logical deduction - Twelve Labours - Prologue

_{This puzzle serves to introduce the ‘Twelve Labours’ series of puzzles. Its solution is standalone and does not contribute to the meta-puzzle at the series end.}

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“Wake up, wake up, wake up! What are you still doing in bed? There’s lots to do – hurry now!”

Hercules groaned as his mother stomped into his room and flung open the shutters to let in the daylight. Rubbing his eyes, he sat up and yawned.

“Mum, it’s six thirty...”

“Then you’re already behind on your chores for the day! Get up, get up, get up!”

Stretching his arms out and yawning yet again, Hercules shook his head, stood up and shuffled towards the bathroom. This was always the usual pattern of his summer holidays when home from university – woken at the crack of dawn and expected to labour all day. He often wondered whether his mother genuinely believed that naming him ‘Hercules’ had bestowed upon him the physical strength of his namesake, such was the amount she expected of him each day.

Splashing cold water on his face to try and wake himself up, Hercules spotted a sheet of paper taped to the mirror. On it was written a list of twelve tasks:

1. Collect dry cleaning (Nemean Iron)


2. Collect stock from wholesalers (Lernaean Hydration)



3. Deliver beer (Golden Hind)


4. Deliver wine (Erymanthian Bar)


5. Help open up the bistro (Eugene’s Tables)


6. Get a haircut! (Stymphalian Beards)


7. Pick up new U7's team kits (Cleats ‘n’ Balls)


8. Submit betting slip (Dark Horse Bookmakers)


9. Pay for reserved belt (House of Hippolyta)


10. Buy beef (Meat Monster)


11. Buy apples (Hesper’s Hypermarket)


12. Collect Cerberus from kennels (Pluto Pups)

Hercules sighed heavily – this was going to take him all day. What’s more, the names of many of the businesses were unfamiliar to him...

“Mum,” he called. “Where even are some of these places?!”

“Turn over the paper – I drew you a map,” came the efficient response.

Sure enough, on the other side of the list was a crudely drawn map of their Athens neighbourhood with several buildings shaded in orange. However...

“None of the buildings have names?!”

“Of course not – that would be much too easy,” smiled his mother. “Read the instructions around the edge.”

Hercules sighed again. He had forgotten how much his mother enjoyed testing his mental agility. He had a horrible suspicion that each of today’s labours would be made more complex than was really necessary...

TASK: Use the map and instructions below to identify the 12 labelled business premises (in orange). Then identify which house (in white) belongs to Hercules’ family.

Note the logical-deduction tag - the solution can be deduced without resorting to guesswork. A complete answer is expected to explain the logical steps required to reach an answer.

Instructions from Hercules' mum:

• Nemean Iron is directly opposite (and across the road from) our house, while Lernaean Hydration is at the end of our row³.


• The symbols for Golden Hind and Erymanthian Bar rhyme⁶ with each other but are not adjacent² to each other.


• Exactly one business shares its starting initial⁵ with that of its symbol.


• Exactly one business shares its starting initial⁵ with that of a road on⁴ which it stands.


• Stymphalian Beards is further north than Pluto Pups, which is further north than Cleats 'N' Balls.



• Eugene’s Tables stands on⁴ the corner of Aristotle Avenue and Socrates Street.


• Hesper’s Hypermarket is not adjacent² to any other business.


• Stymphalian Beards and Dark Horse Bookmakers have two adjacent² neighbours in common.


• Completing the list in order, you will need to cross the road to the building directly opposite on exactly 3 occasions.


• In any given row³, each business has a different starting initial⁵.


• There are no businesses due west of House of Hippolyta.


• Building Mu is Pluto Pups if and only if Building Delta or Building Eta is Monster Meat.

<sup>Clarifications (if in doubt):

1 All buildings shaded orange are business premises. All buildings shaded white are houses. A building is either a business premises or a house (never both).
2 ‘Adjacent’ buildings share a wall boundary and are immediate neighbours either horizontally or vertically, not diagonally (e.g. Zeta and Eta are adjacent as they share a wall boundary; Zeta and Kappa are not adjacent as they meet only at a corner). Buildings with a road in between cannot be considered adjacent.
3 A ‘row’ is a horizontal line of 5 buildings stretching between Socrates Street and Hippocrates Hill.
4 A building is ‘on’ a road if one of its walls is adjacent to that road (e.g. Alpha is on both Socrates Street and Aristotle Avenue; Beta is only on Aristotle Avenue).
5 The ‘starting initial’ of a business, symbol or street is the first letter in its name.
6 For the purposes of this puzzle, only ‘Beta’, ‘Zeta’, ‘Eta’ and ‘Theta’ are considered to rhyme with each other. ‘Gamma’ and ‘Lambda’ do not rhyme, no matter how much you want them to...</sup>

Answer

The completed map:

Method:

1. Rules 5 and 8 make Stymphalian Beards and Dark Horse Bookmakers as a pair be Epsilon and Theta.

2. Rule 6 makes Eugene's Tables be Alpha, since Epsilon is already filled from step 1.

3. Rule 2 makes Erymanthian Bar and Golden Hind be in Beta, Zeta, or Eta. But since they can't be neighbors, one must be in Beta. By rule 10, Beta can't be Erymanthian Bar because Eugene's Tables is already in the same row. So Golden Hind must be in Beta, and Erymanthian Bar must be in either Zeta or Eta.

4. Rule 9 says that 3 times in the path, the next location is directly across the street. One of these times is from the house to Nemean Irons. The other two must occur in Alpha-Epsilon, Iota-Mu, or Delta-Eta. Using rule 12, assume that Mu is Pluto's Pups, which implies that Monster Meat is Delta or Eta. In order for two of the three to be consecutive locations, two of these must be true: Epsilon is Stymphalion, Iota is Hesper's, or Delta/Eta is either House of Hippolyta or Hesper's. However, Iota cannot be Hesper's (which is not adjacent to other's businesses). The two other conditions can't simultaneously be true because both require businesses with the same starting initial as their street, which rule 4 limits to one. Therefore, we have learned that Mu is not Pluto's Pups and Monster Meat is not in Delta or Eta.

5. Continuing to look for ways to satisfy rule 9, only Eugene-Stymphalion and one or two of House-Meat-Hester-Pluto could work (I abbreviate House of Hippolyta as House, not to be confused with Hercules' house). But we know from step 4 that Monster Meat can not be in Delta-Eta. Neither can Pluto because it is north of Cleats. This means that Delta-Eta can not be used. This makes Stymphalian Beards be Epsilon and Dark Horse Bookmakers be Theta. It also makes Iota-Mu one of House-Meat, Meat-Hester, or Hester-Pluto.

6. Since Stymphalian is on Socrates Street, Pluto can't be Iota or Mu (both on Plato street). This leaves only House-Meat and Meat-Hester as possibilities for Iota-Mu. Iota can't be Hester (has neighbor) or House (has business due west), so Meat Monster must be Iota.

7. If Mu is House, then Hesper (no neighbors) has no place to go because the only no-neighbor locations are ruled out: Lamba (two in the same row start with H) and Delta (street starts with same initial). Therefore, Hesper's Hypermarket is Mu. House is now either Gamma or Kappa.

8. There is not yet a business with the same starting initial as the location, and the only choices are Lernean-Lambda and Erymanthian-Eta. It can be shown that putting Lernean in Lambda leads to a contradiction, because that would make Nemean be Kappa, House be Gamma, and Erymanthian be Zeta, leaving Delta and Eta to be Pluto and Cleats, but Pluto is north of Cleats so it doesn't work. Therefore Erymanthian Bar must be Eta.

9. Lernean can now only be in Kappa or Delta since it is at the end of a row. But putting it in Delta leads to a contradiction because Nemean becomes Zeta, forcing Pluto and Cleats to be Gamma and Kappa, but that leaves House nowhere to go. So Lernean must be in Kappa. This puts Nemean in Lambda and the house across from Lambda.

10. House of Hippolyta must now be in Gamma, Pluto's Pups in Zeta, and Cleats N Balls in Delta.

logical deduction - Four coins (plus one) and a balance

Someone gives you 4 apparently identical golden coins.

You know that among them there could be exactly one fake coin, and that a fake coin hasn't the same weight of a golden one.

In your pocket you have a fifth - surely genuine - golden coin, and in front of you a balance with two pans.

What is the minimum number of weighings to determine if there is a fake coin between the four you've been given, and if it's heavier or lighter than a golden one?

Answer

How many weighings you need:

2

Call the four coins A,B,C and D, and the true gold coin G. You start by weighing

AB | CG

1. 
   

   
   

   If this balances, then it is only possible that D is defective. Weigh D against G to find out for sure.

   
   

   
   


2. 
   
   

   
   

   If it tips towards AB, there are three possibilities: either A or B is heavy, or C is light. To figure out which one, weigh A against B. If they don't balance, you learned which of them was heavier. If they do balance, you learn C was light.

   
   

   
   


3. 
   

   
   

   If it tips toward CG, this is very similar to the previous case. Just switch the words "heavy" and "light" in that paragraph.

   
   

   
   

Furthermore, it can't be done in just one weighing. There are nine possible situations (each coin could be heavy or light, or all could be same), and a weighing has only three outcomes. By the pigeonhole principle, there will be some outcome which would result from from two different situations, so a weighing could not distinguish between those situations.

quantum mechanics - Does this photograph portray double muon impact with nanogold atoms?

1PHOTO 1: Macro-photograph of an NIH/FDA TEM of a nanogold dark stained biological sample projected onto Silver Halide (AgX) photographic gel paper.

On June 10 I questioned if PHOTO 1 depicted an electron-positron particle annihilation caused by TEM generated electron impacts. My engineering instincts led me to think that the simultaneous red and green on the left and right hemispheres might be related to a positron (which Dirac likened to an electron moving back in time) was associated with what I thought to be a collision of particles that generated the separation of the two green areas that appear to be undergoing some type of explosive event.

I analogized the image in PHOTO 1 to the characteristics of the Roman era glass cup, the Lycurgus Cup which can be viewed in Wikipedia, reflects green and transmits red light depending on surface which the light photons initially strike.

Anna.v (10-June-14) kindly explained that the energy required to generate an annihilation (511 Kev) would not be present in a 200 KeV TEM and suggested that such energies could be found during collision with a random muon but an accelerator would be needed for verification of nuclear emulsion type material which recorded the event. Unfortunately, that would be the film from which PHOTO 1 was generated which is not readily available, so I began a search for other explanations beginning with the definition of muon, a lepton, one of the four elementary particles - it contains no quarks or sub particles.

The moons shadow from cosmic rays is depicted in Wikipedia under the definition of muon and the greens associated with the rays appear consistent with the greens in PHOTO 1.

However, in the original image of PHOTO 1, the green areas appear to engulf dozens of smaller particles and in the thousands of photographs taken of these TEMS, some type of interference pattern projects from "particle emissions" generated by what I suspect to be electron impacts with nanogold atoms used to dark stain the targeted biological. No others throw off this distinctive green coloration.

The imaging of PHOTO 1 probably inversely projects the green from the bottom to the top based on the angle of the light to the camera and the composition of the photographic gel paper (PHOTO 5).

Now I question if PHOTO 1 may be the image of a double muon impact, separated by the nanometer sized leg of the biological triangle. (PHOTO 2) If so, the deceleration of the muon would produce Bremsstrahlung radiation and decay into anti-particles. In this respect note the black streaks in PHOTO 1 which intersect at about 90 degrees above the two green areas which also align at approximately 90 degrees.

PHOTO 3: Screen photograph of enlarged areas of impact. Above the bottom vertices of the triangle and left of the right leg a circular particle emission spirals upwards possibly imaging expanding electron bands that have been pushed out from the impact area center. (Assuming the Rutherford orbital model is correct). Almost horizontally to the right of spiral and right leg another gold region appears to have been impacted.

Could Bremsstrahlung radiation from the decelerations of the muons provided the shorter wavelengths needed to image an EM wave? In point is the picometer size waveform that transits from the lower vertices towards then away from two areas of impact on the right side of the right leg of the triangle in PHOTO 2. Might this be an errant displaced electron seeking a new home - it appears to traverse towards two spots of glowing gold that repel it - an indication of reversal of ionic attraction.

PHOTO 4: Circular bands of color projected from suspected areas of particle emission/collision/expansion in PHOTOS 2 & 3.

PHOTO 5: PHOTO GEL COMPOSIITON: Photographic paper is comprised of four layer, a paper base coated with a layer of baryta/gel and photo gel, then about .5 MIL (12,500 nm) AgX (silver halide) and topped with a gloss coat which serves as somewhat of a projection screen with ellipsometric photography. This image portrays the reflection and "depth" and inverts the gold, possibly a reflection of gold dopant in the gel, from bottom to top.

Some insight into this inquiry can be gleaned from the following project posted on the web.

“Measurement of the Lifetime of Muons and Pions”

“In competition to muon decay negative muons can be captured by nuclei in a similar fashions as electrons from the K-shell can be captured by a nucleus. The probability for a muon capture increases with the fourth power of the nuclear charge number. Muons can be captured into the K-shell of an atom forming a muonic atom.

Since the rest mass of muons is larger by a factor of about 200 compared to electrons their orbit is approximately a factor of 200 closer to the nucleus. Consequently, there is a non-vanishing overlap of the wave function of the muon with the wave function of the nucleus and this can lead to a capture of the negative muon by the nucleus.

For Z = 40 [Au=79] the Bohr orbit is already inside the nucleus. The capture probability for this condition is very close to 1.”

(Advanced Laboratory Experiments, Universität Siegen, Dr. I. Fleck).

Coherent suggestions, input and answers would certainly be appreciated.

Thank you for considering these issues.

Walter Kyle 29-Jul-2014

classical mechanics - Does the speed of a wave travelling through a chain vary based on the size of the links?

If you have a single, linear, physical, steel chain, and you initiate a wave motion that travels through this chain, how would the speed or momentum of the wave vary based on the thickness and/or length of each link in the chain?

Assume that the force applied to the wave, the total length of the chain, the total weight of the chain, and all other characteristics remain the same -- the only variables are the thicknesses and lengths of the individual links in the chain.

Similarly, how would the speed or momentum of that same wave be different if there were no links at all in the chain, but rather a continuous steel cable of the same weight and length, but just in cable form vs. in chain link form.

enigmatic puzzle - Twisted Connections

_{This is an entry into the Fortnightly Topic Challenge #36}

The following is an Only-Connect wall seen in the BBC TV series. You have to divide the 16 words into groups of 4 each satisfying a specific condition.

Now, before you complain, the 16 words aren't given. Instead, you have been given Cryptic Clues that clue the words.

Next, after getting the words, you find out the groups of 4 and the relations between the words in the respective groups. So, you should get 4 words/phrases as your answer(this is where the game normally stops).

However, I can't stop there. So, next, using those 4 words/phrases, you have to find out another 4 words/phrases(How, you ask? That's the real puzzle)

Your final job is to create another only-connect wall with those groups of 4 words/phrases as the connections.

Answer

Here's my working so far, for the clues I've been able to figure out. I'll keep plugging away and update as I go.

Row 1:

THICKENS (Stiffens: anagram of KITCHENS)
REVIEW (analyse: English ditch=REEN, six=VI, gives REVIEN, then left (West) instead of up (North)). Thanks to Clarkey for the final part of this one
HANDED (Helped: Hotel=H, AND, education=ED)
BACK (to sponsor: the rear of something is the back of it)

Row 2:

LINE (railway track: Portuga-LINE-urope)
MASTER (leader: victory=MATE, small=S, real=R (currency))
STEVEN (Spielberg: anagram of EVENTS)
GRAVE (double straight). Solved by Persona

Row 3:

TAPES (piece=PAT, written left=TAP, is in Spanish=ES). Solved by Stiv
TEMPERED (initial letters + PER + bEDs). Solved by aPaulT

PLAN (design: quiet=P, unending street=LANe)
SCOTLAND (S + COT + LAND). Solved by aPaulT

Row 4

DOCK (COCK [pile of hay] with D replacing C). Solved by aPaulT
PARTY (celebrate: or participant gathering, double-straight)
TWIST (TWISTER [swindler], remove ER). Solved by Stiv
FLOW (current: no hesitation in river=FLOWer)

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Part 2: Connections

EVEN: Flow, Tempered, Handed, Steven(s). Solved by aPaulT
YARD: Dock, Scotland, Back, Grave
PLOT: Thickens, Line, Plan, Twist
RAVE: Master, Party, Review, Tapes. Solved by aPaulT

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Part 3: Using the Connections

Arrange the four 4-letter words into a square and read down the columns to reveal four new words. Solved by aPaulT.
PLOT
RAVE
EVEN
YARD

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Part 4: A new wall

nuclear physics - Is there something like "forced / induced electron-capture"?

I understand how electron-capture (also known as epsilon-decay) works and why it happens.

As far as I know we are able to do "induced fission", i.e. in nuclear reactors.

My question is: Is there a way to do (with our technology) something like "induced electron-capture"?

If so, how does it work? And if not, why isn't it possible?

I've been looking for an answer, but I couldn't find one on the internet.

logical deduction - Variants to blue eyes puzzle

Here are some variants to the Blue Eyes puzzle. All the rules are the same, but there are differences in (a) the number of people of different eye colors and (b) the oracle's statement. It is not necessary that everyone leaves - there could be a deadlock as well.

Variant 1: (a) 50 blue, 50 not blue (b) "At least one, but not all of you are blue-eyed."

Variant 2: (a) 100 blue, 50 not blue (b) "There are more blue-eyed people than those who are not."

Variant 3: (a) 100 blue (b) "At least 10 of you are blue-eyed."

Variant 4: (a) 50 blue, 50 not blue (b) "There are at least 10, but not more than 75 blue-eyed people."

Answer

Variant 1:

The oracle saying "not everyone has blue eyes" does not add any information here. Consider the case where there are only two people on the island, one of whom has blue eyes and the other having brown eyes. Being told that "not everyone has blue eyes" gives the blue-eyed person no information - seeing a brown-eyed person, the statement could have been trivial (i.e "not everyone has blue eyes because nobody has blue eyes") or not, and the blue-eyed person has no way to tell the difference. The brown-eyed person, seeing a blue-eyed person, does get some information about their eye color - it is not blue. Because they still do not know what color their eyes are (only a little bit about what they aren't), this statement, by itself, is not enough to cause anyone to leave.

With the first part of the statement being "At least one person has blue eyes", the second half of the statement is redundant. Following the same logic as in the original problem, all the blue-eyed people will leave on

day 50

. Then everyone who remains will know that they do not have blue eyes, therefore not everyone had blue eyes. Again, all of the people who remain will now have no clue as to their eye color other than that it is not blue.

Variant 2:

"There are more blue-eyed people than those who are not."

Consider the case of $N+1$ blue-eyed people and $N$ non-blue-eyed people (we'll say they're all green for simplicity). On the first day, each B sees $N$ B's and $N$ G's. In order for there to be more B's, they must be B. So on day 1, all blue-eyed people leave. If there are $N+2$ B's and $N$ G's, each blue-eyed person sees $N+1$ B's and $N$ G's. They must have blue eyes for there to be a greater number of blues than greens, so again all B's leave on day 1.

Now consider larger gaps - $N+3$ B's and $N$ G's. Each B cannot tell the difference between the real scenario and $N+2$ B's and $N+1$ G's. Same with $N+4$ vs $N$ not being distinguishable from $N+3$ and $N+1$. However, when nobody leaves the first day, the other possibilities are removed, so all the blue-eyed people leave on day 2.

So for both odd and even numbers of total people, the induction able to start with the halfway point being considered day 0. With 150 people, if there were 76 B's they would leave on day 1. If there were 77 B's, they would leave on day 2.

Following this pattern, all of the blue-eyed people will leave on

day 25

Variant 3:

"At least 10 of you are blue-eyed."

This simply starts off the induction a few steps down the line. 10 B's would leave on day 1, 11 B's would leave on day 2, ... so 100 B's (everyone) will leave on

day 91

Variant 4:

"There at least 10, but not more than 75 blue-eyed people."

As with variant 4, this starts off the induction a few steps down the line so all the blue-eyed people will leave no later than day 41. However, what about the second half of the statement?

Consider the case where we have 1 B and 1 G, and the oracle says "There is not more than 1 blue-eyed person". This statement could be said even if there were not any blue-eyed people, so the blue-eyed person does not know if their eyes are blue or not. The green-eyed person sees blue eyes and so knows that they do not have blue eyes. Again, knowing a single color that their eyes are not not does not reveal what color they are. If there were 2 G's and the oracle said "There is not more than 1 blue-eyed person", neither would know any new information about their eyes. If there are 2 G's and 1 B when the oracle says "There is not more than 1 blue-eyed person", the two G's both know their eyes are not blue, while the B knows nothing new.

If there are 2 G's and 1 B and the oracle says "There are not more than 2 blue-eyed people", nobody will know anything new. The oracle could say that for 2 B's and 1 G, 1 B and 2 G's, or 0 B's and 3 G's. They will each know that everyone else knows that there are not 3 B's, but the only information that can give anyone is what color their eyes are not, which is not actionable. Therefore nobody else's actions (or lack thereof) will give them any information to further distinguish between the possibilities.

So the second half the oracle's statement in this variant does not help give any information, so all of the blue-eyed people will leave on

day 41

waves - Whats behind light?

Once the light is emitted as a pulse, and once that pulse has passed, what's behind it? Is light a continuous wave, or only a wavefront? What is left in the wake?

Uncertainty Principle tricked - so why not Newtonian Determinism?

Recently I read that some results are obtained in directions of tricking the uncertainty principle. The relevant link is here: http://www.caltech.edu/content/tricking-uncertainty-principle , and the paper is on the Arxiv.

Now the problem I have is that if we can trick the uncertainty principle, doesn't this mean that the uncertainty principle is just a restriction on our current ability to make measurements rather than a fundamental limitation?

If so, can we assert that actually the universe does have certain physical knowledge of each particle's momentum and position simultaneously, and the interaction of the rest of the universe with that particle is not uncertain. But isn't this simply Newtonian Determinism?

So were is the catch?

pattern - What is an Organizable Word™?

_{This is in the spirit of the What is a Word™/Phrase™ series started by JLee with a special brand of Phrase™ and Word™ puzzles.}

If a word conforms to a special rule, I call it an Organizable Word™.
Use the examples below to find the rule.

$$\smash{\lower{28px}\bbox[yellow]{\phantom{\rlap{rubio.2017.01.30} \Rule{18px}{20px}{0px}\begin{array}{|c|c|} \textbf{Organizable Words }&\textbf{Not Organizable Words }\\\end{array}}}}\\ \begin{array}{|c|c|}\hline\phantom{\Rule{0px}{20px}{0px}} \textbf{Organizable Words }^™&\textbf{Not Organizable Words }^™\\\hline \text{ BEINGS }&\text{ PEOPLE }\\ \hline \text{ BOOTS }&\text{ SHOES }\\ \hline \text{ CASE }&\text{ CHASE }\\ \hline \text{ ELBOW }&\text{ KNEE }\\ \hline \text{ GOTHS }&\text{ MOTHS }\\ \hline \text{ RIFTS }&\text{ GIFTS }\\ \hline \text{ SADDER }&\text{ LADDER }\\ \hline \text{ SHILL }&\text{ SHELL }\\ \hline \text{ SHIP }&\text{ BOAT }\\ \hline \text{ SLOGS }&\text{ LOGS }\\ \hline \text{ SLOT }&\text{ SLOW }\\ \hline \text{ TAPED }&\text{ TIED }\\ \hline \text{ TON }&\text{ POUND }\\ \hline \text{ WONK }&\text{ NERD }\\ \hline \end{array}$$

In case you want it in CSV:

Organizable Words™,Not Organizable Words™
BEINGS,PEOPLE
BOOTS,SHOES
CASE,CHASE
ELBOW,KNEE
GOTHS,MOTHS
RIFTS,GIFTS
SADDER,LADDER

SHILL,SHELL
SHIP,BOAT
SLOGS,LOGS
SLOT,SLOW
TAPED,TIED
TON,POUND
WONK,NERD

The puzzle relies on the series' inbuilt assumption, that each word can be tested for whether it is an Organizable Word™ without relying on the other words.

These are not the only examples of Organizable Words™, many more exist.

Answer

An Organizable Word is

one that remains a word when alphabetized.

Below is a list of the example words that shows how they are Organizable:

BEINGS => BEGINS;
BOOTS => BOOST;
CASE => ACES;

ELBOW => BELOW;
GOTHS => GHOST;
RIFTS => FIRST;
SADDER => ADDERS;
SHILL => HILLS;
SHIP => HIPS;
SLOGS => GLOSS;
SLOT => LOST
TAPED => ADEPT;
TON => NOT;

WONK => KNOW;

optics - What is exactly an ‘virtual object’? (From the point of view of lens maker’s formula)

Here’s an image from my textbook

• 

It shows, how an image is obtained from a convex lens. The second and the third images, shows in depth , that how a convex lens behaves.

They say, that suppose ( in figure 2 ) the light ray enters the lens from the interface N1 B C. This forms a virtual image I1. Now this virtual image acts like an object for the second interface, and thus forms a real image at I ( figure 3 ) .

What is this virtual object? How can rays come from this virtual object towards the lens, from the right , and again the image is formed at the right hand side. This is possible only in concave lenses and in convex lenses when the object is between P and F. But neither of the cases apply here. What is it actually?

WHY do we treat this as an ‘object’ even when it does not exist?

PS- I know we do this with plane mirrors and others too, but thats reasonable there but here, the object and the image are on the same side! So whats it?

quantum information - Extract single qubit state from combined state in QuTiP

I would like to extract the state of single qubits from combined states, using QuTip (the Quantum Toolbox in Python. Is there some QuTiP function that will extract a single qubit state from a combined state?

Example

For example, suppose I begin with two qubits, $|0\rangle$ and $|1\rangle$. I can create these in QuTiP (as q1 and q2) as follows:

q1 = basis(2,0)
q2 = basis(2,1)

This gives me:

$q1 = \left[\begin{array}{&} 1\\0\end{array}\right]$ and $q2 = \left[\begin{array}{&} 0\\1\end{array}\right]$.

I can then combine q1 and q2 using tensor multiplication:

q1q2 = tensor(q1, q2)

This gives me:

$q1 \otimes q2 = \left[\begin{array}{&} 0\\1\\0\\0 \end{array}\right]$

Now, suppose that I apply a Hadamard gate to q1, and Identity to q2. I can do this as follows:

H = hadamard_transform(1)

I = qeye(2)
HI = tensor(H, I)
result = HI * q1q2

This returns $H = \left[\begin{array}{&&}0.70710678&0.70710678\\0.70710678&-0.70710678\end{array}\right]$ and $I = \left[\begin{array}{&&}1&0\\0&1\end{array}\right]$,

with result:

$\begin{align*}(H \otimes I)(q1 \otimes q2) &= \left[\begin{array}{&&&&} 0.70710678&0&0.70710678&0\\ 0&0.70710678&0&0.70710678\\ 0.70710678&0&-0.70710678&0\\ 0&0.70710678&0&-0.70710678\end{array}\right]\left[\begin{array}{&} 0\\1\\0\\0 \end{array}\right]\\ &= \left[\begin{array}{&} 0\\0.70710678\\0\\0.70710678 \end{array}\right]\end{align*}$

By reading off the values from this matrix, q1 is in the state $\frac{|0\rangle + |1\rangle}{\sqrt{2}}$, and q2 is in the state $|1\rangle$. Put another way, if we write q1 in the form $\alpha|0\rangle + \beta|1\rangle$, we get $\alpha = \frac{1}{\sqrt{2}}$ and $\beta = \frac{1}{\sqrt{2}}$. And for q2 we get $\alpha = 0$ and $\beta = 1$.

Extraction

Given a combined state, finding the values of $\alpha$ and $\beta$ for single qubits are what I mean by "extract a single qubit state from a combined state." I would like QuTiP to find those values for me. Of course in this simple case, I know I can do this by inspection; but in more complicated cases, I would like QuTiP to do it for me.

Answer

Before we talk about how to get QuTip to do what you're asking, we should make sure you understand what it means to "extract" one of the qubits" states. In general, such "extraction" is not possible.

Combined quantum systems

The question, we have four dimensional vectors representing the states of the combined two-qubit system. Note that when you write a vector in terms of components, there's an underlying assumed basis for the vector space. For example, when the question writes $$\frac{1}{\sqrt{2}}\left[ \begin{array}{c} 0\\1\\0\\1 \end{array} \right] \, ,$$ this really means $$\frac{1}{\sqrt{2}} \left( 0 |00 \rangle + 1|01 \rangle + 0 | 10 \rangle + 1 |11 \rangle\right) \tag{$\star$}$$ where in each ket the first number refers to the state of $q_1$ and the second number refers to the state of $q_2$. The notation used here has the tensor multiplication implicitly, but if we write it explicitly we get \begin{align} & \frac{1}{\sqrt{2}} \left( 0 \left( |0 \rangle \otimes |0 \rangle \right) + 1 \left( |0 \rangle \otimes |1 \rangle \right) + 0 \left( |1 \rangle \otimes |0 \rangle \right) + 1 \left( |1 \rangle \otimes |1 \rangle \right) \right)\\ =& \frac{1}{\sqrt{2}} \left( \left |0 \rangle \otimes |1\rangle \right) + \left( |1\rangle \otimes |1\rangle \right) \right) \tag{$\star \star$} \end{align} where here in each term the first ket refers to $q_1$ and the second ket refers to $q_2$. anyway, the point is that we're using a basis set $\{|00\rangle, |01\rangle, |10\rangle, |11\rangle \}$. This is a very common choice of basis, but note that others are possible. The state $(\star \star)$ can be factored into $$\frac{1}{\sqrt{2}} \left(|0\rangle + |1\rangle \right) \otimes |1\rangle \, .$$ Because we can factor the combined state into a product of single individual qubit states, in this case it is meaningful to talk about the individual states. However, this is not generally the case. Consider, for example, a Bell state $$ \frac{1}{2} \left( |0\rangle \otimes |0\rangle + |1\rangle \otimes |1\rangle \right) \, .$$ There is no way to factor this into a product of individual qubit states. States that can't be factored are called "entangled".$^{[a]}$

How describe a sub-part of a combined system

Of course, if you make an entangled state, you can make measurements on just one of the qubits, so there must be some way to talk about the information contained in a single qubit without the second one even if the state can't be factored. To do this, we have to go beyond the usual state vector description of quantum systems and use a density matrix. Given a regular state $|\Psi\rangle$, the associated density matrix $\rho$ is $$\rho \equiv |\Psi \rangle \langle \Psi | \, .$$ If you have a regular (i.e. classical) probability distribution over a bunch of quantum states $\{|\Psi_i \rangle \}$ with probability $p_i$ for each one, then the density matrix is $$\sum_i p_i |\Psi_i \rangle \langle \Psi_i | \,.$$

What's the density matrix for the Bell state $(1/\sqrt{2})(|00\rangle + |11\rangle$)? Well, following the recipe, it's \begin{align} \rho_\text{Bell} =& \frac{1}{2} \left( |00\rangle \langle 00| +|00\rangle \langle 11| +|11\rangle \langle 00| +|11\rangle \langle 11| \right) \, . \end{align} Great, but what do we do with it? The Bell state is not factorable, which means that we cannot find a state vector for each qubit individually. However, we can find a density matrix for each qubit. This is done by taking the trace over the qubit you don't care about. A trace of a matrix is where you sum the diagonal elements, i.e. $$\text{Tr}M = \sum_n \langle n | M | n \rangle \, .$$ In our case, tracing over the second qubit picks out elements of $\rho_\text{Bell}$ where the state of the second qubit is the same (try it yourself to see why). The result is $$ \rho_\text{Bell, first qubit only} = \frac{1}{2} \left( |0\rangle \langle 0| + |1 \rangle \langle 1 | \right) \, . $$ Remember when we said that a density matrix can be a sum of probabilistically weighted quantum states? This is a case of that: we have a (1/2) probability that the first qubit is in state $|0\rangle$ and a (1/2) probability that the first qubit is in state $|1\rangle$. These are not probability amplitudes. These are normal probabilities. This is an example where when you lose part of an entangled system (i.e. by ignoring the second qubit), you lose quantum coherence and wind up with classical probability distributions. This is intimately related to what happens when you measure a quantum system and get so-called "wavefunction collapse".

Answer to the question

QuTip supports combined systems, including the partial trace.

---

$[a]$: Note that there are bases in which the Bell state factors, but the states in those bases mix the two qubits. For example, a Bell state is factorable if you use the Bell states $$\{|00\rangle + |11\rangle, |00\rangle - |11\rangle, |01\rangle + |10\rangle, |01\rangle - |10\rangle \}$$ themselves as your basis.

gravity - Why is the Sun almost perfectly spherical?

Relatively recent measurements indicate that the Sun is nearly the roundest object ever measured. If scaled to the size of a beach ball, it would be so round that the difference between the widest and narrow diameters would be much less than the width of a human hair.

I do appreciate that above result is just one measurement , and I looked for confirmation of the result. However, Wikipedia accepts its validity:

By this measure, the Sun is a near-perfect sphere with an oblateness estimated at about 9 millionths, which means that its polar diameter differs from its equatorial diameter by only 10 kilometres (6.2 mi).

Two questions on this subject:

• 
  

  To me at least, it is a completely counter-intuitive result. Can anyone explain from what causes this symmetry emerged? Is it a combination of a slow rotation rate combined with a highly isotropic central gravitational field? I  thought there would be an equatorial bulge, even though the rotation rate is slow.

  
  


• 
  
  

  Does this result, for just one ordinary, as far as I know, star indicate that asymmetrical stellar collapses are much less likely than may have been previously envisaged? Admitted, it is just one star out of countless billions, but on the other hand, as it is a random sample, it may well be indicative of many more similar "extremely" (if I can use that word) spherical objects.

  
  

Answer

The symmetry of the Sun has got very little to do with any symmetry in its formation.

The Sun has had plenty of time to reach an equilibrium between its self gravity and its internal pressure gradient. Any departure from symmetry would imply a difference in pressure in regions at a similar radius but different polar or azimuthal angles. The resultant pressure gradient would trigger fluid flows that would erase the asymmetry.

Possible sources of asymmetry in stars could include rapid rotation or the presence of a binary companion, both of which break the symmetry of the effective gravitational potential, even if the star were spherically symmetric. The Sun has neither of these (the centrifugal acceleration at the equator is only about 20 millionths of the surface gravity, and Jupiter is too small and far away to have an effect) and simply relaxes to an almost spherically symmetric configuration.

The relationship between oblateness/ellipticity and rotation rate is treated in some detail here for a uniform density, self-gravitating spheroid and the following analytic approximation is obtained for the ratio of equatorial to polar radius $$ \frac{r_e}{r_p} = \frac{1 + \epsilon/3}{1-2\epsilon/3}, $$ where $\epsilon$, the ellipticity is related to rotation and mass as $$\epsilon = \frac{5}{4}\frac{\Omega^2 a^3}{GM}$$ and $a$ is the mean radius, $\Omega$ the angular velocity.

Putting in numbers for the Sun (using the equatorial rotation period), I get $\epsilon=2.8\times10^{-5}$ and hence $r_e/r_p =1.000028$ or $r_e-r_p = \epsilon a = 19.5$ km. Thus this simple calculation gives the observed value to a small factor, but is obviously only an approximation because (a) the Sun does not have a uniform density and (b) rotates differentially with latitude in its outer envelope.

A final thought. The oblateness of a single star like the Sun depends on its rotation. You might ask, how typical is the (small) rotation rate of the Sun that leads to a very small oblateness? More rapidly rotating sun-like (and especially more massive) stars do exist; very young stars can rotate up to about 100 times faster than the Sun, leading to significant oblateness. However, Sun-like stars spin-down through a magnetised wind as they get older. The spin-down rate depends on the rotation rate and this means that single (or at least stars that are not in close, tidally locked binary systems) stars converge to a close-to-unique rotation-age relationship at ages beyond a billion years. Thus we expect (it remains to be proven, since stellar ages are hard to estimate) that all Sun-like stars with a similar age to the Sun should have similar rotation rates and similarly small oblateness.

riddle - Where Is My Friend?

A friend of mine, one of those world-traveler, rugged outdoorsman, live-out-of-a-tent-and-backpack sorts, has just sent me an interesting email:

  From: whatchootalkinboutGrylls@hotmail.com
  Subject: Check out this pic! Guess where I am?!
  To: rubio@[redacted]
  Date: 04/08/17 11:33 AM
 
  Dude, check out this epic pic:
 
 

 
  It's pretty awesome, isn't it? I think we captured the scene perfectly.
  That's obviously just a thumbnail; for full resolution, open the attachment!

  I'll see you in a couple months, man - keep it real!
    — WG
 
  [Attachment(s):fullrespic.doc]

---

Where the picture should have been I just saw a blank space, but I know sometimes embedded images don't work all that well; good thing he provided the attachment. I figured I was just a click away from seeing an amazing photo in all its full-res glory. But on opening the attachment, instead of finding an image I found ... well ... I'll just show you what it said. Since one of you apparently is partially to blame for this, it's only fair you help me figure out where my friend is!

________

Yo, I know you like puzzles. One of the guys I'm with out here is a puzzler himself — he's even on that puzzle website you were telling me about. Anyway, he helped me put together a bit of a puzzle for you. See if you can figure out how to get through my trail of clues to find my picture and figure out where I'm at!

The trail begins with some clues for you to figure out....

1: One thing leads to another

1. I heard you sing Flying High Again? (5)


2. Weird old camel’s place in Maryland. (2*,6)


3. 904


4. IIX857, a synthetic hormone, makes a heavenly body. (####,8)


5. Unusual array: song, nose, tongue, ear, eye, or skin, maybe. (1,7,5)


6. Perhaps a stout tankard made of hot and heavy metal with an edge. (3,5)



7. A measure of temperature just above zero? (3,6?)


8. (½)(12)=6 (4,5)


9. A vehicle full of blind drunk denial heads west on Rodeo Drive for date, maybe. (8,4)


10. Live! Exit around center aisle! (5)


11. Strike her head suddenly, then bug-eye; then, almost unendingly wrong, start sprinting thither. (6’1,10)

2: Looking for connections

Those weren't just simple standalone clues.
There are sets of answers that go together. Figure out how, or your trail ends here!

  1 & 4 & 11
  2 & 5 & 7 & 10
  3 & 8
  6 & 9

Do you have a hunch how these are connected?
Blaise de Vigenére says: C M P X V G V P V :)

I'm going to call the way these are connected, LINKINGS.
When you're sure you know how LINKINGS work, proceed to ...

3: Follow my lead(s)

Taking your final answers from Part 1, collect all the letters from all the answers together. Sort them alphabetically, discarding duplicates. You should end up with 13 letters. They are your key to unlock the next step:

  $\small\texttt{LFCHA A̕DBJZVZB TP WRZLHRZ O}$

If you've got the right answers from Part 1 ...
If you've correctly figured out the LINKINGS for Part 2 ...
If you've deciphered the code above ...
If you're reading me Loud and Clear ...

Then you should be able to figure out what to do with the following leads:

• You're going to need 5 specific letters, and I've just given you what they are.


• You're going to need a grid to put them in, and I've just given you what it looks like.



• There's only one way the letters can go into the grid.


• Make sure ALL your letters make it into the grid!

When you've got everything up to here squared away, proceed to ...

4: How to unlock the solution, from A to Z

You've filled in a grid, more or less. Time for the next step.
Take the grid's outermost letters and sort them alphabetically, discarding duplicates.
They're now your key to unlock the next clues:

Ypw'pt becsvzlz yevf gc jjpt ujiaq nzgcybfvmkmyzp amr zqyz ajb jvliua. :)

• Q: Infhipqe ugbbnbcb vzovv e abvqv dq fjtwb-obok kyiov #255,153,051? (4,7,6)


• V: Xmeh fic cmwi aqise


• E: Boxp wwggv Dzkpikw ugfwt. (5)


• J: Nasv svq pzlz (5)


• A: Ao gbbuaqk Nmlojsnqa qipqtbeiiz qwjxmvftil ua Sdzpy, Ppxidvc. (5)


• B: Kyru pyunrf fic cmwi vgzpvm dovt


• X: Uaactpjas gcmevuyo dos cr Iyrfzxyn nkrs ecstdcs. (5)


• W: Wrsqzdk Nsnecc Aoucfg wovvl tqnrvmq fpnpwivbx hc. (5)



• Q: Nu Hmdry yvq a eqrsql’g mjgcf. (5)


• W: Tidg hyz disux, kxhs kcc njpxp


• U: Swinr pbtx, kxhs epkbft (3)


• D: Mjvgkn: gn fzmafrbtz. (2)


• N: Sjzxp oyiv jd pbtx wzr


• V: Umsnlgr bqrbrbcr’t didqeoxz. (5,3)


• B: Io Tsuq, n blhzes qj ugervm aatgw umawwzqtff mv hvcczlcf krqfnzct. (5)


• Z: Cbuwqgf Qcvw bpzia QFDE NnosvwKqahvm jebfize nbu Amrnwpi Aas ivaes Liiz bf Xdsljcrw. (5)


• M: Pcnvpdma gzk bik dl pbkr. (6)


• O: Oyiv ouo gtsu bnfk jle



• T: Rezf V qcpc VJKM


• N: evunvgtjpjw() ieogkcr dtiifrr joycl qzmdszfr. (2)


• C: Tig zmdl tzmqt dnym


• I: Swinr pbtx, bqahy xjuf


• N: Rmugvvm wet qv va, csicypt. (5)


• A: Tidg cez alvg itqise


• I: Kebpmvs vb "Rtc-azg, Wqd!" (3)


• K: Owocr QT gpmauvpn, cppjcevbx kjadg fmonav xmhftivf. (5)

Bv, fic mpti btvbx. Tmu xqr'b wacn rfau vs la jwkc rhju cmf, oik D'k psqzqpvbx dr npy.

  mushf: k-C-L-Y-t (rro)

Whew! What comes next?

5: Things aren't always black and white

Four of your answers from Part 4 are, well, colorful.
It's time to add some color to those LINKINGS! Here's how:

Sort your colorful answers alphabetically by the index letters of their corresponding clues.
Then apply these cryptical directions* to the answers in their sorted order, to find which clue (and thus what color) goes with each LINKING. $$\def\T#1{\small\text{#1}}\begin{array}{rcl}\hline\T{first clue}&~~~~&\T{painlessly behead}\\\T{second clue}&&\T{no novice in speech}\\\T{third clue}&&\T{cut to the core}\\\T{fourth clue}&&\T{remove hind end; make bland; remove hind end}\\\hline\end{array}$$                     ^{* They're cryptic-like more or less, but not necessarily entirely fair. Deal with it. :)}

The fifth LINKING also has a color; its color is: SILVER.

When you've found the colors for all five LINKINGS, proceed to ...

6: Hex marks the spot

For an effective hex, one must put the right things together in the right way.
It's time to use something I gave you in Part 4.

There are numbers, 1 through 5:

• Each has a location.


• Each location identifies a LINKING.


• Each LINKING identifies a hex code.



• Each LINKING also has a color, which itself identifies a hex code.


• XOR the two hex codes together.

Find the values for 1 to 5.
Put them together to find an image that should look very familiar
... but with some things missing, and some things new.

Useful URLs (among plenty of possibilities):
    http://www.colorhexa.com/
    http://xor.pw/

One more thing:

My friend wants me to tell you this message before you continue. He was going on about some Lord of the Rings quote — something about a phial, and light, and Eärendil’s star, and that his message was like that only the opposite? He's weird sometimes.
Anyway, he says, "King DEDEDE says OOOOOOh when he drinks his C0C0A0." I have no idea what he's talking about, but he made me promise to put that in here so there you go.

Moving on!

7: Being hexed may leave you feeling down

Part 4 gave you answers from A to Z.
Part 5 used four of them.
Time to use five more.

As you look down at your notes you should notice that some of your answers to Part 4 seem familiar from Part 3: some exactly, and some in part.
Now would be an excellent time to put what's missing into the image found in Part 6.

You'll find, as you write things down, that you're able to fill about 4½ columns using 5 answers.

That's another five answers down ...

8: Feeling down might lead to a cross word or two

Cross six more answers off the list.
(Do I really need to say more?)

By the way, remember the theme of this puzzle you found way back in Part 2?
The six entries you just crossed off? Yeah....

9: Be of good cheer — the end is near!

Eleven entries left. They spell out the key that will let you decipher the final text below.
If you're not sure how to find that key, it may help to remember where they came from and what I've told you about these entries previously.

(Tfh "cldc ld", en'h g tzqy!)
(Qsq "txcdra", irl dkxxn)
Zkzoky ^^ klmt lnljr.

Iy sfy'de kyrhunz nyme, yho'mi som nyi dizbk oqy — viekdamocefihhj!

Ef tacj taign psg sailpp htpv juleyu mz tay xvud B arzq yho zr Bakn 6 tsypeykixy, xrtibt yii sze xggxk sjorvq ig nyi pptx. Klmt xggxk sjorvq rxjiieegnj xte ictxgrx C gvambmvh kon... ueh zop, uk pmsm, C'd vqaws ks feef psg wayii ut bm.

Csak uuto mt Iuix 7. Defydfqr mbfwq fbpv ezspyiw? Uf mbv kdiw C xehe ril ams t nptuctf tvaslqfvp pntqpq gkcu, xtoly nsglw vv xte thjaqrl zfv fhx Xfaz ceovw zufvvvqd hhv xtrhoxl rioy. Z amnm sfy fo plzxq, ig iihqr, mbv E-L egnic xemnvve fhl veoh hz klasx zzzq agmnids, uok qmkx nyi adw hlqnekyu iztkcvw xopyi gmsx, ueh fhx ymiz nngsidew yexdixm ltbek wrwq - sh ck paodm cmwe qRoBj Tay iieuen zw, brxxzgfaufp, e bnz cdkgr mux. M fhbhb cau'ef rkdex nyi tizbvv delicyfihh zw m ghiu mypkimiyegn fzqr mbv xtufveeul B jlx un mbv iyabf sspy.

HHV PMSM NYMZG:

Bz psg'rx uj snsxmjmhe tm Z ey, tauk iypms jtat bh klq diuu qgsm vv hdiocek kon HLXE.
Fhlxiftbhx enonn klq ckijwiokx rrp chhjmpekcek vuln klmt wjrh, ihtn ciftxl xsqs bh klmt xggxk siuti, mnw qyc?

Postscript

Hope you enjoyed the puzzle, and the picture.
Happy trails, dude!
  — WG

________

So, I couldn't even get the first bits done!

I could really use a hand, and since my friend had *ahem* some help... it seems only fitting that I get some too.

Can you help me figure out:
What is this a picture of?
Where is my friend?

_{Part of a metapuzzles entry: I, for one, welcome our new ...}
  ^{Metapuzzles answer: What word does the very last letter you solved for represent?}

Answer

This answer is half a record of my solution path and half the "intended" solution path.

Here is the spreadsheet used to solve.

Thanks to Silenus and Sp3000 for the help solving.

Part 1

The clues solve to USING, MD LOCALE, 904, 1714 ASTEROID, A SENSORY ORGAN, PUB ORDER, ONE DEGREE?, HALF DOZEN, CALENDAR WORD, EXIST, and HITHER'S ANTITHESIS.

These are all

clues to more words. For instance, 1714 ASTEROID is SY, EXIST could be BE, and HITHER'S ANTITHESIS could be YON. But what about the rest? There seem to be too many options. Hm... let's continue on for now.

Part 4

Wait, part 4? Yes. A couple educated guesses (armed with the knowledge that the Vigenère key is alphabetical) reveal the key to be abceimnorvy. That reveals a ton of clues, but let's not worry about those just yet. We know that the key from 3 is also alphabetical, and a superset of part 4's key.

Part 3

Time for more educated guesses! Just adding two more letters (L and S) gives the message LEADS P̕ROVIDED TO UNRAVEL X. Well, it seems like the "leads" will be helpful if we want to do any "unraveling"... and we need five letters... let's take LPTUX! (Er, LP̕TUX. Apparently P brought a diacritic along with it.)

The message mentions 5x5 grids... and we need 5 letters... Hey wait, those letters are pentominoes!

And this is where the puzzle finally starts to pick up. Suddenly, we can do a lot of things!

Parts 2 and 3 (again)

We can decode the part 2 message with the keyword PENTOMINO(ES) to get the confirmation, NICE HUNCH.

We can assemble those pentominoes into a 5x5 grid into exactly 1 way (up to reflection/rotation).

We can solve the rest of the part 1 "subclues" and fit the answers into the pentominoes crossword-style.

(And we get the message "MAYBE CLEAR" for confirmation, which currently describes our thoughts on the puzzle!)
We can decode the part 3 and 4 messages the "right way". Speaking of part 4, it's time to solve some clues...

Part 4 (for real this time)

We have 26 clues to solve. Luckily for us, 11 of them are from part 1. The rest are regular (non-contrived (usually)) cryptic clues. (Details on the sheet.)
We also have an image.


This'll be important later.

Part 5

Four of our solutions from the part 4 clues have colors: YELLOW, GREEN TEA, BLUES, and DEEP SAFFRON PRIMER. The instructions tell us what to do with them: remove Y/OW, remove GREEN and take a homophone, take the centre letter, and remove DEE/R from the edges and SAFFRON from the middle.

This gives us letters: ELL, T, U, and P (prime). That's four of our pentominoes! We can now assign colors to them.

Part 6

Look up the colors in each location based on the letter of the pentomino there. They look slightly different from the actual colors of the pentominoes. XORing them together gives us ASCII values kh30K, which lead to this image:

Part 7 and 8

By this point, you may notice that CLEAR and MAYBE are solutions to two part 4 clues. And so are MCMIV and ERYON, which read downwards. We can cross off those words, then add letters in the edges of the X to cross off 7 more words: YES, MESSY, IS, SO, VISON, ALESI, BASSO. (All the added letters are S, incidentally.) This completes the grid (and gives confirmation for missing part 1 answers, if you're missing a clue or two).

Part 9

We've used all clues from part 4 but the ones copied from part 1. Those clues' letters are EGIKOPRSUWY, which anagram to KEY IS GROW UP.

Decoding part 9 with GROW UP gives a helpful message that GROW UP is a cluephrase, not the literal key. So we try MATURE... and it works!
It gives us a confirming message:

Part 9

So, we do what the message says. Take the "down" words, find their clue letters, make them alternate capital and lowercase, and we get this:

...Oh.
Well, I guess I should've expected that.

Now what?

Let's see, what information have we not used? There are the three hex codes DEDEDE, 000000, and C0C0A0. There's the centre square of the d-pad.

And... there's the steganography tag. The image isn't all the same shade of white. Coloring the image in the colors provided gives this nice picture:


Of course! It's a polar bear blinking in a snowstorm, and that's why we couldn't see it. And polar bears live at the north pole, which is Rubio's friend's location.

One Last Thing

So, what do we do with the D-pad? Ignore everything else, as the note says. We have four arrows, all of which have an S. This must be a sort of "compass rose", though of course every direction points south. Meaning the center would be the north pole itself, with the letter N. That means north is the answer we need for the upcoming metapuzzle.

statistical mechanics - Why can't the density difference between the liquid and solid be an appropriate order parameter for liquid-to-solid transition?

The order parameter $\mathcal{O}$ in the case of a liquid-gas transition is the density difference $\mathcal{O}=\rho_{liq}-\rho_{gas}$.

But in the case of a liquid-to-solid transition, the order parameter $\mathcal{O}$ is not taken as the density difference $\mathcal{O}=\rho_{sol}-\rho_{liq}$. What is the reason? Is it just because the density difference is too small to measure?

Answer

There is no unique definition of "the" order parameter. However, we would like to use an order parameter that exhibits the full symmetry breaking pattern of the transition. If the transition corresponds to some symmetry $G$ breaking to a smaller symmetry $H$ then we want the order parameter to transform non-trivially under $G$, and exhibit the residual symmetry $H$. In the liquid-gas transition there are no continuous symmetries, and using the density is fine. In liquid-solid we break translational symmetry to a crystallographic symmetry. This requires a more complicated order parameter, like the F-trafo of the density correlator.

optics - If a lens focuses all incoming light to a point, how do we get 2D images?

How do lenses produce 2-dimensional images, if a lens bends all incoming rays of light to intersect at the focal point? Shouldn't this produce a single dot of light on a screen placed at the focal length?

This is basically the standard diagram that always shows up in textbooks:

I know this doesn't happen in real life--I used to use telescopes pretty frequently for work. The most in-focus image would be the one with the smallest diameters for the stars, which, of course, we think of point sources at infinity. But--the light from all stars (despite being at infinity) is not all focused at a single point. Instead, each star's light is focused at its own $(x,y)$ point on a 2D image.

I can't reconcile the theory as I understand it with any of my real-life experience with optics. What am I missing?

Answer

...if a lens bends all incoming rays of light to intersect at the focal point? Shouldn't this produce a single dot of light...?

(In your diagram, the source image is at infinity. I will continue the analysis along that idea.)

It is true that all rays parallel to the axis focus to that single dot. Not all rays, however, are parallel to the axis:

Rays coming from different angles focus to different points. That is how an image is formed.

special relativity - What do physicists mean by "information"?

On the question why certain velocities (i.e. phase velocity) can be greater than the speed of light, people will say something like:

since no matter or "information" is transferred, therefore the law of relativity is not violated.

What does information mean exactly in this context?

It may help to consider the following scenarios:

If a laser is swept across a distant object, the spot of laser light can easily be made to move across the object at a speed greater than $c$. Similarly, a shadow projected onto a distant object can be made to move across the object faster than $c$. In neither case does the light travel from the source to the object faster than $c$, nor does any information travel faster than light.

Read more: https://www.physicsforums.com/threads/phase-velocity-and-group-velocity.693782/

Answer

In the case of relativity, "information" refers to a signal that enforces causality. That is, if event A causes event B, then some signal must travel from A to B. Otherwise, how would B "know" that A had occurred?

Some examples:

• Light (signal) from a candle (A) hits your eye (B), causing you to see it.


• Electricity (signal) flows from a connected switch (A) to a light bulb (B), turning it on.


• Your friend (A) throws a wad of paper (signal) that hits you (B) in the back of the head, causing you to turn around to see who's trying to get your attention.

In all of these cases, the effect (B) comes after the cause (A) because there must be some signal from A that interacts with B to cause B to happen. The technical term for this is "locality." Over the centuries of studying how the universe works, scientists have found that all causes are local to their effects; nothing happens at a distance without something (light, sound, matter, etc.) acting as a go-between. [1] If you want to interact with some distant object (a friend, a planet, an enemy target), you either have to go there yourself or send something in your place (a letter, a satellite, a missile).

Let's consider the case of a laser beam swept across the face of the Moon. Let's further imagine that there are two astronauts, Alice and Bob, on the surface of the Moon with a large distance between them. The laser spot sweeps across the Moon and falls upon both Alice and later Bob, with the spot moving at faster than the speed of light. So, the question is, does that spot constitute a causality signal from Alice to Bob? The answer is no, because nothing Alice does will affect how the spot moves or when it moves or even if it moves. The cause of the light is on Earth and is not local to Alice. Nothing Alice does will change the spot that Bob sees.

There is a way that Alice can use the spot. She can hold up a mirror and reflect the laser beam towards Bob. The reflected laser beam is a causality signal because its origin is local to Alice. Alice can choose whether or not to reflect the beam at Bob. But, notice that this signal travels at the speed of light. It will arrive after the laser spot sweeping across the surface.

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[1] This is why Einstein and others objected to quantum entanglement weirdness. It looks like signaling at a distance, but it's really not. Various mathematical and experimental discoveries show that not even entanglement's "spooky action at a distance" can transmit information faster than light. Quantum teleportation has been demonstrated in the lab, but there must be a slower-than-light signal between the sender and receiver to make the system work. There's far too much detail to go into here.

acoustics - How to calculate the sound pressure at a given distance of a point source?

Consider a point source in a free field. The source is emitting spherical waves. I know the relationship ${p_1\over r_2}={p_2\over r_1}$ that lets you convert the sound pressure $p_1$ at a point at a distance $r_1$ to the point source the sound pressure $p_2$ at a point at a distance $r_2$ from the point source.

But how can you calculate the sound pressure at a given distance from a point source if you are given the sound pressure at the point source? The corresponding distance would be zero...

homework and exercises - Variation of Maxwell action with respect to the vierbein - Einstein-Cartan Theory

I'm using the reference "Differential Geometry, Gauge Theories and Gravity" by M. Göckeler and T. Schücker and I am having trouble to vary correctly the lagrangian

$$ \mathcal{L}_M=\dfrac{1}{2g^2}F \wedge *F $$

with respect to the vierbein $e^a$ in order to find the Energy-momentum of the Maxwell action.

By doing $e\rightarrow e+f$ in the lagragian above I find $$ \mathcal{L[e+f]}_M-\mathcal{L[e]}_M=\dfrac{1}{g^2}f^c\left(\frac{1}{4} F^{ab}\epsilon_{abcd} e^d \wedge F+\frac{1}{2}F_{cb}e^b \wedge *|_eF\right). $$

However, the right answer, present in the foregoing reference is

$$ \mathcal{L[e+f]}_M-\mathcal{L[e]}_M=\dfrac{1}{g^2}f^c\left(\frac{1}{4} F^{ab}\epsilon_{abcd} e^d \wedge F-\frac{1}{2}F_{cb}e^b \wedge *|_eF\right). $$

(the only difference is the minus sign in the expression inside the brackets)

I worked a lot, but couldn't identify my mistake. So, does anyone know anything that is not trivial in the treatment with the forms in this case?

(In this case the signature used is lorentzian)

The whole calculation that I did was the following:

Since

$$ \mathcal{L}_M[e]=\dfrac{1}{2g^2}F \wedge *F=\frac{1}{2g^2}\left[ \frac{1}{2}F_{ab} e^a \wedge e^b \wedge \left(\frac{1}{4}F^{\alpha \beta}\epsilon_{\alpha \beta cd}e^c \wedge e^d\right) \right] =\frac{1}{16g^2}\left( F_{ab}F^{\alpha \beta}\epsilon_{\alpha \beta cd} e^a \wedge e^b \wedge e^c \wedge e^d \right) $$

Then, by doing $e \rightarrow e+f$, and neglecting terms quadratic in $f$, we are lead to

$$ \mathcal{L}_M[e+f]-\mathcal{L}_M[e] =\frac{1}{16g^2} F_{ab}F^{\alpha \beta}\epsilon_{\alpha \beta cd} \left(f^a \wedge e^b \wedge e^c \wedge e^d + e^a \wedge f^b \wedge e^c \wedge e^d + e^a \wedge e^b \wedge f^c \wedge e^d + e^a \wedge e^b \wedge e^c \wedge f^d \right) =\frac{1}{16g^2} F_{ab}F^{\alpha \beta}\epsilon_{\alpha \beta cd} \left(2f^a \wedge e^b \wedge e^c \wedge e^d + 2e^a \wedge e^b \wedge f^c \wedge e^d \right) =\frac{1}{8g^2} F_{ab}F^{\alpha \beta}\epsilon_{\alpha \beta cd} \left(f^a \wedge e^b \wedge e^c \wedge e^d + e^a \wedge e^b \wedge f^c \wedge e^d \right) =\frac{1}{2g^2} \left(\frac{1}{4}F_{ab}F^{\alpha \beta}\epsilon_{\alpha \beta cd}f^a \wedge e^b \wedge e^c \wedge e^d + \frac{1}{4}F_{ab}F^{\alpha \beta}\epsilon_{\alpha \beta cd}e^a \wedge e^b \wedge f^c \wedge e^d \right) =\frac{1}{2g^2} \left[f^a \wedge \left(\frac{1}{4}F_{ab}F^{\alpha \beta}\epsilon_{\alpha \beta cd}\right) e^b \wedge e^c \wedge e^d + f^c \wedge \left(\frac{1}{4}F_{ab}F^{\alpha \beta}\epsilon_{\alpha \beta cd}\right) e^a \wedge e^b \wedge e^d \right] =\frac{1}{2g^2} \left[f^a \wedge \left(\frac{1}{4}F_{ab}F^{\alpha \beta}\epsilon_{\alpha \beta cd}\right) e^b \wedge e^c \wedge e^d + f^a \wedge \left(\frac{1}{4}F_{cb}F^{\alpha \beta}\epsilon_{\alpha \beta ad}\right) e^c \wedge e^b \wedge e^d \right] $$

And then, we have $$ \mathcal{L}_M[e+f]-\mathcal{L}_M[e] =\frac{1}{2g^2} f^a \wedge \left[ \left(\frac{1}{4}F_{ab}F^{\alpha \beta}\epsilon_{\alpha \beta cd}\right) e^b \wedge e^c \wedge e^d + \left(\frac{1}{4}F_{cb}F^{\alpha \beta}\epsilon_{\alpha \beta ad}\right) e^c \wedge e^b \wedge e^d \right] =\frac{1}{2g^2} f^a \wedge \left[ F_{ab}e^b \wedge *|_e F + \left(\frac{1}{2}F^{\alpha \beta}\epsilon_{\alpha \beta ad}\right) F \wedge e^d \right] =\frac{1}{2g^2} f^a \wedge \left[ F_{ab}e^b \wedge *|_e F + \left(\frac{1}{2}F^{\alpha \beta}\epsilon_{\alpha \beta ad}\right) e^d \wedge F \right] $$

Relabeling the indices, we finally have

$$ \mathcal{L}_M[e+f]-\mathcal{L}_M[e] =\frac{1}{g^2} f^c \wedge \left[\frac{1}{2} F_{cb}e^b \wedge *|_e F + \left(\frac{1}{4}F^{ab}\epsilon_{abcd}\right) e^d \wedge F \right] =\frac{1}{g^2} f^c \wedge \left[\left(\frac{1}{4}F^{ab}\epsilon_{abcd}\right) e^d \wedge F +\frac{1}{2} F_{cb}e^b \wedge *|_e F \right] $$

This is not in accordance with the reference due to the plus sign (it should be a minus) and, again, I could not identify where I have done something wrong.

Answer

I understand the notation in the book this way \begin{eqnarray} e^a \wedge e^b \wedge \star |_e F &\equiv& e^a \wedge e^b \wedge \star F\;\\ &=& F \wedge \star (e^a \wedge e^b)\;\\ &=& \star (e^a \wedge e^b) \wedge F\;. \end{eqnarray} Now in the more accurate definition of Hodge dual (in the proof from definition there is a step of flipping the indices for more symmetric looking result, but we not flip them here, at first) we have \begin{equation}\boxed{ \star (f^a \wedge e^b) =\frac 1 2 \epsilon^{abcd} f_c \wedge e_d }\;. \end{equation} In the case of $\star(e \wedge e)$, there are no problems for flipping the indices \begin{equation} \frac 1 2 \epsilon^{abcd} e_c \wedge e_d = \frac 1 2 \epsilon^{ab}{}_{cd}\, e^c \wedge e^d \;. \end{equation} But in the case of $\star(f \wedge e)$ we have \begin{eqnarray} \star (f^a \wedge e^b) &=& \frac 1 2 \epsilon^{abcd} f_c \wedge e_d \\ &=& \frac 1 2 \epsilon^{ab}{}_{ef} \eta^{ce} \eta^{fd} f_c \wedge e_d \\ &=& \frac 1 2 \epsilon^{ab}{}_{ef} \eta^{ce} f_c \wedge e^f \\ &=& - \frac 1 2 \epsilon^{ab}{}_{ef} f^e \wedge e^f\;. \end{eqnarray} This because $e$ obeys the transformation laws \begin{eqnarray} e^a &\longmapsto& e^a+f^a \\ e_a &\longmapsto& e_a-f_a \\ && \eta^{ab}f_a=-f^b\;. \end{eqnarray} With \begin{equation} f^a \wedge e^b \wedge \star |_e F = \star(f^a \wedge e^b) \wedge F =- \frac 1 2 \epsilon^{ab}{}_{cd} f^c \wedge e^d \wedge F \end{equation} we can obtained the same result as shown in the reference.

differential geometry - Conformal transformation/ Weyl scaling are they two different things? Confused!

I see that the weyl transformation is $g_{ab} \to \Omega(x)g_{ab}$ under which Ricci scalar is not invariant. I am a bit puzzled when conformal transformation is defined as those coordinate transformations that effect the above metric transformation i.e $x \to x' \implies g_{\mu \nu}(x) \to g'_{\mu \nu}(x') = \Omega(x)g_{\mu \nu }(x)$. but any covariant action is clearly invariant under coordinate transformation? I see that what we mean by weyl transformation is just changing the metric by at a point by a scale factor $\Omega(x)$. So my question is why one needs to define these transformations via a coordinate transforms. Is it the case that these two transformations are different things. In flat space time I understand that conformal transformations contain lorentz transformations and lorentz invariant theory is not necessarily invariant under conformal transformations. But in a GR or in a covariant theory effecting weyl transformation via coordinate transformations is going to leave it invariant. Unless we restrict it to just rescaling the metric?

I am really confused pls help.

Can we see ALL of the observable universe?

Reading this wiki page about the observable universe, (and visible universe), implies that we humans, on this rock in the milkyway are seeing a static representation of the universe that is still effectively in the past. They're also saying that the observable universe is about 93 billion light years to observe it, because of Hubble's law, (I think), which implies that other objects far away in the universe are moving away from us faster than the speed of light (SoL). Now if we're moving slower than SoL then it would infer that we would never, from Earth be able to see the entire universe. In fact our sun is expected to consume us long before this would even be conceivably possible.

So, if we are seeing all of the observable universe when it was 13.75 billions years old and there is actually more "universe" to observe beyond this static reference, will we actually see more of it ? Or are we doomed to observe the universe at a specific place in time ?

Can we have physical quantities which have magnitude and direction but are not vectors?

I am not able to understand how to approach the question. Vectors are defined as quantities having magnitude and direction, then how is it possible?

Please explain.

quantum mechanics - A tutorial explanation of decoherence?

Is there a tutorial explanation as to how decoherence transforms a wavefunction (with a superposition of possible observable values) into a set of well-defined specific "classical" observable values without the concept of the wavefunction undergoing "collapse"?

I mean an explanation which is less technical than that decoherence is

the decay or rapid vanishing of the off-diagonal elements of the partial trace of the joint system's density matrix, i.e. the trace, with respect to any environmental basis, of the density matrix of the combined system and its environment [...] (Wikipedia).

Answer

The Copenhagen interpretation consists of two parts, unitary evolution (in which no information is lost) and measurement (in which information is lost). Decoherence gives an explanation of why information appears to be lost when it in actuality is not. "The decay of the off-diagonal elements of the wave function" is the process of turning a superposition: $$\sqrt{{1}/{3}} |{\uparrow}\rangle + \sqrt{{2}/{3}} |{\downarrow}\rangle$$ into a probabilistic mixture of the state $|\uparrow\rangle$ with probability $1/3$, and the state $|\downarrow\rangle$ with probability $2/3$. You get the probabilistic mixture when the off-diagonal elements go to 0; if they just go partway towards 0, you get a mixed quantum state which is best represented as a density matrix.

This description of decoherence is basis dependent. That is, you need to write the density matrix in some basis, and then decoherence is the process of reducing the off-diagonal elements in that basis. How do you decide which basis? What you have to do is look at the interaction of the system with its environment. Quite often (not always), this interaction has a preferred basis, and the effect of the interaction on the system can be represented by multiplying the off-diagonal elements in the preferred basis by some constant. The information contained in the off-diagonal elements does not actually go away (as it would in the Copenhagen interpretation) but gets taken into the environment, where it is experimentally difficult or impossible to recover.

nuclear physics - Predict decay chain of a radioactive element

I know there are tables of decay chain of radioactive elements. Is there a way to predict the whole chain from the first radioactive element?

How does quantum mechanics and quantum field theory explains discrete energy levels of particles?

Please give me a brief explanation as to how qm and qft describe and explain the energy level that exist in an atom.

I understand that in QM energy state is quantized but does not offer an explanation as to why, does qft solves this phenomenon?

lateral thinking - Nobody can solve this puzzle

Yes, literally, nobody can solve this puzzle.

While every answer might get upvotes, nobody's answer will be accepted.

Nobody gets the green check.

In this sense, the puzzle is fair to everyone.

Let's see if the puzzle gets solved!

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...Although it doesn't matter that much in the end.

Just for fun! (:

Answer

Yes, literally, nobody can solve this puzzle.

My name is Nobody

While every answer might get upvotes, nobody's answer will be accepted.

This is my answer

Nobody gets the green check.

I should get the green check when my answer is accepted

In this sense, the puzzle is fair to everyone.

Anybody can change their display name to "Nobody"

enigmatic puzzle - A pirate's treasure-hunt

My dear puzzle friends, I need your help!

An old acquaintance of mine has recently passed away and mentioned me in his will. I was quite thrilled, as the man was rather rich due to some clever investments during his lifetime. Unfortunately, he was as, hmm, unconventional as he was rich. He always imagined himself as a pirate and spent a good deal of his time roleplaying his phantasies. Oh, how I hated all his Arrr-jokes and endless adventure stories told at countless evenings at the hotel bar... Anyway, I'm afraid he also chose to have the last laugh on me, and I can only inherit his fortune when I play along with his last pirate story.

When I met the solicitor, he handed me over the three things pictured below:

• A parchment with a map and some notes to me


• A small square cardboard coaster which has been scribbled on


• A leather-bound book, supposedly a "Pirate Captain's logbook" of sorts

That's all I have to go on, my friends! Please help me. The solicitor told me, I can take my time, but I must not guess. A single attempt for a correct solution is all I have!

---

(There are 5 pages in the journal: Page 1, Page 2, Page 3, Page 4, Page 5)

---

This is a long, laborious, multi-layered treasure hunt and cooperation is strongly encouraged. Solving individual 'layers' of the puzzle hopefully provide enough internal feedback to encourage you in your thinking.

The story above is just flavour. The three images above are the sole starting point. Potential spelling or grammatical errors are non-intentional.

I have tried to cross-check this puzzle multiple times over the last few weeks of building it, but mistakes and errors are possible. If you think you have spotted a potential flaw, please comment. Below is a list of known errors or corrected things.

Errata and correction log

• 31th May 2016: Replaced log book page 3. (Incorrect use of 'nordwards').
  Does not change puzzle. Removed inconsistency. Found by Leppy64.


• 31th May 2016: Replaced log book page 4. (Last turn after 3 not after 4 miles.).
  This changes Journey 7, last leg.
  
  Found by Leppy64. \


• 31th May 2016: Added log book page 5 which was omitted on original posting.
  


• 4th June 2016: One symbol on one of the crosses is rotated by 180 degree. The original image can not be changed for specific reason. The replacement image is found here. The community wiki posts to the correct image.
  This fixes an inconsistency. The puzzle would be solvable without this fix.
  


• 7th June 2016: Fixed hint in journal page 3.
  

Answer

For those who want to edit my base map without the paths:

Coaster

Corners:

The turns in the journey occur on a letter in the grid. When combined together, each is a link to the "stacked" imgur. There are nine more images. 6 pieces of the map, the silver ring, and the silver die. And one other image.

Journey 1 - Top Left of Page

Corners = M7xBm

Journey 2 - Top Right of Page

Corners = 2jK8M

Journey 3 - Bottom Right of Page

Corners = StWHV

Journey 4 - Middle Left of Page

Corners = lWlvC

Journey 5 - Top Left of the Page

Corners = YqUSJ
The ring

Journey 6 - Bottom Right of the page

Corners = HLy4j

Journey 7 - Top Left of the page

Corners = Fmvr0

Journey 8 - Top right of the page

Corners = NccF6

Journey 9 - Top Right of page

Corners = xU9Ke The die

Overlay of all Journies:

Random Notes to remember:

Ring and the Graveyard - This is the next step 4x4 Grid things (Page 1)
F.N. - West / S.N. - North (Page 3)

The Map Pieces - Thanks to Wesley Situ for this merged map pieces

The Crosses - Thanks to Wesley Situ for this pic of the crosses

Page 1 - Left

Uwe Halfaname

Page 1 - Right Top
Navigator

Page 1 - Right Mid
Sick Harming - Doc has a peg-leg
Wil the Joker

Page 2 - Left Top
Wil N'thing
Sick Harming
Navigator (Thickhead or Darmwind)

Cook (Thickhead or Darmwind)

Page 2 - Left Bot
Gorn - Brings a cat aboard

Page 2 - Right Bot
Ron (Darmwind or Fartling)
Eugen (Darmwind or Fartling)

Page 3 - Right Top
Gorn
Grinning Cat (G.C)

Page 4 - Left Bot

Thickhead (Navigator or Cook)
Knut Tattenshoe
Maurice Nutter
Darmwind (Navigator or Cook)
Wil Ironfist (as we find on Page 4 Right)

Page 4 - Right Top
Marice Nutter

Page 4 - Right Mid
Wil Ironfist
Wil N'thing

Page 5 -Right Top
Gorn

Conclusions:
The names on the stones have all been tracked except for Alexander Leffie, Svem Orm, and the blank one.

Two names referenced in the story that are not on stones are (Ron or Eugen) Fartling, and Gorn. Fartling is likely the blank stone due the farting story and the reference to his "wind".

Gorn is either Alexander Leffie or Svem Orm. I lean towards Gorn being Svem Orm but I can't prove it.

That leaves mister Alexander Leffie. Leffie spelt backwards is Eiffel. There is an Eiffel Tower in the mountains of the map.

Alexandre Leffie - "You will not be forgotten your work is central"
- Not mentioned in the journal at all
- Leffie spelt backwards is Eiffel.
- Alexandre Gustave Eiffel is the man the Eiffel Tower is named after
- There is an Eiffel tower on the map

Maurice Nutter - "right down right down right down back you are" - Possibly a reference to Maurice Koechlin - Engineer on the Eiffel Tower

Sick Harming - "A peg for a leg Tock-Tock"
- This man is the doctor on the ship
- Tock-Tock seems to imply a clock, but clocks say "Tick-Tock"
- Morse code on the side of tombstone says "City of Love" associated with Paris

Knut Tattenshoe - "born march 1889, ascension 1710"
- The Eiffel Tower was opened in March 1889
- The Eiffel Tower has 1710 flights of stairs

The right side of the ring says "Quo~Vadis". Translated from Latin this says "Where are you going". The squiggle matches up with the squiggle on the dies faces. When you orient the faces correctly they follow the paths on page 1 of the journal. The straight lines on the die faces seem to point to the trees on the North and West sides.

Using the clues, FN - North and SN - West. FN is the first intial and SN is the second initial of the names on the tombstones.

There are a lot of i's to do and t's to cross to sum up this section. Now to fill them in, across the top we get:
A M R W E U G S K
Alexandre
Maurice

Ron (This could have been Eugen, but Ron works here, Fartlings tombstone sums this up I'm sure)
Wil (Both Wils are in the same column)
Eugen (See Ron)
Uwe
Gorn and Grinning
Sick and Svem
Knut

Down the side we get:
H D N L O F I T C
Harming

Darmwind (This could have been Nav's last name, but Nav shares a row with Tattenshoe)
Nutter and N'thing
Leffie
Orm (Halfaname's surname starts with O)
Fartling
Ironfist (N'thing shares with Nutter)
Thickhead and Tattenshoe
Cat

Following the paths and lines from layers 1 through 6 as laid out by Wesley Situ we get the message:
THE_FAIR_METAL_T

OWER_LAT_MINUS_L
ONG______THEN___
_______TAKE_____
_GRAD_MIN_SEC_AS
__THREE__ANGLES_

"The Fair Metal Tower" that has been alluded to is the Eiffel Tower. It lies at 48°51′29.6″N 2°17′40.2″E accoridng to wikipedia.

Rouding those a little and subtracting yields: 46 43 50

Degrees is Grads in German, apparently, so no conversion needed.

Looking at the map (wow it's been a long time away from it) and finding the three angles. We look around and don't see a 50, but we do see K. Earlier it was found by WesleySitu that K = 50!

The treasure is hidden here:

Great puzzle! We left a few rocks unturned (and man one of them was actually really beautiful) but we made it!