@SillyGoose not sure what you are trying to say, because it is a standard thing treated in the EM textbooks that multipole expansion includes higher order terms, and only the first non-vanishing term is origin-independent.

I think it is misleading to describe (-q) --> (+q) as a canonical example of a dipole

it gives the impression that a dipole requires at least two charges

On page 59 physicsbook.ir/book/… (page 81 of the pdf), why is it written above equation 3.110, $u^s(\Lambda \tilde{p})= \Lambda_{\tfrac{1}{2}}^{-1}u^s(\tilde{p})$ instead of $u^{'s}(\Lambda \tilde{p})= \Lambda_{\tfrac{1}{2}}^{-1}u^s(\tilde{p})$

Note the prime above u on the LHS of the equation that I have written

how does one formalize electromagnetic boundary conditions? In textbooks such as Griffiths and etc. they are derived via some hand-wavey-seeming limits. Is there a more abstract way to view the matter?

Is there an imprecision in the definition of $u$ on the wiki page for Poynting's theorem?

I believe that $u$ should be the energy density of the electromagnetic fields + the energy density of the particles (kinetic and non-electromagnetic potential energies)

hm nevermind

@SillyGoose ???

@SillyGoose no, the usual description that is helpful is when you have, truly far away, a field pattern that is dipole-dominant (maybe with some light radiation) and that cannot be done with a monopole alone, because the monopole term would dominate the field pattern.

In Qmechanic's answer here, balu writes a comment asking why focus on the complexified Lorentz algebra $\mathfrak{so}(1,3)_\mathbb{C}$ in qft. I had always thought it is for convenience (e.g., it allows you to reuse the representation theory of $\mathfrak{su}(2)$). however, Qmechanic mentions that this trick does not always work. What do they mean by this?

Has anyone seen defining a QFT described in this way?

@SillyGoose the easy equivalence of the representation theories is only for the finite-dimensional representations

@SillyGoose it's just lattice QFT and the question of the existence of the continuum limit, also tied to RG flow concepts. Look up "triviality of phi^4 theory".

ncatlab.org/nlab/show/funny+tensor+product

hilarious

Any mods around? Voting ring is forming, those two users are the same person accepting and upvoting own posts: one, two. @ACuriousMind

Good luck nuking it, hopefully before they get too much rep and cause damage.

I red flagged some posts, Smoke Detector also reported and some posts nuked, but the user might still create more.

good morning

Noon

@ShadowWizard Well, many of us are voting against that now, and it should be soon that the pair will be downvoted to oblivion without mod input.

Also flagged and downvoted

m i a o m i a o ~ ~

@ShadowWizard wow, now they are spamming/trolling.

I really don't understand why this happens here on a physics site... I mean... every now and then there is someone doing these kind of things

but what is more annoying are the people who ask ChatGPT for a ToE and post it here :')

cringe

but maybe they're a 13yo kid who knows

yep

it continues...

🤔

Other people same to have the same questions eg. physics.stackexchange.com/questions/107999/… here

in the comments here physics.stackexchange.com/questions/87575/…

Or does anyone else know

@naturallyInconsistent to follow up on your comment on my post: 1) i see that in the lecture notes there indeed is an expression $\epsilon(\omega) = \epsilon_b(\omega) + \frac{4\pi i \sigma(\omega)}{\omega}$ so what im thinking is that this is the permittivity for the case of a conductor. when we look at insulators, the conductivity term is zero and we return $\epsilon_b(\omega)$, which is the normal dielectric constant. [...]

[...] now, the key piece of information would be that, while $\epsilon_b(\omega)$ can never be zero, $\epsilon(\omega) = \epsilon_b(\omega) + \frac{4\pi i \sigma(\omega)}{\omega}$ actually can be zero. by the maxwell equations, this means conductors do permit longitudinal waves, but insulators do not. am i getting this right?

i find this satisfying so i hope it's correct

have any of you read Fradkin's Quantum Field Theory textbook? It is relatively new.

@DIRAC1930 I don't know how often I can repeat that the confusions around this topic run so deep that I cannot resolve them in a few chat messages for someone who does not know the machinery of modern differential geometry.

But this is a simple question that doesn't need any of that

If you view it from the physical point of view

Otherwise all field theory books would start with a discussion on differential geometry

this extends to the language of the "primed" and "unprimed" quantities - without a precise and proper definition of the nature of the relationship between these quantities, you can essentially write arbitrary equations between them and argue they are correct within your convention

@DIRAC1930 no, my opinion is merely that they should

In nearly all modern qft books, primed means that you transform the field in your reference frame

So it is a different field

e.g. if I have a magnet, it will produce a magnetic field say $\mathbf{B}(x)$. If I move the magnet, the magnet will produce a magnetic field $\mathbf{B}'(x)$

Thats all

So it is obvious what P&S is trying to say

The question is whether it is incorrect or not

Which might be the case since several people have asked the same question with no answer

@Relativisticcucumber I dont see how that could help you get the result you want, because all that means is that in a conductor, the dielectric function turns complex; that makes the wavevector pick up a complex part, but how are you hoping for it to give you longitudinal waves? Instead, in accordance with the meme that you have identified, I have just told you that waveguides can have longitudinal parts in the free space...

@DIRAC1930 since it is so obvious, why are you even asking?

Because there is most likely a mistake in the book

... is it an obvious mistake?

No

Because there might be some principle that is not explained making it correct

If you are in doubt about your understanding of the equation, why are you so sure of what it is that ACM is trying to dissuade you over?

Because it will not clear up the question

Maybe when it comes to GR or something, then it is useful to think of things in a different way

Well, AFAICS, ACM is clearly the one who had fully understood the exact topic in question and not you.

@DIRAC1930 I would argue the lack of answers is precisely because the topic is so unexpectedly subtle instead of it being obvious. I say it one last time: Actually fixing the mess that physics texts make around coordinate transformations, symmetries and the active/passive distinction is not simple and I do not know any way to do it without being several degrees more mathematically sophisticated than just acting as if the meaning of things like "coordinate change" or "translation" is obvious.

But it is

I don't see the use in overcomplicating the real world

If it is, you should stop having this question

What are you on about?

The question is about if there is a mistake in the book

@DIRAC1930 Don't you see how you're contradicting yourself? If this topic is so simple and does not require sophisticated mathematical machinery, what are you even asking? Either you understand the correct answer and know it is simple and obvious, or you don't and cannot judge whether it is simple or not.

No

Im saying that there is nothing wrong with the way transformations are done in most QFT textbooks

It is obvious

then I cannot help you, please stop pinging me about this topic

The question is if there is a typo or mistake or some physical principle that needs to be invoked in a textbook

Well you're not much help anyway

one more reason to stop pinging me, one would think :P

I'm not allergic to the active, passive transformation argument that is treated in physics texts, quite fond of it, to be honest, but it is so bleedingly obvious that ACM is correct in this case.

This literally has nothing to do with my question

Well, it is clear that you cannot recognise the correct answer to your exact question when staring it in the face

No part of that is irrelevant

Because noone can hence the other questions I linked to on physicsstackexhcange which currently have no answers

Other people have asked the exact same question

Has it ever occurred to you, that, maybe, just maybe, people who know the answers, do not owe the answers to the people who have the questions?

No, the question is closer to Fermat's Last theorem and you are insisting that just because Fermat couldn't fit a proof within the margins and that he was working with 17th century maths, it couldnt be that difficult

@naturallyInconsistent nah, differential geometry is much easier than the machinery required for Fermat's last theorem :P

I mean one is a theorem and one is a whole field of mathematics

So it's a bit hard to compare

@Slereah I'm comparing the entire subfields of algebraic geometry they had to invent for the proof to differential geometry, not the theorem itself!

I'm actually more reminded of the fact that polygamy with underaged girls is still a thing in USA and how the "wives" would often clash with police to try to get back to their "husband". The conviction with which that they proclaim how correct they are, and the pity that we feel for their plight.

Like, maybe if these people would try, maybe, a good book written this century instead.

@naturallyInconsistent Uhhh...this comparison invites a lot of questionable implications, let's not go there, shall we?

They invented differential geometry for a nobler goal (measuring Germany)

@SillyGoose I have it at my bookshelf, but only skimmed through it

@Slereah lol

@DIRAC1930 well they probably should

@TobiasFünke oh i see. I just found out about it. It seems pretty neat. I might order a copy. Although, i’ve rarely seen a good book from princeton press

haha

yeah, I mean I like what I've read so far...but the last 5-6 chapters are above my pay grade lol

at least for now ;)

Hehe

I am wondering why the active passive transformation cannot be handled simply. Shouldn’t it all be settled with a precise definition of “field transformation under a Lorentz transformation”?

E.g., $\Phi(x) \to \pi(\Lambda)[\Phi(x)]$

I mean that's essentially what it is

But the trick is the details course

people typically want coordinates

yeah, I've never really understood the debate lol

hm okay so the tricky bit is placing the coordinates into the general expression?

yeah the representation can be complicated

@SillyGoose what do you mean?

i am imagining we have the base $n$-dimensional smooth manifold (spacetime) $(M, \tau)$. then we have coordinates (or charts) $\phi: U \to \mathbb{R}^n$ for $U \in \tau$. Then, to write a coordinate free expression like I wrote above in terms of coordinates one needs to place the coordinates judiciously.

aha

er not judiciously, but in a mathematically appropriate way i guess. i am unfamiliar with the details.

anyways turning the map $\Phi: M \to V$ into (locally) a map $\Phi: \mathbb{R}^n \to V$ I imagine. Then appropriately modifying the $\pi(\Lambda)$ if necessary

here's a sneak peek on that topic:

@SillyGoose It is handled simply. Thats the whole point. There is no confusion about that whatsoever

Its like in standard Newtonian mechanics, one has a particle at $x_0$. If one wants to make it move, they just write $x(t) = x_0 + v t$ or something. Theres no reason to say that cannot be understood without differential geometry or something

If someone asked "why do we add $vt$", you would explain why instead of saying that understanding that requires an extensive overview of differential geometry and the question makes no sense to people who don't know any differential geometry

But anyway, my original question wasn't about any of this, it was about a specific computation in book regarding a step and ACuriousMind and NaturallyInconsistent wrongly thought it was about all of this

The switching between different pictures: schroedinger, Heisenberg or interaction one, depends on the case considered ?

does anyone know if anywhere has a list of foundational problems of classical field theory? they can be physical or mathematical.

@SillyGoose what do you mean by "foundational problems"?

a (made-up) example would be: "the theory of electromagnetism in media has no mathematically precise description".

I don't think people are much in doubt over the mathematical formulation of classical field theory

or another (made-up) example would be: "there is a logical contradiction implied when combining the mathematical structures of a classical particle theory and a classical field theory"

yeah, I don't think there's anything like that in classical physics :P

or: "here is an example of a relevant classical field theory that does not fit under the action paradigm but the implications of this are not well understood"

@ACuriousMind your long form writing?!

i think the only actual example I might know is there is some dispute over how to define electromagnetic energy in media, or something like that.

The Abraham-Minkowski controvery

@ACuriousMind no crazy unsolved puzzles of classical physics :P?

@SillyGoose I don't know if it counts/ I don't know much about it but I heard about someone working on the gravitational self force and related topics

@SillyGoose en.wikipedia.org/wiki/… ?

@qwerty thank you for the reference. this seems pretty in line with what i was asking for examples of. i am reading an arxiv thing by wald on the topic now.

@fqq i suppose there is this indeed. i was thinking stuff that is a little more pedestrian :P

Does anyone here know Rickayzen, Green's Functions and Condensed Matter?

I'm looking for a reference discussing RPA with only Hartree diagrams (no Fock) as in section 3.8 of that book, which has some points that I can't understand well

@naturallyInconsistent well i have two separate responses to this message. regarding "waveguides can have longitudinal parts in the free space" -- well im not really sure what this means since the whole discussion is about em waves in materials?

2) regarding how the message i sent would resolve my confusion. well, in the notes linked, the way we sus out which waves are permitted is by looking at the maxwell equations. and we see that the waves must be transverse if $\epsilon(\omega) \neq 0$ so really the culprit here seems to be whether $\epsilon(\omega)$ can be zero or not, no?

and third of all lol