word - Piece de Resistance - Queen Machina, my Majesty, Time for a Hearty Supper!

Queen Machina, my Majesty, Time for a Hearty Supper!

_{This puzzle is part of the "Piece de Resistance" series. Go back to Part 1 (Ace) for the story.
Ace Two Three Four Five Six Seven Eight Nine Ten Queen King... Jack}

No more images... (for the time being), back to words...

$\mathfrak{\text{- rosin control}}$
$\mathfrak{\text{- escape talk}}$
$\mathfrak{\text{- current bond}}$
$\mathfrak{\text{- smallest direction}}$
$\mathfrak{\text{- unpleasant garbage}}$

Rest Assured

1) No obscure words

2) No P.SE usernames involved

3) The answer is a word. The title will make some sense when the answer is found.

Hints

Level 1

"Unpleasant garbage" can be replaced by "unpleasant mess".

Level 2

As some of you have suspected in the comments, this puzzle has nothing to do with rhyme.

Level 3:

@PiIsNot3 has already correctly identified the logic. 2 more to go!

Level 4:

"Unpleasant garbage" can be replaced by "strict code".

Answer

Here's an attempt to tie this up... As already established, the clues here refer to:

Pairs of words whose spellings differ by the removal of a letter.

rosin control

RE[S]IN - REIN

escape talk

CH[E]AT - CHAT

current bond (formerly 'spray heart')

TI[D]E - TIE

smallest direction

WE[E]ST - WEST (found by @PilsNot3 - the two of us both discovered the same solutions to the first two clues independently but I had favoured [L]EAST - EAST here until attempting to spell the linking word out...)

unpleasant garbage (or 'strict code' in hints)

HA[R]SH - HASH

Taking the 'unshared' letters from each pair suggests a solution of:

SEDER - a Jewish ritual service and ceremonial dinner, which would link with 'supper' in the title. Not sure on the 'Queen Machina' reference - isn't she a Power Rangers baddie?!

quantum field theory - Ward identity for 'general' operator and current diagrams

This is actually about two related doubts and I hope is appropriate for a single question (if not, I will happily divide it). So, my problems are related to the analysis and calculation of chiral anomalies in QFT (and are independent of a specific scenario).

(i) Reading a paper of Gross and Jackiw on the topic, it is said that was the chiral current $J_{5\mu}$ to be conserved also in the quantum case, then the ward identity wouldn't be spoiled (which I understand), but then proceed to write this statement explicitly as

$$ p^\mu\int d^4x \;d^4y \; e^{ipx}e^{iqy} \langle0|T{J_{5\mu}(x)J_{5\nu}(y)J_{5\alpha}(0)}|0\rangle=0. $$

What I don't understand is why the currents' Green function (and not the fields') also obey the same kind of ward identity and have never seen it be derived. Why is that? Is it actually a consequence of the 'rightful' Green function Ward identity?

(ii) He then goes on to show that this is actually false, implying that the chiral current is not conserved. He does it in the normal way, that is, setting the above fourier transform of a correlator equal to the triangle diagram (and permutation). This is a basic thing I have never understood: how can we write a diagram for a current expected value? Is just brute force, that is, writing e.g. $J_{5\mu}(x)=e\bar{\psi}(x)\gamma_\mu \gamma_5 \psi(x)$ and evaluating the expected value by expanding $J_{5\mu}(x)J_{5\nu}(y)J_{5\alpha}(0)e^{-iS_{\text{int}}}$ and applying Wick's theorem? By the way that (literally) every author I've read so far made it look as if it was obvious I feel like there some intuition I'm missing.

I suspect a lot of this is related to the Schwinger-Dyson equation and probably points to my poor understanding of the Ward identities, but I cannot pin it down. Any related insights are also greatly appreciated.

riddle - Could you *be* any more metaphorical?

The following, apparently random list of seemingly random nonsense is a metaphorical interpretation of something that exists in the real world. What is it? As always, go line by line. But once you figure it out, you'll have everything you need to write the solution.

The current endeavor is not a genre, it is mass murder

Your were torn from the respirator

You were brought up to date

Like half of a starlet or a whole amputee

You wore vestments

And played Dracula

You scurry down the lane

Defenseless, unhidden

Concealed canine hunters for profit

The statues pursue you with murderous intent

I shall pacify an acquaintance

Hasten hither, my child

Veni, vidi, cessavitque

Take leave of the flora

Lest you fall ill 'mongst the murk

They're mutts, bejeweled

John Samhain is, at present, a snow leopard

Who dwells above a vault

Lift in disrepair, hemp brings him down

So he prowls the macadam with Cheetah

A tartlet nears with The Specials

She wears no grin, but Salvador's accessory

Nostalgic, chocolate chip cookies

Destroyed and quadriplegic, yet the boxers are unwavering

Annus belbus in the dog days of summer

Strut next to the beach; hasten to the trench

"In the Year 2525", a hipster goes solo

Pain is inevitable, try to live through the night

Minor hint:

Rather Orwellian, isn't it?

Enormous, gigantic, probably ruins everything hint:

The riddle's solution is best friend to both sexes.

Answer

This is an interpretation of

The David Bowie song, Diamond Dogs (btw the probably ruins everything hint was the key) https://en.wikipedia.org/wiki/Diamond_Dogs_(song).

The current endeavor is not a genre, it is mass murder

This ain't rock'n'roll. This is genocide!

Your were torn from the respirator

You were brought up to date

As they pulled you out of the oxygen tent
You asked for the latest party

Like half of a starlet or a whole amputee

You wore vestments

And played Dracula

With your silicone hump and your ten inch stump, Dressed like a priest you was, Todd Browning's freak you was

(Tod Browning directed the 1931 Dracula film with Bela Lugosi
https://en.wikipedia.org/wiki/Dracula_(1931_English-language_film))

You scurry down the lane

Defenseless, unhidden

Concealed canine hunters for profit

Crawling down the alley on your hands and knee, I'm sure you're not protected, for it's plain to see, The diamond dogs are poachers and they hide behind trees

The statues pursue you with murderous intent

Hunt you to the ground they will, mannequins with kill appeal

I shall pacify an acquaintance

Hasten hither, my child

I'll keep a friend serene

Oh baby, come unto me

Veni, vidi, cessavitque

Take leave of the flora

Lest you fall ill 'mongst the murk

They're mutts, bejeweled

Well, she's come, been and gone.
Come out of the garden, baby
You'll catch your death in the fog

Young girl, they call them the Diamond Dogs

John Samhain is, at present, a snow leopard

Who dwells above a vault

Lift in disrepair, hemp brings him down

So he prowls the macadam with Cheetah

The Halloween Jack is a real cool cat
And he lives on top of Manhattan Chase
The elevator's broke, so he slides down a rope

Onto the street below, oh Tarzie, go man go

(Samhain is another term for Halloween, Jack is often what John's are called, Tarzan was friends with a Cheetah )

A tartlet nears with The Specials

She wears no grin, but Salvador's accessory

Nostalgic, chocolate chip cookies

Destroyed and quadriplegic, yet the boxers are unwavering

Meet his little hussy with his ghost town approach
Her face is sans feature, but she wears a Dali brooch
Sweetly reminiscent, something mother used to bake

Wrecked up and paralyzed, Diamond Dogs are stabilized

The Specials had a hit with "Ghost Town". Salvador Dali.

Annus belbus in the dog days of summer

Strut next to the beach; hasten to the trench

"In the Year 2525", a hipster goes solo

Pain is inevitable, try to live through the night

In the year of the scavenger,the season of the bitch
Sashay on the boardwalk, scurry to the ditch
Just another future song, lonely little kitsch

(There's gonna be sorrow) try and wake up tomorrow

(Belbus is Latin for Hyena (http://latin-dictionary.net/definition/6268/belbus-belbi). In the year 2525 is a song about the future.)

Minor Hint:

The song has a post apocalypic setting, reminiscent to some of the novel: 1984. According to Wikipedia this was the intention of Bowie on the Album of the same name: "Thematically, it was a marriage of the novel Nineteen Eighty-Four by George Orwell and Bowie's own glam-tinged vision of a post-apocalyptic world." https://en.wikipedia.org/wiki/Diamond_Dogs

Major Hint:

A __ is a man's/girl's best friend.

The genuine article:

http://www.youtube.com/watch?v=36lWAcY9IXE

mathematical physics - Other processes than formal power series expansions in quantum field theory calculations

I am not sure if this question is too naive for this site, but here it goes. In QFT calculations, it seems that everything is rooted in formal power series expansions, i.e. , what dynamical systems people would call Lindstedt series. However, from what I heard, this series (for QFT case) is known to have zero radius of convergence, and it causes tons of difficulties in theory. My question is, are there approaches that start with an iterative process that has a better chance of converging (e.g., a fixed point iteration), and build computational methods for QFT from there?

In other words, when there are so many approaches to approximate the exact solution of, say nonlinear wave (and Klein-Gordon, Yang-Mills-Higgs-Dirac etc) equations on the classical level, why do we choose, when we quantize, only a couple of approaches, such as power series, and lattice regularization (the latter essentially a finite difference method)? Note that it is milder than making QFT completely rigorous, it is just about computing things a bit differently.

quantum field theory - Supersymmetric Localization (Mirror Symmetry)

I'm reading Chapter 9 of Mirror Symmetry book. As you can see in eq. (9.30) his model for SUSY is

$$\begin{align} \delta_\epsilon X &=\epsilon^1\psi_1 + \epsilon^2\psi_2\\ \delta\psi_1 &= \epsilon^2\partial h\\ \delta\psi_2 &=-\epsilon^1\partial h. \end{align}\tag{9.30}$$

I understand that if $\partial h\neq 0$ it is possible to see that partition function vanishes, in particular, there is a total divergence:

$$ \int \mathrm{d}\hat{X}e^{\frac{1}{2}(\partial h(\hat{X}))^2}\frac{\partial^2 h(\hat{X})}{(\partial h(\hat{X}))^2}. \tag{*} $$

The implication that a total divergence yields vanishing integral, it is a bit strange, because, since every function that admit a primitive is a total derivative. What is the meaning of that implication? Then why $(*)$ vanishes? It seems to me that, in general, $$ \int \mathrm{d}(h')\frac{e^{\frac{1}{2}(h')^2}}{(h')^2}\neq 0. $$

Then, again, when he claims that if there is a $x_0$ such that $h'(x_0)=0$ we can use again the argument of the total divergence, why doesn't he keep in account that, removing a small neighborhood of $x_0$, give us some contribution at the border of that neighborhood (two points)?

gravity - How are the 4 km arms of LIGO measured so accurately?

The arms of the LIGO interferometer are 4 km long. Now, LIGO functions by measuring the phase difference between two beams of light coming (as in a Michelson interferometer) to a sensitivity of $10^{-18}\: \mathrm m$. Now, we know that if the path length $2d_2-2d_1$ ($d_1,d_2$ being the length of the paths from the mirrors to the partially silvered diverging mirror) is anything other than $0$ then there is a phase difference between the two waves. So, how are the mirrors placed at exactly $4$ km apart and not even an error of $0.000\ldots 1\:\mathrm{mm} $ might have crept in which could have caused a phase difference, and false results creep in?

How are the measurements so accurate?

Answer

It is a misconception that LIGO is a very accurate instrument, it has an uncertainty in calibration which is on the order of 10%. This means that the measured strain amplitude of GW150914 of $1.0 \cdot 10^{-21}$ could easily have been $1.1 \cdot 10^{-21}$. Note that this is just a scaling error.

LIGO is however extremely sensitive, it can measure relative length variations on the order of $10^{-22}$, but only in a bandwidth between 10 and 2000 Hz. At lower frequencies, the measurement fluctuates by several orders of magnitude more. You need to do band-pass filtering to reveal the actual signal.

As already mentioned in Chris's answer, a Michelson interferometer can only measure incremental changes in the path length difference. It does not say anything about the absolute length of the arms, and not even about the absolute difference in arm lengths. For a perfect Michelson interferometer, the resulting power on the photodiode is$$P = \frac{P_0}{2} \left(1 + \sin\left(4 \pi \frac{L_1 - L_2}{\lambda}\right)\right)\,,$$which will only tell you how much the difference in the arm lengths changes over time.

Still, there are reasons why you want to have the long arms as equal as possible. For a simple bench-top interferometer with a crappy laser diode, the path lengths need to be reasonably similar, otherwise you run into problems with the coherence length. This is not an issue for LIGO, they use Nd:YAG lasers which already have a coherence length measured in kilometers when running alone. These lasers are further pre-stabilized by locking them on ~16 meter suspended cavities, and finally the laser frequency is locked on the average length of the two 4 km long arms. The resulting line-width of the laser is on the order of 10 mHz, so a coherence length larger than $10^{10}$ meters ...

You still want to make the length of the 4 km arms pretty equal, since any imbalance would couple residual noise of the laser frequency to the differential length measurement. With standard GPS-based surveying methods, the mirrors are positioned with an accuracy on the order of millimeters. There is no need to do this much more accurate, since there are other sources of asymmetry that can couple frequency noise to the differential measurement, such as the differences in absorption and reflection of the mirrors used in the two arms.

enigmatic puzzle - An unexpected storyteller

Quietly, we entered right through yesterday's unit.
In our place, all seemed... distant... forgotten... gone.
How Jack knew Lucy zoned, Xavier could verify, but not match.

What's "unusual" about the story?

Answer

The first letter of each word corresponds to the qwerty keyboard layout, there are 26 words in the story.