The h Bar

imbAF

00:13

Anyone in here knows what FORM is and can use it?

Arjun

Dayum hbar been cookin' some wild s#!t lately lol

qwerty

05:06

@Slereah btw how do you use IRC? i tried to try it out last month after you mentioned it a while ago but it seemed kinda dead

Slereah

06:18

@ACuriousMind Well trying to do it with one of the classical way (interaction with EM field) might be tricky

Since they accelerate when they lose energy and radiate infinitely IIRC

Tobias Fünke

07:17

@qwerty yeah

morning

qwerty

hey Tobias :)

Tobias Fünke

Hi :) How are you doing?

qwerty

I'm okay - a bit unmotivated for no good reason. but that's okay, it happens :) yourself?

Tobias Fünke

07:40

unmotivated in regards to what? :d I am fine so far, enjoying the weekend, although today I have to tidy up and so on :/ I hate it hehe

qwerty

oh - I'm supposed to be learning R for work, and separately there's my personal physics writing

one day we will outsource all tidying to robots

Tobias Fünke

07:56

I see. Yes, sometimes there is no motivation, but that's normal indeed

@qwerty Hopefully :d I really hate it. aaargh haha

Ryder Rude

"one wants to live in a world where robots do the dishes and humans do the art, not the other way around"

assuming it is non sentient robots ofc

Slereah

08:12

It must have been pretty weird to read Lawvere's papers in the 70's before the whole trend of Hegelian categories raw.githubusercontent.com/mattearnshaw/lawvere/master/pdfs/…

Ryder Rude

have you read Hegel's work?

@Slereah really interesting

"unification of geometry and logic"

Slereah

I have tried

It is a bit dense

Ryder Rude

oh

i think of logic as a language, while geometry can be one of the things that the language of logic refers to

e.g. logic can refer to a set universe. and within that universe, one can have geometrical entities like manifolds

but I think Lawvere is instead talking about a connection between the study of logic itself and the study of geometry

one always uses logic as a meta-language. but logic itself can be studied using logic. i think Lawvere wants to relate the study of logic to the study of geometry, instead of relating the language of logic to geometry

Herr Feinmann

08:45

@TobiasFünke It has to

qwerty

@HerrFeinmann va

Herr Feinmann

ahahahah

Tobias Fünke

hehe

Herr Feinmann

I love to deliver "va" with a straight face, which apparently stuns the other person

Tobias Fünke

how are you doing, fratello?

@HerrFeinmann not yet

Herr Feinmann

08:48

Oh God, I turned Tobias into a member of an Italian gang

Tobias Fünke

hehe

si

Herr Feinmann

I don't want to sound French now, but it's "sì"

> bel paese là dove 'l sì suona

Tobias Fünke

oh haha ok sorry

Herr Feinmann

Oh, no, just nitpicking :P

Tobias Fünke

didn't you learn that you should never ever disagree with your boss :p

lucky you that I am only a boss intern/trainee

Herr Feinmann

08:54

@TobiasFünke I will have a hard time understanding that in the future

I have a not-so-small problem with concealing disagreement

My face twists in a way I cannot control: I don't do it to be dramatic, but I've had soem lecturers stop, asking why I was glaring at them :P

Tobias Fünke

haha oh man

qwerty

@TobiasFünke i have heard from my german friend about certificates of employment. they sound scary

Tobias Fünke

@qwerty isn't this normal? I mean, isn't this everywhere in the western world the same?

qwerty

@HerrFeinmann one of my friends told me i give people judgement eyes IRL >.> i had no idea i even did that at all

@TobiasFünke I don't think so? we have reference letters and such but thats about it

Herr Feinmann

@qwerty Oh yeah, I can feel that. I have a RBF, so it happens to me :P

I just realized that because of the exchange-correlation hole

qwerty

08:58

like there's internal things similar with progress reviews but afaik that's not used for job applications for example

Tobias Fünke

@qwerty I thought you meant reference letters

qwerty

"It's a German thing... you don't give a reference that people can call, but after each employment you get a certificate and the next employer asks for it"

Tobias Fünke

would you mind elaborating what it is about? I probably know it, but I don't see right now what you mean

ah yes

qwerty

that's all i was told yes

Tobias Fünke

I mean I was not employed too often now haha. But that sounds "reasonable" for my german mind

"Arbeitszeugnis", if I don't misunderstand you again

qwerty

09:01

yes, i suppose so; i guess it depends... i suppose if you have "reference letters" you have more control who you ask to write them, i guess.

Tobias Fünke

employmentlawworldview.com/…

@qwerty yes

the text closes with "All these issues show why employers active in Germany should never underestimate the “Arbeitszeugnis” and the trouble it may cause"

Herr Feinmann

It took me a month but I got the hang of second subharmonics, yay

qwerty

haha >.<

Tobias Fünke

kanzlei-vonpreuschen.de/blog/arbeitsrecht/…

maybe ACM knows more, since is has a real job lol

Herr Feinmann

ACM writes the Arbeitszeugnis for his boss

qwerty

09:06

hehe

Herr Feinmann

I found a nice review of gauge symmetry breaking in superconductivity

Tobias Fünke

would you mind sharing it? ;)

Herr Feinmann

It's not advanced, actually it's so basic that it has skippable subsections about second quantization and so on, but it has Weinberg in the refs so I trust it. Here

I mean, it's book-level stuff

Tobias Fünke

thank you

Herr Feinmann

Useful for me now

Incidentally, Weinberg has a short discussion of SC in the second volume of his QFT book and there is also this paper (both in the references of the review above)

There is an available link but it's too long

>Superconductivity for Particular Theorists*

Tobias Fünke

09:16

do you know where to find the paper?

the crow won't give it to me

nvm, got it

Herr Feinmann

For anyone here, you can legally find a version of it just googling

Tobias Fünke

yes that's the way I got it haha

Usually I don't try it that way, because it won't work for the most cases. But yes, sometimes Google is your (my) friend

Herr Feinmann

Neither do I :P

Herr Feinmann

09:45

I think I'm disappointed by that review :P

Loong

@TobiasFünke I never got one myself.

Well actually there was one almost like this, but not in Germany.

The German ones have a certain code how to express things.

Like "his achievements met with our satisfaction" sounds good but it's actually the lowest passing grade like a D.

Herr Feinmann

10:17

@ACuriousMind I will appeal to your wisdom. I've just read something... unique (in case I'm not clear enough, I'm talking about eq. (48-49) here). The author describes, what is usually known as the unitarity gauge, as not being a gauge transformation (for reference, it looks a little different than the usual Higgs lagrangian because he is neglecting fluctuations of the modulus of the order parameter)

It seems that the author is confused and he's handwaving improper motivations to say that if something is gauge-fixed, then it is not gauge-dependent

Tobias Fünke

10:29

@HerrFeinmann I haven't read it yet, but it might be of interest to you: physik.uni-wuerzburg.de/fileadmin/11030100/Paper_Martin/…

Herr Feinmann

10:42

Judging from the title, it is. It's the reason I started reading that other paper

Although what I'm talking about is possibly a weird way to say things (for which I'd like to have ACM's opinion). I think that the author above should have said that - given that $A$ and $\phi$ are both gauge-dependent, and the unitarity gauge is a specific gauge, the transformation $(48)$ is of course a different gauge transformation if we are in a different gauge. $\tilde{A}$ is the unitarity gauge EM field, so of course it's gauge invariant

In other words: $\tilde{A}_\mu:=A_\mu-\frac{1}{2e}\partial_\mu\phi$ is the obviously gauge invariant of the unitarity gauge potential; what I dislike is saying this is not a gauge transformation. It's precisely the transformation that takes an arbitrary gauge field $A_\mu$ to the unitarity gauge

Okay, I think I convinced myself :P

ACuriousMind

12:13

@HerrFeinmann We've already said that the literature around the Higgs mechanism is often confused, but I think this is actually a careful presentation of it: Eq. (48) is indeed not a gauge transformation, it's a redefinition of a new dynamical field $\tilde{A}$ in terms of the dynamical fields $A$ and $\phi$. Eq. (49) is the correct statement that $\tilde{A}$ is invariant under gauge transformations.

@qwerty It's just a pre-emptive reference letter, meant to ensure that you have a reference even if your previous place of work goes out of business or would otherwise refuse to give you a reference. It's also meant to ensure that references are factual, which you cannot control if they are given personally over the phone or in private communication. It's an official document where you have legal recourse if the employer makes claims about you that are untrue.

This means the only thing those letters contain are facts and praise; a letter that contains only facts and no praise means you were the worst employee they ever had :P (This sounds like "He worked here. He fulfilled the minimum requirements in his job.")

qwerty

12:32

i see. my impression was based on my friend was saying something about a colleague unfairly receiving a poor grade for one of the criteria, and that not much could be done. since i didn't know anything about it, I just nodded in sympathy

ACuriousMind

I mean what level of praise exactly someone should receive in those letters is of course not objective and certainly some employers might write you a letter you feel is unfair to you, but I don't think that's any worse than a less formal reference badmouthing you in any other country

Herr Feinmann

@ACuriousMind Yes, I do remember our previous discussions. I agree that such quantity is invariant, as it is the field in a specific gauge. What I meant is that one may also read it is as the way to get to the unitarity gauge being in a given gauge. In this perspective I call it a gauge transformation (well, actually it's a different one for each gauge), but I agree about the invariance

I probably just don't like the phrasing, I don't know

qwerty

@ACuriousMind yes, fair. but you can choose you ask to write a letter for you, which seems not to be the case for this cert?

ACuriousMind

@qwerty It's "the employer", in larger companies my understanding is that HR essentially generates this from formulaic building blocks based on feedback from your manager or other people you worked with. There's not really a personal aspect to this letter in the way e.g. academic reference letters are personal

@HerrFeinmann I actually like this formulation better, and the author is 100% correct that it's not a gauge transformation in the way he phrased it. But you are correct that there is a way to tell the story with essentially the same equations and results where it is a gauge fixing.

Herr Feinmann

Okay, thanks for your help!

Loong

12:45

@ACuriousMind Yes, I have recently checked with our HR. As a manager, I only give grades like in school for something like seven different aspects; and HR translates that into standardized text blocks.

In smaller companies, the boss might say "you can write it yourself and I will sign it".

Ryder Rude

13:12

@HerrFeinmann i think that this gauge transformation depends on $\phi$. so one can identify it with an element of the gauge group only after one has fixed a $\phi$

as in, if u have two different field configurations on two different spacetimes, then this transformation is identified with different elements of the gauge group on the spacetimes

so in this sense, this transformation is not a "gauge group element"

also, by "gauge group element" I mean the infinite dimensional group whose action is given by $A_{\mu} \to A'_{\mu} -i\partial _{\mu} \alpha (x)$. this should be contrasted with the associated finite dimensional groups like U(1), SU(2), etc

also, one can do this kind of gauge fixing only when one knows the $\phi$. contrast this with other forms of gauge fixing which only rely on the knowledge of $A$

i thought mister nobody were talking about the meta post deletion. but they were talking about the stackexchange account deletion. i think they didn't accept the apology

Ryder Rude

13:54

it was false accusations. so there should've been a direct apology

Tobias Fünke

I think we should stop discussing this incident. Everyone expressed their opinion already.

We should instead learn from this situation, and establish (or re-establish?) a better discussion culture.

Ryder Rude

yeah..

ACuriousMind

14:14

@RyderRude Everyone in this conversation is aware of this. There is in principle no difference between a gauge condition that depends on $A$ and a gauge condition that depends on $\phi$. The gauge condition here in the gauge fixing interpretation is just $\phi = 0$.

Ryder Rude

14:26

@ACuriousMind what do u mean by this being the gauge condition? $\phi$ transforms to $\phi (x)e^{i\phi(x)}$ after this transform. i think this is not equal to 0

Herr Feinmann

If you had read my document, you would know that here the field is $\Delta=\lvert\Delta\rvert e^{i\phi}$, so that's not the field. It's the phase

Ryder Rude

@HerrFeinmann oh

but still, i would say that this transform is slightly different from a gauge transform, as it can be identified with an element of gauge group after fixing a $\phi$

but for a given $\phi$, it is a gauge transform, yes

but still, the usual gauge fixes are not dependent on $\phi$. this differs from the usual gauge fixes in that sense

it can still be called a gauge fix in the general sense though. i see no reason gauge fix conditions should only depend on A

ACuriousMind

That's what I already said.

Ryder Rude

also, it makes sense that this is the phase. i should've seen this. sorry

@ACuriousMind yes

i just found this transform peculiar and was trying to distinguish it from the usual gauge transforms

Slereah

14:47

The whole "super coordinates thing" I wonder how much you could describe it properly because like

The coordinates are patches of $\mathbb{R}^n \times \Lambda^p$

But $\Lambda^p$ is the algebra, not the spectra of that algebra

but de Witt still manages to get "coordinates" out of it

would it be like using as "coordinates" of $U$ the set $C^\infty(U)$

Silly Goose

what does the expression have to do with $S^1$?

this just looks like (part of) the definition of a group representation

ACuriousMind

@SillyGoose They're just spelling out the definition of a group homomorphism

Silly Goose

what does it mean to homomorph into the circle $S^1$?

ACuriousMind

@SillyGoose It's implicit that you are treating the circle with its canonical group structure that you get e.g. by treating it as the unit circle in the complex plane

Silly Goose

so i can just say a character is a unitary representation $\chi: G \to U(1)$?

ACuriousMind

14:58

@Slereah I'm not sure what you mean here: $\mathbb{R}^n$ is also an algebra, in coordinates you multiply different real numbers with each other all the time

@SillyGoose It's a group homomorphism. To be a representation there would have to be an associated vector space here on which that $\mathrm{U}(1)$ is the unitary operators.

Slereah

@ACuriousMind Yeah but they're on the wrong side of each other

ACuriousMind

A representation is a group homomorphism into the linear group of a specific vector space

Slereah

$\mathbb{R}^n$ is the space of probes for the sheaf of the space, $\Lambda^p$ is the space of values that the cosheaf maps into

Like they are different types of objects in this context?

The actual "space" for superspaces is just a single point

Although how much of this can be generalized I have no idea since non-commutative geometry is wildly different from the commutative case

Ryder Rude

are you learning Lawvere's idea about uniting logic and geometry

what is this idea in non category theory terms

Slereah

Maybe I should reread dewitt's book on the topic, perhaps it will make more sense with that angle in mind :p

Ryder Rude

15:07

Is it some kind of homomorphism?

Slereah

There was a lot of process to apply to Grassmann algebras before applying it to manifolds

I think a lot of it relates to nlab's modality business in this case, ie there's a lot of sorting Grassmann numbers into even and odd decomposition (the bosonic and fermionic modality) and the body and soul part is just the reduced and infinitesimal modality

Herr Feinmann

Is "gauge away" a legit phrasal verb? :P

Slereah

15:24

English doesn't have a regulating body for grammar

So gauge away I say

We have it here, though

Council of sleepy grampas

Ryder Rude

16:57

suppose i think of the universe as a set with some structure on top. like the set is the set of points. the structure is topology, etc and the functions on it which are fields

if the above is the set theoretic view of looking at the universe, what would be the category theoretic view

i think we would have to prioritise morphisms somehow instead of points

i am trying to see how to view a platonic mathematical object from A. Set theory perspective B. Category theory perspective

i am used to thinking of sets with some structure on top. how to switch to category theoretic perspective of platonic mathematical objects

and does the question make sense

Silly Goose

17:41

the responses to the meta post have made me lose a lot of faith in the stack exchange community...

Allie

I don't know if this

nevermind

Tobias Fünke

18:20

@SillyGoose Would you mind expanding why?

Bml

19:43

Hi everyone. In [this answer](https://physics.stackexchange.com/a/669614/376364), it is said: "If the equilibrium temperature when $\xi$ is put in thermal contact with $n^\textrm{th} $ auxiliary body is given by (2), then its equilibrium temperature when it is put in thermal contact with $(n+1)^\textrm{th} $ auxiliary body, that has temperature $T_i + (n+1)(T_f-T_i)/N$, is given by the average value:
$$\begin{equation}
\frac{\left( 2^{n+1}(N-n) -1 \right) T_i +\left( {2^{n+1}}n+1\right) T_f }{{2^{n+1}} N }. \tag{5}

ACuriousMind

@Bml If you put two identical bodies with temperatures $T_1$ and $T_2$ into contact and let them equilibrate, the result is that both end up at the average of their temperatures, i.e. $\frac{T_1 + T_2}{2}$.

Bml

20:06

@ACuriousMind Yes, I am aware of this. But this result is obtained by bringing two bodies into contact at what temperatures? I don't understand where this value comes from.

ACuriousMind

@Bml The answer tells you:

> and imagine to put subsequently ξ (initially at temperature Ti) in thermal contact to these bodies (first to the body at temperature Ti+(Tf−Ti)/N, then to the body at temperature Ti+2(Tf−Ti)/N, etc.)

you'll have to be more specific what about this description is unclear

Bml

@ACuriousMind The answer says: "its equilibrium temperature when it is put in thermal contact with $(n+1)^\textrm{th} $ auxiliary body, that has temperature $T_i + (n+1)(T_f-T_i)/N$, is given by the average value...". One temperature is $T_i + (n+1)(T_f-T_i)/N$, what is the other value?

ACuriousMind

@Bml It's the temperature the body $\xi$ has after being put into contact with the previous $n$ auxiliary bodies. Which is equal to eq. (5) by induction as the answer proves.

Bml

@ACuriousMind If we substitute $n+1$ for the formula you want to prove, we get eq. $(5)$, but this is not enough to prove the formula. Where do we get equation $(5)$ from?

@ACuriousMind Perhaps you mean equation $(2)$ instead of equation $(5)$?

ACuriousMind

@Bml The induction is just the elegant way of proving it. If we call that value from eq. (2) (not (5), yeah, sorry) $T(n)$, you can just directly (but laboriously) compute it recursively: $T(1) = \frac{T_i + T_i + \frac{T_f - T_i}{N}}{2}$. Then $T(2) = \frac{T(1) + T_i + 2 \frac{T_f - T_i}{N}}{2}$. Then $T(3) = \frac{T(2) + T_i + 3\frac{T_f - T_i}{N}}{2}$, etc. Plugging in recursively you can compute as many of the $T(n)$ you like to see that they do follow the formula given.

This is how you usually get the formulae we prove by induction: You look at the first few elements of the sequence and guess a formula that you then prove by induction.

(if the proof by induction fails the guess was wrong and you try again :P)

Herr Feinmann

20:30

Hello people

Or rather, Guten evening

Bml

@ACuriousMind OK, I got it. Thank you once again :-)

PM 2Ring

Greetings. I've been sick chat.stackoverflow.com/transcript/message/57854012#57854012 so I haven't been posting here or even lurking for ~1 month. I'm a lot better now, but still recovering, and don't have much energy.

Silly Goose

@TobiasFünke i think it was stated previously that related discussion should be contained in the meta post...and i do not wish to do a write-up at the moment, so i refrain from describing in more detail for now.

Herr Feinmann

@PM2Ring I'm so sorry to read that. I hope you fully recover soon and join us again. :)

ACuriousMind

@PM2Ring Oh, that sounds like it sucked. Good to hear you're doing better again, though!

Herr Feinmann

20:40

I read more about your experience in the message you linked, so as I understand, they cured your problem but you're still fatigued and need some time to recover

PM 2Ring

Thanks, @HerrFeinmann I miss doing Stack Exchange stuff. OTOH, I haven't had the brain power to do anything more complicated than listening to music for a while. :)

Silly Goose

does anyone know an instance of the generic fourier transform being used in physics? that is, for a finite group $G$ we have $\hat{f}(\rho) = \langle f, \rho \rangle$ where $\rho: G \to GL(V)$ is an irrep? Or if any of usual quantum mechanics has been framed in this way.

PM 2Ring

Oh, I'm still very short of breath, and my "toilet issues" aren't totally back to normal either... I was on oxygen for ~4 days.

Herr Feinmann

@PM2Ring I'm sure you'll have plenty of time for that. 3 years ago I was hospitalized suddenly, underwent a surgery and stayed in the hospital for more than a week (and it was during the pandemics). I didn't feel like doing physics - even if I was obsessed back then - and I took a break from study for a while. Later on, I recovered my will and energies, so I promise it will get better

For now just rest your mind, listen to music, watch some show. Don't overwork yourself, you'll have time for that

PM 2Ring

Thanks. I'm feeling better every day, and feel fairly positive, but I have no intention of pushing myself. I'm definitely taking it easy for as long as it takes. :)

See ya later

Herr Feinmann

20:47

See ya!

Silly Goose

is there a connection between flat connections and harmonic analysis?

flat connections on a $U(1)$-bundle over $X$ are characterized by representations $\rho: \pi_1(X) \to U(1)$. in physical examples, $X$ can be the reduced configuration space of two identical particles in $\mathbb{R}^3$ or $\mathbb{R}^2$. These have fundamental groups $\pi_1(X) = \mathbb{Z} / 2\mathbb{Z}$ and $\pi_1(X) = \mathbb{Z}$, respectively.

But then, flat connections are precisely in correspondence with characters of $\pi_1(X)$.

qwerty

@PM2Ring hey gosh that sucks. I hope you continue to be on the mend and feel better soon

@HerrFeinmann good morning

Herr Feinmann

21:08

@qwerty hello, being from the future

Ah, I've already used this line too many times

Tobias Fünke

@PM2Ring Sorry to hear that. Get well soon!

Herr Feinmann

21:33

I was thinking about something not-so-smart, probably. Gauge $\mathrm{U}(1)$ symmetry entails also transformations where the phase is constant $\alpha(x)=\alpha_0$. These transformations do not seem to differ from global $\mathrm{U}(1)$ transformations, so I was wondering, on the one hand the gauge symmetry carries no physical meaning and it's an artifacts of the constraints between our variables;

on the other hand, the global symmetry is a genuine symmetry. I'm a little confused about the idea that you could see the global symmetry as a subcase of the gauge symmetry, having said all that... Am I badly confused?

Should I see that as just a "coincidence"?

ACuriousMind

@HerrFeinmann Note that for the free Maxwell field, $\alpha = \text{const}$ is not a transformation at all. Since $A\mapsto A+\mathrm{d}\alpha$, there is no actual transformation that corresponds to constant $\alpha$. Remember that a gauge transformation is supposed to connect two physically equivalent states, but for constant $\alpha$ you're just doing $A\mapsto A$, there's no information about the gauge redundancy contained in that

So the global U(1) that can act on the objects/fields charged under the local U(1) gauge symmetries is just that - a distinct global symmetry, not a gauge transformation

In general this is why the center of the gauge group appears as a meaningful global symmetry - because the gauge transformations that are constant functions into the center do not change the gauge field at all

Herr Feinmann

Ok, that edit clarified it. I was confused by the "no transformation" at the beginning, but you just meant that it's trivial

The answer to my question is the one above, saying that the center of the group is a meaningful global symmetry

Okay, so my mental picture of breaking the residual global $\mathrm{U}(1)$ after gauge fixing starts to make sense

Slereah

Isn't the set of constant 0 forms part of the resolution of the de rham complex

ACuriousMind

21:58

@Slereah I don't see the relevant to the conversation :P

Slereah

Was wondering if it was

ACuriousMind

I mean the reason that the constant functions are not "gauge" is that they're in the kernel of $\mathrm{d}$

you're right to bring deRham into this in so far as for higher form symmetries where you have $T \mapsto T + \mathrm{d}\omega$ where $T$ is not a 1-form but some higher $p$-form, you get similarly global symmetries as those with $\mathrm{d}\omega = 0$, i.e in those in the kernel of $\mathrm{d}$ at that $p$

Herr Feinmann

22:17

Is there physics where this thing with higher forms comes into play (lighthearted question)?

I think I've only seen 1-forms (at most Lie-Algebra valued) learning QFT

Slereah

Yeah

For an n dimensional object, the connection is an n form

For curves (particles) it's a one form

Strings have a 2-form

Allie

22:36

is it ontopic to ask if my answer to a textbook problem is correct?

like to ask the question

on the SE website

qwerty

33

A: Should any check-my-work questions be made on topic?

Check my work question should always be off-topic. Those that can be rephrased should be rephrased. "Am I right?", "Is this correct?" or something else is always only of use to people who did the exact same derivation, and this is definitely too localized. To answer the bullet points in order: ...

Allie

hmmm

well i think my question is quite general

it doesnt have any details specific to the problem

it was pretty much just "calculate the diffraction pattern of a 1d lattice"

i see how it would be off topic to ask a specific physics question about, say, the trajectory of a ball with a given initial velocity and angle. but also i think the problem i worked on is quite general

qwerty

I don't know why you're debating it here: if you think it can be phrased in a general way, then ask away. if not, then you can still ask - just don't be disappointed by any downvotes or votes to close

Allie

I'm not debating?

I'm just clarifying

if you still think that's off topic then tell me

chatting on the internet is always so hostile

qwerty

there was no intended hostility, just genuine confusion.

Allie

22:47

Okay, sorry that i construed hostility. i just read from the question that "Check my work question should always be off-topic" but then you said "if you think it can be phrased in a general way, then ask away" so I wasn't sure

from the question it sounded like a hard no, from you it sounds like "if it's general enough, yes". I'm gonna post it, downvotes won't hurt me

Herr Feinmann

23:23

Generality typically means that you have to ask your question as a conceptual question which in principle can be interesting to people who do not deal with your specific problem

Also, the more specific you are, the less likely you get an answer

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