The h Bar

12:55 PM

It's just an exponential decay on two sides rly

1:32 PM

Ugh

Trying to write something about thermodynamics of magnetic system, but I'm not 100% sure what is what

Because it's fucking thermo and as usual they're shy about using k-forms

So they just use all the weird deltas

2:21 PM

I suspect they avoid talking about forms because then they'd have to properly define what space these forms live on :P

Physics Meta

2:34 PM

-1

Q: I Hate this website

It is clear that I cannot ask any problem related questions. I think it is good to ask homework or exam questions as the connection between theory and application can be made clear. I know you lot have made it off-topic which I consider ridiculous. Oh well, I am done venting.

2:47 PM

I love this: damtp.cam.ac.uk/user/tong/tasi.html

It's actually step 3 in How to Become JamalS: The Definitive Guide

aka his recommended resources

3:22 PM

> " Instantons, monopoles, vortices and kinks; These are a few of my favourite things."
-- Maria von Trapp (deleted scene)

Something I've never really understood is in what context you would want to act with a representation of some symmetry group/algebra on spacetime and when you'd want to act with it on the operator fields on spacetime

As in, in what context the objects $\phi'(x')$, $\phi'(x)$ and $\phi(x')$ are useful respectively

because it seems like transforming both spacetime and the field would not be a super useful thing to do

maybe to give an example of what I mean, in DG you either talk about an active transformation being a diffeomorphism of a manifold or you would talk about a passive transformation being a smooth coordinate change, but it would be strange to do both at once

silverrahul

@PhysicsMeta This is true. This website is not very welcoming to newbies. Newbies by the very definition are not very knowledgeable. So, naturally their questions would be a bit all over the place. Just shutting down these questions does not help them much

fqq

4:03 PM

@silverrahul they're literally complaining about not being able to copy and paste homework problems, without any work or conceptual question.

@fqq While in this particular case I agree with you. But in general physics stack exchange can be intimidating for the noob :P

4:16 PM

Ill have to agree to that

this site most definitely comes across as harsh to someone whos 1)new to this site and 2) not at a decent level in physics.

much as I would like for it to be different, I think it's a necessary protection mechanism - we need to filter out people that just want someone else to do their homework or else we'll be overrun with such questions (just look at math.SE, which doesn't consider such questions off-topic and as a consequence consists almost entirely of them)

@ACuriousMind I actually love math.SE I dont see a problem with it?

Also I was being more generic than homework problems

@Charlie physicists are very bad at distinguishing between "coordinate change" and "diffeomorphism" - in mathematical notation, $\phi'(x')$ under some diffeomorphism $f$ is just the pushforward $f_\ast \phi$ and neither $\phi'(x)$ nor $\phi(x')$ are really meaningful

@ACuriousMind On that note have you ever considered writing a short list of "essential math books a physicist should read"?

no because as I keep saying I haven't read all that many books :P

4:31 PM

@ACuriousMind Some work to genius others are simply born :P

I learned most of the math I know from lectures, not books, so I cannot in good faith tell others how to learn them from books

@ACuriousMind Got lecture notes :)

?

for most of them not, no

@ACuriousMind Sniff and here I thought I asked the most important question on PSE

also I don't think the written essence of the notes would be equally good - an essential part of lectures are the problem sets, after all

4:33 PM

:'(

@ACuriousMind do you know anything about irreducibility of polynomials over the integers?

(positions himself to ask math question)

I know some things about them, yes

@ACuriousMind So lets say I have a polynomial $P(x)$. I claim if $P(0)$ is odd then $P(|P(0)|)/P(0)$ has $z$ prime factors and $\Deg P(x) > z$ then $P(x)$ has irrational roots

I'm pretty sure a mathematician wouldn't check for irreducibility like this

I don't quite understand what you wrote - there are two "then" but only one "if" there

@ACuriousMind So lets say I have a polynomial $P(x)$. I claim if $P(0)$ is odd and $P(|P(0)|)/P(0)$ has $z$ prime factors and $\Deg P(x) > z$ then $P(x)$ has irrational roots

@ACuriousMind Fixed

I was doing the Feynman technique of first having a shot at the problem and then looking up how others did it. When I made this

what exactly is the problem you're trying to solve here?

4:43 PM

@ACuriousMind How to find $P(x)$ has irrational roots

Uh, $P(x) = x^2 - 1$ has $P(0)$ odd and $P(P(0)) = 0$ has 0 prime factors, which is certainly smaller than the degree of 2, but it has the perfectly rational roots 1 and -1

Also, what if $P(0)$ is even, or if the degree is not larger than $z$? Are you saying no polynomial with even constant term can have irrational roots?

@ACuriousMind Ah yes, $\Deg P(x) > z > 1$

@ACuriousMind No Im saying my condition only works for odd polynmials

If $\Deg P(x) < z$ then $P(x)$ may have irrational roots or rational roots

then that doesn't solve the problem, does it? :P

@ACuriousMind But its progress :P (and interesting) :P

I'm now more confused about the terminology surrounding symmetry in physics than I was 10 minutes ago

"Internal symmetry" and "local symmetry" mean the same thing?!

4:53 PM

@Charlie Suggestion read a mathematicians book (Woit)

(Though personally havent gone through the entire thing myself)

A mathematical book would help understand the actual concept, not the wild language used in physics :P

@Charlie He actually learned language from both sides physics and math

As our sacrificial lamb you can tell the rest of us if the book was any good :P

I've seen parts of woits thing, I assume we're talking about the same book

@Charlie Yes

The reason I asked about $\phi'(x)$ and $\phi(x')$ above was because I was under the impression that acting "on the fields" (what I used to think was an "internal transformation") and acting "on the underlying spacetime points" were the two ways to act with some symmetry

4:55 PM

I saw a bit of it .. Found it dry

@MoreAnonymous Peter Woit? You're recommending Peter Woit's book?

@Charlie which book you studying from?

@JohnRennie Yes why?

None at the moment, I was just thinking about it

@Charlie no, "internal" symmetry just means a symmetry that is not a spacetime symmetry

it just so happens that these symmetries are usually gauged, hence local

@MoreAnonymous He has only published one book, Not Even Wrong, though he does have various online books in beta form. And I wouldn't recommend Not Even Wrong to anyone trying to learn QFT.

4:58 PM

@JohnRennie No he has one on QM as well

and QFT

but it's easy to write down theories with non-gauge internal symmetries, e.g. the theory of $N$ free scalar fields has a global & internal O(N) symmetry

springer.com/us/book/9783319646107

Ok in that case I am very confused now

Ah, OK.

Woit's mathematical write-ups are usually very high quality

5:02 PM

@JohnRennie why though? I have heard he is quite renowned in the community...

He is of course a bit opinionated on a few topics, but he's a hell of a lot more level headed than, er, someone with the opposing viewpoint

@NiharKarve Did you read it? I read the intro and I was like its too dry for me

*a bit more than the intro

@NiharKarve I believe this is called "damning with faint praise" :P

@NiharKarve I wonder who this opposing person could be :P

@ACuriousMind also since we are on the note of symmetries

@satan29 because Not Even Wrong is not a textbook :P

5:04 PM

@MoreAnonymous The representation theory book? I thought it was great, and the extended version he released recently looks even better

doesn't matter how good someone is at physics, an opinion piece from them is not gonna teach you QFT

@NiharKarve Really gotta have another look at it then

@MoreAnonymous Believe me, there are drier textbooks

@Charlie can you be more specific about your confusion or do you need a while ;)

@ACuriousMind What do you make of Woits proposal:

The conventional argument has always been that the Coleman-Mandula no-go theorem says you can’t combine internal and space-time symmetries in a non-trivial way. Coleman-Mandula does not seem to apply here: these symmetries live on PT , not space-time. To really show that this is all consistent, one needs a full theory formulated on PT , but I don’t see a Coleman-Mandula argument that a non-trivial such thing can’t exist.

Relevant link

math.columbia.edu/~woit/wordpress/?p=11899

?

Also what does Coleman-Mandula no-go theorem apply only for flat manifolds? (I'm confused about this)

5:06 PM

@ACuriousMind ah, in that sense.

@NiharKarve I've become quite rusty at chess up for a game sometime

?

@ACuriousMind I'm trying to pick out a specific point of confusion, I am probably going to arrive at the conclusion that I'm unhappy with what it means to act "on the field" instead of "on the spacetime points"

@MoreAnonymous I have no idea what this is specifically about, but whatever Woit's doing there is not a quantum field theory in the usual sense so Coleman-Mandula doesn't apply

Because I have seen the phrase "acting on the field instead of the spacetime points" called an internal transformation. And while it makes sense to talk about the action of a symmetry group/algebra in some representation on the vectors (aka states), I'm still a bit uncomfortable with the idea of acting with operators on other operators.

Although I now understand that this is something about the state-operator correspondence, and it's just a slight change of viewpoint which involves the commutator rather than acting on a vector

Which then leads on to thinking about the situation where you have both operators and spacetime, i.e. an operator valued field on spacetime. Which is where I've seen the notation $\phi'(x)$ and $\phi(x')$ used to denote the "transformed field" evaluated at $x$ or the "untransformed field" evaluated at $x'$

I'm about to go for a walk so I'll think about it

@Charlie "acting on the field" means it's a transformation on the indices of the field. E.g. if you have a non-Abelian field strength $F_{\mu\nu}^{ab}(x)$, then both the non-abelian gauge symmetry and the usual spacetime symmetries act on the field (via the $ab$ and $\mu\nu$ indices, respectively). The gauge symmetry however is internal in that it doesn't do anything to the $x$.

5:14 PM

I don't really know enough about gauge theory to know how to work with the ab indices

@ACuriousMind Does Coleman-Mandula no-go theorem apply only for flat (Lorentzian) manifolds? (I'm confused about this) Quite often I hear it in the context of quantum gravity but we know the metric is not flat ...

The exact context of where I'm coming from is scalar conformal field theory, although my confusion is more general

Most of my upsetting confusion is an artefact of the fact that I've had to cram a lot of physics in the last 2 years and not really be able to dwell too long on certain basic topics which is now coming back to bite me

@Charlie The transformation behaviour of fields is a priori purely classical. It's one of the Wightman axioms that the quantum representation of symmetries (e.g. the Lorentz symmetry) on the space of states is such that $\rho_C(g)\phi(\Lambda(g)x) = U_Q(g)\phi(x)U_Q^{-1}(g)$,

where $\rho_C$ is the classical representation on the target space of the field, $\Lambda(g)$ is whatever the symmetry does to the coordinates (if anything) and $U_Q$ is the unitary representation on the quantum space of states

the problem is I kind of do get that, if $\phi$ is a scalar field then $\rho_C$ is the trivial rep, $\Lambda(g)$ is the fundamental rep and $U_Q$ is the infinite dimensional unitary rep on the field operators

@MoreAnonymous Wiki pretty clearly says you need a manifold with Poincaré symmetry, no?

5:19 PM

@ACuriousMind Then why do they claim this has something to do with quantum gravity?

I guess if I had to be unhappy with something there, it would be the fact that you act with $\Lambda(g)$ on the spacetime points on the classical side and don't act on the $x$ in $\phi(x)$ on the operator side

I'm going to think about it for a bit, I don't want to waste your time slowly talking my way though the entire topic until I find something specific I don't get :P

Hey all!

@RewCie yo

sup bro?

OwO

@RewCie Do you play chess by any chance

?

5:23 PM

@MoreAnonymous you'll have to ask them that :P

Yes, I do

but, busy.... Engineering Assignments

and all

that sucks

@ACuriousMind Is it a valid physics stack exchange question?

@ACuriousMind What you are doing?

Might just type it down

rather write it on paper and post the photo as question :P

5:24 PM

@MoreAnonymous if you can nail down a specific place where this is done, sure

@ACuriousMind The Woit post for example?

Who among here has the best handwriting here?

@RewCie Not me

need a handwritten note on something XD UwU

@MoreAnonymous Lol, My whole handwriting are short signatures for banks

reserved! XD ~(0^0)`

d(^_^)b

@MoreAnonymous I don't see where in the post Woit claims anything about C-M applying to arbitrary manifolds

5:27 PM

@ACuriousMind He's mentioning it in the quantum of quantum gravity

@Charlie sure, no worries

Today, I tried peanut Butter with bread, let me tell you, it sucks, I'm feeling inflamation in my chest since then... I guess I'm allergic to peanut butter

:-3

@RewCie No the problem is you didn't add jam

@MoreAnonymous Jam with peanut butter?

Yes good ol classic

5:28 PM

Actually I use jam, but no refregrators allowed in hostel

*i

so, tried something new that should work

but, that I don't like it

anything else to be tried with bread?

@MoreAnonymous So? It's a throwaway mention in a blog post, I wouldn't equate that to the claim you seem to be reading into it

@ACuriousMind I'll find a better resource

in particular since this is just supposed to be motivation for his personal ToE :P

(8->)

XD :-) :-D

please stop posting random emoticons

5:31 PM

@ACuriousMind Wasn't this the reason for the rise of supersymmetry? (I might be wrong)

@MoreAnonymous Not really - supersymmetry does evade C-M, but why would that be motivation for its "rise"?

@ACuriousMind Found it

mathi.uni-heidelberg.de/~walcher/teaching/sose16/geo_phys/…

This means that for a unified theory of Gravity and Gauge interactions, we are forced to look
at a loophole of this theorem which is to consider superalgebras or general graded algebras. So
as long as we do not want to give up field theory as a framework, we must consider supersymmetry. Because of this, the theorem was the starting point for physicist to try and implement
SUSY into their theories and the general study of SUSY from a mathematical perspective.

I think this is an ahistorical claim

@ACuriousMind Either way it is a claim of quantum gravity requiring this no go theorem

SUSY arose in early string theory, and people also observed that supersymmetric QFTs had interesting properties, in particular with cancellation of certain diagrams, long before people were doing either of these as "quantum gravity"

5:35 PM

I think there is no dedicated Chat SE Room for Cheems, pepe and Doge Memes, it is a big requirement here.

@MoreAnonymous Well, it's saying a unified QG needs to be a theory for which C-M doesn't hold. That's true. It doesn't claim what you said it claimed, namely that C-M holds on arbitrary manifolds

@ACuriousMind But obviously C-M should be irrelevant in QG since the manifold is not flat

I'd say unified QG should be a theory that also "works" in approximate Minkowski space, so the escape via other manifolds would simply be not enough

I've a question to ask, won't disturb now, please ping me once that conversation ends..

remember, if you talk about C-M at all then you're searching for a quantum field theory

if you're doing string theory or loop quantum gravity or whatever, then C-M doesn't even begin to apply in the first place (it only applies to the low-energy QFTs approximating your non-QFT quantum gravity theory)

5:41 PM

@ACuriousMind I see thanks ...

11:11 Happiness all around

@ACuriousMind What are your views on AI Moderating chatrooms? OwO

What do you mean?

11 moderating chat

AI

What's confusing here?

repeating the phrase does not make it any clearer :P What sort of AI? How would you get it to "moderate" chat? Why would you do that?

Whatever does a moderator does, suppose SE Chat AI is able to replicate that

I mean moderation tasks

5:49 PM

That doesn't really answer my questions - is this a sci-fi AI or one of today's neural networks we glorifyingly call "AI"? what has it been trained on?

bit of Sci-fi, but is a bit implementable using ML Methods

6:14 PM

How do you write the first law of thermodynamics so that it's not a mess for magnetic fields

The thermo people use $$\delta U = T \delta S - p dV + \frac{1}{4\pi} \int_V H \cdot \delta B dV$$

But if I try to use differential forms to make it into not nonsense, I can't just replace everything trivially

The magnetization isn't a real value, it's a field

It's gonna be one of those Legendre bundle vector or something

I'm more concerned what the $V$ on the integral there is

just the full volume of the system?

6:53 PM

@Slereah $dU = T dS - P dB + H dM$ why are you integrating in the last term when it's a differential expression and $dV$ is already in the middle term it's not even a valid differential doing that no

@bolbteppa that's the standard theorem!

The integral is over space, not the Gibbs space

I guess it's a bit confusing to call both V

$dV$ in the middle is $dV$, maybe there should be a prime on the last $dV$ and you are working with averages like $\overline{H}$ but even then it's like an average value over a differential volume because the whole thing is a differential

I think it's the total magnetic work over the thermodynamical system

Magnetization

In classical electromagnetism, magnetization or magnetic polarization is the vector field that expresses the density of permanent or induced magnetic dipole moments in a magnetic material. Movement within this field is described by direction and is either Axial or Diametric. The origin of the magnetic moments responsible for magnetization can be either microscopic electric currents resulting from the motion of electrons in atoms, or the spin of the electrons or the nuclei. Net magnetization results from the response of a material to an external magnetic field. Paramagnetic materials have a weak...

Even the $E$ in $dE = TdS - P dV$ is really $\overline{E}$ maybe you're just doing the analog of writing $d \int E dV = T dS - P dV$

7:13 PM

@bolbteppa :0 magnetism and thermodynamics? thats a weird crossover...

In a sense a part of the origins of the standard model trace back to considering magnetism in thermo :p

silverrahul

@ACuriousMind When i was talking about newbie questions, i was not talking about homework questions. I was talking about questions , where a newbie asks about some topic and the question is along the lines, of how does this work or what does this mean . But obviously, the questions are not of a very high level and there are some flaws in the question. But just closing down these questions, does not help them at all.

@silverrahul but we don't close questions for being "too low level"

silverrahul

That is not the phrasing used, but that is what happens in effect lots of timees