The h Bar - 2018-01-09 (page 2 of 4)

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1:03 PM

Does anyone know the right definition of a exactly marginal operator in a QFT? Is an operator whose coupling is exactly marginal(namely, beta function=0)??

I do not

hey @JohnRennie, quick note about this one

Hi Sharon. I suspect you would get a better response if you asked this on the Engineering Stack Exchange. — John Rennie 20 hours ago

we occasionally get people that are (disproportionately) upset about users cross-posting, and comments that can be seen as encouraging it

since then I try to supplement them with stuff along the lines of

... and if you think that's a better home for your question, please flag this for a moderator to migrate it instead of cross-posting. — Emilio Pisanty 1 min ago

there's no harm in adding them and it defuses a potential problem

1:36 PM

The notation looks dazzling. — Ng Chung Tak 2 mins ago

good comment

Anonymous

@BalarkaSen Do you know the proof of why crystals can't have $D_5$ symmetry or $D_n$ (where $n>6$) ? I tried looking for it on the net, but they only state the fact and don't prove it.

Anonymous

Interestingly $D_6$ is possible

@Blue You have to tell me what a crystal is

@Blue Isn't it just related to the possible tesseletions by regular polyedrons

That's what I was thinking

You can't tessellate the plane with pentagons...

(You can do it with hexagons)

Anonymous

1:40 PM

Wait a bit...what is tesselation?

Well slightly more complicated because 3 dimensionals

Tesselation is a tiling

Like how you can make a grid with squares

Or triangles

Or hexagons

But not with any other regular polygon

@Slereah I don't think it's a three dimensional fact.

Anonymous

3

A: Why are there no crystals with 5-fold symmetry?

Your question is much more complicated from mathematical point of view than it seems to be. First, I'll start with a nice photo: (source: Wikipedia). What you see is really a photo and it is almost a mono-crystal. The only problem is that we all know that it cannot be a monocrystal since it canno...

@BalarkaSen regular tesselations are boring

Anonymous

I think they're talking of crystal in this sense ^

1:42 PM

A quasiperiodic crystal, or quasicrystal, is a structure that is ordered but not periodic. A quasicrystalline pattern can continuously fill all available space, but it lacks translational symmetry. While crystals, according to the classical crystallographic restriction theorem, can possess only two, three, four, and six-fold rotational symmetries, the Bragg diffraction pattern of quasicrystals shows sharp peaks with other symmetry orders, for instance five-fold. Aperiodic tilings were discovered by mathematicians in the early 1960s, and, some twenty years later, they were found to apply to the...

@Blue The accepted answer there seems to give my logic

@Emilio Oh yeah those are fucked

@BalarkaSen a.k.a. "interesting"

anyways, here's an answer to that age-old question: what happens if you run out of badges you can track on your profile?

stackoverflow.com/users/501557/templatetypedef?tab=topactivity

stackoverflow.com/users/771848/alecxe?tab=topactivity

answer: nothing much

also answer: currently only a thing for two users on Stack Overflow

Anonymous

@BalarkaSen Yeah, but I'm looking for the proof of that thing: "Note that it is impossible to produce a regular arrangement of unit cells to produce a pentad (order 5 symmetry)". Also for n>6

Anonymous

You could just point out a source which I can read up

@Blue that's normally known as the Crystallographic Theorem

Crystallographic restriction theorem

The crystallographic restriction theorem in its basic form was based on the observation that the rotational symmetries of a crystal are usually limited to 2-fold, 3-fold, 4-fold, and 6-fold. However, quasicrystals can occur with other diffraction pattern symmetries, such as 5-fold; these were not discovered until 1982 by Dan Shechtman. Crystals are modeled as discrete lattices, generated by a list of independent finite translations (Coxeter 1989). Because discreteness requires that the spacings between lattice points have a lower bound, the group of rotational symmetries of the lattice at any point...

that and/or references therein should have a proof

probably a range of different proofs at different levels of rigour

Anonymous

1:49 PM

@EmilioPisanty Thanks!

no worries

hallo

The proof is very easy. If you tile your plane by regular $n$-gons, the interior angle of that $n$-gon has to be a multiple of $2\pi$.

btw @ACuriousMind you're third place on the trackable badges scoreboard

surprisingly enough, @JohnRennie is seventh place

'Cuz at a vertex of such a tiling a bunch of $n$-gons are sitting side-by-side

But the interior angle of a regular $n$-gon is, what, $2\pi(1 - 2/n)$?

I mean $\pi(1 - 2/n)$

So you want $2n/(n-2)$ to be an integer, I think

$n = 3, 4, 6$ are the only options. RIP

@Blue

1:56 PM

what about 18?

yeah that's the proof in 2D

wait I',m dumb

bye

I quit

There should be group theoretic proofs too

Hissss

Mister Euclid was going geometrical proof when group theory wasn't even the dream of a madman

Leave euclidian geometry alone

stop being a euclidean shneck

the group theoretic proof might generalize is the point

good luck classifying regular hyperpolyhedron suckers

there's like 28 of them bikhes in dimension 4

2:02 PM

> All isometries of finite order in six- and seven-dimensional space are of order 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 15, 18, 20, 24 or 30 .

that kind of statement?

Yeah

Exactly

hmmmm

kinda light on the references, that section

oeis.org/A180042

maybe?

[citation needed]

@BalarkaSen yer welcome to step in and add them references

doi.org/10.1107/S0108767385001180 this will probably do

> The crystallographic restriction in higher dimensions
> H. Hiller
> The crystallographic restriction in dimensions two and three is generalized to arbitrary dimensions. It is shown that m can occur as the order of an element of the point group of an n-dimensional space group if and only if [Phi](m) [less-than or equal to] n where [Phi] is an additive version of Euler's totient function. A table of these allowable orders in dimensions [less-than or equal to] 23 is provided.

Good paper!

2:06 PM

What about in 11 dimensions

CC @Blue

What are the string theory crystalline structures

@Slereah presumably obtainable from others in that table?

3 is missing

So I guess the isometry group of tessellation of $\Bbb R^2$ by regular $n$-gons is generated by translations + $D_{2n}$

I need to see what the relators are

is there a Thing in string theory

About closed strings of different homotopy classes

2:08 PM

yes, if a dimension's missing you just take the next lowest on the table

@Slereah 11, 22, 35, 45, 48, 56, 70, 72, 84, 90, 120

Like is there a specific thing for closed strings that are contractible versus the ones around the compactified dimensions

Do they have Important Behaviours

Prolly $aba^{-1} = b^r$

I'm guessing contractible closed loops are probably like proper gravitons

but what of the non-contractible ones

Think you just need to account for winding numbers

It's all linked to t-duality superstringtheory.com/basics/basic6a.html but still need to write it up

I still don't know about t-duality uwu

Haven't been far into interacting strings yet

I probably should reread some perturbative QFT first

2:18 PM

Holy fuck I just looked up a deadline that was like in two weeks but that was for international students

I'm a bit rusty on s and t channels

Had a fucking heart attack good lord

Need some heroin now

@BalarkaSen why do you need measure theory

shoot it boyo

dynamics

It’s not something a Jedi would tell you

@BalarkaSen I learned from the old Halmos book

it's pre-Bourbaki for what it's worth

fairly light, most of the stuff is in 150 pages

2:21 PM

Did you ever hear of the story of Lebesgue the wise

Oh sorry

the tragedy of Lebesgue the wise

(the tragedy is measure theory)

there's some stuff on LCA groups, probability, and some really abstract measure stuff that you probably won't need

I am reading from Royden

looks super dry

:S

measure theory is super dry

it's the most technical math there is

[citation needed]

Bourbaki really really messed up measure theory with their functionals and ignoring probability

2:24 PM

@BalarkaSen Federer :)

What kind of person defines an integral before a measure

@bolbteppa Riemann

4

Q: What exactly is a pomeron?

The term 'pomeron' was apparently important in the early stages of QCD. I can't find any reference to it in modern QFT books, but older resources sometimes refer to it offhand, and I've yet to find any explanation of what it actually is. Old theoretical sources throw up a wall of math and seem ...

quantum-field-theory particle-physics quantum-chromodynamics

I want to post a pomeranian photo

Will this land me in troubles

@DavidZ

@Slereah why stop there?

you should post a photo of a pomeranian eating a pomegranate

pomegranate doesn't sound very much like pomeron, though

2:39 PM

@Slereah i'd accept that answer if the pic was cute enough!

@0celo7 Did you see my ping to you

About flattening?

use a mallet

Yep

@BalarkaSen so it works with my modified rippling lemma?

2:50 PM

I am positive that it does

Do you want me to explain the picture?

did you ask ted about the rippling lemma

we must be missing something in the wording

there's no way the authors missed that their picture doesn't do what they say it does

I don't think this is Ted's cup of tea

ok

give me the picture

let me find the paper

arxiv.org/pdf/math/0003026.pdf

So remember that $M^n$ is embedded compressibly in $Q^q \times \Bbb R^k$ where $q \geq n+1$ and we want to isotope $M$ by a small amount so that the tangent bundle becomes almost-horizontal

wow

I go on PSE

Suddenly, 45 points

Rare treat!

2:55 PM

@BalarkaSen they say that Kuiper (small world, quoting him in my thesis) used rippling in the 50s

To do a different thing, though

yes but maybe the lemma is the same?

It makes sense to use rippling to $C^1$ isometric embedding

mm maybe

user image

^ that's kind of why you use rippling

(it's a C^1 isometric embedding of the flat torus in R^3)

the second looks like a you know what

pls, no 12 year old jokes

2:57 PM

@BalarkaSen huh

@BalarkaSen I am just substituting for Dr. Rennie

poor james damore getting support from the alt right

testicles joke the best jokes

Apparently the big contribution of Nambu is from a conference (unpublished)

>:|

enemy of my enemy

I need a time machine to attend this conference now

@Slereah it was an impromptu talk in the men's room

3:00 PM

[40] Private communication overheard in the mens' room at a recent conference, but we didn't see who it was because we didn't want to get up.

2

im looking up the kuiper papers

just in case

Kuiper had some fun ideas. He had a cool sketch of the sphere eversion before anyone else had it

Namely, take the tautological line bundle on $\Bbb{RP}^2$ (a twisted normal bundle)

@BalarkaSen he showed that any simply connected LCF manifold can be conformally immersed in the sphere

It's unit $S^0$-bundle is $S^2$

it's the start for a potential theoretic proof of the positive mass theorem for LCF folds

Now contract it fiberwise until it gets flipped and passes through itself to the unit $S^0$-bundle again

That's sphere eversion

(where the RP^2 is immersed as Boy's surface in R^3)

@0celo7 What's LCF

3:04 PM

@BalarkaSen locally conformally flat

isee.jpeg

@BalarkaSen it means that the Weyl tensor vanishes identically

iseewhatyoumean.png

(which in turn implies you can take an atlas in which all transition functions are conformal maps)

ah

so it's like the Nijenhuis tensor

3:05 PM

yes

but the formula is worse

rip

$$W_{ijkl}=R_{ijkl}+\frac{R}{(n-1)(n-2)}(g_{jl}g_{ik}-g_{jk}g_{il})-\frac{1}{n-2‌} (R_{ik}g_{jl}-R_{il}g_{jk}+R_{jl}g_{ik}-R_{jk}g_{il})$$

@BalarkaSen it's also the part that contains info about gravitational waves

Pretty cool

Is it, tho

@BalarkaSen ok so flattening?

3:09 PM

Can't you have gravitational waves in a medium even if it's conformally flat?

@Slereah like any GR meme, it's only partially correct

I dunno

@0celo7 Alright, so look at the projection map $M^n \to Q^q$. That's an immersion

@BalarkaSen making eggs, so send me an essay and I'll read

Alright, alright, jeebus

3:13 PM

what is

the egg manifold

is it $B^3$

So remember that $M^n$ is embedded compressibly in $Q^q \times \Bbb R^k$ where $q \geq n+1$ and we want to isotope $M$ by a small amount so that the tangent bundle becomes almost-horizontal (that means for any neighborhood $U$ of the subspace $H$ of $Grass_n(Q \times \Bbb R^k)$ consisting horizontal $n$-plane distributions you can produce such an isometry s.t. $TM$ is restriction of an element of $U$). Look at the projection $f : M \to Q$, which is an immersion. Ripple this immersion by the immersiotopy $f_t : M \to Q$. Define the isotopy of $M \subset Q \times \Bbb R^k$ to be $i_t(x) = (f_…

Basically, we ripple the horizontal coordinate of $M \subset Q \times \Bbb R^k$, keeping the vertical coordinate the same

So if $x, y$ are two (sufficiently far apart) points in the horizontal projection $f(M)$, the distance $d_{f_1(M)}(x, y) \gg d_{f(M)}(x, y)$

That means $d_{i_1(M)}(f^{-1}(x), f^{-1}(y))$ (bad notation; let's just assume it's injective) is also very large compared to $d_M(f^{-1}(x), f^{-1}(y))$

@Slereah I couldn't find the 2 early Nambu articles

Looked a few times

yeah apparently it's a conference

So only the 5 people still alive who attended know what that means

But the vertical components are the same; that forces the slope to die out

i.e. you get closer and closer to a horizontal tangent bundle

It's totally the same spiraling staircase picture from Eliashberg-Mishachev before

@Slereah Wow Schwarz arxiv.org/pdf/hep-th/0007118.pdf says they are in amazon.com/Broken-Symmetry-Selected-Scientific-Century/dp/… which even the amazon comment hints at!

3:25 PM

"goto" is a good name btw

That guy made up his own mechanics

Nambu mechanics

Because then I can make programming jokes about his name

like homepages.cwi.nl/~storm/teaching/reader/Dijkstra68.pdf

GOTO ACTION CONSIDERED HARMFUL

@BalarkaSen I never saw that picture

Is the classical Polyakov string action solvable, anyway

Like is there an exact solution of the EoM

Scherk sets up and solves the wave equations

I.36 onwards

3:43 PM

Good old Shrek

My problem is his second non-covariant (lightcone gauge) solution, if you can translate that into sense that would be help

(it is also the classical solution used for cosmic strings!)

Physics doesn't make sense

But yeah this is some paper taking on Nambu-Goto directly

it's all a lie

How do geodesic work in affine gravity, anyway

Torsion doesn't affect geodesics since the geodesic equation is symmetric in the lower indices of the connection

But on the other hand, non-metric connections should affect the geodesic equation

BUT

The Nambu-Goto action doesn't lead to the geodesic equation, in this case

what is the dealio

3:59 PM

ah fuck I have a class right after calc of variations a mile away

15 minutes to get there

might have to get a bike

Terry Bollinger

4:16 PM

So the new bi-weekly chat structure is for the chat room to empty out completely at 4:00 PM UTC sharp? Heh! Bye!

0

Q: Geodesics and actions in affine gravity

It is a well known result in Riemannian geometry that geodesics are also the curves that minimize length (or extremize it anyway, I know what you're gonna say @0celo7), with a very similar result in Lorentzian geometry, where for instance the Nambu-Goto action $$S = -m \int_{t_0}^{t_1} ds = -m ...

general-relativity geodesics

plz halp

@TerryBollinger Bad idea since it's 5:20 here

I'm getting pretty close to going home time!

But if you have chat discussion that can hold in 10 minutes I'm here!

@Slereah wtf is that comment

Because I know what you're gonna say if I say "geodesics are the shortest curve"

Oooh this isn't true if the manifold is incomplete

they don't even extremize

etc etc

Like a big baby

4:25 PM

they're critical points

isn't that what an extremal curve is

no

extremal usually means max or min

not all critical points are minmax

Oh does extremal not cover saddle points

Or whatever they're called

4:36 PM

no

@JohnRennie yo

dusted the pc last night

@0celo7 Afternoon

the amount of dust in the GPU was pretty comical

I just took some compressed air and put it on the fans, and a big cloud came out

It is amazing how much dust gets swept into a PC when you leave it running 24x7.

That why you always lock servers away in a clean room.

@JohnRennie yeah it was running for a month straight

dusty boi

user image

Have you used the GPUs for any numerical work (apart from gaming :-)

4:46 PM

@JohnRennie woe is me if I have to use anything more involved than matlab or mathematica

Weren't you doing some Monte Carlo calculations a little while back?

yeah but it was CPU based

Using the GPU for computations isn't too hard

It's hard for graphics tho

user image

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