The h Bar

Silly Goose

12:16 AM

okay this is my fast and loose summary of the discussion on differential forms over a differentiable manifold :P

Relativisticcucumber

1:27 AM

@ACuriousMind what does this mean

@SillyGoose u eldritch being

Silly Goose

2:17 AM

u shellshermichael @Relativisticcucumber

lucabtz

3:02 AM

@SillyGoose you said you define the exterior derivative but you didn't

Anyhow what's the name of the font

Silly Goose

@lucabtz oh yes i wrote "we define the usual..." so i didn't have to write down the definition XD but i should add that in...

let me check about hte font

im not sure it might be called "libertine"

lucabtz

@SillyGoose thanks

Silly Goose

this latex template also uses this package which seems to affect the font: \usepackage{mathptmx} % d. ...in a simple roman face except where indicated below (§3).

naturallyInconsistent

5:07 AM

@SillyGoose you can always go back to old Newtonian mechanics with every single particle in physical space instead of the combined configuration space~

Silly Goose

meow

naturallyInconsistent

@SillyGoose KathyLovesPhysics covers the original papers of E&M. Extremely amazing stuff.

@SillyGoose M I A O ~ ~

@SillyGoose It cannot be, because the completion of putting together Maxwell's equations were all done without the conception of SR nor QM. Of course, if you want a modern account that emphasises how SR and QM are actually really needed for Maxwell's equations to be as they are, you might find one, but it is not how we got here.

Also, ACM's reply to you there is undoubtedly correct, but he omitted an important aspect, probably because everybody talks about this: The spin-statistics theorem says that the spin one of photons makes it so that photons obey Bose-Einstein statistics, and this amplifies the collective wave aspects, so that the classical wave aspects are salient. Contrasted with electrons obeying Fermi-Dirac statistics, which amplifies individual particle aspects.

@SillyGoose burn this. Either define the 1-form with a minus sign, or define the 2-form with the opposite sign. Nobody should be taking the negative of a exterior derivative as a convention.

John Rennie

I have a probably really simple question about an example in Lancaster and Blundell's QFT book:

In 1.37 how does the derivative of $\tfrac12 (\partial_\mu\phi)^2$ end up as $\partial^\mu\phi$ rather than $\partial_\mu\phi$?

Presumably it is related to the fact that $(\partial_\mu\phi)^2 = (\partial_\mu\phi)(\partial^\mu\phi)$?

Obliv

5:25 AM

Is it possible that I have finally found a moment where I could help John Rennie?

naturallyInconsistent

@SillyGoose horrifying mistake, but it is also the kind of mistake that would be completely prevented if you had concrete applications of all the various entities involved, instead of an abstract-only learning. It is not just the metric that is a symmetric tensor. There are plenty of others in physics.

@JohnRennie yes

Silly Goose

@JohnRennie i think if you write it out as you did and then use the metric to lower the index of the second term and then differentiate using product rule you will get your wanted result

Obliv

I think $$\frac{\partial}{(\partial_{\mu}\phi)}\frac{1}{2}(\partial_\mu\phi)^2 = \frac{2}{2}\frac{\partial}{(\partial_{\mu}\phi)}(\partial_\mu\phi)(\partial^{u}\phi)$$

naturallyInconsistent

In particular, it is that $\frac12(\partial_\mu\phi)^2=\frac12\partial_\alpha\phi g^{\alpha\beta}\partial_\beta\phi$ so that the derivative denominator that is $\partial\partial_\mu\phi$ with the two partials, converts first just one, and then the other, into a kronecker delta. Then the two parts are the same thing, and cancel away with the half.

@Obliv why do you have a dangling $\partial^\mu$ at the end?

Obliv

woops

naturallyInconsistent

5:30 AM

@Obliv the new edit is straight up false...

Obliv

welp too late to edit it now

I shoulda put a bracket around those two terms

it's meant to be partial of those last 2 terms which should just be $\partial^u\phi$

Silly Goose

Obliv

or whatever that var. is in the superscript

Silly Goose

writing index notation in latex is tiresome XD

naturallyInconsistent

@SillyGoose This package defines Adobe Times Roman (or equivalent) as default text font, and provides maths support using glyphs from the Symbol, Chancery and Computer Modern fonts together with letters, etc., from Times Roman. It supersedes both the original times and the mathptm packages.

Silly Goose

5:32 AM

the bottom half of the page is the correct calculation (im pretty sure)

i obtain $2$ times the correct result because i left out the factor of $\frac{1}{2}$

Obliv

@JohnRennie I have no idea what's going on physically/mathematically but it seems to me if you plug in that identity and then take the derivative, the term $\partial_{\mu}\phi$ vanishes and we have $(\partial^u\phi)$ kinda like $\frac{\partial}{\partial x}xy = y$ where $x = \partial_{\mu}\phi$ and $y = \partial^u\phi$

Silly Goose

@naturallyInconsistent oh

naturallyInconsistent

@SillyGoose it is. It skipped the important step, though

@Obliv no, this is not what is happening

John Rennie

Of course, yes, silly me :-)

Thanks everyone :-)

naturallyInconsistent

yay

Obliv

5:34 AM

I mean that's literally what's happening @naturallyInconsistent

naturallyInconsistent

It is not a usual day that we get to help Rennie

@Obliv and I am telling you that you are wrong on that

Obliv

In what way?

Silly Goose

@Obliv $(\partial^\mu \phi)$ is not an independent variable

from $\partial_\mu \phi$

naturallyInconsistent

@Obliv The correct reason is already written by the wacky fowl

If you want it in more detail and LaTeXed goodness:

Silly Goose

if you wanted to do the calculation systematically, you should lower all indices via the metric tensor and then apply usual calculus to take derivatives

Obliv

5:36 AM

where did the $\eta$ come from

what da dog doin

John Rennie

Minkowski metric

It's being used to lower the index.

Obliv

I like my explanation better because I can understand it, even though it's wrong :D

naturallyInconsistent

$$\begin{align}\frac{\partial\ \ }{\partial\partial_\mu\phi}\frac12(\partial_\nu\phi)^2&=\frac12\frac{\partial\ \ }{\partial\partial_\mu\phi}\partial_\alpha\phi\,g^{\alpha\beta}\,\partial_\beta\phi\\&=\frac12\left[\delta_\alpha^\mu\,g^{\alpha\beta}\partial_\beta\phi+\partial_\alpha\phi\,g^{\alpha\beta}\delta_\beta^\mu\right]\\&=\partial^\mu\phi\end {align}$$

@Obliv that's a paddling

Obliv

so what do the subscript/superscripts represent

naturallyInconsistent

@Obliv contravariant v.s. covariant indices

Obliv

5:42 AM

can I learn this power

naturallyInconsistent

@Obliv you should

Obliv

I thought my intro abstract algebra course would touch on concepts like vector spaces,modules, etc

i guess technically we did touch on vector spaces via fields

but not explicitly

I'm just gonna learn on my own what manifolds are and tensors etc

all the important stuff in physics

didn't even know this was a subject of its own

meow

naturallyInconsistent

@Obliv technically, none of it is of any importance. Plenty of physicists get by with none of these. But it is very helpful to know some of these maths

M I A O ~ ~

Obliv

I guess you're right, but it feels like I've come this far, so might as well put in the extra effort to learn a few more abstract concepts

plus then I can decipher what is going on with these scribbles in the h-bar

naturallyInconsistent

yes, these are very helpful stuff to learn, but unfortunately many coverage are just bad

Obliv

5:49 AM

it looks like truly nonsense from my PoV :P

@naturallyInconsistent do u know of a good resource for deriving the ensembles in stat mech

Schroeder served me well in like the first couple chapters but I'm starting to see the shortcomings

it's probably still my best resource cuz the curriculum follows it pretty closely

naturallyInconsistent

@Obliv You should try Ian Ford Statistical Physics, an Entropic Approach. If you want a lot of maths, do Callen.

Callen does a lot of the physical arguments properly.

Obliv

Okay, will check those out. Thanks :)

ACuriousMind

@Relativisticcucumber incomprehensible utterances are a hallmark of eldritch beings a la Lovecraft - once you understand them, you typically go insane, i. e. become a physicist

Obliv

oh no the mods are no longer asleep. shoos the animals out of the bar

naturallyInconsistent

M I A O !!!

Obliv

6:00 AM

We have geese just chillin on campus snacking on the grass and they're so docile/used to students just walking near them.. but I have to remind myself that they're actually just cobra chickens..

no offense @SillyGoose

John Rennie

I wonder if I should admit we had roast goose for the big Easter meal.
It was very tasty!

Silly Goose

6:14 AM

i will admit that i think duck is one of the greatest foods ever

@Obliv omg i wish we had geese

we have squirrels and lizards

Relativisticcucumber

6:32 AM

@ACuriousMind lol

my supervisor believes that nobody really understands physics and that people just say random things that they think are related to what is being discussed

i think thats a him problem

i have begun to dig too deep into his endeavors. i have found that 30% of his citations are from himself. on average, 70% of the citations of his papers are by him

or "on median" im not sure which i should say. selecting papers at random, examining who they have been cited by, 70% of it is him is what i mean

Silly Goose

zoinks

naturallyInconsistent

@Relativisticcucumber omg that is something that I would consider for reporting to the university. Can you, like, pivot to another supervisor or something? Very very scary

Relativisticcucumber

im leaving in august anyways but yeah i didnt realize how bad this guy was until recently

some events have made me think quite negatively of him and i began to realize more and more bad things once i lost my initial respect for him

sadly i have come to realize many figures in science are corrupt :( and it seems that trying to do anything ab this just gets u blacklisted :(

naturallyInconsistent

6:50 AM

Sigh

That is also why niche parts of science deserve scrutiny. It is much more difficult to make shit up in hotly contested experimental areas, whereas if you write shit in a niche area that nobody really cares about, you can hide quite well

Relativisticcucumber

yeah well this guy is quantum info but hes not that big. it just bothers me bc he refers to himself as someone who is big in the field and its just annoying bc i wanna be like bro i know u r just a citation machine with urself who r u fooling

but it does make me sad that science is corrupt. but its just smth to learn to navigate i guess : ////

Silly Goose

@Relativisticcucumber MIT be like: 0.0

Relativisticcucumber

but ill just learn as much as i can and then onto bigger and better things in august !!!

naturallyInconsistent

@Relativisticcucumber omg so many horror stories of QC. Very very annoying

Relativisticcucumber

@SillyGoose MIT be like "lets name this new building after a pedophile!"

lets visit sex offenders in jail!

still better than stanford

there the offenders arent even sent outside county lines!

nuff said

naturallyInconsistent

6:57 AM

When miao miao jumped out of QC to join this company, the physics people who heard what miao miao was doing in QC, all agreed that miao miao dodged a bullet. And then the people who touched QC before, added their own QC horror stories

Relativisticcucumber

@naturallyInconsistent god

Silly Goose

this notation makes me sad ;_;

naturallyInconsistent

yeah, eyes are not happy with that

@Relativisticcucumber anyway, your latest question about BEC, im not sure what you want more than the fact that Bose-Einstein statistics favour a LOT of particles sharing the same state.

Silly Goose

7:21 AM

i wish there was a mathematics for theoretical physics course

naturallyInconsistent

@SillyGoose actually, miao miao benefitted a lot from one

but even that is not at all enough. There are just way too much maths that can be applied to physics

Silly Goose

true dat

i really wish i knew diffe g...but maybe i can learn enough over this summer

123

7:36 AM

Hello Everyone..

Ryder Rude

hi

123

What is the gravitational field equation in 2D in newtonian mechanics?

Slereah

Depends how you derive it :p

123

At which i can find field at point in space as vector field by magnitude and direction.

Ryder Rude

depends on the context

123

7:38 AM

like the case of earth gravitational field

Slereah

You can just pick it to be the same as the Newtonian one, the Laplace equation : $\Delta \phi = \rho$

In which case since it is in 2D the field drops off as $1/r$ instead of $1/r^2$

123

$g = \frac{GM}{r^3}\vec{r}$ is one dimensional

Slereah

The force I should say

The field drops off as $\ln r$

123

how to do this in two dimensional in Earth case

Silly Goose

@123 what context are you working in? are you trying to describe the force of gravity on an object close to the earth's surface?

and what do you mean by 2D? are you looking at the earth as if it were a circle drawn on the page and an object drawn above it?

123

7:40 AM

@SillyGoose I just want to create Earth gravitational field nothing else.

Ryder Rude

@123 this is a vector equation. why do u say it's 1d

this eqn is for 3d

123

@RyderRude because it has only $\vec{r}$ not angle

what is the description of "r"

Ryder Rude

$r$ has 3 components, so 3d

it has the vector r, not the length r

123

@RyderRude Can you pls share the equation of "r" which shows 3D

Silly Goose

$r = x^2 + y^2 + z^2$ as measured from the center of the earth (most likely but it is hard to tell because i am not really certain what sort of context you are working in)

Ryder Rude

7:43 AM

@123 can u pls first describe what this formula means to u

123

@SillyGoose Thanks. What do you mean by context. Can you pls tell few example of different context.

Silly Goose

@123 if you are writing down this formula then it sounds like you are doing a problem in which you are considering the earth and another object in space and considering the force one mass exerts on the other

alternatively, you can be at the surface of the earth and considering the gravitational force acting on an object near the surface of the earth like you do when you consider a baseball flying through the air

123

@SillyGoose No , i just want to draw vector gravitational field of Earth.

Silly Goose

okay so are you just considering earth in space

123

@SillyGoose Yes

Silly Goose

7:47 AM

@123 then the formula you wrote seems appropriate

for 3D

123

@SillyGoose Let me animate vector field using this formula. Then i share you.

Silly Goose

i think you are missing a negative sign--the gravitational force should be attractive

123

I do not have problem with minus sign. but $\vec{r} = <x^2 , y^2>$ doesn't work

$\vec{r} = <-x^2 , -y^2>$

It is not looking spherically inward.

Silly Goose

$\vec{r} = (x, y)$

123

What is the problem with my gravitational field

Silly Goose

7:54 AM

if you are looking at earth in 3D space, it has gravitational field $\vec{g} = -\frac{GM_e}{r^3}\vec{r}$ where $r^3 = (x^2 + y^2 + z^2)^{3/2}$ and $\vec{r} = (x, y, z)$. If you are looking at just a plane in space (so specializing to 2 of the 3 dimensions), just set $z = 0$ and let $\vec{r} = (x, y)$.

123

but scalar "r" doesn't help in animating field.

Pls share equation of $\vec{r}$

Is it okay $\vec{r} = <-x , -y>$

Ryder Rude

the force shud increase as we go toward the origin

it is undefined at the origin

123

@RyderRude Aaah okay.. let me again animate

naturallyInconsistent

You should be trying $$\vec F_G=\mu\begin{pmatrix}\frac{-x}{(x^2+y^2)^{\frac32}}\\\frac{-y}{(x^2+y^2)^{\frac32}}\end {pmatrix}$$

123

@naturallyInconsistent Aaah Thank... This is what i am searching for.

@naturallyInconsistent Perfect it worked thanks a lot

Ryder Rude

9:09 AM

the militaries of modern countries are something out of sci-fi when compared to the greatest militaries from hundreds of years ago

air force alone makes it a completely different thing

but in the movie Avatar, they show the old war techniques of horses and arrows winning over modern military tech in a war

but they didnt use nukes in the first movie maybe because of public backlash back on earth. in the beginning of the second Avatar, they use a nuke and win the war

Avatar: The Way of Water - RDA returns to Pandora (HDR - 4K - 5.1)

►

Silly Goose

let $M$ be a smooth manifold. let $x \in M$. i want to define the tangent space $T_xM$ as the space of all derivations over smooth functions at $x$ (with appropriate addition and scalar multiplication defined on the derivations).
I am a bit confused because wikipedia states that a derivation as $D: C^\infty(M) \to \mathbb{R}$, so it sends a smooth function $f: M \to \mathbb{R}$ to a real number.
Is that right? I would have thought that it should send the smooth function evaluated at the point $x$ to a real number, so $f(x = x') \mapsto Df(x) \lvert_{x = x'}$

Ryder Rude

9:33 AM

u can think of $D_{x'}$ as the map $D_{x'}(f(x))= Df(x)|_{x=x'}$. the input has to be the function $f(x)$, not the function evaluated at the point $f(x')$ becuz the latter loses the neighborhood information required to differentiate @SillyGoose

to be more precise, the input is a germ of functions

becuz the equivalence class of all functions, which hav the same neighborhood information around $x'$, get mapped to the same real number

Silly Goose

i see

Ryder Rude

10:16 AM

Oxygen started out as poison for most of the life on earth

life persisted despite that massive catastrophe

now life has come to a point where we might start terraforming other planets after another 1000 years

lucabtz

10:44 AM

@SillyGoose it is in fact wrong. You should see it as a map $D: C^\infty_{x^0}(M) \longrightarrow \mathbb{R}$. $C^\infty_{x_0}(M)$ is usually called the germ of $C^\infty(M)$ functions at $x_0$. It is the quotient of $C^\infty(M)$ by the relation $f \sim g$ if there exists a neighbourhood $U$ of $x_0$ such that $f|_U = g|_U$

this is just telling you that the derivations at $x_0$ should only depend on the local behaviour of the functions at $x_0$

glS

@naturallyInconsistent sorry, stumbled about this conversation and am very confused. I know who you're talking about in MIT, but who is "miao miao" here??

Mr. Feynman

12:39 PM

@glS when naturallyInconsistent writes "meow" or "miao" he means "me" or "I"

user 85795

I always thought "miao" was Italian for "meow." :P

Mr. Feynman

It is

It's pronounced the same way

user 85795

> “Meow” is an established spelling in English. Rossini spells it “miau” in his famous duet for two cats - but he would; he was Italian. “Miao” is the preferred French spelling.

re: Quora

cc @Slereah

Now I'm wondering about the German spelling 🤔

Mr. Feynman

12:55 PM

@user85795 "Züfefrysvrfhrbden"

user 85795

Thanks pal.

lucabtz

1:54 PM

What's a Cartan connection

Slereah

2:20 PM

A connection that has a "translational" component to it

lucabtz

3:10 PM

@Slereah it seems related to solder forms somehow

lucabtz

3:38 PM

Ahh the translational part IS the solder form

Ohh it's called affine connection because of this translational bit

Vincent Thacker

3:56 PM

8 latest consecutive questions all closed

Longest chain I've seen so far

Ryder Rude

hi

there was a spacecraft which recorded Saturn

Real Images From Saturn: What Cassini Actually Saw There

►

these planets are hellish rubbles

i really want to be in different places of the universe, but our tech doesnt allow it

people get to be in the universe only to roam a point sized volume

bolbteppa

4:23 PM

He is indeed the king

@SillyGoose all the big language of bundles and manifolds and derivations wasn't very helpful when it came to actually computing a simple derivative, there is a very deep lesson in this

Mr. Feynman

5:13 PM

I love that bolbteppa is magically summoned as soon as there is a possibility of insulting bundles :P

Confess @bolbteppa you dream of bundles in your most gruesome nightmares

naturallyInconsistent

@Mr.Feynman oh, im sure he is admitting guilty of something

Loong

@user85795 miau

Sir Cumference

5:28 PM

anyone else get bugged out when people write the $dx$ before the integrand in integrals?

like I know what they're going for by making it look like an operator, but it just feels weird

ACuriousMind

I had a phase where I preferred that notation :P

nowadays I don't really care

nickbros123

real ones dont write the $dx$

Sir Cumference

@nickbros123 from what I recall in my shoddy calc on manifolds class, it does actually matter in some cases

ACuriousMind

@nickbros123 well, you kinda have to write it when you're about to do substitution or potentially work with more than one measure

Sir Cumference

what's really weird is this derivative notation

$d^2$ in the denominator

like again I get what it's going for, but just keep the notation conventional

ACuriousMind

5:33 PM

but sure, I, too, prefer the pristine efficiency of notation like $\int_V \omega$ integrating a form over a volume

nickbros123

speaking of the subject matter, @SirCumference, I recently bought the hard-cover version of spivak calc on manifolds :)

Sir Cumference

@nickbros123 I'm so sorry

nickbros123

@SirCumference xD

Arjun

@nickbros123 go study biology lol

nickbros123

sad lyf

Vincent Thacker

6:02 PM

@SirCumference Yes, it should be better written as solid angle

The power of two looks ugly

bolbteppa

@SirCumference One becomes an adult when this becomes their default ;)

Sir Cumference

@bolbteppa I want to stay a kid forever then :P

it's just nice to have it clearly written where the integrand ends

also just looks more appealing tbh

Obliv

only disadvantage is $\int (\dots)+(\dots)dx$ looks like it's enclosed by the integral and dx whereas $\int dx (\dots) + (\dots)$ need to use a bracket/parentheses to enclose what you're trying to integrate

Sir Cumference

6:17 PM

there ought to be an international mathematical committee that decides the convention for a lot of the notation

not that anyone would need to follow it, but we could deem any deviation from it as nonstandard

well then again, I doubt physicists would follow the conventions mathematicians establish in any case

Obliv

Don't you think standards detract from the beauty of the art that is mathematics? :P

Sir Cumference

I do not, in fact :P

Obliv

If one is to believe beauty is immutable and the way one perceives it doesn't matter 🤔

when one "understands" something in mathematics is like any other evocation :O

I was kinda tongue in cheek about it because I don't really think of math as "art" like most mathematicians do but now Idk

like how a piece of art can evoke deep feelings in some regions of the brain, math can do the same just in a different way..but one is emotional and the other is logical. I'm not convinced you can call them the same thing (calling math art)

bolbteppa

$\int dx f(x)$ has always looked more badass, when you've suffered through qft you've earned the right to write this notation

Obliv

I agree. But when everyone starts doing it, it becomes less cool like most things..

Mr. Feynman

6:39 PM

@bolbteppa I realized I am a man by now when yesterday I wrote an integral with the measure after the function and I barred it to have the measure first :P

It's also much easier to read, so you know the variable first. It's similar to the Spanish ¿ (although we don't put the measure at the end too :P)

I can't let myself do that now though, not until I take measure theory. My prof would definitely fail me if I wrote $\int d\mu f$ :P

bolbteppa

I remember purposely not doing it until I got to qft

Mr. Feynman

What is astonishing to me is that somebody once mentioned it in the MSE chat and they were like "no one uses that notation"

I got a similar feedback when I used $\frac{1}{A}$ for the inverse operator and there was no way to have them accept it :P

Sir Cumference

@bolbteppa I don't trust the quantum physicists with notation after they started putting arrows on top of norms

or in some case, boldface

like wtf sakurai

well actually that's a bad example

but i swear it's something he does and it drives me nuts

Mr. Feynman

I don't understand the problem :P

ACuriousMind

what's wrong with that notation

bolbteppa

6:45 PM

Physics notation is just efficiency, when they write $|x|$ you know they mean a 4-vector (or an $n$-vector depending on the context)

Sir Cumference

@ACuriousMind I chose a terrible example in the heat of the moment

this one's better

Mr. Feynman

I love the way you panicked right after sending that pic

Sir Cumference

like why is $J^2$ bold and yet $J^2_z$ isn't

$J^2$ isn't a vector

Mr. Feynman

@SirCumference oh you mean $\vec{v}^2$ instead of $|\vec{v}|^2$?

Sir Cumference

yeah

ACuriousMind

6:47 PM

again, nothing wrong with that

since $\vec v\cdot \vec v = \lvert \vec v\rvert^2$

Mr. Feynman

Because $\vec{v}^2$ is shorthand for $\vec{v}\cdot\vec{v}$

Sir Cumference

it's just that it's bizarre to write norms (scalars) in the same fashion as vectors

ACuriousMind

and of course $\vec J = (J_x,J_y,J_z)$ is a vector (operator)

Sir Cumference

Like this isn't a thing that most people do outside of QM anyways

Usually in classical mechanics people would just write $J^2$ without the boldface

Mr. Feynman

So, if you want to nitpick: $|\vec{v}|^2$ is by definition $\vec{v}\cdot\vec{v}$, while $\vec{v}^2$ is just a shorthand for the latter

Sir Cumference

6:49 PM

and more generally mathematicians don't even use arrows or boldface when writing vectors in linear algebra

@Mr.Feynman Most people would just write $v^2 = \vec{v} \cdot \vec{v}$

I mean I guess it doesn't really matter, but it just feels weird to see arrows on top of scalars

Mr. Feynman

@SirCumference in that case they mean $|\vec{J}|^2$ of course but it's potentially confusing to use square without either $|\cdot|$ or $^2$

Sir Cumference

@Mr.Feynman well that is true

ACuriousMind

@SirCumference that's because in classical mechanics there is always a number $J$ - the QM intro text is emphasizing that while the operator $\vec J^2$ is a scalar operator, there is no scalar operator $J$ that you could square

Sir Cumference

@ACuriousMind I mean I feel like that really only needs a paragraph of explanation

not modifying the whole notation in the book

ACuriousMind

sure there's other ways to do that, but it's just a choice of convention

Sir Cumference

6:52 PM

I guess so

Mr. Feynman

I love also that the discussions about notations are the most heated up :P

ACuriousMind

there's nothing wrong about doing this, and it is internally consistent

Sir Cumference

like most things, how we feel about convention is arbitrary

but some things just stick out like a sore thumb to me

@Mr.Feynman that is true lol

Mr. Feynman

All physicists know that conventions are conventions and yet they'd be ready to kill to defend their favorite one

4

Sir Cumference

quite so

ACuriousMind

6:54 PM

@Mr.Feynman for me it's more that I rag on physicists for not doing math properly all the time, so I feel some need to defend them when they're just being a bit silly about notation :P

Sir Cumference

I wonder what physics would look like if they followed all the advice of mathematicians

well I guess just mathematical physics

but I'm curious if it'd really make the major topics that much harder to learn

Mr. Feynman

@ACuriousMind we kind of are on the same boat, yet

it is time to torture ACM now. Let's switch to passive vs active stuff

bolbteppa

Active vs passive can get you into real trouble unfortunately

Mr. Feynman

@SirCumference still studying simple pendula

@bolbteppa Most definitely, I hate discussing that stuff

Every time I manage to get myself confused

Silly Goose

7:52 PM

@bolbteppa true it’s not so useful for computing things…but one doesn’t really need to know much at all to compute things. The “abstract nonsense” is my preferred way of organizing information in my head :P

It is hard to visualize the abstract holonomy of a connection…and i prefer to think in words anyways. “Abstract nonsense” also lets you recognize things in the future with little additional effort

it also just answers questions that people maybe are okay with glossing over but which seem pretty fundamental. like why should the connection 1-form be lie algebra valued? and the lie algebra of the structure group no less. well the answer is answered immediately if one just constructs it, and the construction is quite natural

@SirCumference i like it :D it makes me not forget to add the measure in the first place lol

a lot of stuff even in the abstract nonsense just seems to be linear algebra anyways...

Obliv

8:16 PM

How does an LLM like chatgpt or any of the other ones correctly analyze problems?

I know at the end of the day it's just a brute forced non-intelligent piece of code but it's weird how it can do moderate computations and problems

DIRAC1930

10:36 PM

@SirCumference Have you tried reading a quantum mechanics book written for mathematicians?

It's impossible to decipher what is actually going on lol

Silly Goose

11:03 PM

what do you mean ;)

DIRAC1930

11:17 PM

lol precisely

Silly Goose

11:47 PM

let $P$ be a principal bundle. define a connection on $P$. i am trying to show the equivalence of the horizontal subspaces defined by the corresponding connection one-form and the horizontal subspaces defined by the definition of a connection.

I should be able to ignore definition 10.4 and from the listed properties in the definitions show that $X \in H_pP$ iff $X \in \ker\omega_p$

Nakahara proves one direction in the text, so I am attempting the other direction and got to this point.

However, I am not sure how to show that $(10)$ vanishes

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