@ACuriousMind i too hate this word :P. Sorry. I shall end this discussion. It's the bounds of language

@EmilioPisanty <-- this mood would just point to this answer

the average philosophical discussion

I hope this room is age limited lol

It's the h Bar but where exactly is the bar

Under the sink in my case

two pies under the sink ^_^

What does the highlighted part mean?

What is 3.166

The covariant derivative acts on Dirac spinors (in this case), why would I define it also for the gauge field?

Looks like covariant derivative equals partial derivative, but i'm literally just translating symbols I recognize from somewhere else without understanding the context

@Slereah the gauge field transformation $A_\mu\to A_\mu-\partial_\mu\theta(x)$

I mean, why would I be interested in defining a covariant derivative acting on the gauge field itself in this context?

Maybe because when you take higher derivatives it comes up?

Like I'm imagining applying an additional covariant derivative to 3.167...

Hello, I am wondering why, when writing the Lagrangian density, we can say $S = \int dt L = \int dt \int d^3x \mathcal{L}= \int d^4x \mathcal{L}$

specifically why we can make the last jump moving to $d^4x$

@Mr.Feynman omg i just used this fact

@Mr.Feynman I think it's that : $\partial ^{\mu} A^{\nu}$, under a gauge transformation, transforms to $\partial ^{\mu} (A^{\nu} +\partial ^{\nu} \alpha) = \partial ^{\mu} A' ^{\nu}$

So $\partial ^{\mu} A^{\nu}$ is a covariant object under the transformation properties of the field $A^{\mu}$

For the other field, $\phi$, $\partial ^{\mu} \phi$ wont be a covariant object under the transformation property of $\phi$ which is $\phi e^{i\alpha (x) }$ @Mr.Feynman

This is why the simple partial derivative doesnt work as a covariant notion of derivative for $\phi$, but works for $A^{\mu}$

It's becuz both fields have different transformation properties under the gauge transformation in consideration

I still don't find a mathematical reason for that remark, though. $A_\mu$ are the components of a connection, it makes no sense to take the covariant derivative of those

It's becuz u need some covariant notion of derivative. I guess u think that the simple partial derivative should just be the default for A. But consider that we always need to check that we are working with gauge covariant objects @Mr.Feynman

@Relativisticcucumber it's just the definition of a 4D integral. The measure is $d^4x=dtd^3x$

Becuz theory wud b weird if we werent working with covariant objects. Becuz gauge choice is supposed to be physically menaingless

@Mr.Feynman so it is worth checking that the simple partial derivative is a covariant object, given the transformation property of A

But yeah, i cant think of any use of this notion @Mr.Feynman as u said, A is the connection of the curvature

@Mr.Feynman okay so this integral is actually an integral over all of space and time then? so if we have a two dimensional situation (one space and one time), this means we integrate over 1D space and then over time? because im trying to figure out why we can throw out the boundary term if we integrate the lagrangian in the schwinger model case

Mh, again mathematically I don't find this compelling. We have the gauge field which makes the components of the connection on a bundle. The only mathematically meaningful operation here is the ordinary derivative, the idea of a covariant derivative of the connection itself - which is not a tensor field - doesn't make sense to me. @ACuriousMind may I ask your opinion on the matter?

@Mr.Feynman uggggh

it means nothing

it's a silly statement that should just be stricken from the text

I think it's one of those physicist explanation of gauge theory

@Relativisticcucumber yes, just in that case you wouldn't write $d^4x=dtd^3\vec{x}$ but $d^2x=dtdx$ but no one uses such notation I think :P

@ACuriousMind D:

The ones that are like "The covariant derivative is the derivative that is invariant under gauge transformation" kind of stuff

it makes no sense to define a "gauge covariant derivative" on the gauge field itself - the gauge field does not transform in a linear representation of the group of gauge transformations

local symmetry I should say

also this fundamentally misunderstands the nature of $A_\mu$ as a form

something like $\partial_\mu A_\nu$ is not something you can do to a form

This is physics @ACuriousMind

$A_\mu$ is four functions

you always need to antisymmetrize, and in that case they're right that in the Abelian case the anti-symmetrized version $\partial_\mu A_\nu - \partial_\nu A_\mu$ is equal to $D_\mu A_\nu - D_\nu A_\mu$

i.e. you could say that the "covariant derivative" is just the normal derivative on the connection form

I could rant about this for a bit longer but I would really just ignore that sentence

don't even commit to memory what I just said, just forget about this

@ACuriousMind A form? Aren't these just the components of the connection?

it's a form $A_\mu \mathrm{d}x^\mu$

I've seen that in the context of covariant classical E&M, but here they are also the components of the connection, and forms transform linearly...?

I think there is one value in this remark : that the partial derivative of A is a gauge covariant object. But that doesnt mean that we should start calling the partial derivative as covariant derivative, yes

It's just worth recognising gauge covariant objects

@Mr.Feynman forms transform linearly under diffeomorphisms. The other transformation of the connection is the gauge transformation

My bad, it's indeed a form. The bundle indexes of the connection components are fixed

I think the term is : "internal vs external transformations". The internal stuff operates on the internal degrees of freedom rather than spacetime ones.

@ACuriousMind so the logical steps are the following: 1) I define a connection as above 2) I acknoledge that by construction the components of the connection i.e. the connection one form are a differential form, that is a section of the cotangent bundle 3) Since they are forms the only sensible operation is exterior differentiation, which I "promote" to covariant exterior differentiation (?)

Point 3) is not true though, I can take covariant derivatives of forms

I have to go, might take a bit before I can return to this discussion

@Mr.Feynman Remember that $\phi$ is also a scalar. But that doesn't we can have the simple partial derivative as a covarianr object

I think of it like this : u gotta look for objects that are covariant wrt all redudancies, whether they are gauge redundancies or diffeomorphisms

@ACuriousMind Ok, if you don't mind I'll ping you in the next days if you are available :)

For the field $A^{\mu}$, the simple partial derivative qualifies as a covariant object wrt all redundancies. This means that the partial derivative is a physically meaningful object independent of the gauge choice

@RyderRude That is the reason to introduce a covariant derivative of a scalar field, sure

and AFAIK, if you take that definition, doing a $\partial_\nu$ on $A_\mu$ is no problem...

Whether that gives a quantity that transforms "nicely" or not is a different story

But maybe things are different in QFT land :) IDK. I am talking from my modest GR land knowledge lol

In GR land often it is emphasized that the covariant derivative also gives back a tensor quantity (not only acts on one). Where tensor again means "transforms nicely"

@Amit In QFT, we have the same ideas but we are also considering Gauge transformations in addition to diffeomorphisms. And we're looking for physically meaningful objects modulo all these redundancies

probably not a good idea to bring up GR in an EM gauge theory discussion

it's not the easiest fit

Okay I thought that 1-forms are kind of universal in that respect

The GR connection transforms very differently under coordinate transform

as you may know

Oh yeah it looks very different. Also, skimming through some equations with the $A$ object, it doesn't always have the same number of indices. Is this cookie or what? :)

$A$ always has the same number of indices, sometimes they just don't show it :p

they suppress the upper one sometimes?

In the case of EM they are suppressed because it's one dimensional

since it's the tangent space of U(1)

Ergo only one indice so it's not important to keep track

Oh I see. Well ok I won't dabble in that anymore, every field has its own favorite abuse of notation :)

it is noted for other gauge groups like SU(2)

So it's not a 1-form

that's why EM has only one particle but the weak interaction has 3 and the strong interaction 8

@Amit It's a Lie algebra valued 1-form

Cool, it's like a decomposition thingy if I understand

@RyderRude youtube.com/watch?v=dslafN9afMg

i dont think we can communicate 100% what we'd like to communicate

i don't really get how you could ever formalize the concept of sending information from yourself to another person

and if you can't even do that then certainly you cannot formalize the concept of sending information from one person to another person in general

You can formalize the process of sending the information, but the process of converting it to some internal representation you can formalize probably only up to a constant, a constant function, or a divergent series...

Formalizing the concept of sending information is what linguistics is all about eh? ^_^

Hm I guess I should have used the word communicate. Since I mean to encompass not only the means of transmission of the information, but also the pre-processing of the information (say by me) and the processing of information (say by you)

Yeah, it's the endpoints which are the tricky ones

:P it just seems impossible to ascertain how your brain processes the information I am sending

And whether it even does

There may be only one consciousness, not your brain and my brain

i suppose that is one way to potentially see the world :)

It makes sense 'cause we're the outcome of a certain process with shared roots

So a lot of what we call consciousness was forced on us

For good and for not so besty

Tea?

black tea today :D how about you

i dont quite resort to meth as was discussed earlier, but when I need to wake up I do some cups of black tea

lol

black for me too!

@SillyGoose wait what

I had to google if "meth" had another meaning I don't know

now is english breakfast a black tea?

XD oh there seemed to be some chatting about amphetamines

chat.stackexchange.com/transcript/message/63285943#63285943 scroll slightly past this msg

@Slereah Slereah, why do your comments always crack me up?

people have been extra funny lately

Slereah is always funny

not in the world i mean, but in hbar

@SillyGoose we need to cook

cook some delicious chicken parmesean i say yes

That was a BB reference :(

hehe

i've only watched a few episodes of BB

@SillyGoose you can't graduate if you haven't watched BB

And then BCS

@SillyGoose Oh sure it is, I think it's a prime example of such

Oh my god, Superconductivity has ruined my life. Now when I say BCS (Better call Saul) I think of superconductors

ah i see so a splash of milk then :D

lol do you watch nilered?

:)

i onoly know of BCS because i watched his video on making a superconductor

@SillyGoose No, but I remember one night watching videos and falling asleep and then a video of this guy started playing

I didn't actually watch it but I remember being in half-sleep and hearing him going on about extracting water from urine...?

oh no for some reason i mixed up BCS which i actually do not know and have not heard of with YBCO which is a material i guess used to make super conductors

that seems like it might be something he would do

No, maybe that was ammonia. Whatever, I was kind of sleeping and I suck at chemistry

looks like he has a video extracting urea from urine

Maybe it's the space station recycling system

@SillyGoose yeah that one

There's reasonably close Related threads, right?

If not duplicates

@EmilioPisanty This is from 2012, I'm not sure what you're asking