Looks like covariant derivative equals partial derivative, but i'm literally just translating symbols I recognize from somewhere else without understanding the context
@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
@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?
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
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
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
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
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
AFAIK, When you have a form usually what is meant by writing $A_{\mu}$ is actually $$ A_{\mu} := \left( A_{\alpha}dx^{\alpha} \right )_{\mu} = A_{\alpha}\frac{\partial{x^{\alpha}}}{\partial{x^\mu}} = A_{\alpha}\delta^{\alpha}_{\mu} $$
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
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? :)
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)
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
Since thrusters have no matter to push off of, how can a spacecraft generate thrust in the vacuum of space?
