back and forth exerting force. This same concept is in the photon Wikipedia article too - but it isn't true. As as anna said, "Virtual particles only exist in the mathematics of the model." So, who can tell me what a real photon is, or refer me to some kind of authoritative informative definition that is accepted and trusted by particle physicists? I say all this because I think it's of paramount importance. If we have no clear idea of what a photon actually is, we lack foundation. It's like what kotozna said:

Photons seem to be one of the foundation ideas of quantum mechanics, so I am concerned that without a clear definition or set of concrete examples, the basis for understanding quantum experiments is a little fuzzy.

I second that, only more so. How can we understand pair production if we don't understand what the photon is? Or the electron? Or the electromagnetic field? Or everything else? It all starts with the photon.

I will give a 400-point bounty to the least-worst answer to the question. One answer will get the bounty, even if I don't like it. And the question is this:

What exactly is a photon?

experimental physics - How can a charge of a particle be determined through its cloud chamber photograph?

Specifically in reference to Carl Anderson's famous cloud chamber photograph providing evidence of a positron. Any insight is appreciated.

Answer

In that experiment there exists a magnetic field perpendicular to the plane of the photo.

The motion of the charged particle in a magnetic field is used , equating the centrifugal with the centripetal force one gets

Bqv=mv^2/R

where B is the magnetic field, q is the charge ( with its sign) v the velocity m the mass and R the radius.

From the incoming direction and the right hand rule one knows the charge of the particle, in this case it has to be positive from the setup.

From the curvature one gets the momentum , thus the energy.

There exist ionization curves for a given particle mass

Ionisation is proportional to the number of ions scattered off as the particle moves through the cloud or bubble chamber, and is a function of the mass.

From the ionization characteristic for the measured energy, it could not be a proton, (at that small momentum the proton would have a much thicker ionization deposit and subsequent energy loss) it was consistent with an electron track which confirmed it as the proposed by Dirac positron.

energy - Doesn't the "Mexican hat" potential give a misleading impression that the barrier height between two vacua is finite?

For $\mathbb{Z}_2$ symmetry breaking in a Classical Field Theory described by a potential $$V(\phi)=\lambda(\phi^2-v^2)^2,\tag{1}$$ there is a finite energy barrier of height $\epsilon=\lambda v^4$ that exists between two minima $\phi=\pm v$ of $V(\phi)$ separated by a maximum at $\phi=0$. This can be seen clearly from the picture of a double-well potential. This finite potential barrier when multiplied with the infinite volume of space gives rise to infinite (potential) energy barrier. To be explicit, the infinite energy barrier comes from $$[V(0)-V(\pm v)]\int d^3x=\epsilon\int d^3x\to \infty!\tag{2}$$

But for $U(1)$ symmetry breaking, the potential energy barrier cannot be seen between two neighbouring vacua from the picture of Mexican hat potential. According to (1), this would mean by the potential barrier between two minima can be finite! But probably changing the vacuum costs infinite energy. If so, does it mean that the "Mexican hat" potential give a misleading impression? Why does the figure incapable of showing the actual barrier?

Answer

Good question. You are correct that, in principle, it is possible to rigidly, uniformly, and adiabatically rotate the entire field configuration by an identical angle in such a way that you always stay exactly inside the degenerate ground-state manifold. This gives the misleading impression that you can tunnel from one ground state to another with zero energy cost, as you suggest in your question.

But in practice, this isn't really realistically physically possible. That's because of the local nature of the thermal or quantum fluctuations. For concreteness, imagine the $XY$ model $$H = J \sum_{\langle i, j\rangle} \cos(\theta_i - \theta_j)$$ on a hypercubic lattice in $D \geq 3$ spacetime dimensions, in which the global rotational symmetry gets simultaneously broken. You can't realistically set up a situation in which every spin in the entire lattice rotates from pointing toward angle $\theta_i$ to angle $\theta_f$ exactly simultaneously, in one giant nonlocal perturbation. (For one thing, such a "perturbation" would require an infinite amount of time to set up, since the "planning message" about what angle to rotate by would need to propagate arbitrarily far before the process actually begins.) In practice, all you can do is first rotate a small little local cluster of spins from $\theta_i$ to $\theta_j$, then rotate that cluster's neighboring spins, then the next neighboring spins, etc. That's because by assumption, in a local QFT (or lattice system), the microscopic couplings are purely local, so nonlocal "update moves" (in Monte Carlo language) are not allowed.

Eventually, the effect of the initial perturbation will have propagated out far enough that a huge region of spacetime will have rotated its angle, so that to a good approximation you're locally in a new ground state. But the point is that as you do this process, at any given instant there will always be a boundary between the old and new ground states (kind of like a domain wall, but less sharply localized). That boundary will need to move and hit every spin in order for the entire system to rotate. So even if the rotation angle is infinitesimal and each individual spin's rotation only costs energy $\epsilon J$, the total energy of the entire tunneling process will still get multiplied by the volume of spacetime and will be a huge $\epsilon J V$.

Thought of another way: you can create a Goldstone mode with arbitrarily low energy by delocalizing it enough, but in practice it's impossible to create a Goldstone mode with exactly zero energy. The difference is subtle, but conceptually important.

mathematics - Hidden in plain sight - By what method do these images hide a prime number?

Note: This question has been severely edited from its original version to better fit the format of this site. Some answers below - including the accepted, correct answer - have been given based on the original question.

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Consider the following two digital images below:

Although they depict two very different things, they share a strong common feature: They hide an identical 3-digits prime number within themselves.

Not only do the hide the identical number, but they hide it by the identical method.

If you apply the method on image one, you'll get the prime number.
If you apply the identical method in the identical way to the other image, you will get the same prime number.

Important hints & restrictions:

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  The answer to this puzzle is not an ambiguous or subjective solution. Subjective interpretation is not involved.
  ("...looks like..."; "...reminds me of..."; "...could be interpreted as...";... are all invalid concepts.)

  
  


• 
  

  The answer lies within the image itself and can not be 'forced' upon it by altering the data.
  ( "...paint everything black and draw the prime onto it...";... is not a valid answer. )

  
  



• 
  

  There is no hidden meta-data to the images.
  (You can right-click & save it from here; or you may print it and then rescan it again; or you may simply take a good picture of you screen and use that; all of this will keep the image suitable for the method in question. However, you do need to use a computer to find the answer -- simply looking at the images does not suffice.)

  
  

Answer

The answer is 101, which you can see

by computing the Fourier transform.

How this works

You can think of an image as a function $F$ of two variables, $x$ and $y$. $x$ and $y$ represent the horizontal and vertical positions of each pixel, and $F(x,y)$ represents the pixel intensity (lightness or darkness). Now it turns out that any periodic function (and with images we only care about a finite range, so we can always just suppose $F$ to repeat once you reach the image boundaries) can be represented as a weighted sum of sine and cosine functions of different frequencies. These sums are called fourier series; the image below (taken from Wolfram MathWorld) shows various sums of sinusoids converging to a few different periodic functions in the one-dimensional case.

Note that in general for this representation to be exact for a continuous function you may need an infinite number of terms in the sum. However, in the discrete case, you only need to sum over as many different frequency values as you have samples (pixels) of the original function (image).

So in our discrete, two-dimensional case, what we have is this: a function $F(x,y)$ representing an image of size M-by-N can be written as a weighted sum of 2D sinusoids with horizontal frequency $u$ and vertical frequency $v$, where $u$ ranges from $1$ to $M$ and $v$ ranges from $1$ to $N$. You can think of the weight of each term as a function, say $G$, of the frequency variables $u$ and $v$. There's then nothing stopping you from thinking of $u$ and $v$ as horizontal and vertical positions of a new image, with $G(u,v)$ determining pixel intensity at each point. This is the process that was carried out to produce the "101" image above.

I've tried to keep this explanation as non-technical as possible; if you'd like to see the gory details of how the coefficients are actually calculated, MathWorld's Discrete Fourier Transform article has a nice overview of the 1D case. You can see the (relatively straightforward) generalization to two dimensions here.

To play around with this stuff yourself you could try ImageJ, a public domain image processing package developed by the NIH. It'll do all the dirty work for you -- just install, open an image, and select Process->FFT (Fast Fourier Transform).

To see a reason (one of the many, I swear) why you'd bother to do this outside of solving puzzles, try this nice, quick example of using FFT filtering in ImageJ to remove periodic noise from an image.

For a deep understanding of why this works, I recommend a course in applied analysis. (Or, I'm sure, any number of good books or online tutorials -- but I don't know what the good ones are, so I won't try to suggest one.)