12:02 AM
@SillyGoose as far as I can tell you're still not writing it out fully, e.g. as
$$\mathrm{tr} (A \wedge B) = A_{\mu}^a B_{\nu}^b dx^{\mu} \wedge dx^{\nu} \mathrm{tr} (T^a T^b) = - B_{\nu}^b A_{\mu}^a dx^{\nu} \wedge dx^{\mu} \mathrm{tr} (T^b T^a) = - \mathrm{tr} (B \wedge A)$$

12:26 AM
@bolbteppa hm i think i see. because the trace operation should really only act on things with lie algebraic indices. so i should actually fully expand the form in coordinate and lie algebra indices

I wish we had more lower level physics people here
It's like all grad lvl + ;(

12:43 AM

true, but I'd rather ask a peer than pester a legit physicist like ACM
or naturallymeow
but asking stuff on here is just my lazy way of not searching PSE/the internet thoroughly
once I get to the hard stuff, I'm sure I'll need all the help I can get

Id have to think about it before replying. Itas too early to do maths

don't trouble yourself lol
it's just me computing a bunch of stuff. although problem 2) i'm pretty sure I did wrong

which is problem 2?

"A circular region of radius R has a uniform electric flux that points normal to the
plane of the circle with magnitude ΦE = kt, where k is a dimensionful constant
and t is the time. What is the magnitude of the magnetic field that is induced as
a function of r, the distance from the center in the plane?"

12:56 AM
thats the thing, it feels wrong but it looks correct

I gotta change the flux thing as ACM pointed out
wait is that really correct lol
that stokes theorem part looks wrong
ill edit it
that is what it should be
EM is weird because once we couple the E and M fields it's no longer two separate phenomenon but interconnected

1:27 AM
@Obliv i think the hard stuff is the first stuff one encounters :D

1:44 AM
@Obliv This is correct. It is also why we often prefer to use the vector potential A, because then we reduce that back to one single multivariable equation
@SillyGoose This I concur. I always feel for the poor students who didnt get a decent introduction, and that causes everything to go extremely painful for them

2:39 AM
is it correct to think of a virtual derivative delta x as being a more general version of dx? so dx is more strict because it doesnt generate a new arbitrary function. but if you have an equation which only involves delta x, then the same equation should also hold replacing delta with d? im thinking if dx in a way being "along some function f(x)" and delta x being "away from the function", dx is a special case of delta x

@Obliv You are using the wrong equations. No wonder I felt something was so off.

2:57 AM
how so

You used $\vec\nabla\times\vec E=-\partial_t\vec B$ but this equation is not applicable here. You should have used $\vec\nabla\times\vec B=\mu_0\vec j+\frac1{c^2}\partial_t\vec E$

that's what I thought..
but I didn't know if I could use that one because I wasn't sure if there was a current or not

3:17 AM
You are supposed to intuit that there is no current, or else the problem would make no sense.

dang, I woulda never figured that out

It is supposed to be "free space" between two capacitor plates that you are connecting to a linearly increasing voltage power supply.

Right, that makes sense
But theoretically.. couldn't it also just be a cross section of a wire where current is flowing and voltage is increasing, since all we're given is $\Phi_E = kt$

do all four maxwell's immediately come out of stationizing the electromagnetic action?

3:33 AM
@SillyGoose i think the inhomogeneous ones do
the homogeneous ones are fufilled automatically when you write the potential formulation iirc

@SillyGoose Not like I know what you're talking about, but I was under the impression these laws came somewhat from experimentation (like the proportionality constants $\mu_0,\epsilon_0$)

(in the differential form formalism) i thought two of the MEs came form stationizing the action and two come from a mathematical property of differential forms. something like alluded to in this post: physics.stackexchange.com/questions/746207/…

@Obliv it can be this. I thought of it too. But then the question would have told you it is a wire if that is the case. The phrasing is meant to evoke that you are supposed to do this in free space

@Obliv i think ultimately indeed maxwells are models fit to experimental data

@Obliv yes all physical laws are fundamentally from experiment, but you once you know them then you can also motivate them from an action principle usually

3:36 AM
yeah plus the context is we just started electrodynamics and coupled the E,M fields

@Obliv no, you are misunderstanding the question, just as you have predicted in advance that you would be

lol
after I'm done with finals I'm going to finally start learning how to read those indices

@SillyGoose more or less yes i think.

@SillyGoose this is correct, as qwerty also asserted that you are correct. The non-homogeneous part is the interesting part anyway, so it is tolerable to say that we can get Maxwell's by stationarising the action.

welp, I got this $$\nabla \times \mathbf{B} = \frac{1}{c^2}\partial_t\mathbf{E} \implies \iint_{S}(\nabla\times\mathbf{B})\cdot d\mathbf{S} = \oint_{\partial S}\mathbf{B}\cdot d\ell = \frac{1}{c^2}\iint_S \frac{k}{\pi r^2}d\mathbf{S}$$

3:43 AM
wait do EM waves exist in 2 spatial dimensions?

I feel like I could literally just leave it like this but whatever I'll explicitly write $B(r) = \frac{k}{c^2}$ I think is what that reduces to
@SillyGoose idk, if u consider propagation vector part of it then no

@Obliv This is close to the answer. You are, however, not given $\vec E$, but rather you were given $\Phi_E$, so that you have some irritating conversion factors.

I skipped a step and computed $\partial_t\mathbf{E} = \partial_t\frac{\Phi_E}{A} = \frac{k}{\pi r^2}$

i am confused about why in some resources like Tong it is stated that the raw Chern-Simons equation of motion (suppose $G$ is abelian) $F = dA = 0$ is uninteresting. But isn't this literally the same eom as with the Maxwell equation of motion in 2 + 1 dimensions?

Ugh, I can't even interpret what $\mathbf{E}(x,y,z,t) = \mathbf{E}_0\cos(kz-\omega t)$, where $\mathbf{E}_0$ is a constant vector, is meant to represent. As in, is it propagating along the z-axis

3:56 AM
@Obliv But this is incorrect. It is $\frac{k\pi r^2}{\pi R^2}=k\left(\frac rR\right)^2$

Ohh
tricky tricky

@Obliv $\cos(kz-\omega t)$ part tells you that it is necessarily propagating along the z-axis. If this is a free space plane wave, then $\vec E_0$ is constrained to be in only the x-y plane

ooh

@Obliv You have not yet extracted the full B field part out.

I must divide it by $2\pi r$?

3:59 AM
i think my question is: how is classical vacuum Maxwell theory in 2+1 spacetime different from $U(1)$ classical Chern-Simons theory?

How come you characterize it by $2+1$ and not $3$
2 spatial 1 time?

Yesh

uh hm does anyone know how to taylor expand something like df/d(x+y) where y is small?

4:15 AM
are those infinitesimal quantities lol
never seen that :O
also @naturallyInconsistent $$B(r) = \frac{kR^2}{2\pi c^2r^3}$$ final answer

yes i mean a derivative and I need to expand in terms of the parameter to 1st o

how much smaller is y from x though, if they're both infinitesimally small..

uh x is just a parameter
you could think of it as df(x')/dx'
and then sub x' = x+y

@Obliv I just gave you that I have $r$ in numerator and $R$ in denominator; why do you have it the opposite?
@Obliv yes

oops
final answer (revision 001) $$B(r) = \frac{kr}{2\pi c^2R^2}$$

4:19 AM
@Obliv Yeah, I think this is correct. I am not going to spend more effort into it, though. The cough medicine is making meow drowsy

Np, and thank you :)
@qwerty I'm not sure but maybe this might be helpful?

the maxwell eom is not $F = 0$
silly me

5:27 AM
silly silly goose

@qwerty I think what you want is $f(x+y)=e^{y\partial_x}f(x)$

@naturallyInconsistent hmm, not sure i see it
no exp's in the final expression either

5:43 AM
@qwerty I think what you are trying to do is that you have a function f and you know all its derivatives at a point x, and you want to find out the Taylor expansion of the function at a slightly different point x+y, in terms of orders of y
The exp is an extremely easy way to remember how that Taylor expansion goes. It is term by term equivalent to the exponential function's Taylor expansion.

@naturallyInconsistent no, I don't know all its derivatives?

@qwerty then, with what sense are you trying to find $\frac{\mathrm df}{\mathrm d(x+y)}$

it's an arbitrary curve f
ok, let me back up

yes, and how is the x+y supposed to be interpreted?

let me use different notation

5:46 AM
sure

have a curve x(t) say
neighbouring curve is x'(t') where x' = x + delta x, t' = t + delta t
now i want to find a first order expansion

ahhh, that Lie derivative shit

(dx'/dt')^2 say, and subtract (dx/dt)^2
yeah
@naturallyInconsistent haha nice spot. i knew it was related but i liked this derivation which actually didnt mention the lie derivative
so yeah, typical perturbation stuff, but the dt' is throwing me

Well, stereotypically, if you have some neighbouring curve shit like that, you break it down into one that the time is not shifted, and the time shifting alone. Some shit like that. This part is just arcane magic

in the expansion
@naturallyInconsistent come again?

5:55 AM
@qwerty I think it is like the partial derivatives shit. You keep one part fixed and just vary the other, etc. Im not sure; it has been a long time since I last touched that part of the arcane

ah that's ok. i'm sure i'll figure it out
how hard can it be, it was a skipped line of working in the paper nervous chuckles

could you show meow where is the starting point and where is the ending point? Because that is always very intriguing

sorry still new to chat, how are people dropping in images all the time?

squints i see no such button...
>cant upload images without 100 rep

6:02 AM
oh lol
you could manually set it up at imgur or something, and then send the link here

@_@ much efforts
lets see if this website works? ibb.co/fXn30T2
never tried it before
(2.71 is just 2.72 without the primes)

Yes, I see it. It works
now imma hatha meditate on it

since i have your interest @naturallyInconsistent , did you see this related thing chat.stackexchange.com/transcript/message/65539254#65539254 up here? i'm like 80% sure it's right but nice to have a second opinion

@qwerty I saw that, and immediately noped myow way outta that. That question needed context: virtual displacements mean different things in CoV i.e. path integrals v.s. other places it might be used.

i seeee
it always came across to me that although the "philosophy" of the delta is different, you really just end up manipulating it the same most of the time
(unless you have d's and delta's in the same equation)

6:18 AM
Kinda like that, but I'm interested in teaching, so I'm always making sure my students have a fighting chance of understanding what it is I'm teaching.
It is much more difficult to understand when all the notations are jumbled up.

well, in this case im not trying to mix the notations per se
its just that if i can show that something is true with delta's then should it also be true with d's?
with the different meanings
of delta being a virtual variation of the path, and d being an infinitesimal "along" the path

6:55 AM
Hello Everyone...

@qwerty That equation is purely first order. All the complications differ at the 2nd order so it should be perfectly fine being glib with this one

@naturallyInconsistent sorry what do you mean?
im aware it's a first order expansion
im thinking they just invert it now
havent checked though
nvm i dont think it works

7:12 AM
@qwerty \begin{align}g^\prime(\dot x^\prime)^2&=\frac{\partial g^\prime}{\partial x}\delta x(\dot x)^2+\cdots\\&+2g(\dot x)(\dot{\delta x})+\cdots\\&-2g(\dot x)^2\frac{\partial\delta\lambda}{\partial\lambda}+\cdots\end {align} where each $\cdots$ are higher order terms that we drop.

is your \dot differentiation wrt lambda or lambda'?
it's just the last term i can't quite see

That's the thing, differentiation between the two variables are the same at first order
Any difference is at least 2nd order and above

where does the third term come from then?

From the denominator of the difference, to first order
The negative sign from binomial theorem to negative order
They are supposed to be $\frac{\mathrm dx^\prime}{\mathrm d\lambda^\prime}$, which turns into $\frac{\mathrm dx}{\mathrm d\lambda}\left(\frac{\mathrm d\lambda^\prime}{\mathrm d\lambda}\right)^{-1}$
And then you subtracted the term proportional to 1, leaving just the $\frac{\mathrm d\delta\lambda}{\mathrm d\lambda}$

7:37 AM
@naturallyInconsistent OH. i havent cranked it through but if i understand you manipulate the numerator and denominator separately?! like you split up $frac{dx^\mu}{d \lambda'} \frac{dx^\nu}{d \lambda'} = (dx^\mu dx^\nu) (d \lambda')^{-2}$ !?
i would have not thought to do that

@qwerty That is not what I am doing. I am kind of doing (multiple) product rule, keeping only one of the terms to properly expand, whereas setting every other thing in the product to be unprimed.

oh right I see where youre coming from now
i didn't realise that was a valid way to do it, i thought you needed to taylor expand dx'/d \lambda' itself
thanks

8:03 AM
Such silly arguments that only work to first order is all over the place in physics. We are kinda very lucky that we only ever care about first order

if you wanted to go to second order do you have any idea how you would proceed?

8:18 AM
I will burn the place down and run away
The issue is that there will be a lot of horrible cross terms $\delta x\delta\lambda$ of all possible combinations coming into play

haha fair

8:42 AM
@naturallyInconsistent tamest Lie group enjoyer:

9:37 AM
i believe i have read that fock states are energy eigenstates and of the number operator. this means that the number of particles is conserved in time, right? but what happens if i create a particle and its antiparticle and they annihilate?
also
HOOOOOOOOOOOOOONK
did anyone else listen to the new taylor album

@Relativisticcucumber H O N K ~ ~

@Relativisticcucumber The Fock states are stationary in the free theory, and in the free theory, particles can't annihilate

@Relativisticcucumber QFT simultaneous eigenstates of the Hamiltonian operator and what we call the number operator is probably not what you think it is. Yes, the number of particles is conserved, but it is "number of electrons minus number of positrons" that is conserved. i.e. 1 electron, and 2 electron + 1 positron, and 3 electron + 2 positron, states, all contribute to the same wavefunctional.

@ACuriousMind oh no why not? is it because there is no way for particles to interact? or am i prevented from creating an electron positron pair at all

@Relativisticcucumber if there is no interaction, then there is also no way to generate electron-positron pairs

9:49 AM
@Relativisticcucumber yes - free really means free, there's nothing going on in that theory

When you learnt QFT properly, you will see what it is what we are drawing in Feynman diagrams.

spoiler: the real world is not free :P

@naturallyInconsistent sadly i think i will never learn qft. im operating off of the qm notation of creation and annihilation operators and what not but my phd program doesnt require qft for my track :,(

d'aww
@Relativisticcucumber with your kind of curiosity, it is inevitable that you will learn it by yourself

but i think i do need to understand how these operators work. but there is nothing inherently related to qft about creation and annihilation operators, right?
@naturallyInconsistent lol i do wish to learn the basics. i can live without all the in depth computations i think xD

9:51 AM
correct. You can learn creation and annihilation operators as ladder operators in angular momenta and in QHO
@Relativisticcucumber quite a lot of HEP experimentalists who only learn up to tree-level diagrams. They do good work

okay so i see that the hamiltonian is different for interacting, so i could say that theres no reason apriori to assume that the multiparticle fock states are still states of the interacting ham. i guess in fact they are not. but if i start my formalism by defining creation and annihilation operators, then i write the hamiltonian, then i write the commutation relations, what stops me from coming to the kind of contradiction i came to above? [...]
[...] it would seem that i should be prevented from even creating a particle/antiparticle pair since this model wouldnt be able to describe the behavior of these two particles?
@naturallyInconsistent i guess what i mean is can you elaborate on this XD
because im thinking for some reason that defining C/A operators lets me make whatever i want in terms of particle types

@Relativisticcucumber The interaction term $\bar\psi\vec A\psi$ in QED is the one that "eats an electron, eats or spits a photon, and then spits back out an electron". With appropriate exchanges if we want to have positrons instead, this is the term that causes creation and annihilation of particles.

10:28 AM
hi

gday

@naturallyInconsistent that arrow is disturbing :P

what r some lesser known interesting fields of mathematical science other than math and physics

@Relativisticcucumber it seems you think that "you" can create/destroy stuff by applying the c/a operators, but that's not how physics works - everything that happens is the result of time evolution via a Hamiltonian, so if your Hamiltonian commutes with the number operator, it will preserve total particle number.

"the branches of philosophy, which become non-vague, become science". this is how chemistry was.born out of philosophy

10:34 AM
that your mathematical formalism has c/a operators is irrelevant as long as there's no "relativisticcucumber creates a particle" term in your Hamiltonian
2

the fetus stage of a science is labelled philosophy
and any science must have a fetus stage. this means philosophy is useful by definition
5

Philosophy, by nature, is the study of questions around which there exists no widespread agreement, shared framework of inquiry or self-certifying means of confirmation. If a philosopher succeeds at systemizing a topic of inquiry to the point it provides consistent, widely accepted, objective res...

10:59 AM
@Mr.Feynman cant write a 4-arrow in chat
Also, there was a small earthquake just now, scale 5
and then the metro was very fun. Along the way, there was a part that was slanted.
Felt like disneyland

@naturallyInconsistent the earth wanted to hug you and failed :eyes: i hope it was small enough to not be nervewracking

it was very small and not at all nervewreacking, despite miao miao being all high at the 9th floor

are you a cat?

where's schrodinger?

11:07 AM
in the box

good, apparently he was a kiddy fiddler

six feet under

lol

imma be out for dinner date miehehehe
bye

@naturallyInconsistent any relation? en.wikipedia.org/wiki/F._D._C._Willard
ciao

11:09 AM

11:25 AM
@ACuriousMind hm interesting

11:57 AM
i am quite confused about the formulation of coherent states. in this paper in equation 2.1 along with this wiki, the formulation of the state itself is not that intuitive to me. on the wiki it says the definition of a coherent state is that it's an eigenstate of the annihilation operator. what is the best way to think about the state itself?
oh no i have made a mistake

links are [text](link)
but you did [text][link]

bah i was looking at the way it was on the main site
curses

I'll fix it

12:02 PM
yay!

it's also [text](link) on the main site, you just there also have the variant [text][1] with the link then being in a "footnote" labeled by [1]

@ACuriousMind bah
@ACuriousMind i see. this answer says "A maximally classical state should have minimum and equally distributed uncertainty in X and P." where does the equally distributed uncertainty come from?

12:33 PM
@Relativisticcucumber for a "classical" state, you want something that becomes "equally quickly" sharp in momentum and position has $\hbar \to 0$ - otherwise it would keep its quantum character much longer when you look at one of these property than at the other

@Relativisticcucumber I usually describe it like this: by being an eigenstate of the annihilation operator, the state that is being described has "infinitely many particles" so that when you remove one, you get back itself. It is kinda a semi-classical EM wave kinda state. Yet still quantum enough to do quantum stuff with. Very fun stuff.

on a different note, being an eigenstate of the annihilation operator is also nice in the sense that the coherent state "doesn't care" when you destroy a single particle in it - it just stays the same, and this is also what you expect from a classical state: It shouldn't care about losing one of its many constituent particles
and at least for light, the nice thing is that the coherent states do map neatly onto the classical waves, see e.g. physics.stackexchange.com/a/89047/50583

thats what miao miao just said...

@ACuriousMind ah i see
@naturallyInconsistent i see that makes sense. thanks and also thanks @ACuriousMind

12:48 PM
@naturallyInconsistent yeah, I started typing that before you posted, I didn't mean to imply I was saying anything different

B A H ~
M I A ~

<_<
>_>
←⁠_⁠←

1:14 PM
But no. Newton was not entirely right. Yes, there is some difference between the stellar mechanics and the earth mechanics. And the galactic rotation curves are so different, simply because that is their nature. So has God created them. There is no dark matter.
:-) Not that am I believing. But I could argument for it...

1:50 PM
@peterh do u argue for a panpsychist god or more traditional god?

@RyderRude panpsychist or panphysist? :-)

2:17 PM
@RyderRude I googled for that. First I thought you are thinking about pantheism but not. I argued for a traditional God, but I think also this panphysicist idea is very interesting.

2:31 PM
@peterh yeah it's really interesting. i dont find the traditional idea very interesting. god is like an emotional human in the traditional idea
@peterh youtu.be/lLc5x3v75ug?si=yBT4yuSq1rTkDIBd this is an okay debate on panpsychism

@RyderRude In our scientific world-view, the laws of the Universe are eternal. Only our comprehension about them changes (grows, mostly). But what if the reality is different. What if both our comprehension and also the laws of the Universe are changing.
What if once the Earth was really flat.

i think it's okay to dismiss these ideas unless there is positive evidence for them
laws of the universe can change tho.. e.g. symmetry breaking
but we can dismiss the idea where earth was once flat

That does not contradict to the concept.

yes. i was agreeing that laws can change
for earth to once be flat despite all evidence, one would need a cartesian demon instead of god :P @peterh

Newton and Galilei lived in a world of two physics. There was the "celestial mechanics", about eternal and always moving things. And there was the terrestrial physics, where the nature of the objects are standing. We all know what happens with anything without any force. Once it stops.
@RyderRude There was no evidence against it at the time. Only now, but not then.
Newton and Galilei unified the mechanics of the heaven with the mechanics of the Earth. They were very fast, maybe even some decades before, the laws of the nature were not yet ready for their discoveries.

2:46 PM
empiricism didnt use to be a big thing before newton and galileao, which is why people believed in false things
modern scientific reasoning is rational. back then, it could be irrational

@RyderRude Empiricism was not useful in a world where the things were just given by a supreme being, and the Being indirectly stated, they are not important.
salvation is important. Galilean relativity or Aristotelan idealism are low-prio
Science says, the Universe is eternal with eternal laws. Solipsism says, the Universe is created by our mind. What if reality is between them.

we cannot grasp fundamental questions like that. so yes, the closest rational answer would be "something in between" (but we cant even grasp what that means)
these questions come under philosophy. there's no objective or satisfactory way of dealing with them

Because our brain is wired very deeply for science since our early childhood. People lived thousand years ago, they were not.

idk about that. going by what they wrote down, they werent cooking

@RyderRude Historians interpret everything as the consequence of scientific or sociological changes. For example, protestantism, "enlightenment" is interpreted by them as the result of that the scientific progression created a wealthy citizenry, and they needed an ideological base to compete the political-military power of the classical nobility.
@RyderRude Because we all can think only in a framework, wired into us, and we can not really leave this frame.
@RyderRude You know that the birth year of many kings simply was not recorded? At the time, it was not an important data. When they were crowned, when they died, these were important. Birth year was not. No one cared...

3:08 PM
it could be, but i didnt know that

@RyderRude Btw, even the years. There was not a year like "476, Fall of Rome". It is a newer invention. Instead, the line of kings came from the infinite past and went into the infinite future, and years were refered like this: "in the 7. year of the reign our Lord, John the Second". There was no universal coordinate system for the time.

@peterh this comes under the problem of induction. theres no proof that this framework works, but we use it becuz it has worked so far
@peterh oh

Now, there is some chance, that things will change again
In many sources, scientists cry, because most today created papers are trash; but they are paid only for creating them and not for real inventions. Even Peter Higgs said, today he could not discover the Higgs boson; they had no time and most importantly, they had not got his academian job.
Today children... seem having trouble even with basic literacy in any language.

@naturallyInconsistent why are you not hissing?
What kind of cat goes "bah"?!

They don't need to read any more, there are educational videos. They don't need to learn foreign languages any more, there is machine translation. And, probably we are the last generation who needed to be able to write programs.
Odds are... that the laws of the Universe will change again
How does it feel, being a fossil?

3:30 PM
@Mr.Feynman a funny kitty
and you thoroughly misjudge how much shakies ass miao miao is getting and how difficult it would be to get angry

1 hour later…
4:32 PM
@Relativisticcucumber bah

4:53 PM
Is it hard to quantize a theory?

Depends what theory and what quantisation method

@SillyGoose it depends upon how you quantise it. If you use Cartesian coördinates and perform canonical quantisation, it is easy. If you can ignore ordering problems and just use path integration, it is moderately easy. If you have to use something worse than geometric quantisation, tough luck

5:22 PM
@Mr.Feynman my friends orange cat yelps it's quite unsettling
i think he is half turkish van half crazy gremlin

@SillyGoose yes..i guess. the functor from classical theories to quantum theories does not exist

5:41 PM
@SillyGoose It sounds like this is one of these weirdly abstract questions where there's actually something more specific that triggered it - why do you ask?

6:36 PM
I am wondering what procedures are available to quantize the classical Chern-Simons theory

@Obliv always the orange cat

hm i see

ACM is now getting an upgrade i.e. customizing the answer according to the features of the asker :P
Truly the best AI on the market
I'd never trade you for chat GPT

Lol

6:52 PM
@qwerty i think i just realized that extremize might be more common than stationize in discussions of the action
is the "physics" of a chern-simons theory purely in the observables?

7:14 PM
hm formally it seems like actually two of maxwell's are extremely different than the other two maxwells. is this accurate? in the sense that two maxwell's are induced by imposing an action and two are mathematical facts

7:25 PM
in abelian Chern-Simons, the equation of motion is $F = dA = 0$. and, a "gauge transformation" is $A \to A + dX$ where $X \in \mathfrak{g}$.
$F$ is gauge invariant since $d^2 = 0$.
Then, the class of gauge equivalent and on-shell $A$ is given by the first de Rham cohomology class $H^1(M) = \ker d^2 / \text{im} d^1$ where $M$ is the base manifold
being in $\ker d^2$ is the condition that $dA = 0$ (closed) and modding out by $\text{im} d^1$ is the condition that two closed $1$-forms that differ by an exact $1$-form are equivalent, i.e. gauge equivalence
Is there an analogous identification for non-abelian Chern-Simons theory?
in which we now have that an on-shell $A$ satisfies $dA + A \wedge A = 0$ and a "gauge transformation" is taken to be $A \to A + dX + A \wedge X$

7:42 PM
@Mr.Feynman chatPPT sux
useless in any capacity at helping me with my physics/math homeworks thus far
but I suppose it's good at feigning intelligence, which is what makes it marketable

@SillyGoose ah yeah but that's incorrect. I always say "make stationary" myself btw

in an ideal world we would have $\ker(d^2 + A \wedge) / \text{im}(d^1 + A \wedge)$, but since $d + A \wedge$ is not a coboundary map (does not square to the $0$ map), we do not even have a guarantee that the image described is a subset of the kernel

@SillyGoose I assume you didn't have in mind GR

yeah i had in mind this Chern-Simons theory stuff

@ACuriousMind and @bolbteppa I thought that maybe there is a scenario in which the manifold is not completely generic but still breaks some SUSY. Maybe some of the conditions are relaxed not because of some physical principle but as a simplifying assumption.

7:56 PM
sorry lol you replied before I typed the obligatory tone marker /s or /jks

oh i see

(also it's about 6am here and I'm barely awake) good morning

trying to figure out how to word something

Anyway, I was asking all these because I met a person who is convinced that there's no low energy SUSY and so there must be such compactifications and said that 90% people of the string people didn't research on compactifications with resulted in $\mathcal{N}=0$ SUSY.
So I wondered what the other 10% did

8:28 PM
i've turned the question i asked in here to a proper question on the main site: physics.stackexchange.com/questions/811420/…

9:24 PM
When a capacitor is charging, current flows to and from each end, but when it's fully charged is it only displacement current that flows
or like while charging is it $I+I_{\text{disp.}}$ and when fully charged we have no current whatsoever
Why do we need to adjust ampere's law with displacement current to begin with? My prof used the capacitor example and choosing an amperean loop in b/t the capacitor plates to demonstrate there was indeed an induced magnetic field but no current I think
But current does flow while a capacitor charges..
Ohh I think he was rushing and didn't draw it properly. If u consider an amperean loop within the plates then yes, no current flows so u need disp. current
but in his picture, ampere's law alone would have been sufficient since the current flow to that end plate induces a magnetic field
but maybe the B field it predicts is still inaccurate

10:10 PM
maybe i am more interested in math than physics :P good golly

10:27 PM
they go hand in hand
as we venture into deeper physics the math gets deeper so it's natural to be interested in that side
but it's important to remind ourselves that math is just logical arguments, physics is meant to reflect the world we interact with or whatever
shrug

11:04 PM
"God created natural numbers. All the rest is the work of man" - Leopold Kronecher
i count natural numbers as physics :)

why only the naturals?

the line is blurred. but natural numbers are more comfortably physics, as they are a discovery
real numbers less so

how do you just discover numbers?
you count things

yes. multiple distinct things exist so numbers exist
the symbols are made up though

yes, but in physics the things are physically relevant things
in maths, just need the numbers by themselves
the idea of a number dettached from an object is maths

11:17 PM
hmmm.. but all physics is math which also happens to model something real
numbers model real things. u can detach the real thing from the study of numbers, but u can do that to any physical theory

@RyderRude physics isnt maths. if physics = maths just because it uses maths then finance and economics is physics
as well as ecology, and a whole host of other subjects.

physics is math which happens to model something real at the fundamental level...now the fundamental level is not well defined. As we go less fundamental, we give it different names like chemistry, ecology and finance
but there is a continuous spectrum from physics to these subjects

there is no continuous spectrum to wall street, although the tools and skills are transferrable
economics is a purely human construct

im just saying that of these can technically be modelled by physics due to reductionism

no, they can be modelled by maths

11:23 PM
they can both. using math is just more practical

the study of numbers as a mathematical object is its own discipline

and every property u discover there carries over to the real world at the fundamental level
so it's just a matter of semantics. their study is extremely close to physics
numbers are everywhere in the universe. u cant say that for finance
@qwerty i agree. if u define numbers as the subject u get after detaching the real world, then it's not physics by definition :) @qwerty
also see this post philosophy.stackexchange.com/a/2634 @qwerty
it explains the technical idea why natural numbers are more of a discovery than real numbers