The h Bar

Slereah

8:00 AM

Mostly I'm just being contrarian :p

user507974

or determinable

Slereah

Well Bohmian QM does have different results from Copenhagen

TECHNICALLY

In practice, I don't think any experiment can, though

Shing

um.... isn't stat mech having different physics states to qft? can't blame it for being total approx... (surely, it has its own approx just like other physics)

Slereah

I'm just being willfully difficult :p

Give me any theory and I'll give you all kinds of reason why it's the worst

Shing

lol

I see

user507974

8:01 AM

E&M

Slereah

EM's pretty good

user507974

what happened to our contrarian

also doesnt it ignore self interactions a decent bit, or approximate on it

Slereah

Depends

user507974

thats the only main physics i have left to take so i dont know and i say Slereah take the wheel

Slereah

I think that was a big debate, before QED was a thing

Like do particles interact with their own EM field

Because then you get weird results if you take that into account

user507974

8:03 AM

i have taken 4 stat mech courses tho...

Slereah

Especially for point particles

Since a point particle has an infinite electric field at $r = 0$

And if a particle tries to move, it will move inside its own magnetic field

user507974

@Slereah I think its normalizable though in 3d

Slereah

Making it acquire an effective mass

@user507974 Well you can describe it with distributions

user507974

well what if the effective mass is the true mass

Slereah

That was the theory!

Part of the Lorentz ether theory was that the electron had no mass

It only had a charge

The mass was acquired by its interaction with its own EM field

And do you know the relation between that mass and the energy of its EM field?

user507974

8:06 AM

the ether theory we are talking about is the same one that got disproved by einstein?

Slereah

That's the one yeah

Although it wasn't disproven

It was equivalent to it

But since it made unnecessary assumptions, it was dropped

user507974

are the unnecessary assumptions it made self interaction

or other things too

Slereah

Oh no that problem still happens in relativity

That is actually the root of renormalization later on

The assumption was that there was a prefered inertial frame of reference

user507974

oh, yea, i can see occams razor tearing that up

a universal preferred frame for everything or a frame for each particle/object thing?

Slereah

universal frame

user507974

8:10 AM

so basically a universal rest frame

Slereah

Yes

it's the frame in which light goes exactly at c in all directions

But, of course

that wasn't what was measured

user507974

yea

Slereah

So they added some things on top to make it so that the frame still existed, but the speed was c in all other frames

user507974

more to be razorred

user116211

Hallo!

user507974

8:12 AM

oyo

how goes your day

user116211

where is @yuggib ;(

user116211

@user507974 going~~~~

user507974

@MAFIA36790 so no outside forces are acting on you? are you in space or something?

user116211

:D

user116211

Maybe the Speed-force ;P

user507974

8:14 AM

and yes i know forces still act on you in space

I just kind of hypothesized the Lorentz ether theory while talking with Slereah about why he said he will say any theory is the worst but decided not to when I said E&M

hey @Slereah , is there REALLY any water vapor in the air

Slereah

Well I could say that it's a classical theory ad not the real theory

But that would be pretty petty

There should be, yes

Unless you are in the sahara

user507974

@Slereah but when you think about it, isnt it probably instead a suspension of a tiny droplets of water together

Slereah

dunno

Is the water bonded in any form in the air?

Not that I know

unless it condenses for fog

user507974

thats the question, is water truly a gaseous phase

you know what discipline could help answer this

Slereah

Is it musicology

Actually

I'm not sure thermodynamics could answer

yuggib

8:21 AM

@MAFIA36790 here I am

user116211

o//

Slereah

Because a thermodynamical phase is only defined for a large enough sample of particles

You can't say a single atom is in a phase

user507974

@Slereah but you could pick a volume and look for particle correlations

actually im auditing a class where I should probably be able to answer this...

damn not having enough time to do the hw

yuggib

@Slereah people do QFT with statistical mechanics

Slereah

Well yes

As an APPROXIMATION OF QFT

yuggib

8:23 AM

no no

they do QFT using stat mech

it is called path integral (in euclidean time)

i.e. the only feasible path integral

@MAFIA36790 \\o

user507974

@yuggib, actually a question i bugged sle with a bit ago

is water in the air really a vapor, or is it just a suspension of a nanoscale glob of water

yuggib

I suppose that at low temperatures/suff high pressure (i.e. common earth conditions), it is the second

user507974

you mean STP?

or colder/higher pressure

yuggib

the empirical curve usually works quite well... www1.lsbu.ac.uk/water/water_phase_diagram.html

@user507974 I meant stp as well

user507974

@yuggib 404 error

yuggib

8:31 AM

www1.lsbu.ac.uk/water/water_phase_diagram.html

Slereah

I wouldn't call euclidian path integrals statistical mech

yuggib

for me it works...

Slereah

sure it's the same mathematical formalism

But it is also used in economics

I wouldn't call QFT economics

yuggib

it is actually our way of calculating partition functions and all

user507974

@yuggib yea, this

yuggib

8:32 AM

they're pretty much the same thing when it comes up to terminology and what you actually do

user507974

wait, so how does the path integral relate to fields in qft

Slereah

Same way as QM?

You integrate the field between two boundaries to find the transition between them

user507974

basically the sum of all possible trajectories?

yuggib

@Slereah for example this is a book of stat mech

but if you read it, you will find pretty familiar stuff

Slereah

Actually

When I did my master thesis

My advisor told me to get the path integral book by Feynman and Hibbs

So I asked for that book from my family, because no money

user507974

8:35 AM

wth, so many people just came online

howdy folks

Slereah

And I got a book on stat mech path integrals with the same title

Because the layman doesn't know the thing about physics books

They all have the same title

user507974

btw, how does renormalization groups come up in qft

im learning about it in a stat mech course currently

Slereah

Well if I'm to believe @yuggib

Nobody knows how renormalization works

So who knows

user116211

I would sound TOO much overambitious if I say this.

user507974

@Slereah RG seems like it makes sense to me

yuggib

8:40 AM

nobody knows how rigorous, non-perturbative renormalization works in reasonable dimensions

user116211

I'm going to study Abstract Algebra.... before my undergrad classes start....

user116211

don't kill

Slereah

1+1D is totally reasonable

It's a line

Totally reasonable

user507974

at least classical stat mech

user116211

First group theory, then ring theory....

Slereah

8:41 AM

Have you ever read the Planiverse, btw

It's a great book

It's kind of a more science-oriented Flatland

2+1D universe sort of thing

user507974

@Slereah whats the 1, time?

Slereah

Yes

user507974

well what about the other 9

user116211

@yuggib: Should I go with Dummit and Foote's Abstract Algebra?

user507974

or is it 11

or monster groups

Slereah

8:43 AM

It's a common notation to use $p+q$ D for p spacelike dimensions and q timelike ones

user507974

@MAFIA36790 you can go for Zee's and learn of the woes of confuzio and try to find me in the book

user116211

@user507974 :))

user507974

@Slereah i know, just was making 100% sure

yuggib

@MAFIA36790 springer.com/us/book/9783540642435

if you're brave enough... ;-P

user116211

The GREAT BOURBAKI

user507974

8:47 AM

@Slereah were you being serious about RG being not understood?

Slereah

Can't say

I don't know much about RG

for me renormalization is like

"Separate the divergent term and sweep it under the rug"

zoop zoop

Currently reading about zeta regularization btw

user116211

@yuggib indeed; thanks man... I owe you ;)

yuggib

@MAFIA36790 I have also this book springer.com/it/book/9780387715674

it seems reasonable

Slereah

I still don' get how the analytic continuation is done

yuggib

and much more introductory than bourbaki

Slereah

8:48 AM

Is it just a limit process?

Like as it approach the branch cut by going around, it has this value?

yuggib

@Slereah it is something you can't do apart if you prove reflection positivity or some other shit

Slereah

the wikipedia article is pretty vague about it

do you have a pithy statement for how to do analytic continuation for the Riemann zeta function

But none of that bullshit using the sum

yuggib

@Slereah I don't care about riemann zeta

Slereah

I know it's the spectral zeta function for QFT yeah

yuggib

I say that in order to recover real-time qft from euclidean time one you need reflection positivity

Slereah

8:50 AM

But one step at a time

isn't that one of the Wightman axiom

Or... OS

yuggib

i.e. to fullfil osterwalder-schrader axioms

Slereah

Osterwalder-Schrader

I never remember that name

yuggib

arxiv.org/abs/1411.2964

user507974

@Slereah is zeta regularization like analytic continuation?

Slereah

kinda yeah

How many regularization procedures exist, anyway

There's zeta

lattice

dimensional

colombeau

regulators

yuggib

8:56 AM

ultraviolet regularization (mode reg)

Slereah

Is that the momentum cutoff one

isn't that equivalent to lattice

Physics Meta

0

Q: What the site will do if a user with little/no knowledge of physics down-votes me when I answer his question correctly?

What the site will do if a user with little/no knowledge of physics down-votes me when I answer his question correctly, but that user assumes that my answer is not helpful-answer to the question posed by him/her?

yuggib

no, they should be different

or maybe I'm wrong...who cares it's all b*******t

s.harp

9:57 AM

Hi, quick question, how would one define a quasi local operator or even a local operator?

The context in which I am asking could be if we have a lattice $\mathbb Z^2$ and on each point of the lattice a HIlbert space $\mathcal H_x$, then I guess a local operator would be something which is acts on $\otimes \mathcal H_x$ via identity on all $H_x$ except for at most $1$.

I'm not sure how this would extend to a continuum model

And I think quasi local could be in such a context that the action is different from unity on a finite amount of $\mathcal H_x$?

But maybe it is something more general?

I would appreciate some help :)

yuggib

I don't know what you're trying to do; however it is possible to have tensor products of Hilbert spaces with a "continuous" index...simply you don't have separability in the end

s.harp

@yuggib I think I remember what I have seen in QFT is that a operator valued distribution is called local if $[\phi(x),\phi(y)] = A \delta(x-y)$ or something like this.

yuggib

that is a different thing

it is an operator-valued distribution, with a notion of support and commutation properties in "local", i.e. time-like separated space intervals

Slereah

obviously not since $\varphi(x)$ isn't defined for a distribution :p

It should be $[\varphi[f], \varphi[g]] = i\int dx f(x) g(x)$

yuggib

see wightman axioms

s.harp

10:08 AM

@Slereah Ok, I guess you are right about a stricter formulation. My knowledge of QFT is limited to the physics blah blah courses where everybody pretends we are in finite dimensional hilbert spaces

Slereah

Physicists just say it's an operator, they don't give a shit

s.harp

What I precisely interested in is happening in a quantum statistical physics setting on a lattice. Is what I wrote what one would usually call a local operator? And what is usually called a quasi local operator?

yuggib

10:21 AM

I am afraid you should take a look at what are called quasi-local algebras

I suspect that the term comes from there

if you want to do that with a statistical physics flavor you may check the books by bratteli and robinson

s.harp

ha, I have those books at home, and they are actually exactly the right flavour of things

when I get back I'll check the index ^^

Slereah

10:46 AM

So

What objects are CSR fields

Is it like

$A_{\mu \theta}$

A vector field with a continuous index

but then what bundle section would that be

I can only find papers about the unitary rep of that

For the wavefunctions

Nothing about the field operator, though

user116211

11:13 AM

@yuggib: just 1 left at Amazon.... I'm buying it ;P

user116211

Also, I downloaded the pdf of Dummit and Foote; they started from set theory ;))

yuggib

@MAFIA36790 it's a nice couple of books...a bit heavy, but lots of informations

also, a purely algebraic approach to qm

user116211

@yuggib not expensive also....

user116211

@yuggib books? o-O

user116211

I've not ordered the Bourbaki ;_;

yuggib

11:16 AM

there are two bratteli robinson books

part 1 is more mathematical

user116211

@yuggib ohh.

yuggib

essentially it is theory of operator algebras

user116211

checking my order

yuggib

the second is more physical, CCR and CAR and then statistical mechanics (free bose&fermi gases, etc.)

s.harp

the books are super nice

I started reading as undergrad, but when they had 6 new topologies on the same space I stopped reading for about 2 years :D

yuggib

11:18 AM

@s.harp yeah von neumann algebras of bounded operators

the good ol' $\sigma(X,X^*)$, $\sigma(X^*,X)$ etc. topologies

ACuriousMind

11:50 AM

@Slereah I think there is no field that would have the CSR associated to it, as we always demand that the target space of field is finite-dimensional, and none of these gives you a CSR

Slereah

Do we?

Can't we have an infinite-dimensional field

I guess it would be weird but would it be haram

ACuriousMind

@Slereah How? Generally, a field is thought of as a tensor field or a form on spacetime, neither of these have infinite-dimensional target spaces

You can think about maps from the manifold to infinite-dimensional spaces, but they just don't arise in any remotely natural way

Slereah

True dat

But could we impose it by fiat

ACuriousMind

So, well, it's not obviously forbidden, but there is really zero motivation to consider them, I'd say

Slereah

Yeah

I'm just curious to know what it would look like

I guess a fiber bundle with an infinite dimensional fiber

That is invariant under some Lorentz rep

ACuriousMind

11:57 AM

@Slereah Yeah, and there the trouble already begins - there is no unique notion of what you want to understand as an "infinite-dimensional manifold"

I.e. what kind of regularities you should or should not demand

Slereah

I guess physicalities would narrow it down

ACuriousMind

@Slereah Only if you have some physics you're trying to explain with these fields in the first place :P

Slereah

Hm

I recall reading that CSR are not causal

ACuriousMind

Since we lack any physical motivation to consider such objects that I could see, I don't really see how one could make physical arguments here

Slereah

But don't you need some observables to make that determination?

And wouldn't you need some field operators for that

Well you don't need objects to be real to make some reasonable demands on them

ACuriousMind

11:59 AM

@Slereah I don't know anything about that, but perhaps it's simply impossible to have any local causal operator on them?

Slereah

Maybe

I should find that paper

Also I recall that the vacuum for a CSR field has infinite heat capacity

hence why they were mostly thrown out

What are some manifolds with dimension $\mathfrak{c}$

yuggib

what the hell are CSR

ACuriousMind

@yuggib "continuous spin representations"

They are unitary representations of the Poincaré group that normally don't play any role in QFT

But Slereah seems determined to use them :P

@Slereah $\mathbb{R}^\mathfrak{c}$? :P

John Duffield

@DanielSank : that's why some IT guys are good at physics.

yuggib

I smell b******t

Slereah

12:03 PM

Well basically

if you look at unitary reps of the Poincaré group

For the small groups with $m = 0$

You have two choices

Either the small group for massless particles that transforms under $ISO(2)$

yuggib

@ACuriousMind $\mathbb{R}^{\mathfrak{c}}=2^{2^{\aleph_0}}$, i.e. the power set of the reals

Slereah

Or the continuous spin representation

ACuriousMind

It's little group, not "small" group!

Slereah

Little, small

They ain't big

ACuriousMind

@yuggib Well, the question is meaningless anyway until he defines what "dimension" is for a not-finite-dimensional manifold ;P

Slereah

12:05 PM

Well I don't want to sound like a physicist, but

I'd say that the index is continuous :p

yuggib

@ACuriousMind I agree..probably a schauder basis of given cardinality

(at least a local schauder basis of given cardinality)

Slereah

An ordered pair $(\theta, A_\theta)$ such that both elements are $\in \Bbb R$

not quite sure how you'd apply a Lorentz transform on that though

yuggib

take your favorite non-separable Banach space and construct its Banach manifold

Slereah

What's a cool non-separable Banach space

yuggib

I'm astonished how you couldn't get into a phd in hep-th

you sound perfectly like an hep-th/string guy

Slereah

12:08 PM

heh

Well the lab I was at was more practical oriented

yuggib

doing random useless stuff that are vaguely physical and non-mathematical just for the sake of it ;-P

Slereah

Mostly particle simulations and such

Parton jets, plasmas, etc

aka boring

I was the student of the only big theory guy around

Probably a mistake, retrospectively

yuggib

@Slereah any non-trivial weyl algebra

Slereah

Probably didn't help when I tried to go for funding

John Duffield

@Slereah : of course they do. Only then you don't call it a photon any more...

yuggib

12:11 PM

@Slereah or $L^{\infty}(\mathbb{R}^d)$

Slereah

Let's go with that one

it sounds nice and physicky

Bounded function bundle

yuggib

@Slereah the usual $L^\infty$ does not consist of bounded functions alone...

John Duffield

@user507974 : that's the guy who wrote this.

yuggib

but of almost everywhere bounded functions (wrt Lebesgue measure)

ACuriousMind

almost everywhere bounded measureable functions

It's a pretty ugly space :P

Slereah

12:13 PM

How do you even make a rep of the Lorentz group on that

yuggib

however also continuous bounded functions are non-separable

and banach with the sup norm

John Duffield

@user507974 : and note this: "There clearly is a frame where the CMB is at rest, and so this is, in some sense, the rest frame of the Universe".

ACuriousMind

@Slereah You might just have found the reason no one does this :P Since these "fields" don't arise naturally, they don't carry a natural representation of the transformations on the base manifold, either

Slereah

Oh bother

Scihub is down

How am I suppose to get things legally

ah, found it here

neurotheory.columbia.edu/Larry/AbbottPR76a.pdf

yuggib

they got it in the end

Slereah

12:23 PM

They do construct field operators for it

Let's see

ACuriousMind

$\theta_n$ is continuous and $\Gamma_{mn}$ acting on it is a "matrix". Sigh

Slereah

Hm

Apparently

yeah

They construct it exactly like how I would have done

Like a filthy physicist

not really any indication of the space used

on the other hand

Might be a good PSE question

yuggib

no

Slereah

Yes indeed, @yuggib

yuggib

it is at most a good theoretical physics SE question, sadly it ceased to exist

Slereah

12:27 PM

you're right @yuggib, I should ask it right away

ACuriousMind

@yuggib Anything that was on-topic there is on-topic here.

yuggib

T__T

ACuriousMind

Whether you find someone to answer it is another question :P

yuggib

metaphysics should not be on topic here

Slereah

I don't have a good track record for getting answers

yuggib

12:28 PM

and there are people complaining about mathy questions

Slereah

Toy models and such are allowed!

ACuriousMind

@yuggib Tell that to the quantum-interpretations and measurement-problem tags :P

yuggib

a toy model for God is still methaphysical

@ACuriousMind T______T

ACuriousMind

Hehe

Yeah, I could live without those

yuggib

you barely missed the discussion about that with MM yesterday

ACuriousMind

12:30 PM

I read it, though

Anyway, I have to go

I'll spend the next few days OUT AT SEA

yuggib

good ol' experimentalists...they are attracted by ultra-finitism as mosquitos by a pool of blood

@ACuriousMind that seems a sensible idea

ACuriousMind

See you all on Tuesday, unless the waves claim me

yuggib

but cold germanic sea or true sea?

see ya ;-)

ACuriousMind

@yuggib North Sea in Friesland, so not really true sea ;P

yuggib

agh...you have the physique du rôle

being german

Slereah

12:52 PM

0

Q: What type of fields are continuous spin representations?

Continous spin representations (infinite dimensional representations of the Lorentz group) are pretty rarely discussed, and usually not in that much mathematical details. And usually it is done in a very "physicist" way. The field operators for a CSR is given as $\hat \varphi_{\theta_n}(x)$, a fi...

quantum-field-theory representation-theory

There u go

Slereah

1:14 PM

I wonder if CSR reps are unitary

@Qmechanic got wild on tagging my question

user116211

@Slereah He is the master of tagging :)

user116211

He even added LMGTFY ;P

user116211

But yeh always a silent diligent worker.

user116211

Kudos

Slereah

He is assuming that people don't know what CSR are

Even though everybody does

user116211

1:17 PM

@Slereah yeh.

Slereah

all the cool kids love CSR

yuggib

2:09 PM

no one loves CSR

Slereah

2:44 PM

"The Chebyshev distance is sometimes used in warehouse logistics,[4] as it effectively measures the time an overhead crane takes to move an object (as the crane can move on the x and y axes at the same time but at the same speed along each axis)."

heh

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