If all theories are diffeomorphism invariant, then why do we say Maxwell theory is not invariant under Galilean transformations? I know the standard derivation, but I am confusing with the point that "physics is invariant under coordinate transformations"

Indeed $F_{\alpha \beta} F^{\alpha \beta}$ seems like a scalar under general coordinate transformations, unless we define it to be a scalar under only Lorentz transformations.

@NairitSahoo what is maxwell theory?

classical em without SR?

@Obliv No I was just meaning full fledged special relativistic classical electromagnetism theory defined by the action $S=\int d^d x -\frac{1}{4} F_{\mu \nu}F^{\mu \nu}+A_\mu j^\mu$ with $\partial_\mu j^\mu=0$ for consistency with gauge invariance.

Oh, well I didn't think SR even accommodated Galilean transformations because they allow $v>c$

oh ur asking why

That is a very weird question.

I think it's kind of an experimental result moreso than derivation. When you couple the maxwell equations to get the wave eq. of light in vacuum u get a fixed wave velocity

But maybe it could make more sense if you specify to higher detail: Equations in physics are Lorentz invariant under coördinate transformations. Galilean coördinate transforms will not preserve the invariance of those equations.

@naturallyInconsistent If we simply change the coordinates in the expression of any equation in physics, we are just describing the same physics in different terms, right?

@NairitSahoo Yes, you have that, but then you don't get to assert that the equations themselves are invariant under arbitrary coördinate transforms.

We are trying to say something with actual physical content in it, not just state the mathematically obvious fact.

@naturallyInconsistent Is the action invariant? I mean does $S'=S$ hold for arbitrary coordinate transformations $x' \to x$?

No, I don't think so.

@naturallyInconsistent So actions of all are not invariant under coordinate transformations?

It is a sad fact of life that we have no choice but to use human languages to communicate, which necessarily incorporates vagueness, but when we are stating these things in physics, we actually often mean something rather physically insightful and remarkable, something that cannot be obtained just by thinking about stuff. We are summarising a lot of experimentally obtained facts forced upon us whether we like it or not, and not trying to utter vacuously true statements.

@NairitSahoo What are you even talking about? Have you not seen some examples of transforms in physics that give extra boundary terms in the action?

@naturallyInconsistent Oh apart from that :| I am not asking about quasi-symmetry vs. symmetry stuff.

@naturallyInconsistent True. I am trying to convert the human language/physics facts to precise math statements

@NairitSahoo But it is a direct contradiction to what you think is being said! You should have immediately realised that your interpretation of the statement must be wrong, and instead looked for a nuanced understanding.

@NairitSahoo Does it even exist? You might need to contend with non-existence. Physics is not maths.

@naturallyInconsistent Atleast for the subtopic I am pondering about, I think it does. it seems this is a confusing and controversial topic

I want to understand exactly what do they mean and how are they consistent?

I see some discussion over here: physics.stackexchange.com/questions/76721 for example

Precise math statements are good for communication, but as Hilbert said (paraphrasing) "when rigor enters, meaning departs" not to mention that one can't even prove consistency/completeness within ones axiomatic system

but idk why I said that because there must be a balance in real experience & abstract thought to form a good theory. It can't lean too hard into either side I think

@NairitSahoo It definitely is confusing. But mainstream physics is not considering any of these to be controversial.

is there an easy way to computationally find simultanous eigenstates of two hermitian matrices?

@naturallyInconsistent People seem to put this under the rug. But even in 2023 papers come up on this, and there's no general concensus.

This should be settled once and for all.

See arxiv.org/html/2405.09703v1 this came out in 2024...

@SillyGoose i learned what a hermitian matrix is today, but idk what an eigenstate is :D good luck

This has particularly no new concept.

I mean, I really don't know what you are doing by reading all of these things in such an order. You jumped from basic Maxwell's equations to Conformal Field Theory to GR in a span of months. I don't think you really understand what it is you are reading, not least judging by the history of your questions.

@naturallyInconsistent Well I am a 2nd year undergrad and ofcourse I really feel that i don't understand a lot of things.

@SillyGoose This is a not just mathematically difficult, but also numerically difficult, so much so that we needed theorems to prove that solutions actually exist so that the search is not futile.

@NairitSahoo Sometimes people say an expression is "invariant" when they really mean that the expression transforms in a prescribed way

get used to that feeling, that's the human condition lol

@NairitSahoo That makes way more sense. All of these things are wayyyyy beyond your current grade

@naturallyInconsistent But the issue I am facing. It is not an easy one I am sure because see even Josh (a 6th year grad student from I forgot: some Stanford or something) got confused by this point.

@naturallyInconsistent err I mean to say for two commuting hermitian matrices

See people out there in that question are challenging giants like Carlo Rovelli

I am trying to generate the eigenstates of total spin operators to circumvent doing some clebsh-gordan business

What do you have against clebsh-gordan business?

@NairitSahoo Correct, it is horrible. To even state the diffeomorphism invariance you need to have learnt standard GR and know what that means in that context, and learn enough differential geometry to convert the GR knowledge into a form that you can start understanding. This is not just one year's worth of study. It is multiple.

so really the question is how to computationally find eigenstates of $S^2, S_z$ where these are total spin operators for $N$ qubits

@Obliv it seems more involved than just finding simultaneous eigenstates

@naturallyInconsistent Well, I don't mean to brag. But I think that I understand it almost except I get confused by notions of diffeomorphism invariance, sometimes...

@SillyGoose I like the name Alfred CLEBSCH it's got a lot of successive consonants

@SillyGoose Then isnt the eigenspace of any one of them (like, not just one single eigenvector in it, but rather that you must have span any single eigenspace) also going to be some (sub-)eigenspace of the other? In particular, if you find a non-degenerate eigenvector of one of them, then it is also an eigenvector of the other

@sillygoose u can also pawn it off to one of the maths people in room 36

as a true physicist would

@NairitSahoo You seem to be approaching these things from maths-first. I would not deny that your understanding, if you can do it quickly, is stable and solid. But you should not be able to understand all these things properly as it is. Your questions here are indicative of confusion due to an unmooring of the knowledge from actual applications and examples.

@naturallyInconsistent No no. What happened was this: sometimes you read some basic stuff, and then some advanced stuff. From the advanced stuff the basic stuff should come out naturally. But it isn't coming out. That's where the problem is.

@SillyGoose Obliv is actually giving you the correct answer. You are not looking for the simultaneous eigenstates as you originally worded the question.

@naturallyInconsistent Ofcourse I "know" as a fact that Maxwell equations are not invariant under Galilean invariance. I even know that the Galilean limit is pesky (there are two Galilean limits!)

@NairitSahoo That is because physics is difficult and such an attempt is doomed to fail. Even in mathematics you do not start by learning category theory. You start by working with merely real analysis, and then working upwards.

oh qutip just has a simultaneous diagonalize command

which will return the wanted eigenvecs

yay

@NairitSahoo sometimes basic stuff is wrong

@SillyGoose how is puncturing time manifolds going? (I barely skimmed that conversation but it looked interesting)

@Obliv dead

High level physics sure does get abstract..

@Obliv have since returned to earth

boo, I mean welcome back.

i think as a start people use the word "invariant" in ways they shouldn't. so it would be nice to know precisely what your definition of the word "invariant" is in this context

@naturallyInconsistent See? That's why I was focussing on the math in this context!

@NairitSahoo galilean invariance doesn't enter the picture because classical EM with SR doesn't allow for them

the people of that time probably believed it was "diffeomorphism invariant" if they had that word in their vocabulary in the late 1800s

but it wasn't the kind they expected until after SR was developed

(this is my own note, so perhaps it could be wrong) but some people refer to (18) as "scalar fields are invariant under lorentz transformations"

obviously that statement is literally not true

under lorentz transformations, scalar fields transform as $\phi(x) \to \phi'(x) = \phi(\Lambda^{-1}x)$

so this whole textbook use of the word "invariant" is (to me) quite confusing. things transform.

what's the big lambda mean? I've seen it before

@Obliv the picture literally explained...

it is (weinberg's) notation for arbitrary (restricted) lorentz transformation (in the fundamental representation)

$\Lambda \in SO^+(1,3)$

ah yes, sorry I didn't catch that @naturallyInconsistent and makes sense we went over SR briefly in the spring

jmurray echos my (and many others') complaint (in more diplomatic terms) + provides some exposition here

@Obliv See these are the things I know, believe and am taught at college. But then this seems to be inconsistent with what I am reading as "extra" stuff cz I wanna enter HEP-th.

@NairitSahoo that's unlikely to help you understand things, not least because you are far out of context and learning things in haphazard order.

Doesn't this theorem follow directly from the definition of inner product

@SillyGoose That's because that is precisely not what is wanted to be meant.

@naturallyInconsistent Well... try me :) I am damn sure that without understanding this, I can definitely understand a lot of say... string theory.

@NairitSahoo ah, string theory. subtracts one from your score

@naturallyInconsistent what do you mean?

@NairitSahoo not sure what HEP-th is but maybe if you point to where you saw someone say classical EM with SR isn't galilean invariant, we can figure out what they meant

@Obliv high energy physics-theoretical

@SillyGoose Ofcourse. Scalar "fields" aren't meant to be taking the same value at each point in spacetime. We are free to choose different values (of the field) at different points in spacetime.

I would imagine it's not a mathematical inconsistency

@SillyGoose When they say that scalar fields are invariant under Lorentz transformations, they don't mean what you are arguing about. They mean something completely else.

@naturallyInconsistent are you referring to an expression like $\phi(x) = \phi'(x')$ or something akin to this

@naturallyInconsistent you know... It's like RG. You don't need to know quantum gravity to describe a glass of water :p

@NairitSahoo That's worse.

@Obliv See physics.stackexchange.com/questions/76721

Also see physics.stackexchange.com/questions/643635 . The OP is saying "as a conformal transformation is a particular type of coordinate transformation, shouldn't any action be conformally invariant?". Similarly, I can say since Galilean transformation is a particular type of coordinate transformation, shouldn't any action be Galilei invariant?

The answer there is specific to conformal transformations: CFTs are simply not field theories invariant under conformal transformations. They are field theories invariant under conformal transformations+rescalings

So that doesn't answer my question for the case of Galilean transforamtions

Consider $\phi^\prime(x)=\phi(\Lambda^{-1}x)$ as is happening inside your screenshot. Assuming that this is a scalar field satisfying the Klein-Gordon equation $\frac1{c^2}\frac{\partial^2\ }{\partial t^2}\phi-\vec{\nabla}\cdot\vec{\nabla}\phi=0$, what we are trying to say is that this equation will be form-invariant, i.e. $\frac1{c^2}\frac{\partial^2\ }{\partial t^{\prime2}}\phi^\prime-\vec{\nabla}{}^\prime\cdot\vec{\nabla}{}^\prime\phi^\prime=0$ with no extra terms popping out

If you applied Galilean transform to the Klein-Gordon equation, the resulting equation will fail to be form-invariant because extra terms will appear.

On top of that, there is also a "spin-"invariance. For example, in spin-half Dirac wavefunction, $\psi^\prime(x)\neq\psi(\Lambda^{-1}x)$ because when you Lorentz transform the Dirac wavefunction, not only is the location of the point being considered being moved (the $\Lambda^{-1}x$ part), but that the spinor wavefunction itself gets a mixture between the 4 components.

In fact, it is this horrible "spin-"covariance nonsense that is the fundamentally important new physics that is always being heavily emphasised and computed right at the start of QFT.

And this is already 2 totally different "invariance" concepts to grapple with when you start down this path.

I haven't started talking about diffeomorphism invariance.

Trying to say that you understand these things right away, is just hubris.

@naturallyInconsistent Now we are talking :p Yeah, so this is precisely the proof that I push within the "basic stuff". I mean this is as obvious as daylight! But just like this, can you by doing math, show that the Maxwell EM lagrangian is not invariant under galilean transformations?

Even Carroll's book on GR says "Any semi-respectable theory of physics is coordinate invariant, including those based on special relativity or Newtonian mechanics"

See Newtonian mechanics!

cant u try it and find out :P

And I definitely see that $F_{\mu \nu} F^{\mu \nu}$ transforms as a scalar under any coordinate transformation $x \to x'$

@NairitSahoo Yes; and it is trivial. However, it is obvious that the Maxwell's equations are not Galilean form-invariant because it is Lorentz form-invariant, and you can only be one or the other. It is a tedious and non-instructive to compute the Galilean form-non-invariance, and it is already rather tedious to directly prove the Lorentz form-invariance.

@NairitSahoo That is simply not true.

@naturallyInconsistent At the Lagrangian level, what does form-invariance mathematically mean?

@Obliv Yes. Okay. So $F'^{\mu \nu}= \frac{\partial x '^\mu}{\partial x^\rho}\frac{\partial x'^\nu}{\partial x^\sigma} F^{\rho \sigma}$. So $F^{\mu \nu}F_{\mu \nu}$ is a scalar because the Jacobian factors cancel out. Usually these Jacobians are taken to be Lorentz transformation matrices $\Lambda^{\alpha}_ \beta$

@NairitSahoo Carroll is trying to mean the trivial sense. Definitely not the form-invariance. In particular, if you perform a non-inertial coördinate transformation, Newton's laws spit out stuff like Coriolis and Centrifugal effects, and that is manifestly breaking the form-invariance, even though under Carroll's point they still count as coördinate-invariant.

@NairitSahoo That mapping is famously troublesome to state and is in textbooks, which is why I have taken so much pains to avoid this. Just study things properly from the textbooks in the proper order. It is basically impossible for you to miss them along the standard curriculum. Why are you being like this?

@naturallyInconsistent Yes yes. exactly! See that's where the confusion is... I know these to be true. I derived these: these extra terms are the extra terms in the Christoffel symbol transformations for example

idk how to read einstein indices but yeah I'm sure you got this :D time for sleepy time

good night naturallyMiao

yay >:D i can just simdiag $S^2, S_z$ instead of worry about some CG business...

i wonder why i haven't just seen this "naive" approach used in practice...maybe it is slow

@naturallyInconsistent You are saying diffeos in general change the form of EOMs. I know form invariance=isometry for metrics. True there, also. Generic diffeos change the form of the metric but isometries don't.

@NairitSahoo You have absolutely neglected the possibility that $\frac{\partial x^{\prime\mu}}{\partial x^\rho}\frac{\partial x^{\prime\nu}}{\partial x^\sigma}g_{\mu\nu}\neq g_{\rho\sigma}$; that this is equal in the case of Lorentz transform is a remarkable fact that is emphasised, but it is not true in the general case.

@Obliv good night miao miao~

@naturallyInconsistent This is not used in the above derivation, is it?

I mean I have to transform the metric too

@NairitSahoo It is. $F_{\mu\nu}F^{\mu\nu}=F_{\mu\nu}g^{\mu\rho}g^{\nu\sigma}F_{\rho\sigma}$

@naturallyInconsistent Yeah ofcourse. So? I would transform the metrics also and cancel the Jacobians.

@NairitSahoo And immediately you can lose the form-invariance. In any case it is a known fact that if an equation is SR invariant, it is not Galilean invariant and vice versa; that means if you think you proved it, you have just made a mistake.

Ofcourse.... I have made a mistake, but I want to know where it is...

@naturallyInconsistent That the Jacobians are Lorentz transformations are something which has to be imposed as "physics input". If I give you a $F^2$ term, you have no way to say by pure math that it is a Lorentz scalar or a scalar under all passive diffeos!

That's correct.

We have specific physics meanings to these things, and you can always insist for people to give you an example of the kind of in-variance or co-variance that they are trying to talk about.

@NairitSahoo How do you know that this is true? How do you know that when $F$ transforms according to a general coordinate transformation, it transforms nicely/homogeneously?

I know that it is true only for Lorentz transformations. If you approach this from a "theory" side, this comes from the definition of your theory: you define the fields to be irreducible representations of the Lorentz group. If you ask "why?" too deeply, at the end of the day like @naturallyInconsistent tried to emphasize, "because experiments told you so".

@Sanjana So what's the conclusion? Is Maxwell field theory invariant under just Lorentz transformations or all diffeos?

@NairitSahoo I am learning physics myself, so I am not sure if I am the right person for making conclusions :p But yeah, I agree with what naturallyInconsistent said.

@Sanjana that is a boulder of humility, coming from ya. You are way deeper into these physics than meow meow

aww... not at all :p

@Sanjana Okay I will take your words with a grain of salt. But still can you say something on this?

When you try to change the coordinates according to a generic diffeo, you get some extra terms in the action. In that sense it is not invariant under diffeos. But in some sense the action is still "invariant" because you don't go to a frame from which you cannot come back. You don't run into singularities.

But you can go to non-inertial frames: It is perfectly fine to describe Maxwell electrodynamics from a non-inertial frame (just harder). In that sense it just signifies "physics is invariant under general change of coordinates".

But then sometimes you don't even want to change the form of the Lagrangian. That is you still wanna have just the $F^2$ terms (or at the level of EOM, keep the Maxwell equations "form invariant"): not some extra weird pieces (like the Coriolis or centrifugal terms which nI mentioned).

hi

For that you can't do a generic diffeo anymore: you need to do an isometry (literally diffeos which keep the metric form invariant!) or a transformation which takes an inertial frame to another inertial frame only.

@Sanjana hi

But what does iso- metry i.e. properties of the metric has to do with fields? That's because you defined your theory with fields which are tensors (or irreps in a rep. theoretic sense) under the isometry group of the spacetime on which you defined the theory. So in that sense your action is truly form invariant only if you choose to transform it by Lorentz transformations.

@NairitSahoo I would recommend confirming everything with ACM who has deep knowledge on diffeo invariance

@RyderRude Hi. How are you? Sorry, I forgot to reply earlier: I didn't watch Deadpool and Wolverine yet.

@Sanjana oh.. i am good. thanks

@Sanjana What about formulating physics on spacetimes with no isometries? How do we define fields there?

@Sanjana i thought it was good

it has captain America too

but he is playing human torch

@RyderRude I have read the wiki page... But, I don't think people here will like to hear spoilers, Ryder!

@Sanjana yeah.. sorry

in non spoiler terms, it is about forgiving urself and sacrifice

@ACuriousMind Is what Sanjana and naturally Inconsistent told correct? Or maybe if it is too long, my simple question is: people say physics doesn't care about choice of coordinates. So all theories are invariant under passive small diffeomorphisms. But since Galilean transformations are subset of diffeos. it means that Maxwell field theory is invariant under Galilean transformations which is not what is being taught in courses.

@NairitSahoo i have a post about this.lemme find

@NairitSahoo see this physics.stackexchange.com/questions/746794/…

The second answer is relevant too

@ACuriousMind If you ask what do you mean by diffeomorphism/invariance etc. I just pray that you just explain what do people mean exactly when they say "physics is invariant under diffeos" and "Maxwell theory isn't invariant under galilean transformations".

@RyderRude Tanks. Let me see

it is not about Maxwell specifically, but it is about Newton's second law's invariance @NairitSahoo

@ACuriousMind can u pls delete the deadpool.and Wolverine spoilers i posted above

@RyderRude So what Sanjana and naturally Inconsistent was wrong? All theories are invariant under all passive diffeomorphisms?

Where by invariant I mean form-invariant

@NairitSahoo i have not yet read those msgs

@NairitSahoo depends on what u mean by "form invariant". e.g. r u allowing the metric to transform or have u fixed it to be minkowski

Only certain transforms will leave the action "form invariant" in a way that the metric is fixed in form

@RyderRude I am fixing it to be Minkowski

These r Poincaire transforms of the Maxwell eqns

@NairitSahoo oh.. then a lagrangian is not form invariant under all transforms, when u define form invariance like that

@RyderRude Is Maxwell equation not "invariant" under all diffeomorphisms?

Depends on how u write it down

if u leave the metric arbitrary, like write it as $g$, then $g$ can transform and that Lagrangian is invariant under all transforms

But if u write the Lagrangian using the Minkowski metric, then that Lagrangian is only invariant under Poincaire @NairitSahoo

@SillyGoose No need to overcomplicate this - the statement that a scalar is "invariant" is just a statement about the target space of the field, not about the field as an element of function space.

@NairitSahoo i think the "active" vs "passive" transform distinction is the most clear

@NairitSahoo See this answer of mine - the invariance of the action is not about coordinate transformations.

@NairitSahoo i just read.. naturallyinconsistent and Sanjana r saying the same thing

for the specific case of diffeomorphism invariance, see this answer of mine for a line of thinking much like yours ("But Maxwell is diffeo-invariant, too!") and this answer of mine for my most recent understanding of the issue of "diffeomorphisms" as symmetries in GR

@ACuriousMind i'm surprised if the authors of that statement think that is what they are saying...because I don't get why you would call the scalar field itself invariant under a transformation if what you mean is that the target space of the field is invariant

@SillyGoose You're a pedant like me. Not everyone is. :P

@ACuriousMind So all theories are invariant under diffeomorphisms?!!!!

"it's a field it's obvious that if it was fully invariant it would just be a constant, obviously that can't be what we mean, stop nitpicking"

^typical response when you bring this up to people who write like this

@NairitSahoo No, I didn't exactly say that; did you read all my three answers in this short time? :P

:66087155 in that case that is obviously what they mean :P

@ACuriousMind I have read two of them this morning before coming here :) I just want to know... what do they mean when they say that Maxwell equations are only Lorentz invariant and not invariant under Galilean transformations?

@NairitSahoo They mean the invariance in the sense of my answer about Noether's theorem

@ACuriousMind Can you dumb it down a bit?] I can't relate.

@ACuriousMind according to them, I shall justifiably write my theorems as such: Let us assume X, Y, Z. It is then simply obvious that result is true.

@ACuriousMind I mean in terms of coordinates...

@SillyGoose I would not simplify the issue into so binary a distinction. We always assume some context when we talk about things; many words mean more than one thing in different contexts, and for authors it is sometimes difficult to judge what amount of context to expect from a reader (e.g. when I talk about a connected space, I am assuming the reader understands I mean the topological and not some colloquial meaning).

Also, there is the inertia of jargon - someone started the terminology like this, and now it's in almost a hundred years of literature. Even if the terminology is inaccurate, to all active practitioners it will be even more confusing if you now start using your own "more accurate" terminology no one else is used to.

using inaccurate terminology is not necessarily some deep sin, not some fundamental commitment to being unserious about one's work, it's not even some sort of principle...sometimes, it just happens

@NairitSahoo Mostly you have to be careful that you don't transform the volume element if $g$ is not a dynamical field (since that is what distinguishes invariance of $\int F^{\mu\nu} F_{\mu\nu}\mathrm{d}^4 x$ under the Lorentz isometries from the ability to change coordinates arbitrarily - the latter will incur some $\sqrt{g}$ from transforming the $\mathrm{d}^4 x$)

but it is also a massive display of "doing what is easy just because". perhaps the deepest of sins :P

that is not necessarily my personal attitude towards it, it is just a sort of observation

i think generally i use my own vocabulary for my own notes and then attempt to translate my vocabulary into the common vocabulary if necessary to communicate with others

and often times the common vocabulary is pretty good to begin with

but an atrocity like macrostate is hard to ignore

@ACuriousMind Just tell me one thing: if I write $S=\int \mathrm{d} ^4 x F_{\mu \nu} F^{\mu \nu}$ for fixed background spacetime which is Minkowski spacetime. Is this invariant under general coordinate transformations $x' \to x$?

@NairitSahoo The problem is the meaning of "invariant" ;)

@ACuriousMind I mean is $S'=S$ if I do $x' \to x$?

Again, Noether's theorem and symmetries are not about coordinate transformations, so really asking whether an action is invariant under any coordinate transformation doesn't make any sense

@NairitSahoo but what do you do when you do $x' \to x$?

that's the crux: You do something subtly different when this is a "change of coordinates" as opposed to a transformation in the sense of Noether's theorem, and that difference is what matters

@ACuriousMind My action is made up of fields $\phi(x)$. I substitute $x \to x'=f(x)$ in place of every field I see in my action and also put primes on the fields because the coordinate transformations induce a change in the fields. E.g. I put some jacobians if I have some vector fields

@SillyGoose Pragmatism and laziness can be difficult to distinguish, but I would not always err on the side of laziness. I don't think I claimed that people do this "just because" - note that solving the problem of jargon inertia is a coordination problem that cannot be solved by a single person no matter how much effort they invest, and also that many students struggle already so much with the concepts that having many different names for them would mostly confuse them further.

@ACuriousMind Then, do I get $S'=S$?

@NairitSahoo what happens to the $\mathrm{d}^4 x$?

@ACuriousMind $d^4 x= d^4 x'$

ah, but that's not what happens during a coordinate transformation!

@ACuriousMind Oh yes. Okay $d^4 x'= |J| d^4 x$

when you change the coordinates in an integral (remember integration by substitution!), you get the determinant of the Jacobian of the transformation in there

@ACuriousMind Yeah sorry. I forgot... I put that back

@ACuriousMind If I do the coordinate transformations and the field transformations which is induced by the coordinate transformations, and the transformation of the volume element--- all of these, do I have $S'=S$ for Maxwell electrodynamics for arbitrary coordinate transformations?

and also you need to consider that $F_{\mu\nu}F^{\mu\nu}$ is really two $F_{\mu\nu}$s contracted with the metric $\eta^{\mu\nu}$

when you do a coordinate transformation, you also have to transform the $\eta$, since the metric changes its expression when the coordinates change

@ACuriousMind Yes I transform every of these (I pointed this earlier to naturally Inconsistent too: that I am already being careful about these)

but since the SR metric is not dynamical in SR EM, it will not change during a symmetry transformation in the sense of Noether's theorem (physical transformations can only act on the dynamical variables)

@NairitSahoo then yes, $S' = S$. But again: This is not the sense of invariance demanded by Noether's theorem, it is only the much more trivial statement that you can change the integration variables in an integral.

@ACuriousMind But when doing Galilean transformations on Maxwell Lagrangian. Since these are coordinate transformations, they induce some change in the field. Do you mean to say that these changes would somehow cancel and make the action invariant?

I don't understand the question

schematically, we have a term like $F^T \eta F$. This is invariant under $F\mapsto \Lambda F$ iff $\Lambda^T \eta \Lambda = \eta$, i.e. $\Lambda$ is an isometry of $\eta$.

@ACuriousMind But this was what Sanjana was saying. You said something different earlier.

when we do a coordinate transformation, we additionally transform $\eta \mapsto (\Lambda^T)^{-1} \eta \Lambda^{-1}$, cancelling these terms regardless of the properties of $\Lambda$.

@NairitSahoo You have to let me finish typing things out, see above :P

The transformation in the sense of Noether transforms only $F$, and hence leaves everything invariant only for isometries $\Lambda$. The coordinate transformation transforms both $F$ and $\eta$, leaving everything invariant always.

note, in particular, that in GR $\eta$ would be dynamical, too, so there a transformation in the sense of Noether is also allowed to transform $F$ and $\eta$, thus "promoting" all coordinate transformations/diffeomorphisms to proper symmetries

@ACuriousMind So in GR whenever we want to talk of transformation in the sense of Noether, we are bound to transform the $\eta$ too?

sure - the metric is a dynamical variable in GR, is it not?

hello guys

@ACuriousMind Yes. I think it is clear now.

@ACuriousMind How to do this for spacetimes in which there are no isometries?

well, then you have no symmetry :P

what's the problem?

@ACuriousMind I mean to do field theory we need to define fields as irreducible representations of the isometry group of the background spacetime.

ACuriousMind is an omniscient AI deployed by the NSA change my mind

@ACuriousMind If there's no isometry group, how do we define fields?

@Davyz2 I don't work for the NSA

@NairitSahoo This claim is completely wrong :P

Mmmmh...that's exactly what a omniscient AI would say...

the notion of vector and tensor fields is perfectly well-defined on arbitrary smooth manifolds, you don't even need a metric (hence you don't even have the notion of isometries)

@ACuriousMind But in introductory QFT, we take fields to be irreps of the Lorentz group, right?

@NairitSahoo sure, but that's because introductory QFT takes place very specifically on Minkowski space

the whole machinery of QFT as we learn it cannot just be transferred to arbitrary spacetimes

in GR you would not even be able to quantize the field theory, because the notion of "space-like" depends on the metric

which in turn depends on a classical stress-energy

a lot of the constructions don't make any sense if you're not on $\mathbb{R}^{3,1}$, starting e.g. with the Fourier transform and the notion of asymptotic past and future

"QFT on curved spacetimes" is its own, more advanced, field

@SillyGoose Note that what ACM said (that you are replying to in your comment) is what miao miao was saying about "spin-"invariance of scalar fields.

in QFT i still don't understand how you normal order the hamiltonian to make a stable vacuum

@NairitSahoo also see topological qfts

my prof used to say "yeah alright we switch the operators" in the usual extremely rigorous physicist fascion

@Davyz2 that's the rabbit hole of what exactly renormalization is

it's normal (and probably unavoidable) to handwave that in an intro if you want to get anywhere by the end of the semester :P

@NairitSahoo what subjects are u interested in

i totally never understood it, to me normal ordering is defined when you can subtract the commutator from the product

but the commutator is not something you can subtract, because it's not defined

(for p' = p in the hamiltonian, ofc)

@Davyz2 I always find this an ugly bit of instruction. What I would have said, is that, yes, we started with position-based operators when we initially defined the Hamiltonian, and so when we Fourier transform to the momentum-based operator form of the Hamiltonian, we find that it fails to be normal ordered and that means the vacuum has infinite constant energy. What makes sense to do at this point is to start from the normal ordered momentum-based operator form of the Hamiltonian

@Davyz2 yes, that's exactly the issue at the heart of renormalization - that stuff like $\delta(x)$ or $\phi(x)^2$ cannot really be functions, but must be distributions, but we continue to treat them like functions, and thus incur as penalty the "infinities" we renormalize away.

don't worry about this too much if you don't want to, many practitioners of QFT don't understand this beyond a superficial level, either

@ACuriousMind good to know. But when you add source terms in general relativity, how do you define the fields if you haven't already fixed the background spacetime. E.g. In Einstein Maxwell theory, what do those $F$s mean where we don't define fields as irreps of Lorentz group?

@NairitSahoo in introductory QFT (to my knowledge) quantum fields characteristically transform under (finite dimensional) irreducible representations of the Poincaré group (and direct sum reps thereof)

and dictate that that is the actual Hamiltonian. By Fourier transforming back to position-based operator form of the Hamiltonian, you will either get a constant term subtracting in the Hamiltonian, which doesn't matter, or that the specific form of the Hamiltonian will look somewhat awkward, just to keep the normal ordering intact in the position form. In either case, the new Hamiltonian will be much more physically sensible.

@SillyGoose Yes and Poincare group is the isometry group of the base space on which your theory is defined.

in a discussion I had some time ago, I pointed out that the problem is the commutation relations of the field operators, because as soon as you make them singular at x=y, the vacuum energy becomes undefined (not infinite, because delta^{3}(0) is like writing "teapot" in the equation)

@SillyGoose Oh are you talking about the Poincare vs Lorentz? I mean yeah... fields transform trivially under translations. SO I got rid of that

@NairitSahoo They're just 2-forms/antisymmetric 2-tensors - again, vector and tensor fields are defined on arbitrary manifolds without reference to a metric.

@RyderRude physics and math

and to me, writing H_{new} = H_{old} - teapot was something of a worry

@NairitSahoo i am making the distinction that "fields are irreps of [blah]" is not literally true

@NairitSahoo i also like philosophy on top of these

@ACuriousMind Oh you mean the "transform" part... Oh yes, sure. Abuse of language :p

@RyderRude Philosophy, yeah. Missed that. That's the toughest for me. I mean according to me we all can handwavingly speak about some ideas but it is more difficult to understand some other person's thoughts and ideas.

@Davyz2 yes, you are perfectly correct - and this is all related to renormalization. If one does all these steps "correctly" from the start, one incurs no infinities - but the approach that does that (Epstein-Glaser renormalization/causal perturbation theory) is so technical that it is not what people use either in intro QFT courses or for practical computations

@ACuriousMind what? why is it not used for practical computations?

@NairitSahoo yeah.. in a philosophical debate they have to fix the framework they use and all the definition, because different people use words differently

This also happens in casual conversations, which is y it's important to define words

I often ask people "how r u defining x"

the problem however is that if you change the commutation relations to be non-singular, maybe by approximating the delta as a limit of functions, the theory becomes non-local, but gives you a stable vacuum

@naturallyInconsistent ...as far as I know, all the phenomenologists at CERN etc. use "normal" renormalization schemes like MS-bar

@ACuriousMind oh, that, the trivial interpretation of your words. It would be difficult for people to use something that they don't even know exists...

@naturallyInconsistent are you also an omniscient AI?

I'm feeling really bad. I will talk later

@ACuriousMind But that would be too general, don't you think? I mean if we started doing intro QFT in flat space just like that then we would start including condensed matter field theories and what not into our definition of field theory. But still we restrict to the class of field theories where the fields transform under finite dim. irreps of Poincare group, when we do "relativistic field theory".

@Davyz2 ACM just talked about a scheme that carefully splits the Dirac delta so that the infinities go away, with the standard effect that mass and charge are free parameters that you define, rather than coming out of the theory.

I am merely asking what analogous thing we do when we want to do "relativistic field theory" on curved spacetime?

@NairitSahoo No, I think you fundamentally misunderstand what the Poincaré reps are doing here.

@Davyz2 No?

There's no "class of field theories where the fields transform under Poincaré irreps"

A field is simply a vector/tensor field in the mathematical sense

@NairitSahoo Why do you think it makes sense for people who know things, to help you run before you can walk?

it transforms, generically, under $\mathrm{GL}(n)$ where $n$ is the dimension of the manifold

@naturallyInconsistent how do you split the dirac delta? do the field commutation relations still describe a local theory after the splitting?

Now, when you have any kind of special isometry group $G\subset\mathrm{GL}(n)$, then of course all the field also transform under that - because it's a subgroup

@naturallyInconsistent Because... I can?

it is not some additional restriction on the kinds of fields or something like that

Narit plz don't make ACM angry he can erase the internet

@Davyz2 as I said, the details are rather technical - it's why we don't do this in most courses. But yes, causality/locality is carefully preserved by the process

@Davyz2 Read the textbook... Yes, it is still local and causal. The requirement that the resulting theory be those is how they found a way to split the Dirac delta. I am still studying the front bits of that text

but what are you referring to when you say "split"? as a general thing I mean

@NairitSahoo The conversation that has happened here made it clear that you do not understand the nuances being shown to you, and you also admitted to not understanding things. Why do you feel like you have to insist that you understand stuff that you do not understand?

@Davyz2 this picture is as much as I understand of the splitting...

@ACuriousMind Yes that is true. But I am saying that there are two kinds of books: the ones which deal with condensed matter systems, and the ones which deals with HEP. The latter are the ones which restrict to the irreps of Lorentz group. I am saying, what to do if I wanna do field theory in curved spacetime and HEP?

@naturallyInconsistent most rigorous statement in a physics textbook

i hate when college tries to squeeze an immensely hard and poorly understood topic in a semester as if it was "how does a pendulum move 101" or something

@naturallyInconsistent It has not shown anything like that. What you have been saying is clearly wrong.

@Davyz2 It is 234 pages of very dense maths just to get to the splitting

@NairitSahoo then, again, that is the topic of "QFT in curved spacetime"

the problem is not defining the fields, the problem is figuring out which of the methods of Minkowski QFT transfer and which do not; famously one result of QFT in curved spaces is Hawking radiation of black holes

@ACuriousMind Yes obvio :p I mean, how do people define fields there? I have not seen a book where they define fields properly like they do in relativistic QFT books. So they deal with generic fields (sections of bundles) or what?

@ACuriousMind oh...

@naturallyInconsistent oh btw, which book does this derivation?

@NairitSahoo didn't I just explain like twice that fields are not defined by the isometry group or anything? If you didn't get what I said about GL(n) and isometry subgroups I don't know what else to say.

I asked this question on stack exchange btw and somebody just commented me "go read books lolz" as if I didn't already cover 4 QFT books

@NairitSahoo Books and articles that talk about QFT in curved spacetimes assume that the reader already are completely familiar with SR QFT. Why do you think a book that starts beneath that point would even make sense?

luckily hbar is not some rep-intoxicated pit of narcisism and people do have sensible answers

@Davyz2 why you no go look up the hints that ACM already gave you? It is Finite Quantum Electrodynamics, the Causal Approach

thanks i'll check it out

@Davyz2 QFT is a topic so dayum horrible that "go read books lolz" is just not sufficient. There is a need for all of 1) holding nose and just following a presentation 2) discuss with people who know better 3) work through stuff 4) read, lots 5) cry a lot

@Davyz2 do note that hbar is also on P.SE...

@ACuriousMind Yeah I got that. So does it mean that if I formulate field theory on flat spacetime and even if I do non-relativistic physics, I will work with representations of Poincare group?

@NairitSahoo you would know this answer yourself if you actually bothered to..

@naturallyInconsistent yes but the difference between hbar and P.SE is that one gives answers instead of intoxicated ramblings

or copy-pasted GPT answers from undergraduates

@naturallyInconsistent 1. No, books do not always assume that you must know everything about field theory in flat spacetime. Do you absolutely need to know about the parton model to understand Unruh effect? Maybe not. 2. I think so because it is essential to have a notion of what fields mean before starting to do field theory. This might be trivial as indicated by ACM, but still it is worth asking.

@Davyz2 the stuff I've seen on P.SE is not all bad...

oh yes of course, but hbar is still better

@naturallyInconsistent You would tell this answer yourself if you actually bothered to know the answer.

Narit do note that naturallyInconsistent is also an omniscient AI like ACM and I would avoid enraging him for reasons explained in sci-fi novels

@NairitSahoo I never said anything about everything. But you should know enough of SR QFT that you understand the basic bits of renormalisation, and of why major parts of even just defining things is already horrible in SR QFT. One learns humility in QFT.

@NairitSahoo PLENTY of introductory condensed matter physics in QFT framework textbooks cover QFT in the NR context. You can please get off your high horse.

If you are this level of arrogant as a 2nd year undergrad, it will not bode well for your future interactions.

@naturallyInconsistent But do you know whether the fields they discuss are irreducible representations of Lorentz group or not? I think not. Then why not? Why do they allow for more generic description of fields (but still within the realm of ACM's geometric definition), whereas in HEP books they restrict to irreps of Poincare group?

In fact, it is often remarked that NR QFT as treated in the standard condensed matter textbooks tend to be easier to follow and understand than the standard introductory SR QFT textbooks. At least for the initial 2nd quantisation bits

@NairitSahoo Why would it mean that?

@NairitSahoo They are NR. I just told you earlier that an equation can only either be Galilean invariant or Lorentz invariant, not both.

As for why SR QFT is interested in irreps of Poincaré group, well, the interesting consequences of SR QFT is precisely coming from the merger of SR and quantum theory, including the all-important spin-statistics theorem and its implications for microlocality.

like, why would you work with Poincaré representations if your isometry group is not Poincaré?

@ACuriousMind In NR QFT, my isometry group is still Poincare, right?

@NairitSahoo no, of course not

what? no right?

why would it be poincare

poincare is the pope of QFT confirmed

One of the most annoying things about all the elitism in science is exemplified in the headache of reading Weinberg. The man derived a LOT of the basic building blocks of QFT, yet he would not explain why he is presenting QFT in the precise order that he is presenting them. He definitely knows why this particular road is the only one that makes sense, yet he would not tell you that if you walk down this alley way, there lies a dead end.

the main problem is pretending that QFT exists in the first place as a theory

@NairitSahoo How is this a statement anything else than showing off how little you understand? Why must you do things this way? How can this be a constructive conversation?

@Davyz2 note that if you take this position, then you would throw all of physics under the bus.

@Davyz2 From a mathematically rigorous standpoint: Sure, that's true. But as a physical theory, it undeniably exists, and it has undeniable success. Don't throw out the baby with the bathwater, physics is - as much as I wish it was - not mathematics.

no you would just throw glue-stichted algebraic schemes like QFT under the bus, or at least not having QFT presented as something that a graduate could understand without 10 extra years of math

@ACuriousMind I mean... suppose we wanna do fluid mechanics. The fields are velocity fields of the fluid. They are $\vec{u}(\vec{x},t)$, right? Now, $(\vec{x},t)$ indicates that the base space contains both space and time labels. So I thought that it is formulated on Minkowski space afterall

but ok, of course QFT is extremely succesful, that I am very happy about

my problem is more about teachers pretending to know QFT

@NairitSahoo but if you actually do formulate the theory on Minkowski space, you're doing relativistic hydrodynamics

but we were talking about NR = non-relativistic theories

the defining characteristic of a non-relativistic theory is that it is not taking place in Minkowski space, but good old Newtonian/Galilean spacetime

the good ol "i can sync my clocks everywhere"

@naturallyInconsistent 1. We all understand little of physics. There might be some mistakes in my understanding but they are not total bullshit from some pop-sci journal. So I insist. I just shared my understanding. I am not atleast saying that vectors are dimensions or Poincare group is a relativistic field or something completely crazy like that...

@naturallyInconsistent 2. I must do things this way because I don't want to waste any time. I believe I can make some progress with some help

Narit what was your original question? it got lost in the chat and I am curious

@naturallyInconsistent That this can be a constructive conversation, can be seen from the ongoing conversation itself!

@ACuriousMind Oh, got it.

See @naturallyInconsistent it wasn't so hard to just be nice and reply?

@Davyz2 I thought that just cz in relativistic field theories fields are irreps of Lorentz group, so fields must always be defined as transforming as finite dim. irreps of some isometry group... then I thought... what about spacetimes which don't have isometry group?

The answer which I learnt is that... well... you don't define fields that way. It just happens to be that way for certain theories automatically.

ok bye everyone.