The h Bar

Relativisticcucumber

01:01

@Allie i think my issue is not the chemistry -- it's the physics. i get the chemistry explanations but they arent really suitable to me, but i think the discussion w acm gave me a picture that makes sense. i think this should be viewed through the lens of mild stat mech.

i mean in general this is smth that i experience. i like chemistry but the approach the subject takes is entirely lacking in any explanatory power to me. this is kind of what drove me to physics. now i think i wanna work my way back through some chem now that i have a physics mentality, hence my condensed matter fetish. i also plan to explore inorganic chem soon but i think ill never grow out of despising organic chem :PPP

recently i am developing a like for metals but my interests ebb and flow so who knows where ill end up

Silly Goose

Can the representation theory of pure translations of the Poincaré group be rigorously purely determined by Schur's lemma? Since $\mathbb{R}^{1,3}$ is abelian? I am wondering why Weinberg does not come out and use Schur's lemma if so.

Relativisticcucumber

01:29

how does doping semiconductors allow for greater control over conductivity? i get that if one introduces an impurity via doping, it introduces a new atom, say per every 1000 atoms in the lattice (or something like this), and if those atoms each have one more valence than those in the mother lattice, they can be added [...]

[...] to the conduction band. but the thing is i dont see how this is "control" or "tunability" of conductivity. the dopants are not something we can turn on or off, they are just there, so it seems to me it's effectively like using a slightly more conductive metal to begin with. so i feel like im missing the point of why doping is so useful?

User1865345

01:59

@Allie 👏🏻👏🏻👏🏻 . Happy holidays. Enjoy your new books.

I love Dover.

Silly Goose

02:10

does Schur's lemma fail in the case I am considering? I think the translations should not be trivially represented...

Allie

@Relativisticcucumber yeah bc ur not a chemist

@User1865345 :3 its a really nice book

User1865345

02:23

@Allie don't break the spine of the book. 😅

This year I learnt a hack how to avoid spine crack of paperbacks:

How to keep your book spines from cracking

►

Btw @Allie, I noted that big fat red book in the snap! Did you read that whole book?! 🙄🤯

I have not encountered such big books in my life though

Anyways happy reading!

Allie

02:46

yes thats one of the books!

im like 30% through it

Silly Goose

03:09

I am confused about finding the unitary irrep(s) of the Poincaré translations. Naively, we have at the algebra level $[P^\mu, P^\nu] = 0$. So, WLOG in any representation space, we can work with the simultaneous eigenstates $\pi_*(P)^\mu \Phi_p = p^\mu \Phi_p$ where $\pi_*$ is a Lie algebra representation (sub-asterik since it is "induced" by a Lie group representation).

Now, suppose this representation is $\pi: \mathbb{R}^{1,3} \to U(\mathcal{H})$. Then, (1) $\mathbb{R}^{1,3}$ is abelian, so (2) $U(\mathcal{H})$ is one-dimensional by Schur's lemma. But, this is contradictory with the actual irrep of translations $\exp(-iP^\mu a_\mu)$, which is infinite-dimensional. And, Schur's lemma has an infinite-dimensional generalization, so I am not sure what the problem is.

Also, how do we prove that the actual irrep is unique (up to isomorphism)?

User1865345

@Allie great job 👍🏻

Now that I remember I had seen Organic Chem books this big. In statistics though, there doesn't exist one 😅.

Silly Goose

03:57

The only thing I can think of is that indeed the irreducible representation of $\mathbb{R}^4$ looks like $\pi(a) \sim e^{a}$, which can be naturally included into $U(\mathcal{H})$ by tensoring with the identity $\mathbb{I}: \mathcal{H} \to \mathcal{H}$.

Jakobian

04:17

@ACuriousMind hi, could you look at this chat as a moderator? See channel description and owner's profile description.

Allie

@Jakobian Clearly someone who is really not mentally well. I hope they find help but this is not appropriate for the site. at least in my opinion

its concerning tbh

Jakobian

04:32

Sure, let's avoid discussing it though

User1865345

Hmm.

Allie

ok i would delete my message if i could

meow hi @User1865345

User1865345

Good that you brought light to this. Hope site mods would take care of that.

@Allie I see no problem with what you said.

@Allie Hey.

End this discussion here as Jakobian noted. We cannot do anything if they have psychological issues. Nor should that get reflected in an inappropriate way directly or indirectly here.

John Rennie

05:46

@SillyGoose I agree with Ghoster that it's unclear what you are asking. If you were specifically asking about phonons then that would make a lot of sense as you'd expect the phonon (quasi) quantum field to reflect the symmetry of the lattice.

You say in a comment that you're asking about "generic quasi-particles" but that's rather vague.

qwerty

06:32

oh no this answer seems to invoke diffeomorphism invariance to explain why GR is sensitive to absolute energies but not QFT physics.stackexchange.com/a/201151/92181

John Rennie

06:49

I thought that was a very good answer.

I still maintain that question is not a duplicate :-)

qwerty

I see you are not in agreement with ACM and many others in this chat about the invoking of "diffeomorphism invariance"

John Rennie

The details of that argument are way above my pay grade :-)
I'm just a colloid scientist who studies GR for fun. I spent most of my career studying soap not spacetime.

(and ice cream - that was fun :-)

3

User1865345

07:06

I have never read such a rigorous description of ice cream. Wow!

John, you would have a nice CV with that quote only: "I'm just a colloid scientist who studies GR for fun."

Only another guy could beat that:

in Ten fold, yesterday, by User1865345

Survived WW1, did maths, defeated Grim Reaper till 110, wrote papers till 103!

Vietoris topology was named after that centenarian, for context.

John Rennie

@User1865345 Unilever makes millions of pounds selling ice cream. Their research budgets would make academics weep :-)

User1865345

Lol. True.

qwerty

I may need a few months to think about it but i have a feeling that a maths phys answer would tie in with math.ucr.edu/home/baez/torsors.html

@JohnRennie I'm reminded of the biscuit dunking formula nature.com/articles/17203

Jakobian

@User1865345 I'd be surprised to learn that Vietoris topology appears in physics, outside of the case when its already covered by Hausdorff distance between sets

(does it? I don't know)

User1865345

No idea about physics; I do math stats :-)

I covered it while studying general topology for stats.

John Rennie

07:19

> “Dear Sir, I think there is something wrong with your biscuit dunking equation. Please send me some biscuits for noticing this. Chao Quan (aged 12).”

That student will go far :-)

User1865345

@JohnRennie 🤣👏🏻

Jakobian

@User1865345 Ah. I do general topology. That's interesting too though, that Vietoris topology is useful for stats

User1865345

Great @Jakobian. You can't imagine where some maths creep up in stats. Lol.

John Rennie

One of my colleagues at Unilever did her PhD on the crystal structure of chocolate.
When I asked her is she liked chocolate she said "I did before I started the PhD".

User1865345

Of course lots of people do stats without proper maths training. No problem. Applied stats is whole different stuff.

Jakobian

07:21

By stats you mean things that deal with actual data, or something like heavy probability theory? I assume the former

User1865345

@Jakobian for me, stats is mathematical statistics. My research area is limited to that.

But colloquially stats mean applied stats.

Practitioners are applied statisticians.

Jakobian

Yeah. How much theory does that actually cover?

User1865345

@Jakobian I have gone to the fringe of differential geometry, combinatorics in particular. Personally I ventured to Lie groups for invariant results.

@JohnRennie lol.

I have also studied analytic sets. This is important for we need selection rules to study minimal sufficient results.

Jakobian

That makes sense, since I imagine you'd be mainly concerned about measures on the standard probability space

so Borel sets on a Polish space, and maybe more further analytic sets may be of concern

User1865345

We are not taught with full force. For example, I cannot say I even know full topology. Never. Not that we needed that in first place.

Jakobian

07:28

yeah, I cannot say that either
> I cannot say I even know full topology

User1865345

I read Munkres till paracompactness. Metrization concepts were more then enough. In fact, Ash's book has enough materials as appendix for our relevance.

@Jakobian I meant to say, you are way more well trained as mathematicians.

Jakobian

aren't you a mathematician yourself

Tobias Fünke

morning

User1865345

Morning @TobiasFünke.

@Jakobian I wish I were. That would have saved time. I am trained as complete statistician.

Jakobian

ah, you mean by training

User1865345

07:31

Every now and then. I do have problems with basic maths stuff. But I can say that I do know a lot maths than say my other colleagues in stats

Jakobian

@User1865345 I can't blame you since, all those properties relating to coverings of spaces combine in complicated ways

User1865345

Hmm

There is also the application of vector lattices when I went to study comparison of statistical experiments.

This is a new field still. Just developed after 2001 or so.

Jakobian

I see, I think that's more what people who do number theory study

User1865345

Point is there is no limit to how maths can come out of nowhere in hardcore stat.

@Jakobian 😅

And then there is category theory. They have some applications in machine learning. But this is way beyond my research area. Even beyond the grey zone.

@Jakobian yes.

Also you have ergodic theory. Those who do semi parametric stuffs as well as stochastics deal with that.

There was this one off opinion by Hopf:

in Ten fold, Dec 9 at 7:08, by User1865345

1

Q: When did E. Hopf say "ergodic theory is statistics and statistics is measure theory"?

In the archived version of Kolmogorov's Foundations of the Theory of Probability, at the very end of the book, p. $84,$ few books have been listed, one being E. Hopf's Ergodentheorie, where it is mentioned Measure theoretic viewpoints are preferred over topological ones throughout because, as th...

I don't agree with him fully though. He never would have been able to see the full potential of statistics beyond measure theory in his lifetime.

Jakobian

That's okay, you shouldn't agree with someone based on being established in a given field alone. Arnold had similar takes about math and physics

User1865345

07:46

Vladimir Arnold?

Can't remember where and how I heard his name first. But he did have some opinions re that. Good, Jakobian, you reminded me of him.

Jakobian

@User1865345 I believe they use category theory in cluster analysis? I'm not completely sure

User1865345

I wish I could provide you some insight here. I could hardly have knowledge of machine learning areas.

I have worked extensively with cluster analysis though.

Mind it. I am not saying category theory is known to machine learning researchers only. We math stats guys also do deal with that (not me in my research though)

Jakobian

@User1865345 I heard of him first from my physics professor at university. He did contribute some cool fields like topological Galois theory though

User1865345

There is literally a monograph by Nikolai Cencov who employed the language of category theory for explaining statistical decision theory and optimal rules.

@Jakobian I see.

Jakobian

@User1865345 One of my professors specialized in ergodic theory. I would say that, a lot of researchers when they make general claims like these, speak from their own experience alone, and should be taken as such - from the perspective of what Hopf studied it appeared to him that ergodic theory is statistics, and that statistics is measure theory, whetever this is true or not is another issue

User1865345

07:55

There you go. There is a discussion at MO re deep learning:

12

Q: Soft question: Deep learning and higher categories

Recently, I have stumbled upon certain articles and lecture videos that use category theory to explain certain aspects of machine learning or deep learning (e.g. Cats for AI and the paper An enriched category theory of language: from syntax to semantics). On the other hand, people have looked int...

ct.category-theory higher-category-theory computer-science machine-learning homotopy-type-theory

Jakobian

there might be some truth to it though in the sense that the fields overlap a lot when it comes to things that Hopf studied or known at the time

User1865345

@Jakobian exactly. Agree completely.

@Jakobian hm.

Madder

12:21

so two comments
1) what happened to the $k_B$ factor at $TS$?
2) so does this mean that the derivation i provided wrong? it still did not answer my question about if the sum is writting as a function of S(E)
3) is my derivation wrong?

User1865345

12:54

I’m a 95 year old mathematician what can I say. Have. 1 year left according to doctors. — Raynard Bond 1 hour ago

in Ten fold, yesterday, by User1865345

We have already a new user following the same path traced by their compatriots using ai, claiming to be "95 year old mathematician" answering convoluted neural network problems in CV and answering geometry problems in Maths.

Sine of the Time

@User1865345 🥺

User1865345

Don't get empathetic for this. I wish it were a reality.

Already unaware users are showing sympathy. 😅

Sine of the Time

flag for mod attention if you suspect it's AI generated

User1865345

Already done yesterday. This is a big phenomenon.

21

Q: Reporting the recent infestation of new users resorting to gen AI targeting STEM communities

Of late, there has been a surge of new users who venture to various communities and answer at a pretty staggering rate with apparently good looking posts, majority of which would be long and heavy with mathjax formatted equations. The initial targets were: Cross Validated, Mathematics, Maths Over...

discussion abuse generative-ai

Not everyone can be Vietoris. 🙂

ACuriousMind

@Madder 1. I did it in $k_B = 1$ since I didn't want to track constants :P 2. & 3. depends on what you think you derived - your sequence of equations is certainly true, but my derivation shows $F = E - TS$ which yours doesn't as the total $S$ (in contrast to the $S(E)$) doesn't even appear in your equations

Madder

13:09

But that means the formula of the Free energy is always given in the units of $k_B=1$?
Second of all, exactly. How would you justify the transition from $S(E)$ to $S$?
beccause that is what the authors did. Honestly that was the crux of my question.

For me,only $S(E)$ makes sense in that situation...

ACuriousMind

@Madder I just used the general definition of entropy (Gibbs entropy) for a probability distribution. (See also this answer of mine for the relation between Gibbs and Boltzmann entropy)

for the units - the $k_B$ would be the factor between $\beta^{-1}$ and $T$. In my units, $\beta^{-1} = T$ but if you have non-trivial $k_B$ it's a factor in there.

Madder

I will take a look.

ACuriousMind

13:26

@qwerty The answer is correct in everything except the reference to diffeomorphism invariance - I've posted an answer explaining that.

Jakobian

14:11

@ACuriousMind hi

I've messaged you above about something that might require moderator attention in case you've missed it

10 hours ago, by Jakobian

@ACuriousMind hi, could you look at this chat as a moderator? See channel description and owner's profile description.

thank you

ACuriousMind

14:34

@Jakobian Instead of a chat ping, you can raise a custom moderator flag on any message in problematic rooms if you think they need moderator attention. Chat pings aren't really reliable (even though I frequent chat a lot).

Jakobian

@ACuriousMind ah. I didn't realize that there is such flag system, I was only aware of the normal flags

I'll do that next time

Slereah

15:07

I think it's time

samuel-lereah.com/articles/Mathematics/categorical-cheat-sheet

bit sick of the lack of proper proof ressources for category theory :p

User1865345

@Slereah It's a cheat sheet, after all.

And you would expand that.

So a handy and useful one.

Slereah

Like most of math category theory is secretly like 10 theorems used most of the times but that people do not write down much

User1865345

Hm

@Jakobian good riddance

naturallyInconsistent

16:20

@LeakyNun you have made many diagrams here. They only demonstrate that you are not understanding the topic you are trying to discuss. Even in the simplest case of strictly linear Hubble stretching by a fixed constant, the worldlines you should be getting are curves. You have all of them straight lines. They are all too wrong to begin discussing.

@Relativisticcucumber The quote is true, and it is a general statement about the validity and nuances of the core v.s. valence approximation. It is basically independent of the metallicity of the system, i.e. the staircase has nothing to do with the validity and general applicability of the statement.

@SillyGoose That's nonsense. You can almost always get a reasonable approximation by starting with classical concepts, and then when it comes to inserting the statistical distribution function, try out both FD and BE cases, and viola, a tolerable semi-classical approximation.

uugh im too tired to reply

sneeppuuu first

Relativisticcucumber

16:43

can i say that D'Alembert's principle is just newtons second law + the definition of momentum or is there smth im missing here?

@naturallyInconsistent well yes i can see that the statement that only the valence matters can be true regardless. i guess it's a seperate question of why this staircase shape is formed? i mean it seems to imply that there is some nuance beyond the valence number, so im curious about that. i would expect it to be a vertical line not a staircase

ACuriousMind

@Relativisticcucumber d'Alembert principle is annoying - it's not meant to be a trivial consequence of Newton's law because it considers only the "applied" forces, not all forces. See e.g. Qmechanic's answer here

fwiw I don't find it particularly interesting to worry too much about it in any case since no one does mechanics in terms of Newton's laws anyway once you have access to Lagrangian or Hamiltonian mechancics :P

Relativisticcucumber

ok ill look thanks

separate q, this ashmerm excerpt says that we need to maintain charge neutrality even in the case of having defects. the specific defect being discussed here is a vacancy point defect i believe. so what i dont understand is why there would ever be a case irl where the defect is that the ion core is missing but the electrons are present. id imagine that it's actually like the entire atom is missing, which is no issue for maintaining charge neutrality ?

@ACuriousMind ok now i see. that is interesting that this principle doesnt always work :P

ACuriousMind

17:01

@Relativisticcucumber I'm not sure what the question is - all the text is saying is that they are imposing as a constraint that they will only consider such cases where the net charge is still zero

if you agree that most practical situations will fall under this constraint, all the better :P

but I think in reality such defects are more variedly produced than you might think, e.g. through doping

Relativisticcucumber

hm well i was concerned that i was misunderstanding the nature of the vacancy defect

but it seems that indeed the entire atom should be missing

bc i think doping isnt a vacancy defect, right? it should be a defect related to adding a different atom? im not sure what this defect is called

maybe insertion

ACuriousMind

sure, "vacancy" means by definition more or less that the atom is missing

and yes for vacancies the text essentially just means that in some sense the "entire atom" has to be missing, or, if only the ion is missing, somewhere else an ion of the opposite valence has to be missing to restore overall charge

Relativisticcucumber

oh but maybe it's possible that you have like two embedded lattices / a lattice w a two point basis, one anion one cation and only one of those is missing

so maybe this is what ashmerm is banning

@ACuriousMind yeah i was wondering tho if its even possible for just the ion to be missing. that would be of interest to me :P

Silly Goose

17:24

Why is the way that $R^4$ acts on elementary particles $\exp(-iP^\mu a_\mu)$?

Tobias Fünke

sorry, what? :d

ACuriousMind

@SillyGoose You will have to be a bit more explicit with your question, since we generically write the representation of any Lie group as $\mathrm{exp}(-\mathrm{i}T^a c_a)$ for generators $T^a$ and numbers $c_a$.

Silly Goose

17:40

@ACuriousMind is this what is being done in weinberg’s exposition of wigner’s classification?

also, isn’t that only valid if $\exp$ is surjective?

ACuriousMind

I do not have Weinberg memorized and I'm not at the location where my copy of it is :P

@SillyGoose you can do it in a neighbourhood of the identity anyway

it's just how physicists do Lie theory, you should've seen this elsewhere already

Silly Goose

But isn’t it wrong?

The exp map for $R^4$ seems supremely not surjective but maybe i am mistaken

ACuriousMind

@SillyGoose it doesn't need to be surjective, as I said, it suffices to do it in a neighbourhood of the identity

Silly Goose

how do you extend it to represent the rest of the group?

ACuriousMind

depends on what statements you prove with it exactly :P

Silly Goose

17:49

well say i want to produce the concrete irreps

ACuriousMind

often it suffices to argue that any neighbourhood $U$ of $g$ can be turned into a neighbourhood $g^{-1}U$ of the identity, then you do the stuff with exp, then you translate back

Silly Goose

ACuriousMind

@SillyGoose There you have the general Lie correspondence that homomorphisms of Lie algebras are in bijection to homomorphisms of the corresponding Lie groups. Physics texts rarely discuss this in terms of Lie's theorems but just by exponentiating the algebra maps, but the correspondence holds in general with no constraints on exp.

@SillyGoose And in this case I see no issue with this at all - definitely you can form the r.h.s. of 2.4.26 for any value of $a_\mu$. What is the alleged problem?

Silly Goose

@ACuriousMind hm but isn't it more subtle than this? for example, one want to work with simply-connected Lie groups as this gives you the bijective correspondence

ACuriousMind

@SillyGoose ...and do you think $\mathbb{R}^n$ is not simply-connected???

Silly Goose

17:54

@ACuriousMind im just saying bc you said "the general..."

ACuriousMind

@SillyGoose I just omitted the "simple-connected". Of course in the non-simply-connected case you have to check if the map actually descends to a map of the Lie group you're looking at - or, as in every single instance in QM it doesn't matter because you're after projective representations anyway.

but again, in your concrete case there is no issue at all, the r.h.s. is well-defined for all $a_\mu$ and so this defines a full map $\mathbb{R}^4\to U(H), a_\mu \mapsto \exp(\mathrm{i}P^\mu a_\mu)$, what is the problem supposed to be?

Silly Goose

18:10

@ACuriousMind i think i do not understand what is meant to be emphasized by writing 2.4.26

Also, can't we immediately write down what the irrep of $\mathbb{R}^4$ is by Schur's lemma?

unless we are not interested in the irrep(s) of $\mathbb{R}^4$, which would be another one of my confusions

ACuriousMind

@SillyGoose Any irrep of $\mathbb{R}^4$, as it is an Abelian group, would be one-dimensional and supremely boring.

Silly Goose

Yes

Slereah

don't you be dissing Abelian groups

Silly Goose

well i think i do not understand what needs to be done to classify the irreps of a semidirect product. i think this might be my problem.

ACuriousMind

You are (presumably) looking for irreps of the Poincaré group, which is a semi-direct product $\mathrm{SO}(1,3)\rtimes \mathbb{R}^4$

an irrep of such a product is not necessarily an irrep of the factors (just like with a normal product)

@SillyGoose well, what Weinberg is doing "by hand" is essentially the theory of induced representations of Mackey

The theory-building necessary to explain in general what's happening and why it works is a bunch of math you will not find in any physics text :P

Silly Goose

18:14

@ACuriousMind is there a directionality with $\rtimes$?

the wikipedia has it the other way round

Slereah

@ACuriousMind do you mean the classification using little groups thing

Silly Goose

or are both directionalities okay but you define a different semidirect product

Slereah

it's in Sourieau's book on mechanics

Silly Goose

@ACuriousMind i do not see how it even works by weinberg's exposition :P. when i first read through the section, i thought it was a proof. looking at it now, it seems more like a sketch of conceivability, not anything near a proof. but is it actually a proof?

Also, is it generically true that the universal cover of a semidirect product is the semidirect product of the universal covers?

Allie

18:41

meow

Tobias Fünke

hello

Allie

19:09

Hi!!!

I have a question that ive been confused about. im sure it would be answered if i continued reading ashcroft which i will, but i do want to know now just so i can understand my research better

So regarding the Brillouin zone, I'm a little confused on how it contains all the info of the system

When we talk about the Brillouin zone we are talking about the fourier transform of the wave function right?

if so, for a periodic system, wouldn't the fourier transform be discrete?

Tobias Fünke

mhmh

Allie

is that a yes to me? sorry

Tobias Fünke

no

Allie

ohh my bad

Tobias Fünke

I am trying to make sense of what exactly you mean

Allie

19:19

sorry!

so backing up, is reciprocal space just the fourier transform of the wave function?

Tobias Fünke

no, no problem

nah, that does not really make sense

the BZ is the Wigner Seitz cell of the reciprocal lattice

Allie

yes that makes sense

Tobias Fünke

good

Allie

but, when they say "all the information is included in the BZ". i assume youve heard something like that before?

Tobias Fünke

yes

okay, give me a minute

Allie

19:21

sorry

meoww

Tobias Fünke

I can do that here only in a rough manner, but the basic idea (at least this is how I see the things) is this:

a) You impose BvK (periodic) boundary conditions, this gives you a set of $k$ vectors. b) you prove Bloch's theorem, showing that the (lattice) periodic Hamiltonian commutes with all lattice translation vectors. The latter are unitary, and the eigenvalues are directly connected to the aforementioned $k$ vectors. c) You want to label the eigenstates of the Hamiltonian in a neat manner. Usually in QM you probably would label eigenstates of a set of commuting operators by their eigenvalues

and a possible degeneracy index. Here, however, you notice that you can choose a set of $k$ vectors which determine the eigenvalues of the translation operators. This set of $k$ vectors is your 1BZ

Allie

hmm im gonna eat lunch and think more about that. thank you bestie!

Tobias Fünke

ok

...let me add: for a solid containing of $N$ unit cells, you need $N$ such $k$ vectors which make up the 1BZ. It does not matter which you choose, however it seems extremely convenient to choose these as symmetric as possible, and with the smallest absolute value. For example, in 1D for a solid of length $L=Na$, $N$ number of unit cells and $a$ unit cell length/volume, all wave vectors are of the form $k=\frac{2\pi}{N}n$, where $n\in \mathbb Z$.

A "good" choice for the 1BZ would be $n=-N/2,-N/2+1,\ldots, N/2-2, N/2-1$.

sure, let me know if still something is unclear or so

Tobias Fünke

19:49

perhaps the condensed matter youngsters here have a better explanation :p

ACuriousMind

@Slereah yes

@SillyGoose the notation isn't unambiguous anyway since the semi-direct product depends on additional data, there is no "the" semi-direct product

Relativisticcucumber

20:10

i have a follow up ab that ashmerm picture. i have found an example where it seems charge would not be conserved unless im misunderstanding. usually when we invoke some constraint, it's a constraint that we expect to be obeyed, but i just dont see the rationale behind this constraint that charge remains neutral. [...]

[...] consider the examples below where in a) Mo substitutes for S and in b) S has vacancies. surely here we have disturbed charge neutrality and these seem to be illustrative of very common defects

Silly Goose

20:21

Hm, I am trying to spell out some details in Weinberg's exposition of Wigner's classification.

I am at the point where I am just computing what an arbitrary homogenous Poincaré transformation does to a simultaneous eigenstate of the $P^\mu$ in the representation space.

$ \begin{aligned} \pi(\Lambda)\pi(\Lambda)^{-1}\pi_*(P)^\mu\pi(\Lambda)\Phi_{p,\sigma} &= \pi(\Lambda)\text{Ad}_{\exp(c_aT_a)}\pi_*(P)^\mu \Phi_{p,\sigma} \\ &=\pi(\Lambda)\exp(c_a\text{ad}T_a)\pi_*(P)^\mu \Phi_{p,\sigma}\end{aligned}$

Here $T_a$ are generators of the Lorentz algebra.

Herr Feinmann

@TobiasFünke you are a youngster technically :P

Silly Goose

Should this computation yield the same result Weinberg obtains?

It seems like I maybe have done something wrong. Because we should get that $\pi_*(P)^\mu\pi(\Lambda)\Phi_{p,\sigma} = \Lambda^\mu_\nu p^\nu \pi(\Lambda) \Phi_{p, \sigma}$

I feel like there should be a simple way to relate what i will call the $\pi$-induced adjoint representation with the actual adjoint representation

bleb wait i have some errors

Tobias Fünke

@HerrFeinmann hehe sure

Silly Goose

Is this an appropriate way to investigate how an arbitrary homogenous lorentz transformation acts on a simultaneous energy-momentum eigenstate?

Since we have the algebra, we can write down concretely $[T_a, P^\mu]$ in general...

Tobias Fünke

20:36

@HerrFeinmann how are you doing? :) have you already had your superconductivity exam? or do you have to study during the holidays? :/

Relativisticcucumber

@TobiasFünke i think you mean do you get to study during the holidays ;)

Tobias Fünke

hehe

qwerty

yawn good morning all

Tobias Fünke

hey

qwerty

20:52

@ACuriousMind thanks ACM! I'll try digest this. I can't help feeling I "could" make it more complicated or deeper, whether or not I "should"...

Tobias Fünke

@Relativisticcucumber the charge neutrality concerns ionic crystals, no? I am by far no expert in this business... is this the case you discuss (the screenshot you attached won't load atm, for whatever reason)?

Silly Goose

21:12

Starting with the Poincaré algebra, is the first screenshot a proof that $P^\mu$ transforms like a four-vector? My logic is that the commutation structure of $\mathfrak{so}(1,3)$ does not allow for $\text{ad}_{J^i}$ nor $\text{ad}_{K^i}$ to act on $P^\mu$ and produce anything outside the vector subspace $c_\mu P^\mu$.

But I sense something is not quite right. Because a four-vector should transform like $\Lambda^\mu{}_\nu P^\nu$ where $\Lambda^\mu{}_\nu := [\exp(c_i J^i+ d_iK^i)]^\mu{}_\nu$, I thought.

Relativisticcucumber

@TobiasFünke i think so because its molybdenum and sulfur, so it is, right?

if i had a dollar for everytime i was confused about smth related to the harmonic oscillator

i could buy at least one vinyl

Tobias Fünke

21:30

@Relativisticcucumber well, I honestly don't know if it is considered a ionic crystal

Wikipedia did not mention it, but rather used the term "covalent bond"

Molybdenum disulfide

Molybdenum disulfide (or moly) is an inorganic compound composed of molybdenum and sulfur. Its chemical formula is MoS2. The compound is classified as a transition metal dichalcogenide. It is a silvery black solid that occurs as the mineral molybdenite, the principal ore for molybdenum. MoS2 is relatively unreactive. It is unaffected by dilute acids and oxygen. In appearance and feel, molybdenum disulfide is similar to graphite. It is widely used as a dry lubricant because of its low friction and robustness. Bulk MoS2 is a diamagnetic, indirect bandgap semiconductor similar to silicon, with a bandgap...

all that being said, I don't know to which extent the rule A&M describe is a "no-go" result

qwerty

@Relativisticcucumber a lecturer once told me that physics is just harmonic oscillators with increasingly levels of complexity lol

Tobias Fünke

"Everything is a harmonic oscillator if you are brave enough"

Relativisticcucumber

@qwerty this is sidney coleman

@TobiasFünke oh yes ur right

hm ok so maybe the logic can be that this criteria applies to ionic crystals and if these atoms are "sticking" to the crystal via ionic bonds, then something without a partner to bind to wont be part of the crystal so therefore we should expect net charge neutrality? and the criteria just doesn't apply to nonionic crystals?

but then it would seem like a crystal like this would be hard pressed to have defects...

but ionic crystals must have defects because that's how they carry current, right?

like by translating the vacancies? so again im lost

Tobias Fünke

check frenkel and schottky defects

Relativisticcucumber

yeah i was looking at these

the thing is

for the frenkel defect i cant understand how this defect would form. this seems to unfavorable to me

i guess the schottky defect makes sense tho

Tobias Fünke

21:44

yeah, sorry, I am really no expert, and the last time I read about this is long ago :/ I cannot help you much, sorry

Relativisticcucumber

thats ok this has been productive. at least the schottky defect gives me an example that makes sense so thanks for that

there must be some info ab this so i will go in search of it

i find these defects quite interesting

Tobias Fünke

yeah. a colleague of mine works on defects for a particular material class. But tbh I've never understood too much haha

Relativisticcucumber

@TobiasFünke :o

wait what do you work on?

Tobias Fünke

I somehow cannot permalink my previous message; wait a sec

Dec 18 at 9:50, by Tobias Fünke

but broadly: (many-body) theory development, DFT, spectroscopy

Relativisticcucumber

wow to me people who can do dft are like hulks

ACuriousMind

21:55

...i.e. people who can barely conceal the anger that's the only thing that drives them?

Tobias Fünke

@Relativisticcucumber have you discussed DFT in some of your lectures already? :)

qwerty

@Relativisticcucumber you're Sidney Coleman ? :p

Allie

meow

Tobias Fünke

OK, I cannot resist anymore. What's the thing with the cats here? :D

Allie

im a cat

qwerty

22:21

@Relativisticcucumber oh you meant it was in a similar vein to "The career of a young theoretical physicist consists of treating the harmonic oscillator in ever-increasing levels of abstraction."

Relativisticcucumber

@qwerty LOL I wish. No, I mean this statement is a SC quote, right?

@qwerty yes yes

qwerty

@Relativisticcucumber xD I got there

Relativisticcucumber

@ACuriousMind lol no I mean they are just so OP they can crush others with their bare hands

@TobiasFünke I wish but sadly no :(

Tobias Fünke

@Relativisticcucumber it is a beautiful theory, and actually very easy to understand

Dec 19 at 22:13, by Tobias Fünke

I really like Jaynes' "Information theory and statistical mechanics" (1957) or Hohenberg and Kohn "Inhomogeneous electron gas" (1964)

Relativisticcucumber

23:09

I went to look up DFT and saw a lecture with this preview and screamed of joy

A man who knows his market

This is the best cat I’ve ever seen

qwerty

@ACuriousMind I think part of the reason I don't feel entirely satisfied with the outline of these answers (I haven't thought deeply about the details) is that it relies on the existence of the Lagrangian formalism. I was thinking that a more "physical" argument could be made that would cover why energy is a torsor in other theories/models, but not in GR, that would make it clear even for situations where we don't have a Lagrangian.

after all, the lagrangian is just a way to get to the equations of motions, right?

ACuriousMind

@qwerty What is your definition of "energy"? :P

@qwerty all of physics is just a model to produce the correct equations of motion :P

qwerty

@ACuriousMind sure, so is there a way to somehow see it from that level?

@ACuriousMind I won't have one that satisfies you :p

ACuriousMind

I'm not sure what you mean

I mean, sure, if you don't accept that heuristically "energy" has something to do with the integrand of the action (particularly because we usually identify the Hamiltonian as the energy) then none of those answers make any sense to you, but then you need to explain what energy is to you

Tobias Fünke

@Relativisticcucumber hehe, enjoy the lecture

qwerty

23:20

Hm, well, I was just thinking about Newtonian physics. we don't need a lagrangian for that; and if you have dissipation or friction there isn't one. yet, we learn in introductory physics that we can measure differences in energy - at least in some forms.

Silly Goose

I am not understanding why in your answer the wanted result is $\text{ad}_A(b) = Ab$ @ACuriousMind

Isn't the wanted statement about $\exp(\text{ad}_A)$?

ACuriousMind

@qwerty well I'm not doing Newtonian mechanics for GR, that doesn't make any sense

and the question was about GR

Silly Goose

Or is it that that action of the adjoint implies $\text{Ad}_\Lambda(b) = \Lambda b$

qwerty

@ACuriousMind well I meant there's still a conceptual difference there

in terms of "what is energy", I guess, to go along with what you're saying

ACuriousMind

@SillyGoose Obviously if the infinitesimal action is $Ab$, then the finite action is $\mathrm{e}^{cA} b$, i.e. the Lie group similarly just acts by direct multiplication on $b$.

I've just shown the statement that it transforms "like a vector" in terms of the Lie algebra

it should be easy for you to see this implies the corresponding statement for the Lie group, again by the Lie correspondence

you seem to continuously suspect something horrible to happen during the passage between the algebra and the group when you ask these finicky questions about $\exp$, but the beauty of Lie theory is precisely that it almost never matters, and that the first-order approximation via the Lie algebra already tells you everything you need to know about the group

Silly Goose

23:30

How does one show this statement without writing down the "model" of the poincaré group?

qwerty

@ACuriousMind also, as you mention, the Hamiltonian does not necessarily have anything to do with the energy at all. So in those cases where it's not related, why should I be convinced that the argument still holds?

ACuriousMind

@qwerty You shouldn't be, and instead realize that there isn't even a good definition of "energy" in the GR context

without a time-like Killing vector you can't really say what energy even is

@SillyGoose How are you supposed to prove anything about the group without knowing how its elements act on each other?

If you don't have access to the affine 5d model, you need to have something equivalent, otherwise you don't even know what group you're talking about

and proving things about a group you don't actually know is of course hard :P

qwerty

@ACuriousMind I have heard this claim in the context of Noether's theorem. However you did not seem to think that the question "GR cares about absolute energies" is ill posed? I also thought this was to do with the energy conditions in GR, and iirc they do not rely on the existence of Killing vectors?

ACuriousMind

@qwerty Do I always have to unravel every question down to its absolute extreme? I just thought Andrew's argument was a neat way to explain the structural difference without generating an entire course on GR, and phrased it in a less confusing way.

yes, of course, the question doesn't really make sense on a formal level - there is no unique notion of "energy" in GR (the energy conditions are usually conditions on what the energy is in a certain specified frame)

but it's pretty clear what caused the original question: It's folklore that QFTs are allowed to renormalize the vacuum energy to whatever they want unless they are gravitational theories, because then the "vacuum energy density" shows up as a cosmological constant (a constant added to the Lagrangian density exactly in the sense of Andrew's and my answer!)

so the question is: Why can GR detect this cosmological constant, and other theories cannot? Andrew's and my answer answer that, instead of going down the rather irrelevant route of whether this constant deserves to be called "energy"

qwerty

mhmm. sorry if I implied anything which irritated you. i wasn't aware of where the question came from, I guess I was thinking about it from a different (non-qft) background.

ACuriousMind

23:46

I'm mean the folklore also holds in a non-QFT version: The cosmological constant is a vacuum energy density at least in some sense, and because it exists you can't just arbitrarily set the energy density of the empty universe as you can in non-gravitational theories because it shows up in expansion/contraction of the universe

qwerty

I guess maybe I should go away and read about Noether's theorem or something else that might discuss the formal issues around defining energy in general and see if it makes it any clearer.

imbAF

Does anyone has any suggestion regarding books about complex analysis ?

ACuriousMind

@qwerty I don't think it will, because the reason we call the cosmological constant an energy density is just because of the structure of the stress-energy tensor: We call the 00-th component of the tensor the "energy density" (regardless of whether or not it corresponds to some globally time-conserved notion of energy) and in a frame where the metric is diagonal, the cosmological constant just directly adds to that 00 component

(and with the inverse sign to the ii-th components, you could just as justifiedly call it a "vacuum pressure :P)

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