The h Bar

qwerty

00:06

@Slereah I think you mentioned some of this stuff recently johncarlosbaez.wordpress.com/2024/10/31/bradwardines-rule

Silly Goose

00:23

in the discussion of macroscopic mawells in jackson, he mentions that the equations vary wildly on atomic length and time scales. However, he only ends up doing a spatial average. What about a time average?

naturallyInconsistent

01:47

@Mr.Feynman that's more meow meow than ACM

@SillyGoose On the chapter that he analyses that, he specifically covered the time average and found that it is not sufficient to solve the problem, whereas for any time slice, spatial averaging suffices.

That whole thing is a mess; even if yes, that is technically an interesting result, that is nowhere near how anybody would be attempting to solve for real physics.

@ACuriousMind miao miao would add that multipole expansion is the harmonic extension of Laurent series, famous from complex analysis, to 3D.

@Mr.Feynman you've put in such incredible amount of effort...

@Slereah I dont know what you mean; I would think that Caesar had to have enough interest in the issue to come up with what we honour him as the Julian calendar.

Slereah

02:08

I don't think he personally made the research

naturallyInconsistent

02:32

That is definitely the case, but there must at least have been people knowledgeable enough to provide a smart enough person to come up with the system

Also, @Slereah, on your blog page samuel-lereah.com/articles/Physics/… , where, ignoring the trivial bit that the exponential of Hamiltonians that you have right at the top is missing the time factor and imaginary unit, and that the interacting Hamiltonian forces you to include the time ordering factor too, the next line is also in violation of BCH. Or rather, you might be interested in Zessenhaus' formula, or Magnus's expansion.

John Rennie

05:36

Mathtoons

4

Ryder Rude

06:57

hi

Ryder Rude

07:14

what r the main defining physical features of the universe in ur view

i have arranged fundamental features of the universe into tiers

tier 1- locality, uncertainty principle

tier 2- principle of least action, local poincaire invariance

tier 3- gauge group U(1)xSU(2)xSU(3), Higgs mechanism

tier 2 should also have General Relativity in it

what would be the tiers according to u

this is in the Non Humean framework of physics. according to non Humeans, the laws of physics have other fundamental laws behind them

think_meaning_builds

08:07

@Slereah would you be interested in working for this company?

Slereah

10:31

I would not

Dannyu NDos

@RyderRude The quirk is, science is about criticizing and editing itself. We surely have not found the true physical laws, and even after we'd found them, we cannot confirm we did.

That's how the string theory rendered to be a fad.

naturallyInconsistent

@DannyuNDos why are you talking facts to someone who is trying to tierzoo physics?

Dannyu NDos

10:48

*Cough*

Mr. Feynman

10:59

Wow, nR MBT is so different from QFT. Green's functions having a physical meaning

I wonder, is there nothing like the LSZ in MBT? Or is it that the S-matrix is not relevant for condensed matter people?

Don't put too much effort in answering these questions, in case you decide to. They're just things I'm wondering now

naturallyInconsistent

@Mr.Feynman The S matrix calculation is relevant to cond mat too. Why would you think it is not? In particular, the Euclidean propagator is the partition function.

Mr. Feynman

The euclidean propagator? Not the euclidean action?

naturallyInconsistent

Yes. It is actually a LOT different. There is no LSZ because it is not about particles coming in from infinity and scattering to infinity

Mr. Feynman

I was wondering that because unlike QFT, in many-body theory (which is my focus, probably it was wrong to say condensed matter thereafter), they calculate observables using Green's functions, which do carry a physical interpretations because of obvious reasons I'm lazy to mention :P

@naturallyInconsistent Since I learned about relativistic QFT first, I was a little bewildered when I encountered the MBT version :D

Ryder Rude

@DannyuNDos yes. my analysis is about physics as we know it, what has been tested so far

naturallyInconsistent

11:06

Instead, you are dealing with "greens functions" that is not between vacuum states, but rather is between ground state. It is about correlation between excitations at different spacetime points

And these correlation function turn up in Kubo relations

Ryder Rude

@naturallyInconsistent what is ur problem??

Mr. Feynman

That's another great point I've raised a few days ago. Even if you can write down an algebra with respect to which the Fermi Sea is vacuum, it is not the Fock Vacuum: it's not the "same" GF hep people would consider

Oh no they're about to flame D:

Mast

Calm down, people.

naturallyInconsistent

There is a LOT of mess because, for example, I was asking a senior colleague about Mahan's assertion that the 2-point correlation function is unphysical, not least because in basic SR QFT the 2-point Green's function is very important, and he was like "textbook authors can write whatever nonsense they want..." and miao miao immediately realised that it would not be useful to discuss that with him

@Mr.Feynman that wasnt going to happen

@Mr.Feynman The problem is, which Fermi sea? The Dirac's one cannot work.

Mr. Feynman

@naturallyInconsistent I'm not sure "Fermi sea" was a correct choice of words; I meant the g.s. of an electron-many body system, i.e. filling the levels up to $k_F$.

naturallyInconsistent

11:12

@Mr.Feynman Yeah, that is an extremely important thing to keep in mind.

Mr. Feynman

@naturallyInconsistent In relativistic QFT it is definitely unphysical in the sense that it carries no suitable interpretation whatsoever: no particles propagating from there to there. There is no position operator, at least not in standard presentations

In MBT they seem to give a proper physical meaning, as discussed in section 7 of Fetter&Walecka; the idea is the same of rQFT book, just this time a position operator is defined

naturallyInconsistent

@Mr.Feynman But you cannot say that. The 2-point GF in basic SR QFT is the single particle propagator that you have to renormalise with all the bubbles. It is extremely important to get correct.

Mr. Feynman

Unphysical is not unimportant; I mean not observable :P

naturallyInconsistent

But a single particle propagator is the thing you use to define the dayum observable.

Ryder Rude

@Mr.Feynman it has a physical meaning in rQFT but in terms of fields rather than particles. u would have to think of fields as observables

then it just becomes an expectation value

there is a post about measuring this

Mr. Feynman

11:16

The LSZ links what is observable (scattering amplitudes) to unphysical objects that are defined mathematically (Green's functions)

Ryder Rude

1

Q: How are general quantum correlation functions actually measured?

These days, the majority of work in theoretical particle, condensed matter, and AMO physics is about methods for calculating exotic correlation functions, of the rough form $$G_{ij} \sim \langle \mathcal{O}_i(t) \, \mathcal{O}_j(t') \rangle$$ though possibly also including time ordering, (anti)sy...

@Mr.Feynman it is not unphysical. fields are the fundamental observables

Mr. Feynman

I'm afraid that's about condensed matter, which if you read is not what I'm currently talking about.

Ryder Rude

yes, i just noticed

naturallyInconsistent

@Mr.Feynman You must first have specific forms of the GF, then you can have LSZ act on them. In particular, if you use the infraparticle way to get the free-field electron get the Coulomb field around them, then the propagator is no longer the usual k^2-m^2 kind, and boom, your LSZ is fucked up

Ryder Rude

i will try to find something about rQFT. im positive that fields are observables

naturallyInconsistent

11:20

I really think the equivalent to LSZ in cond mat is just Kubo relations. You can just extract so much info from the correlation functions using Kubo just like how you use LSZ on GF.

Mr. Feynman

Fields are not even hermitian...

@naturallyInconsistent Ah. More to read about it?

Ryder Rude

@Mr.Feynman their real and imaginary parts are hermitian

also, electric field is def an observable

naturallyInconsistent

@Mr.Feynman Infraparticle is the search term. I really really don't understand the whole structure, only that it is such a headache to work with.

Mr. Feynman

I'll keep in mind for the future D:

naturallyInconsistent

@Mr.Feynman why do you even want to read from him? The amount of nonsense being spewed is just beyond refutation; I'm low-key holding back from labelling them as inappropriate.

Mr. Feynman

11:22

I reply if I'm pinged typically

naturallyInconsistent

@Mr.Feynman You might want to low-key forget about them. ACM also said that he isn't into that; apparently, the winning move is to ignore those; the usual formalism with infrared renormalisation is sufficient to get good answers.

Mr. Feynman

Oh, it's definitely not on my top priority list :D

naturallyInconsistent

:D

Ryder Rude

13

A: Is the S-Matrix the only quantum field observable?

Correlation functions of local operators are the other observables that field theorists talk about most of the time: things like $\left<{\cal O}_1(x_1) {\cal O}_2(x_2) \ldots\right>$. But there's no shortage of observables in quantum field theory. The situation where you'll hear people say that ...

Mr. Feynman

I learned a Chinese word yesterday

Ryder Rude

11:25

this answer says that correlation functions r observavles but he doesnt give any sources

naturallyInconsistent

I'm kinda amused that my tierzoo comment got starred. Like, I've specifically avoided casting any vote on the complaint that pinged meow too. If someone else was wanting to flag it, it would be funny.

@Mr.Feynman which is it?

Ryder Rude

in general, note that observables must be local smearings of fields or one would violate causality

Mr. Feynman

@naturallyInconsistent 奶

Milk

Ryder Rude

this is the idea in Algebraic QFT. S-matrix is an approximate idea

naturallyInconsistent

@Mr.Feynman ... lots of fun ...

Mr. Feynman

11:27

@naturallyInconsistent Context

naturallyInconsistent

@Mr.Feynman of course, the fun is all in the context miehehehe

Mr. Feynman

Funnily enough, it is pronounced like the Japanese word ない, which means "there isn't"

naturallyInconsistent

heh

@Mr.Feynman also, there is a lot of inappropriate (racist) jokes just awaiting exploitation

and miao miao can say it; you likely cannot

Mr. Feynman

@naturallyInconsistent I won't indulge in such mundane pleasure 💀

naturallyInconsistent

@Mr.Feynman just yesterday miao miao sent a new vocab to my friend: prurient

Mr. Feynman

11:30

Ah, it means lustful

naturallyInconsistent

turned up on the court case that qwerty sent here: the "I know it when I see it"

miao miao iz just like, hmm, fun vocab, snatch

qwerty

@Mr.Feynman i've heard stories of anglophones going to france/italy, trying to order a caffè latte by asking for a "latte" and ending up with plain milk

naturallyInconsistent

@qwerty lol yay~

lack toast in toblerone~

qwerty

@naturallyInconsistent that's a very good bone apple tea

naturallyInconsistent

miahahaha

anyway, if you have more questions, send them quickly. miao miao is about to have to go for a massage at the gym

how are the two of you?

M I A O ~

Anyway, @Mr.Feynman in RR's latest link, there is an answer that Ron Maimon gave a comment that goes into slightly more detail: namely that even though an S-matrix itself is not an observable in one sense of the word, it is an observable in another sense.

qwerty

11:58

@naturallyInconsistent I'm doing well, thanks :) remedial physics/maths reading :p

Ryder Rude

12:27

This Single Rule Underpins All Of Physics

►

it is about the historical development of the principle of least action

qwerty

"how many prefixes can we attach to `-morphism'?" -- Mathematicians, probably.

i've been told to understand DG I need to first read category theory

Mr. Feynman

@qwerty Well, yes. Although abroad "latte" is caffè latte (I guess), here it's just the word for milk.

Similarly, if you step into an Italian bar and order a caffè (coffee), you will get an espresso

ACuriousMind

@qwerty I...don't think that's true in any defensible sense :P

(and I like category theory)

Mr. Feynman

@qwerty Bullshit. Knowing category theory is good and helps but it's far from being fundamental

Ryder Rude

@qwerty i think category theory is suposd to be learned after u have learned all other branches of math

qwerty

12:39

@ACuriousMind ok, I misrepresented, I started reading DG and writing down my List of Definitions including you-know-what, and I got suggested to read category theory

Mr. Feynman

And even then, for basic DG, all the category theory that would be useful (not fundamental) is like the very basics: understanding what morphisms and functors are, so you understand that you're really defining the "same" things over and over

Ryder Rude

this is what mathematicians told me

ACuriousMind

@qwerty still a pretty weird suggestion, imo

Ryder Rude

one should learn linear algebra, abstract algebra, topology and differential geometry stuff before learning category theory

Mr. Feynman

Like, with category theory you will understand that (in a given sense) there is no difference between a diffeomorphism of manifolds, an isomorphism of bundles, a vector space isomorphism: they're all isomorphisms in the sense of the given structure

Concerning functors, you may grasp a better idea of what a pullback is and what a pushforward is, but that's about the amount of category theory that I've encountered in DG

@RyderRude You can arguably get away with the basics of the first three (i.e. just the definitions of the structures)

Ryder Rude

12:43

@Mr.Feynman i didnt mean this as in pre-requisites for category theory. but more like : after u have learned this stuff, category theory helps u see connections between them

Mr. Feynman

oh, okay

qwerty

@ACuriousMind the culprit mathematician: "In my defense, I know qwerty and I know how she tends to understand things"

Mr. Feynman

So your real name is qwerty :P

When all's said and done, the suggestion is like "you need to study Lie theory before learning QM"

qwerty

probably I'm not going as deep as you all think :P I'm just reading wiki and SEP right now

Ryder Rude

that is a good analogy. lie theory only enhances Qm understanding after u have already learned qm

Mr. Feynman

12:46

I know it's a good analogy, I'm never wro-HELLO @Relativisticcucumber

Slereah

@qwerty My favorite one is itchyomorphism for a Poisson preserving morphism

@ACuriousMind That's some New Math shit

Surely the children will understand addition better if they are taught set theory

Relativisticcucumber

@Mr.Feynman 奶茶！！！

Mr. Feynman

Slereah talks about math like Jesse Pinkman talks about meth

@Relativisticcucumber !!!

qwerty

@Mr.Feynman I do like putting things in boxes. Actually I like real physical boxes too. I started bookbinding once and ended up making boxes once or twice... I may not like cats much, but they have the correct take on boxes...

Relativisticcucumber

@Mr.Feynman こんにちは

did i get it right? i know no japanese

qwerty

12:50

@Slereah that's a punny name lol

Slereah

It's not a very commonly used one, but I decided to spread it more

Mr. Feynman

@Relativisticcucumber こんにちは。キュカンバーはどうですか。

Yes

Relativisticcucumber

wow wait i just seem to have learned ciao is hi AND bye

Slereah

$$\mathrm{Poiss} \overset{🐟}{\longrightarrow} \mathrm{Poiss}$$

Mr. Feynman

@qwerty You may not like cats but you are a cat deep down

Relativisticcucumber

12:52

everyone must like cats

Slereah

Quantomorphism is also a good one because it sounds fake

Mr. Feynman

@Relativisticcucumber Why is everyone so astonished that ciao is also a way to say bye?!

Relativisticcucumber

0.o i was shocked that it means hi actually

i think theres an impression that it means only bye

for example, theres a famous song that says "ciao adios im done"

Ryder Rude

@Fermion hi

qwerty

"Bella Ciao" - Italian Anti-Fascist Song (Rare Version)

►

Mr. Feynman

12:53

Incidentally, the italian verb for "say hello" is "salutare", which also means "say goodbye". I'm very triggered by the fact that in English there is a single verb for "say hello", i.e. "greet" but not one for "say goodbye"

"farewell" is a bit too much :P

Relativisticcucumber

@Mr.Feynman wow never thought of that

Fermion

hi @RyderRude

Relativisticcucumber

yeah i guess "bid farewell"

Ryder Rude

@Fermion did u just join the chat

Relativisticcucumber

even then tho bid cant stand on its own so this is a valid problem

Mr. Feynman

12:55

I mean, yes, but it's kind of dramatic :P

Slereah

George Carlin had a whole bit about how you could beef something up but not beef it down

Mr. Feynman

Farewell is also a verb on its own, by the way

I think

Fermion

@RyderRude yes

Mr. Feynman

No, it's not D:

Then "bid farewell" be it

Ryder Rude

@Fermion we discuss physics, math and sometimes philosophy

and other casual things too, like rn

Relativisticcucumber

12:56

wow this q has been asked precisely reddit.com/r/words/comments/w9idvk/…

Mr. Feynman

SO FAREWELL CAN BE A VERB! YEAH!

Relativisticcucumber

yesh

Slereah

dictionary.cambridge.org/dictionary/english/see-off

Relativisticcucumber

ok i migrate to condensed matter ,,,, ciao 朋友们

Mr. Feynman

Bye bye, friend

I recognize only the second hanzi from Japanese :P

@Slereah It's probably the best way, yeah. I always forget about it, though

Not a single word though :P

Slereah

12:59

it is lexicalized :p

Mr. Feynman

How do I get the right to make up a word? :D

Slereah

Just do it and see if other people go with it

Mr. Feynman

From now on "bye" is a verb

Slereah

Worked for Shakespeare

Mr. Feynman

I guess I'm ancient

Ryder Rude

13:03

anyone see The Exorcist? is it still good

Relativisticcucumber

@Mr.Feynman i couldnt resist:

Slereah

Sorry as far as movies with Linda Blaire go, I only like

How to Get... Revenge! [1989] [Edutainment Videoguide]

►

Relativisticcucumber

4

qwerty

@Relativisticcucumber top tier meme

Mr. Feynman

@Relativisticcucumber I don't know if you know Liam Carpenter. He makes stereotypes videos about Germans (he's an Englishman living there)

And at the end of the end of his videos, Germans typically make a slight smile, probably more of a sneer.

That's how I imagine the scene unfold.

Ryder Rude

13:08

@Slereah nice..

Slereah

She had a rocky career after the Exorcist

Ryder Rude

i also saw a movie called Hausu

it's the most bizzare thing ive seen

it is about cats

@Slereah what happened

Slereah

just the luck of the draw I guess?

Ryder Rude

oh

Slereah

not everyone gets to have a splendid career after making one big movie

Linda Blair

Linda Denise Blair (born January 22, 1959) is an American actress and activist. Her portrayal of Regan MacNeil in the horror film The Exorcist (1973) established her in popular culture and as a scream queen, earning her a Golden Globe and an Academy Award nomination. She reprised the role in two sequels: Exorcist II: The Heretic (1977) and The Exorcist: Believer (2023). Blair has starred in several television films, including Born Innocent (1974), Sarah T. – Portrait of a Teenage Alcoholic (1975), and Stranger in Our House (1978). Her role in the musical film Roller Boogie (1979) brought her a...

Depressing career

Ryder Rude

13:11

i read that people died making the Exorcist. so maybe she was affected too

@Slereah sad..

it seems she did a lot of horror movies

Repossessed

Mr. Feynman

@Relativisticcucumber like the one at the end

qwerty

im now veering quickly towards team "category theory isn't helping" lol

Mr. Feynman

What about "it helps but it's not essential"?

qwerty

im discussing the definition of a morphism and being offered a choice between standard pedagogy which apparently is "lies to children" and "the correct rigorous answer"

Slereah

Category theory is good but it's more of an end of learning bit than a beginning

qwerty

13:34

@Slereah insert pun about arrows and arrow of time

the concept of arrows are vexing indeed

ACuriousMind

@qwerty "Morphism" is just a fancy name for function

qwerty

@ACuriousMind "lies to children", apparently.

Slereah

@ACuriousMind the categorists will not approve of this

ACuriousMind

I mean...from the viewpoint of category theory, yes, kind of, but if you're not explicitly doing things in the categorial viewpoint it's just true

qwerty

>"a morphism is a function that preserves some kind of structure" ---- this is the heuristic, non-mathematically rigorous definition that was used historically and that we tell students

>"a morphism is an arrow in a category" --- this is the modern fully rigorous definition. It turns out that to make the concept general enough, we are forced to consider things more general than functions

ACuriousMind

13:42

@qwerty I know of no "morphism" in the context of introductory differential geometry that's not just a function

Ryder Rude

i doubt these two ideas are related. As in, the morphisms in Cat theory do not need to preserve any structure. They're just functions

but in non cat theory contexts, whenever we use morphism, we have to preserve some structure i guess

e.g. homomorphism, isomorphism

ACuriousMind

sure, if you're doing the abstract kind of differential geometry like the nLab does where you want to have generalized smooth spaces instead of just manifolds, then this becomes relevant because many of the non-manifold smooth spaces are not "sets with extra structure" so the morphisms involving them are not maps on sets obeying some extra properties

but manifolds are sets with extra structure and the morphisms between them are functions between those sets obeying certain properties

If your goal is "I want to learn differential geometry", then I absolutely think it's not useful to bother with category theory at the start

Slereah

question is, can you do GR on a non-concrete smooth space

Although I'd suspect that any finite dimensional smooth space is concrete

Ryder Rude

i think homomorphism is the general term for structure preserving functions. other structure preservations r special cases of homomorphisms

but diff geo instead uses the term homeomorphism for some reason

maybe cuz homeomorphism is a special kind of structure preserving map. So they invent a new name

Slereah

homeo is not homo :p

Ryder Rude

13:48

in general, one can also have homotopy maps

@Slereah what is the difference??

ACuriousMind

@RyderRude Homeomorphism is not a differential geometry term, it's the name for a isomorphism of topological spaces.

Ryder Rude

@ACuriousMind yes. That's what i meant

ACuriousMind

it's not what you said

not at all

Ryder Rude

Why don't they call this a homomorphism

@ACuriousMind sorry. I thought topology was part of dif geo, sort of

qwerty

@ACuriousMind culprit mathematician: "the problem is that I can't tell qwerty the traditional pedagogy because she will complain I'm lying if I give anything other than the rigorous definitions". I reserve my own opinions, I'm just going to read and make notes I guess >.<

Ryder Rude

13:50

but i see ur distinction. Maybe u use the term diff geo only when u have an atlas

ACuriousMind

@qwerty Interesting reply, but the category-free version is perfectly rigorous. There's a difference between not being rigorous and not using category theory :P

Slereah

Differential geometry is when you have a differential structure

As the name implies

Ryder Rude

yes. I wasn't precise enough. But my question is : can we say that homeomorphisms is a kind of homomorphism for topological manifolds

ACuriousMind

@RyderRude You're just throwing around the words here and it's evident you don't understand their respective definitions. Homomorphisms are not isomorphisms, in general.

Homeomorphisms are the isomorphisms in the category of topological spaces

It's the continuous maps that are the (homo)morphisms in that category

qwerty

the two definitions feel totally contradictory, but apparently this is ok >.<

Ryder Rude

13:53

@ACuriousMind but that's what I said i suspected. Homomorphism seems to be the general term for structure preserving maps. And homeomorphism a special case of structure preserving maps. In general, one can also have homotopy of topological manifolds

Slereah

If you want a homomorphism that isn't a homeomorphism in topological spaces, just pick a projection

Like the map from a space to a point

Ryder Rude

I wrote my suspicion here chat.stackexchange.com/transcript/message/66554463#66554463

ACuriousMind

@qwerty which two definitions?

Ryder Rude

is homotopy the most general kind of homomorphism for topological manifolds? @Slereah

Slereah

no that is another word yet :p

Apr 19, 2020 at 6:41, by Slereah

Between homology, homothetie, holonomy, homotopy and homomorphy

It's a tricky thing

qwerty

13:55

@ACuriousMind whether a morphism is a special kind of structure-preserving function; or an abstracted generalisation of a function

Ryder Rude

@Slereah i thought homotopy is a structure preserving map..

ACuriousMind

@qwerty It can be both, depending on the category!

qwerty

this seems like overloading terminology lol

ACuriousMind

In the category of sets, the morphisms are just functions between sets. In any category of "sets with structure" (like manifolds), the morphisms are special kinds of functions preserving the structure. In a category that's made out of things that are not "sets with structure", the morphisms are more generalized objects

Ryder Rude

but homotopy only seems to preserve homotopy groups. I'm not sure if it preserves topology in any sense

while homeomorphism preserves topology

Slereah

13:56

Math uses a lot of similar sounding words

u must learn them

ACuriousMind

@RyderRude It's not. Please stop throwing around random terms.

Ryder Rude

i have learned these... I'm just trying to organise them

@ACuriousMind it preserves homotopy groups

Slereah

the typical example of morphisms that aren't functions in category theory are when the category is a partial order

ACuriousMind

@RyderRude No, what you mean is a homotopy equivalence, not a homotopy.

Slereah

Then the morphisms are just an order relation

Ryder Rude

13:57

@ACuriousMind yes

i understand the distinction

homotopy is defined between curves on the same space

while homotopy equivalence is between different spaces

Slereah

Like you can treat $\mathbb{Z}$ as a category, in which case the presence of a morphism between two elements means that they are ordered

Ryder Rude

i meant the latter

naturallyInconsistent

@Slereah one can always insist not to learn them.

Ryder Rude

@Slereah yes

Slereah

@naturallyInconsistent yes, but then you'd be @RyderRude

qwerty

13:59

@ACuriousMind thanks, I'm copying this to my notes xD

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