i have a follow up to this physics.meta.stackexchange.com/questions/14564/… i mean if i search up "xxx model" on my computer this question does not come up AT all, actually nothing physics related comes up, so how does the accepted answer actually explain the phenomena here?

LOL

"Those who are ready to find ecstasy in the desiccated arms of Heisenberg are the true pioneers of pleasure."

btw what do u all think of my new mascot miffy -- I LOVE HER

awww

@Relativisticcucumber hahahah

@SillyGoose this was so f*** cool hahaha

@Relativisticcucumber I don't know for sure, but maybe the traffic was so high because the question was new and thus popped up more often? Does that sound reasonable?

and one comment also said that it could be that many site views are due to bots or so

@Relativisticcucumber yes, but Ditto was cute, too :d

hi

hello, can someone tell me why my recent question has been marked as dupe?

thanks in advance!

@AntonioDeAngelis Hi Antonio :-)

I didn't vote to close the question so I can only guess what the people who voted to close it were thinking.

But it does seem as if you are using the Boyle's and Charles's laws to get the equation of state PV = nRT and that is covered by the previous question. That's probably why it was closed as a duplicate.

(that's the density in moles per cubic metre)

In the feynman diagrams for bhabha scattering, just by using the feynman rules to write down the matrix element, it is impossible to argue the minus that appear in one of the 2 diagrams

is that accurate?

@JohnRennie Hello! I understand, but I do think it is true that the scope of my question differs from the post mentioned duplicate.

@AntonioDeAngelis I have voted to reopen your question, but it needs two more votes to get it reopened.

@JohnRennie Thank you!

do u all believe that, in principle, it should be possible to re-write physics so that all the entities involved are discrete instead of continuous

like, spacetime being a lattice and all the fields taking values in a discrete set

sure

We can't do a continous amount of measurements

some approximation by a discretization is always possible

yes, that is what I expect. some subtle difficulties do arise but I believe it should be possible

i somewhat want to be able to view reality as a platonic mathematical object

like, a mathematical model that interprets the axioms of physics

but I came across this post curtjaimungal.substack.com/p/… which mentions a lot of difficulties that arise when trying of think of reality as involving the Platonic real numbers

@imbAF no, of course you should have formulated "the Feynman rules" for spinors such that they include this minus sign. But usually we just derive the rules for scalar and then have to add the fermionic minus ad hoc

it is a short article. can you tell me if discretisation of physics solves the problems presented in the article

the current axiomatisation of physics involves the concept of a "manifold". now, according to Loweinheim Skolem, no axiomatisation of a manifold can pin down a unique platonic mathematical entity

What's the problem with the minus sign?

@HerrFeinmann no problem, just for fermionic lines there's this additional minus sign if they "cross" an odd number of times (at least I assume that's what imbaf is talking about)

It comes from the fermion time ordering, doesn't it?

Probably this is what imbaf is asking (?)

Hello again! I'm trying to reopen physics.stackexchange.com/questions/840071/… but I am unsure if it is being reviewed.

And BTW thank you for reopening my post :)

is it enough to note that $C_v$ (molar specific heat) and $\Delta T$ are extensive and intensive, and give units of energy when multiplied, to say that these are conjugate variables?

im just slowly going through this article en.wikipedia.org/wiki/Conjugate_variables_(thermodynamics)

sorry, just specific heat, not molar, i was thinking of the quantity $n * C_v$

got the answer from therein, "The intensive (force) variable is the derivative of the internal energy with respect to the extensive"

Trying to prove that the diffeomorphism group of the infinitesimal disk is GL(n), but it is challenging

Easy to do on the corresponding Weil algebra, but otoh I have to prove that it is true on the dual too

good evening

hello

is it convention for fermionic c/a operators to be written like $\{c^\dagger, c \} = 1$?

I am asking becuase this is opposite of bosonic c/a operators often written with teh annihilation operator first: $[a, a^\dagger] = 1$

@SillyGoose ...why does it matter? $\{-,-\}$ is symmetric in its arguments.

oh right

yeah it does not matter, but I use the same convention as in the bosonic case to not get confused lel

reading this little square here: upload.wikimedia.org/wikipedia/commons/thumb/f/fb/…

it gives the wrong sign for internal energy, do you think it's just convention or smth else?

oh no, the sign only affects coefficients

this is a typo right? $C_m^\dagger$ should have $i \mapsto -i$?

but they did not change the order of operators; does this affect the result?

what do you mean?

the adjoint would swap the order of all operators, no?...

it would swap it to $\sigma^-_m[\exp(-\pi i \sum_j \sigma_j^+ \sigma_j^-)]$ right? and then $\sigma_m^-$ commutes with the exponential since they are different sites for all $j$.

yeah, but what about the sum in the exponential?

I am just asking, I did not think through it. I just noticed that all orders are the same, which I'd guess could affect the sign

$(\sigma_+ \sigma_-)^\dagger = \sigma_-^\dagger \sigma_+^\dagger = \sigma_+ \sigma_-$

ok. then it is a typo

okay cool

@TobiasFünke the important thing is to always use {} for anticommutators and [] for commutators

$[\cdot,\cdot]_\pm$ people are evil

haha

Well, I've used the latter some times :/

it is much easier if you deal with fermions and bosons at the same time

don't judge

@TobiasFünke I've had to :P

hehe

@HerrFeinmann You just use $[-,-]$ for both as the bracket in a super-Lie algebra :P

In that case I'm more into using the same symbol and defining it as $[A,B]=AB-\varepsilon BA$

@ACuriousMind I knew, my gut instincts never lie to me

then you should put an epsilon also as a subscript on the LHS

i.e. write $[]_\epsilon$

or \varepsilon, if you are such a person

:d

I won't abide by such cruelty

Oh, are you one of those $\epsilon$ people?

wait. why does it show 10.1k below my name in the chat? Oo

I'm so disappointed, Tobias...

@HerrFeinmann hehe I swap every now and then

I always considered $\epsilon$ users in books as not caring enough about what they do

ah, is it the cumulated rep from all sites? or what?

xD

Oh, yeah. Here you see it all

thanks

Probably except meta (?)

let me check: $\epsilon$ vs $\varepsilon$

I don't know, I don't use meta

mhmhmh

I think on Meta you have the same rep as on the corresponding main site

My way to go is: $\varepsilon$ over $\epsilon$, $\theta$ over $\vartheta$ and $\varphi$ over $\phi$

@TobiasFünke I'm not sure

okay I agree with you except for the epsilon :d

@TobiasFünke total sum of rep across all sites

@TobiasFünke No, the $\epsilon$ is a function of $A$ and $B$ - $\epsilon(A,B) = (-1)^{\mathrm{deg}(A)\mathrm{deg}(B)}$, where bosonic operators have degree 0 and fermionic operators have degree 1.

@HerrFeinmann per-site metas do not have their own rep

only Meta Stack Exchange has its own rep

Why did you study supermanifolds too?! :P

you don't need to have supermanifolds to have super-Lie algebras

(but yes I know about supermanifolds, too :P)

@ACuriousMind okay, why not :d

i have 8k total rep

no, 9k

why is reality describable using math

@RyderRude Gg

@RyderRude it is unreasonably effective... so to say

@ACuriousMind now answer!

oh that was a genuine question and not an expression of frustration, I misunderstood that, sorry :P

I needed to understand what all the "superfields" business was about, you can't avoid it if you study string theory

@ACuriousMind it was both, my friend

But now I have a reason why!

@ACuriousMind what do you mean for scalar ?

@imbAF I (he) mean(s)... Scalar fields

You learn the basics of QFT with scalar fields before using spinors

@ACuriousMind where did you learn about sueprmanifolds? Papers?

Did you read De Witt? 💀

@HerrFeinmann probably? I don't really remember, I found the concept more of a bookkeeping device than of intrinsic interest

Oh, in a similar manner to category theory?

In the sense that it helped you see the bigger picture

no, not really

the "bookkeeping" nature of category theory is interesting because once you grasp it it applies to a lot of stuff; the applications of supermanifolds are rather specific and few :P

@HerrFeinmann I tried a few times but the book sits in a really weird space where it wants to be more mathematical than the usual treatment but then spends all of its time being not quite rigorous. I don't like it, but I haven't fully read it.

De witt's book is kind of the worst way to do supermanifolds

It's the more traditional way to do manifolds but it's a bit of a hassle

Fundamental reason being that anticommuting coordinates are fake

There's no real space involved

Well, there's a weird thing where "geometry" has taken on multiple meanings in modern mathematics

if I asked you to explain what's "geometrical" about algebraic geometry over arbitrary rings, would it be any more about "real space" than supermanifolds?

the "coordinates" are generalized to being some kind of local trivialization (or local isomorphism to some object the geometrical object is "locally modelled on"), and I think in that sense the anticommuting "coordinates" aren't fake - but they no longer reflect what we used to mean by "coordinate" in a more naive sense, i.e. real variables that describe patches of of space(time)

it's a battle about semantics that doesn't really have an unambiguous solution

Apparently some people call bogolons "Bogoliubons". Tch.

@HerrFeinmann arguably more correct, but 100% less funny

The associated space to the Grassmann algebra has a single point and no other region than that point

Bit short for a geometry

@Slereah The universal moduli space of n-forms is the same

does that mean n-forms are not geometry :P

@ACuriousMind probably not

:)

@ACuriousMind well, bogolons are everywhere, Bogoliubons just on Tinkham

What language are you two even talking? What does "are not geometry" mean?

@HerrFeinmann that's exactly the point of contention :P

you could call it a vibes-based conversation about synthetic differential geometry if you wanted to be all nLab about it

I mean, can you explain in layman terms what "being geometry" means? That the space of n-forms is not determined by the geometry of a manifold or something else?

Well you can generalize some things in geometry to rather extreme measures

By taking advantages of some equivalences

For instance you can always build a space from an algebra

But that space may not have much interesting structure, topologically speaking

A lot of the weirder ones are built from a combination of a normal algebra that gives you a normal space and a weird algebra that gives you just a single point, but that point has attitude

@HerrFeinmann I just responded to Slereah calling the space associated with the Graßmann algebra a "bit short for a geometry". I wouldn't use that phrasing but I understood what he meant in the sense that we say "supermanifolds" are some kind of product between normal $\mathbb{R}^n$s and Graßmann algebras locally and that doesn't make a lot of sense as geometry.

But the notion of "point" he used was the notion of point of some form of synthetic differential geometry (clearly as a set the Graßmann algebra has more than one element)

It's the uuuh

Not quite the spectrum

Well it is the spectrum, but in the categorical sense

As some Kan extension of the Yoneda embedding

If I wanted to explain to you what the notion of "point" used here means or why the universal moduli space of differential forms has similarly only a single point (or perhaps none? I don't remember) I would have to explain all the idea of these generalized smooth spaces which I'm not up to right now, sorry

The same way that the algebra of R is also a single point spectrum wise

All the smooth spaces have at least one point

That is part of the definition of cohesive spaces

One point per connected component at least

yeah, the trivial map, makes sense (for whatever value of "sense" :P)

I mean there are some topoi where it's not true :p

They're just not cohesive

A good idea to look at it is to think about how you can make algebras from spaces

For instance the map that takes a manifold and outputs an algebra of functions on that manifold, ie M → C(M)

There is an inverse process to this

It is called the spectrum of an algebra

And you can do it for a lot of weird algebras

@Slereah you left out the crucial part where you have to say what kind of functions ;)

because this construction behaves a lot differently for continuous, smooth or analytic functions

Same principle tho :p

How can those heads of yours have so much math inside? :P

I sound like a normie now D:

Basic weird example is that you can do it on the integers as a ring

In which case it's a space of mostly discrete points for each prime numbers, except at 0 where its closure is the whole space

@HerrFeinmann is there something wrong with de witt

normie is crazy

@SillyGoose nothing, just that books take more time to read and they pile up

I mentioned De Witt because it's the standard reference about it

MIFFY

wonder what sound a bunny makes

meow