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

@ACuriousMind what does this mean

@SillyGoose u eldritch being

u shellshermichael @Relativisticcucumber

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

Anyhow what's the name of the font

@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"

@SillyGoose thanks

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).

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

meow

@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.

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 $(\partial_\mu\phi)^2 = (\partial_\mu\phi)(\partial^\mu\phi)$?

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

@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

@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

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

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?

woops

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

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 $\partial^u\phi$

or whatever that var. is in the superscript

writing index notation in latex is tiresome XD

@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.

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 $\frac{1}{2}$

@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 $y = \partial^u\phi$

@naturallyInconsistent oh

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

@Obliv no, this is not what is happening

Of course, yes, silly me :-)

Thanks everyone :-)

yay

I mean that's literally what's happening @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

In what way?

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

from $\partial_\mu \phi$

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

If you want it in more detail and LaTeXed goodness:

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

where did the $\eta$ come from

what da dog doin

Minkowski metric

It's being used to lower the index.

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

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

@Obliv that's a paddling

so what do the subscript/superscripts represent

@Obliv contravariant v.s. covariant indices

can I learn this power

@Obliv you should

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

@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 ~ ~

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

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

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

@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.

Okay, will check those out. Thanks :)

@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

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

M I A O !!!

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

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

@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

zoinks