2:01 AM
a priori equal probability distributions are so sus
i mean for practical purposes it's useful but feels so dirty
feels like classical mechanics was just a flavor of physics, but nothing I learned was fundamental in any sense. I guess learning how to learn and think systematically was beneficial..

the principle of least action is a sort of "fundamental" tool :D

can you explain like im 5 what the principle of least action is? we veeery briefly went over it in my mechanics class
$\delta S = 0$ means..?
The wiki page says it's the integral of the lagrangian w.r.t. time for some interval so I guess as that interval becomes infinitesimal the "functional" is 0?

2:17 AM
Consider classical mechanics and a particle system in it. Let $L(x, \dot{x})$ be the lagrangian of your system. Then, the action is the integral of $L$ over a time interval. The equation $\delta S = 0$ puts a constraint on $x(t)$ and $\dot{x}(t)$. This constraint is called the equation(s) of motion. Solving them gives you $x(t)$ and $\dot{x}(t)$, i.e., the dynamics of the system
in a course on classical mechanics, you will find under certain assumptions that $\delta S = 0 \implies \frac{\partial L}{dx} - \frac{d}{dt}\frac{\partial L}{\partial \dot{x}} = 0$, which is called the Euler-Lagrange equation, which is the general form of the equations of motion under these assumptions

is the constraint that $T = V$?

hmm okay yeah he never derived the E-L equations for us but we did work with them a lot.
he said don't worry about it, just know that it works lol.

the constraint is a differential equation in two variables: $x$ and $\dot{x}$
^ if your system consists of a single particle

so the action is the integral/functional of the lagrangian over some interval of time, if we vary it infinitesimally, we find that when it's "stationary" i.e when $\delta S = 0$, we can derive the EoM for the system

2:24 AM
the stationary conditions gives us the EoM yes

how come in my class we just derived the lagrangian directly based on the system, then just plugged into the E-L equations without worrying about the action?

the E-L equations come from stationizing the appropriate action

does it follow directly from $S = \int_{t_1}^{t_2}L(q(t),\dot{q}(t),t)dt$?

the procedure in general would be 1) write down your action, 2) stationize it, 3) see what EoM is, 4) solve EoM. In a textbook in classical mechanics, we write down a general action (subject to some assumptions), stationize it, and find the E-L equations. So we do steps 1) and 2) in a more general setting so that we don't have to do it every time for particular systems. With E-L equations in hand, we can more immediately write down particular EoMs by just knowing the Lagrangian
@Obliv after stationizing it and assuming a couple of things, yes

what does the process of stationizing look like
Is it a calculus theorem
I'm so noob D:
oh boy it's just the fundamental theorem of calculus isn't it
nvm maybe not

2:32 AM
en.wikipedia.org/wiki/Euler%E2%80%93Lagrange_equation u might already be on here, but a derivation is written out here

ooh ok thanks

it looks mostly like like calculus, but (at least for me) it was hard to get used to what a variation is :P

the problem (and good thing) about learning from wiki is the language is very precise so there are a lot of math/physics terms I have to look up and I get so sidetracked because of it :P
but it's good to get an overview on the different terms
definitely going to go on a wikipedia binge tomorrow

if you would like a resource that would stick to physics (more than math) for a derivation of E-L equations, then John R. Taylor's classical mechanics seems pretty decent. if you want a flavor of the maths as well, then the supposed classic text is Vladimir Arnold's Mathematical Methods of Classical Mechanics
or if you want to go full maths you can read something about symplectic mechanics :P

someone here was posting problems from taylor and unfortunately the ones he's posted had some typos/misprints so I'm wary of him now :P
but yeah I'll check those out to see what I missed out on..
I need to buff up my knowledge of hamiltonians too because we baarely covered that.
and it's supposedly very useful in qm

2:43 AM
i personally do not find classical mechanics to help with quantum mechanics. but i do (so far) find classical field theory to majorly help with quantum field theory
my personal bias is that if you would like to attack both classical and quantum mechanics at once: learn linear algebra and lie theory and representation theory :P

I'm thinking about just following ted shifrin's multivariable mathematics
he covers a lot of useful math
I think he does linear forms & tensors towards the end

also using only usual methods from calculus, if you treat $\delta S$ as taking the differential $dS$, then you can obtain the E-L equations

so the total derivative of the lagrangian within the integral. hmm ok

@Obliv right

classical EM using maxwell's equations can be a classical field theory :P I don't know how much gauge invariance matters in deriving the equation of EM waves
other than it makes the derivation easier by choosing a specific gauge (lorenzian one i think?)

2:50 AM

I can't imagine how complicated QFT gets without the simplicity of assuming smoothness in functions, symmetries etc.

this is what it looks like in any case
in the final step if you assume that the variation of $x$ vanishes at the boundaries of the time interval, the second term vanishes.
Finally, the remaining eqn is true for all variations $\delta x$, which allows you to obtain the E-L eqns

thank you ! I will digest this
eats the paper

Just for confirmation, it's $$\delta \int_{t_1}^{t_2}dt L(x,\dot{x})=0 \implies \int_{t_1}^{t_2}dt \delta L(x,\dot{x}) = 0\implies \int_{t_1}^{t_2}dt \left[\frac{\partial L}{\partial x}\partial x + \frac{\partial L}{\partial \dot{x}}d\dot{x}\right] = 0$$

2:55 AM
right
i guess the quirk is that $dx \to \delta x$ i don't really know what it means :P, but I think of $\delta x$ as a variation in $x$
and recall that here $x$ is really a function

uh... I probably shouldn't be thinking of the term variation to mean change
is it bad I think variation, curl, derivative, change are all synonymous :'D?

probably depends on who you ask :P
at the least the curl is a particular type of derivative, so I wouldn't make it synonymous with the other terms you listed

so I guess I'll leave it as $\delta x$

3 hours later…
6:11 AM
Hello Everyone...

@SirCumference Indeed, Sir Crackpot has gone dormant

2 hours later…
8:03 AM
@Obliv In a sense they are, but if you get technical you have a quite different meaning for each of those
They all describe the way something changes, the difference is what

monday is the deadline for phd commitment
BAH

@Relativisticcucumber BAH

2 hours later…
10:07 AM
what is the meaning of finite extension?

@SillyGoose that $\pi_0(G)$ is finite

is it an extension in the sense that we start with $1 \to G \to \pi_0(G) \to 1$ and we are adding in another group $G_0$ with the inclusion map
I am also wondering: (1) Is (1.2) just referencing the definition of a compact topological space? I.e., that every open cover of a topological space has a finite subcover. (2) Where does (1.3) come from :P is this some sort of classification of finite covers or something?
(3) In (1.2), what is the role of the abelian group $A$? (4) Why are we writing all of this out in terms of sequences?

10:29 AM
@SillyGoose you can't "start with" $1\to G \to \pi_0(G)\to 1$ because if that were an exact sequence, $G\cong \pi_0(G)$
the name "extension" comes from the idea that $G$ is an "extension" of $G_0$ by $\pi_0(G)$, cf. group extension
@SillyGoose no, and no
in (1.2), $\tilde{G}_0$ is the universal cover, and this uses that the fundamental group of Lie grous is Abelian

@ACuriousMind hm does the torus in (1.3) not make the total space not simply connected? I am unfamiliar with how "simply connectedness" interacts with taking direct products of spaces

in (1.3), this is the classification of simply connected Lie groups derived from the classification of Lie algebras

@ACuriousMind oh so we need to throw in the normal subgroup before $G$ in order to write an exact sequence and have $im f_1 = G_0 = ker f_2$

ah, sorry, this is not the universal covering

instead of $\text{Im} id = \{1\} = \text{Ker} f_2$, which would happen if I left out the normal subgroup in the sequence

10:47 AM
What's happening here is this: Since the representations of compact Lie groups are unitarizable, they are in particular completely reducible, so in particular the adjoint is completely reducible, and so the Lie algebra is reductive, i.e. it is a sum of an abelian $\mathfrak{t}$ and a semi-simple $\mathfrak{k}$.
This, in turn, means that there is a simply connected semi-simple Lie group $K$ such that $T\times K$ has the same Lie algebra, and hence $\tilde{G}_0 = T\times K$ is a cover of $G_0$
replace $T$ by $\mathbb{R}^n$ and you have the universal cover

okay and so the semisimple $K$ splits into a direct product of the corresponding lie groups of the summands in the direct sum of the semisimple lie algebra?
so we end up with $T$ times a product of simply connected, simple lie groups?
nooo not again ;_;
maybe i should read the first bits of a more differential geometric lie theory text :P
is there a special meaning to the brackets around $\theta \wedge \theta$?
If $\wedge$ is an antisymmetric product, then I would think $\theta \wedge \theta = 0$, but that it is a special equation suggests that is not the case :P
er I guess maybe it is because $\theta$ is $\mathfrak{g}$-valued that it does not necessarily vanish?

11:11 AM
@SillyGoose yes, the brackets are intended to be a reminder that you get commutator terms $[\theta_i,\theta_j]\mathrm{d}x^i\wedge\mathrm{d}x^j$

okay i see
how would i write $\theta$'s codomain? from looking at other resources it sounds like $\theta$ should assign to each $g$ a map $f: T_gG \to T_1G$, so the codomain is like $\text{Hom}(T_g G, T_1 G)$..?

sure, a $V$-valued 1-form on a manifold $M$ is equivalently a map $T_x M \to V$ at every point; I don't know why you'd insist on writing $T_1 G$ instead of $\mathfrak{g}$, though :P

i wasn't sure if it was safe to do so XD but okay i see

11:49 AM
@ACuriousMind is this related to reductive groups?
Like the lie algebra of a reductive group is reductive

yes

@ACuriousMind also the opposite? Like a lie group is reductive iff it's lie algebra is?
I never understood the definition of reductive group, but this one for algebras is simpler. I guess kinda like with Borel subgroup/algebra

yes
@lucabtz "reductive group" is usually defined in a more general context of algebraic groups, where not everything is a Lie group

@ACuriousMind yeah I noticed
But that seems to require algebraic geometry which as you know I'm pretty scared of
I'd honestly would study some if I had the time too
and there are so many stuff id like to study better
such as theory of Riemann surfaces, algebraic topology, functional analysis are all things id like to study better
and then there is algebraic geometry which i find scary mostly because i see a lot of theory of rings is used there and i basically known nothing about them
oh also Lie theory and representation is something id like to study better right

2 hours later…
2:29 PM
hi

3:09 PM
there's two different notations for irreps of the Lorentz group. I'm familiar with the $(m,n)$ notation, but how do i translate the notation $\mathbf{k}$ to this? Is it the case that $\dim(\mathbf{k}) = k$? So, $\mathbf{1} = (0,0), \mathbf{2} = (1/2,0),\ \mathbf{3} = (1,0)$?

@notmyrealname the $\mathbf{k}$ notation is not specific to the Lorentz group, and yes, it just denotes the dimension of the rep

okay, thanks. What other groups is it used for?

note that there are multiple reps of the same dimension, so this notation isn't unambiguous and you usually fix conventions in the context
It's common to use it for SU(N), specifically SU(2) and SU(3) in the Standard Model. But since it's just the dimension you might use it for any group

On Wikipedia it's claimed that $3 \oplus 1$ is the adjoint representation of $so(3,1)$. This doesn't make sense to me: The dimension of $3 \oplus 1$ should be 4 but that of the adjoint representation should be 6, no?
The Poincaré group, named after Henri Poincaré (1906), was first defined by Hermann Minkowski (1908) as the group of Minkowski spacetime isometries. It is a ten-dimensional non-abelian Lie group that is of importance as a model in our understanding of the most basic fundamentals of physics. == Overview == A Minkowski spacetime isometry has the property that the interval between events is left invariant. For example, if everything were postponed by two hours, including the two events and the path you took to go from one to the other, then the time interval between the events recorded by a stopwatch...

that section of the Wikipedia article is complete nonsense
(note also that it does not reference a single source)

3:26 PM
strange. Is it at least true that $2 \oplus \bar 2 = 3 \oplus 1$?

you mistyped a $\oplus$ for an $\otimes$, but that's where the nonsense already starts - for representations of $\mathrm{SO}(3)$ or $\mathrm{SU}(2)$ this is true, not for the Poincaré or Lorentz group.
$(1/2,0)\otimes (0,1/2) = (1/2,1/2)$ is irreducible as a representation of the Lorentz group

thank you!

3:47 PM
(just so that no one gets confused by the Wiki link above pointing to no specific section: I've submitted an edit to remove that section)

I thought we had a "No AI" policy regarding both answers and questions. When did that change? physics.stackexchange.com/help/ai-policy
3
19

Over on Meta they recently announced a new option for individual sites to have a "AI-generated content" policy banner, offering several options: We will initially offer two banner text options that all sites in the Stack Exchange network can opt-in to. Those options are the following: Reminder: ...

@PM2Ring note that that post argues (explicitly at the end) that we indeed don't have any specific "no AI" policy, since undisclosed usage of LLM output violates already other preexisting policies on plagiarism and academic honesty, which is also what the help center page you linked says
you can post LLM output, if properly attributed, and this would not be grounds for suspension (but you'll probably get a lot of downvotes)

Ah, ok. ;)

hm
That's surprisingly weak.

OTOH, attributing a chunk of text to some LLM doesn't tell you what the real source of the information is. It merely identifies the text as a piece of generated text. Which is better than not knowing that it's generated. But it's certainly nothing like the usual chain of references that science & mathematics is based on.

4:02 PM
@Loong honestly, no one ever posts LLM output properly attributed - really the only people who use it are those trying to disguise it as their own words so they can farm rep from it
@PM2Ring sure - which is why if you just cite a chunk of LLM text you will probably be downvoted, and note that answers that "consist entirely of copied text" are also subject to deletion regardless of where the text is copied from, cf. the end of physics.stackexchange.com/help/referencing

@Loong I agree.
hm

I'm also concerned about questions based on AI output. The OP has asked the LLM some question, they get an answer that looks right, but with maybe some puzzling aspects. Then they ask us to verify that answer. That can get messy, especially if the OP has some common misconceptions baked into the question.
Sure, we get that sort of thing all the time anyway, eg twin paradox, one-way speed of light. But when LLM is involved the OP can get into much deeper rabbit-holes than usual.

I mean, that's true, but again the worst of these questions are already off-topic by other rules (e.g. just asking whether some nonsense a model generated is correct is often off-topic as non-mainstream)

and those deep rabbit-holes are full of word-salad for the rabbit

the thing that makes the question bad is the content, not necessarily the involvement of an LLM

4:15 PM
or rabbits :P

Nevertheless, I still don't think that all the users who upvoted
actually meant
"Generative artificial intelligence (a.k.a. GPT, LLM, generative AI, genAI) tools can be used to generate content for Physics Stack Exchange, but this content must be properly referenced"
4

^ This

I feel this is splitting hairs: What kind of answers that disclose which parts are computer-generated and that are not deletion-worthy by other pre-existing rules are you worried about here?

@Loong why does your profile say "Member for 3 years, 4 months" but also "Moderator from 2015 "

Yes, technically there is a difference between forbidding the undisclosed use of these models and all use of these models; in practice I have not seen any disclosed usage at all except for a few questions in the vein of what PM2Ring just described
the big stink about the network policy (cf. physics.meta.stackexchange.com/q/14438/50583) was because it was interpreted as an attempt to remove moderators' ability to punish the undisclosed use of the models; I'm fairly confident that is what most people are really concerned about

4:26 PM
I suppose so. If it's disclosed content, we can safely ignore it. ;)

@ACuriousMind I am not concerned over any particular answer. If something is clearly written in bad faith, you may act on this even if the particular case is not mentioned in the help center at all.
For me it's more a matter of principle. If you really don't want A, don't say "you may do A but you also have to do B" and hope that users will trip over B. Just say "don't do A".

Ah, but SE Inc have made it a bit tricky to moderate people doing A, unless they've disclosed they're doing A. So it's simpler to use the "trip them up with B" tactic.

and add in C, D, E,... at their discretion :P

At least we don't have the level of AI use that happens on SO. I believe there's still a huge backlog of "suspected GenAI" flags on SO. And some of the mods are still on-strike.

@PM2Ring Okay, if that's the real problem, I don't mind a little detour.

4:33 PM
Wow, still?

@Loong well, but at least I personally don't know that I don't want A - maybe there is, now or in the future, some legitimate use for including LLM generated content into an answer that also contains some human-written bits. While I personally doubt it, that may be a limit of my imagination. What I know I don't want is people copy-pasting ChatGPT output into the answer box to farm rep, and that's what the current policy succeeds in preventing

some are still waiting for Monica to return :(

stackoverflow.com/users/366904/cody-gray-on-strike stackoverflow.com/users/6296561/zoe-is-on-strike Zoe is still active on the MSO site, but I haven't seen Cody in ages.

yeah, SO is a jungle

I wouldn't mind people using language models as a super spelling checker / grammar checker / style booster. But with current LLMs there's a huge risk that it will change the author's intentions. And the author may not realise it, if they don't have enough expertise, or aren't scrutinising the LLM output sufficiently. It's too easy to develop a false sense of security because of the indefatigable self-confident tone underlying the LLMs utterances.

5:08 PM
I made a Horizons script for comparing angular diameters of Solar System bodies. astronomy.stackexchange.com/a/57419/16685 You can use it to see what bodies are capable of causing eclipses, etc. Eg,
Apollo 11 LRRR is the Laser Ranging RetroReflector placed by Buzz Aldrin.

5:36 PM
My kind of playlist:

2 hours later…
7:11 PM
Hi, everybody.
I haven't been around for a while. What's new?

wb pal

Is there a scalar product suitable to invert the mode expansion in ST? Something like $$a_p\propto\int d^3 x e^{ikx}\overset{\leftrightarrow}{\partial_0}\phi$$ in QFT
Sorry, that's $a_k$
The mode expansion is $$X^\mu=x^\mu+\alpha' p^\mu\tau+\sum_{n\neq0}\frac{\tilde{\alpha}^\mu_n \mathrm{e}^{-in\sigma^+}+\alpha^\mu_n\mathrm{e}^{-in\sigma^-}}{n}$$
I tried $$\int_0^{2\pi} d\sigma e^{im\sigma^\pm}\overset{\leftrightarrow}{\partial_\tau}X^\mu$$
Where $\pm$ determines if we take left movers or right movers
Since both exponentials are written with the same sign, it doesn't seem to work and I should replace $\overset{\leftrightarrow}{\partial_\tau}$ with "symmetric" version with a plus

7:36 PM
@Mr.Feynman It's just the formula for Fourier coefficients after you've split the $X$ into $X = X_L(\sigma^+) + X_R(\sigma^-)$
but it only holds on-shell

@DanielSank not much, how is the quantum ai lab

in that way the ST expansion is closer to a true Fourier expansion than the mode expansion of the free field

@ACuriousMind Yes, you're right. This avoids derivatives too
Yet, left and right movers are so similar to positive/negative frequencies I hoped there would be some analogy
Well, I always wondered why in the free field expansion of QFT we end up with that "weird" scalar product

it's because the "modes" are a mixture of the Fourier transform of the field and its canonical momentum, and the canonical momentum involves a time derivative
there's no canonical momentum involved in the mode expansion of ST, hence no derivative

wait what $\log$ is a trig function?

7:50 PM
no?

Log is a piece of wood
2

weird, maybe they meant trig function in the sense that it's taught first in a trigonometry class or something
(not that we have trigonometry classes anymore where I'm from)

@ACuriousMind Uh, I never thought about that bit about the canonical momentum, makes sense. I always justified that product with its appearance in the Noether charge
Good insight
@SillyGoose legendary on Clash Royale

XD
If it’s called topological quantum field theory why do i need all this algebra 🤔 lol jk

are you a $\sum\limits_{n=0}^{\infty}\frac{1}{n!}$ simp or a $\lim_{n\to \infty}(1+\frac{1}{n})^n$ chad
@SillyGoose dude at that level you need everything

7:57 PM
Does anyone know of a textbook which covers chern-simons theory? or even a good text which would contain the math preliminaries to learn CS theory? I am currently trying to work through a set of notes but i feel i would benefit from the level of thoroughness that a textbook may provide :P

@SillyGoose that's the level at which you don't find "textbook", you find "monographs" :P

@Obliv hopefully not analysis ;D
Blebs well if that’s the best one can do

I think i had a nightmare I was trying to construct $\mathbb{R}$ using something called "dedekind cuts" which I don't know what they are but my subconscious tried to imagine what they were
I think I thought they were sections of $\mathbb{R}$ that you just string together with some rope

you will probably not find a more elementary discussion of CS theory than in Witten's original paper

like a finite interval plane that you tie string around the numbers :D

8:02 PM
Is Quantum Field Theory and the Jones Polynomial this the original paper?

yes

it makes me sad how difficult it is to draw $\mathbb{Q}$ irl
compared to the other Mbb characters

Just write a Q and do a line on it

like on the right side?

I do on the left

8:12 PM
what's the canonical way
does the pope do it on the left or right, or maybe both?

I mean you don't really hand-write $\mathbb{C}$ etc. either, do you?

in order to distinguish it as the set/field isn't that something people do? I rarely write in general (I just write up assignments using overleaf)

you just draw a C and then a straight line inside it, also usually to the left
i.e. the "draw a line somewhere on the left" is in fact how people write most of these symbols

yeah $\mathbb{C}$ isn't so bad, $\mathbb{Z}$ was kinda tricky for a while but $\mathbb{Q}$ was just atrocious looking.. like a grade schooler's attempt at graffiti D:
I understand $\mathbb{C}$ is complex, $\mathbb{Z}$ is zahlen, but wtf is $\mathbb{Q}$ anyway
quotient?

it's quotients

8:15 PM
so $\mathbb{Q}/A$ is a quotient ring of the quotient ring

9:03 PM
@Obliv It's very good. Lots of progress.

there is a neat site called mathpix which turns images of latex into latex that you can copy and paste. useful for if you are typing up notes but using well known expressions that are in other resources :P (or if you're doing topology homework...)
@DanielSank cucumber and i are committing to phd programs this monday :D

@ACuriousMind on a second thought, I'm not sure I agree on this statement. If you expand the conjugate momentum $\Pi_\mu=\frac{1}{2\pi\alpha'}\dot{X}^\mu$ you are in a similar situation. The string modes are a combination of the field and the conjugate momentum Fourier transforms too
Sure, it works using the Fourier decomposition of left movers and right movers as you mention, but it's not wrong to proceed like in QFT apparently (mutatis mutandis in the scalar product)

@SillyGoose What PhD programme have you decided on?

im not sure yet :0

9:20 PM
you should decide on the best one
imo

Did you say you wanted to do condensed matter?

Yes
i think topological matter stuff would be cool 0:

Sounds like interesting stuff

9:39 PM
In an ideal world i think id like to work on axiomatic qft or something similar:P

you should put that on your profile
in an alternate universe you work on axiomatic QFT :D
but in this one you work at a goose company

In an alternative universe, I would work on alternative universes

h o n k
I wonder what los alamos's interest in decoherence is :P
since it seems usually the theory department of a national (in the US) lab is still on like nuclear physics or materials physics or something like that

There are so many interesting areas of physics
But I always end up getting drawn back to the same things

what are those things for you

10:42 PM
@Obliv what if in that alternate universe Silly Goose is a swan?

10:55 PM
or a penguin :D