@Mr.Feynman sure, this is one of these definitions that only start feeling natural after you've seen how useful it is

Oh, ok so it is ad hoc. I never doubted it was correct but I thought there was something I was missing to make it natural

I mean you could present it in a way that is not ad hoc

but really the point of this expression is just that it is convenient notation for following computations, not that it's some great insight into the structure of Lie groups or whatever, so I don't think anyone bothers

@ACuriousMind For a matrix Lie group yes, for a general Lie group I can't see how right now

@WaveInPlace u shud also look up "Fourier Analysis" to see how signals with boundaries r sums of plane waves

But good to know, I can now stop thinking about it

And keep on writing my self-study booklet regarding gauge theories and principal bundles

Which became necessary as soon as I was introduced to the physicists' way of writing down the QCD lagrangian

@RyderRude, I'm somewhat familiar with Fourier analysis. Most of my experience is with 1D work though, so I think of it more as a sum of sines.

:)

@WaveInPlace the 1D stuff can also express the same idea. U can make pretty much arbitrary functions $f(x)$ with a boundary by superposing plane waves that stretch to infinity

But yeah, for electromagnetic field in a box, u need higher dimensional Fourier-transform

@WaveInPlace I also want to say that u cannot have half a wavelength of some plane wave inside a box. For a box, only those fields r allowed that r zero at the boundaries becuz in physics we assume continuous functions

I think this may b relevant becuz u were asking about a box containing a "chunk" of the wave

It shud make sense that only continuous electromagnetic fields exist in nature

Depending on the continuity, sure. I don't think you could have like 0.75 wavelengths of something. Definitely not in a box, and probably not free-standing.

Kind of analogous to sound waves, you could have a wave with a defined beginning and end though. At least in principle.

Yes. U can have fields with boundaries inside an enclosed volume. Their values vanish at the boundaries becuz of continuity. And the divergence is zero everywhere including at the boundaries @WaveInPlace

I mean if no charges r present

EE18 says that he does not know hwo to access the chat. Can we summon him somehow?

(good night. I really gtg)

Where is Slereah

Weird question, but seems like something people here may know. Does anyone know of an electronic component that would be suitable for measuring pressure inside of an espresso machine? Say some pipes with steam at ≤150C and ~0-15 bar

@Amit You should try to summon him

You'll need some drops of blood and an old GR book

@Mr.Feynman I'm not sure I am able to summon anyone yet. It reminds me of how I like to say... "I made many assumptions in my life, but I've never made an Ansatz!"

@Mr.Feynman lol, sounds about right

Oh come on, I got a downvote because of two evident typos

I corrected that instantly and it was removed but I feel that's excessive

@Mr.Feynman What? Where?

An answer I posted on the site

Ah! I saw that question and liked it! :)

Which is rather poor because I didn't feel like giving a long answer but the typos were just a misplaced "with" and saying that "flat=*non*-zero curvature". I mean, after typing non-zero several times one can imagine that's a slip

I'm overreacting I know :P

ahah :) It's cool, I am reading through your answer and I have a question

Yes, tell me

maybe it really is time to summon Slereah? lol

I think that may happen, yeah. In the case of polar coordinates I just knew it is not the case (and it's pretty straighforward to see)

yes, it's just that you extended it to this idea and it struck me as non-trivial

I wondered it too while writing

But I had no examples for it and I decided to settle it with that spherical coordinates example

It will be interesting to see how such a metric looks like, if one exists. Anyway, if you want you can remove this part imo because it's not essential to the correct point you're conveying -- the christoffels depend on the choice of coordinate chart

@Amit what exactly do you mean by a "constant metric"?

@ACuriousMind doesn't depend on the chart variables... or more simply put, numerical factors only :)

@Amit but that is not a coordinate-independent statement

the standard Euclidean metric on $\mathbb{R}^n$ does not depend on the coordinates in Cartesian coordinates, but it does depend on the coordinates in spherical coordinates

@ACuriousMind yes! you're very right, I stand corrected. I should have written, it would be interesting to know what kind of coordinate chart will have a non-constant metric (in the sense I mentioned) but still zero connection coefficients

also, the Christoffels are not coordinate independent, as that example also shows: They are zero in Cartesian coordinates but non-zero in spherical coordinates

@ACuriousMind Does the reformulation of my query make sense?

@Amit zero Christoffels in a region (and not just at a point) exist if and only if the metric is flat in that region

@ACuriousMind Oh I know that zero Christoffels imply no curvature, but non-zero Christoffels do not necessarily imply curvature

so I don't really understand the question

I was just wondering, how would a coordinate chart look like in which the metric takes a non-constant form but the christoffels still vanish identically

For some reason the first thing that comes to mind is "stretching proportionally to the coordinate value"... just as a simple idea on how to make those partial derivatives vanish identically, if you see what I mean

I'm not sure there are any such coordinate systems but I also don't really understand why this is an interesting question

Ahh, mainly because what Mr.Feynman wrote in his answer made me think about it :) I also instinctively associated a non-constant metric with non-zero Christoffels, it was interesting to realize it's not necessarily so

Why would we Kerr? :P

I used that today @Amit

@Mr.Feynman :D It's fine, it's under copyleft... what was the response though?

My GR Prof. was considering to do some QFT in curved spacetime but then he decided it was not appropriate because he didn't want to "torture astrophysicists, they don't deserve it"

So I said "Why would I Kerr?"

But did you clarify the spelling?! :D

Also, I didn't think you'd communicate in english...

lol, torture astrophysicists huh... OTOH, isn't that basically the only field of physics in which QFTCS is actually the closest to observations... I'm way out of my depth here, but I thought, BHs, Hawking radiation and all that...

@Amit I wrote it on a piece of paper to be precise :P

@Mr.Feynman lmao... good move!

@Amit well, the lectures are (should be) in English but even then an English joke was viable

I see, nice