by the way, why is it okay to keep up to first order terms and still claim that $\omega$ is truly antisymmetric as in pg 59 eq 2.4.2 in weinberg? or is this like approximately antisymmetric

@SillyGoose the fact that "to first order" already captures everything infinitesimal about Lie groups is one of the wonders of Lie theory

if you want to really understand why this works you'll have to learn Lie theory from the mathematicians

also is there a name for a general procedure from starting with the data of a Lie group and obtaining its Lie algebra's commutators. In particular, does Weinberg's pg 59-60 follow this general procedure?

@ACuriousMind next semester i will :D

@SillyGoose The general procedure for this are Lie's theorems, and the Lie algebra is the algebra of left-invariant vector fields on the manifold with the Lie bracket being the Lie bracket of vector fields

oh :0

wait so in this land of Weinberg's relativistic quantum, the hamiltonian always commutes with the momentum operators (as an example) as a consequence of the lorentz condition?

I am asking because this seems a big change from say textbook quantum mechanics in which you can have a system that is not translationally invariant, so momentum and hamiltonian do not have to commtue

@SillyGoose do you have any example for such a system?

hm not in mind right now

but i guess i just thought translationally invariant systems were special because i thought things like lattices were special because they are translationally invariant

how is a lattice translationally invariant

well then i must have misunderstood xD

a lattice is translationally invariant if you translate it by some integral multiple of its lattice vectors

but not under general translations

oh so then a lattice is an example

but also i am thinking in classical physics a system that you do work on breaks the condition that $\frac{dp}{dt} = 0$ so maybe some quantum analogue of that? but im not so sure

I'm not sure what you mean

well i suppose if your system is a billiard ball then you apply an external force the momentum has gone from $0$ in the initial instant to some momentum value induced by the external force

so momentum is not conserved if you just look at the system

what

you're really overcomplicating things

the simplest system that does not have momentum conservation is just the harmonic oscillator

it also doesn't have momentum conservation classically

oh

so before you start claiming that somehow non-conservation of momentum is a feature of quantum mechanics, you first need to sort out your thoughts about the classical harmonic oscillator

okay well i can see that i think the classical harmonic oscillator is a specific case of the general idea that applying a force to a system makes the system's momentum won't be conserved

@SillyGoose but momentum, overall, is conserved, once you consider the momentum of the spring!

QFT is a "theory of everything" with no external parts

so naturally momentum should be conserved when you consider "everything"

hm i see

so QFT allows no notion of environment (relative to a system)?

sometimes it does

err idk i think i don't know enough about what QFT looks like as a theory

hm how interesting well i should see what an actual hamiltonian will look like later on in the book

does totally antisymmetric mean that any exchange of two indices introduces a negative sign. so $\epsilon_{123} = +1 \implies \epsilon_{213} = -1$?

what else would the definition of antisymmetric be?

just making sure :) i was not certain about what "totally" should be taken of.

now this i can thank weinberg for. i recall trying to figure out what the levi cevita symbol was and nothing made sense (though perhaps i was just being silly). but now this is as simple as can be

Levi civita is dolce vita

so the result from the content surrounding (2.2.24) is that abelian lie groups have commuting generators, which implies vanishing structure constants, which implies being able to represent finite transformations of the lie group as exponentials of the generators, at least to my understanding

So now on page 60-61, don't we already know that the generators of the group of pure translations commute (2.4.14) on pg 60? thus, we do not have to go through the step of 2.4.25 which considers the parameter additivity?

or perhaps i am confused

oh nevermind. i think maybe we don't know that P is the generator of pure translations

but these steps show that it is

@ACuriousMind i'm having a hard time accepting this, given our current "infinite time" formulation of QFT. I do think that some formulation of QFT would be a "theory of everything", but the current partition function stuff only talks about infinite time

The way I think about the current QFT is as a "black box" theory. QFT describes what wud happen if u measured some property of the black box.

But a non-black box understanding of qft remains a mystery. I wud like to think of the entire universe as a black box with quantum fields everywhere, but then, u can no longer measure a property of that black box as a whole. Becuz we wud need omnipresent measurement devices for that

So I think what we r really measuring is some local integral of energy density, instead of the total energy of the black box

I am trying to do a bit of a write up on the poincare group; does this way of describing Wigner's theorem seem sufficiently accurate :0

given the principle of relativity, is the linearity of the lorentz tranforms only formally derivable using differential arguments as in here: physics.stackexchange.com/questions/12664/…?

@SillyGoose what would be a non differential argument? linearity after all is a property of a derivative too right?

BTW thanks for linking that question, never saw that before

(but did wonder about it!)

@SillyGoose scrolling through the answers, this one may be what you're looking for

@SillyGoose this link is not using the "principle of relativity". It's using the Minkowski metric to derive stuff. Assuming the Minkowski metric is pretty much equivalent to assuming Lorentz transformations themselves

Becuz Lorentz transformations preserve that metric

@RyderRude Look at the link I mention :) great answer that uses only linear algebra, proves a very general theorem and then applies it to SR

Lemme see

This answer is also assuming the Minkowski metric in the las paragraph @Amit

Yeah, but it's not clear that only linear transformations preserve the metric

Also note even there, he has to assume that the transformation is surjective. This implies there can be other non surjective transformations that preserve the metric

Yeah. If u assume that metric, u can derive this. I'm just saying that this is more than assuming "Principle of Relativity"

There r other derivations of this that do not start with the metric

But i think they assume linearity itself

Or perhaps they justify linearity using "homogeneity of spacetime", which is an idea tied to the "Principle of relativity"

i don't remember how you derive the spacetime interval

U can derive lorentz transforms instead using four assumptions. And then those transforms give u the metric becuz they preserve it

yeah

The justification for linearity on wikipedia is : "Now since the transformation we are looking after connects two inertial frames, it has to transform a linear motion in (t, x) into a linear motion in (t′, x′) coordinates."

This justification is closer to "principle of relativity"

but you know this proof i've linked to, it doesn't really assume a specific metric. it only assumes the metric itself is bilinear

it would have worked even for a Riemannian (non pseudo) metric

in the end he does explicitly state it works for taking $g=\eta$ but that's not necessary

Yes. Becuz rotations and Galilean transforms r also linear

I still think that assuming a metric is pretty much assuming the full theory. But I philosophically prefer this approach too

but that's cheating too maybe lol, maybe there's a non linear metric that preserves the interval

Maybe idk

no I mean, everything he did was 100% general apart from assuming the metric itself is linear, he didn't specialize it to the minkowski metric (aka spacetime interval)

When i was trying to rule out Aristotle's relativity, I took this same approach. Assuming the metric gives u the relativity of constant velocity frames, rather than constant rest frames

So this philosophically prefer this metric approach

note also it implies we take for granted inner products are linear

weinberg calls this condition the principle of relativity

Yeah. U r right @Amit . There cud b any metric if we wanted.

althought im not sure how to show that that mathematical statement implies that experimetnal results at a fixed point in space-time are identical from the perspective of two different inertial reference frames (weinberg's def of principle of relativity)

I read that answer, I think it is actually the same one as the first answer (with the Christoffels) only that the christoffels are in disguise :D

This stuff is kicking the can down the road from "linearity of transformations" to "linearity of metric"

Wikipedia explains linearity of transformations using principle of relativity itself

Pls see this en.m.wikipedia.org/wiki/…

The section is "Group theory"

> "Now since the transformation we are looking after connects two inertial frames, it has to transform a linear motion in (t, x) into a linear motion in (t′, x′) coordinates. Therefore, it must be a linear transformation."

This approach is pretty good too becuz it's using some natural philosophical assumptions like Principle of Relativity. But I too like the metric approach becuz it's shorter @Amit

@Amit This uses the principle of relativity of "constant velocity frames"

isn't that equivalent to saying that the derivatives wrt to time of both $x$ and $x'$ identically vanish after the same number of differentiations?

I mean maybe this assumption can be formulated concretely, then use it to proceed with the "blind" math lol

It just says that lines get mapped to lines. This is justified by principle of relativity in either of Euclidean, Minkowski or Galilean theories

And their final result is consistent with any of those theories

oh for inertial frame it would just be the second derivative right. so it's linear... it's equivalent to saying it's linear but from a physical argument

Yes

The physical argument being that "constant velocity frames r equivalent". This is y i recommended this approach becuz @SillyGoose wanted an approach using principle of relatovity

But i myself dont find this a justified assumption in itself. Becuz it rules out Aristotle for no reason. This is y I prefer the metric approach. It's shorter and rules out Aristotle @Amit

If u find the "equivalence of constant velocity frames" a justified assumption in itself, then this is a good approaxh

physical arguments are always hard to pin down :)

it would be crazy if i see you having constant velocity but you see me accelerating right

Yeah

but there's no formal mathematical way to explain why that's crazy, or is there?

This is assuming more than that. It's assuming that any straight line gets mapped to a straight line

if i map your motion to a straight line, yeah? why it means you have to map mine also to a straight line? 'cause it has to be invertible? inverse of a linear map is linear?

@Amit i dont think so. Principle of relatovity is the "dead end" in this approach

@Amit since no inertial frame is "preferred by the universe", if i am moving at v wrt to u, then u r moving at -v wrt me

v wrt u lol

The situation must b symmetrical becuz of principle of relatovoty

Lol

i also think it may just have to do with functions being invertible. a linear x'(x,t) implies a linear x(x',t'), etc.

Yes

Inertial frames must remain inertial wrt every inertial frame. This is what justifies that any straight line gets mapped to a straight line. This is what allows wikipedia to assume a matrix representing Lorentz transformations @Amit

Like, if i am in an inertial frame, and i see u as inertial and theres a third observer that's inertial, then the third observer also must see u as inertial

@RyderRude Did you see this? arxiv.org/abs/1604.06491

More of a philosophy paper than physics

@SillyGoose Here is a more intuitive way to see it. Lorentz transformations are coordinate transformations $x'^\mu=x'^\mu(x^\nu)$ such that the induced transformation (which is linear by definition as it's merely a change of basis in each tangent space) on rank 2 tensors leaves the metric tensor components $\eta_{\mu\nu}$ invariant. The Minkowski space is flat though and you can identify it with its own tangent space, so up to translations (with which I'm including Poincaré transformations)

the transformation is the same

The real reason behind the last sentence is that $\partial x'^\mu/\partial x^\nu=\Lambda^\mu{}_\nu$ does not depend on $x^\nu$, so you can integrate it and find $x'^\mu=\Lambda^\mu{}_\nu x^\nu+ a^\mu$

@Amit But this is saying that GR brought back the notion of "natural motion vs forced motion". This notion has been present in physics since Newton.

i didn't really think the paper makes any great point

I think it misunderstands Aristotle. Aristotle just said that the natural motion is to be at rest wrt some absolute space. This is the part that no physics theory agrees with.

yeah but maybe if people still misunderstand aristotle it means he wasn't so clear too

@Amit yeah. It is a minsunderstanding of GR. idk how he even brought QFT into this lol

lol, he tries to draw an analogy between the divide aristotle made and the divide today between forces and geometry

Ooh

Except they have nothing in common other than being divides lol

right :D lol

I guess it's not so hard to get a philosophy paper up on arxiv

What is the divide Aristotle made tho? Is it same natural motion vs forced motion divide he is talking about?

I take the paper's word for what it says about Aristotle yeah. I don't think there's any factual errors... just.. pointless

Yeah. Philosophers misunderstand physics way too much. Feynman said philosophers thought special relativity was trivial lol

"natural motion" and "violent motion" is what he apparently called it

he didnt need to bring GR for natural motion and QFT for forced motion analogies. Newtonian physics already has natural motion and forced motion

You mean "Philozophers" in Feynman's pronunciation lol

According to Feynman, Philosophers conflated the physics notion of relativity with the everyday notion of relativity that "left and right is relative" or something

So philosophers were suprised that acceleration was not relative in the physics notion

There was a criticism at the beginning of the lectures about relativity in FLP I iirc

I think that was the early days of pop sci. Someone probably gave a popular talk about SR to a Philosophy class.. it was probably quite good but was misunderstood. It went downhill from there, lol

It was like lecture 14 (?)

@Mr.Feynman yeah. He mentioned philosophers told him "no shit, everything is relative" :P

@Amit even today, philosophers do this stuff i guess

@RyderRude lmao

And today we are at the point that "electrons shake between parallel universes" a la Kaku

Lol

Mwi is a great bedtime story

some philosophers still push for "relational spacetime", even tho GR doesnt support it

Einstein abandoned this "relational spacetime" idea aftr he discovered GR

Tbf, depending on phrasing, relational spacetime can survive in GR

Like, u can just say that the fields exist on top of each other but the manifold doesnt exist

These existence questions wud b pointless in "logical positivism"

But philosophy is full of such discussions

There are only very very few metaphysical ideas that are fruitful in physics apparently

But this relational thing, maybe Penrose likes it. 'Cause his Cosmology I think presumes that in the absence of matter the whole thing somehow changes its character

At the point when only radiation is left

Penrose says there is a meaningful distinction between matter and radiation? @Amit

In QFT, these boundaries go away. U just hav boson particles and Fermion particles

in the sense of what happens when the universe evolves far in the future yes. he has some argument that since mass is a clock (equivalence of mass and frequency) then without mass something weird happens to time. that's as far as I could understand his popular'ish explanation

because his whole thing is that he needs to create a singularity of some sort once it is all radiation, and he claims another cycle (big bang event of sorts) happens then. his model of cosmology is cyclic

Oh. So he wants time to not exist in the asymptotic future. For this, there shud b no mass

And then big bang happens all over again (but this is just the popular explanation)

> "Penrose's basic construction[2] is to connect a countable sequence of open Friedmann–Lemaître–Robertson–Walker metric (FLRW) spacetimes, each representing a Big Bang followed by an infinite future expansion."

Woww

sounds technical and inscrutable to me lol

So he is defining "matter" as "mass-ful stuff". He is ruling out even radiation as long as it has mass

This is indeed a concrete way to define "matter" and distinguish it from "non-matter"

> "In his Nobel Prize Lecture video, Roger Penrose moderated his previous requirement for no mass, beginning at 26:30 in the video, allowing some mass particles to be present as long as the amounts are insignificant with nearly all of their energy being kinetic, and in a conformal geometry dominated by photons."

In standard relativity, the manifold cannot become a singularity just becuz it only has massless stuff. So i think he has made a different theory

He has made the manifold topology dependent on the fields. Idk.

Like, in SR, it is perfectly fine to hav a non-singularity universe which contans just light

that makes also some intuitive sense

Idk y people dont want the metric to just be its own independent stuff. Even weirder r people who want to reduce time to entropy. This shit is way too popular in pop sci

I think Rovelli has made a theory like this. He wants to reduce time into temperature

what do you mean the metric being independent stuff?

Like, it shud just b its own field just like Dirac or Maxwell field. It shudnt b an emergent phenomenon. Def not emergent from statistical mechanics lol

@Mr.Feynman is the line of reasoning leading to this in carroll?

@SillyGoose I skipped the first 3 chapters, if you tell me the page I'll check

@RyderRude It's not a usual field

Just the page, I have Carroll's book on my desk rn

oh i am not sure about naything in carroll. the claim comes from weinberg vol 1, but i was wondering if carroll shows linearity of the LT

or teaches the diffe G to show what you wrote

Lemme check

The idea is the same but he follows a more intuitive approach defining the distance between two points (of course it's here that the metric comes into play) and requiring that it is invariant

hm okay maybe ill try to grasp the intuitive (or motivational) way and then try to approach the more precise way

>:D my goals for these notes have been set

But he assumes the coordinate transformation is linear from the beginning

oh

I don't know, I find the way I wrote more appealing

wait now i am confusing the usage of linear i think

an equation like mx + b is linear because it has a first order variable term and a constant

which is what the lorentz transforms look like

that would actually be affine due to the translation term

but there is also the sense of linear as in by wigner's theorem a lorentz transform can be represented as a lienar unitary on hilbert space

That would a generalized Lorentz aka Poincaré transofrmation

oh i see okay so affine

wait now so affine implies linear?

like an affine equation must be linear. or is it a linear affine equation

@SillyGoose No, affine is more general

okay i see

Affine is basically linear transformation+translation

oh okay okay yes i was wondering if the non translation bit had to be linear for the thing to be affine

In a vector space your origin is fixed, namely the $0$ vector. Linear maps take $0$ into $0$. An affine transformation would shift it

@SillyGoose Yes

@SillyGoose I've found this answer by joshphysics that is basically the same argument as mine, maybe it is explained better

there are nonlinear reps of the Lorentz group IIRC

@Mr.Feynman this is also a very good argument. It neither uses the metric, nor the equivalence of inertial frames. It just uses equivalence of translated frames.

So this one is much simpler philosophically becuz it only says there's no preferred origin

But it gets us to affine transformations instead of linear

wait does wigner's theorem + lorentz transformations are Wigner symmetries (ray symmetrices that preserve experimental results) tell us for free that the lorentz transforms are linear? and all we have to prove is that they are affine

No, Wigner only tells you specifically about linear reps

What are we trying to do here?

@SillyGoose wiger's theorem tells u that representations of symmetries on the Hilbert space r either unitary or anti-unitary operators. This does not tell u anything about the Lorentz transformations that we apply on spacetime co ordinates

@SillyGoose And what do you mean by "the Lorentz transforms are linear"? In the context of quantum theory, we must distinguish carefully between a group and its representations: The Poincaré group acts on Minkowski space not linearly, but merely affinely (because translations shift the origin). But Wigner's theorem says that the representation of every group in QM has to be by (anti-)unitary linear operators on the Hilbert space.

@Amit pls also see this. It uses the best physical argument : "Non-preference of any origin"

@Amit I cud think of one way to justify Penroses's reasoning. Let's say u hav a minkowski universe of just light and u draw its spacetime diagram. Then the time labels of this diagram do not refer to the time experienced by any object in that universe

So i guess penrose cud say that time doesnt exist in that universe. But idk. The math works out anyway. It's just differential equations

Also, the spacetime labels r mostly meaningless co-ordinates. They mostly need not refer to time- experienced by anyone.

oh i see lots of moving parts

@RyderRude that argument justifies the linearity when we are already dealing with that kind of transformations. The fact that Lorentz transformations are of that form descends from the requirement on the metric

@ACuriousMind i have two confusions, 1) how is weinberg's mathematical statement of the principle of relativity (pg 55, 2.3.2) equivalent to the qualitative description of PoR (from two different inertial ref frames, experimental result at fixed space-time point predicted by laws of physics is same). 2) how to prove from weinberg's mathematical statement of PoR that Lorentz transforms are affine

@ACuriousMind I'm pretty sure @SillyGoose was referring to why, ignoring translations, the coordinates transform linearly (and so why one would want it to be affine in the first place)

@SillyGoose ad 2: This is just the standard proof that isometries of Minkowksi space are exactly the Poincaré transformations, see e.g. math.stackexchange.com/a/4485120/143136

I think that josh's argument together with the fact that the Minkowski metric is constant is more straightforward though

@Mr.Feynman i think homogeneity can also b justified using principle of relativity. Josh's argument doesnt need metric, but it only gets u to affine transformations instead of linear

also just to see if i have these bits straight. There is the poincaré group which 1) acts on Minkowski spacetime, which is $\mathbb{R}^4$ endowed with an inner product written in terms of the Minkowski metric (?), and 2) via Wigner's theorem as linear unitaries on Hilbert space

Yes but it's about a generic transformation. Lorentz transformations are defined in terms of the metric...

Yeah. But Josh's argument does apply to Lorentz transformations too. Josh's argument is very general. It applies to any theory following principle of relativity. This is y Josh's argument can b used to derive the affine nature of Galilean transforms too @Mr.Feynman

Again, yes but what is a Lorentz transformation?

I personally love this homogeneity approach. I myself used this approach in detiving Lorentz transforms a few years ago

@SillyGoose ad 1: How you arrive at the statement that transformations between inertial frames should be isometries of the Minkowksi metric depends on how you think about relativity. The constancy of the speed of light means that $c^2t^2 = r^2$ for a light ray in all frames, and so $c^2 t^2 - r^2 = 0$ needs to be preserved by the transformations between frames. The l.h.s. is just the Minkowksi metric, so transformations between frames are isometries of it.

@Mr.Feynman to get to full Lorentz transforms, u again dont need the metric. There r four other very natural assumptions along with Josh's argument that can get u there

@SillyGoose Yes

what is the proper mathematical object to call Minkowski spacetime?

@SillyGoose pseudo-Riemannian manifold

It's worth picking up a book on manifolds if you want a good grounding for GR, a lot of them are quite good. Things feel a lot less ad-hoc if you have learned a bit beforehand

@Mr.Feynman Pls see "using group theory" section : en.m.wikipedia.org/wiki/…

@RyderRude I didn't understand the part when he claims the order of epsilon vanishes 'cause he picked the standard basis

I was trying to derive Lorentz transforms a few years ago. And homogeneity just came to me naturally

@Mr.Feynman If you're talking about this I agree with the most recent comment: This derivation doesn't really seem to derive anything. If you start by saying that Lorentz transformations preserve the Minkowski metric, you don't need to add "homogeneity" at all, it's just a mathematical fact that the only isometries of that metric are the Poincaré transformations, so I'm confused why you consider adding superfluous assumptions more convincing.

Becuz homogeneity follows frm principle of relativtit

Either way it seems like it's a game of choose your assumption... if you drop one you have to start from another

because several of those are probably equivalent mathematically. especially since it's inherently linear algebra that has a lot of nice relationships

Yeah. I like the metric approach becuz it's the shortest. It's not worth deriving Lorentz transformations from four assumptions

I'd say you can view a Lorentz transform in two ways basically

okay so one route of relativity is 1) God decrees speed of light is $c$ as a physical law, 2) Principle of relativity implies that the speed of light should be $c$ in all inertial frames. 3) Any transformation between inertial frames must preserve that the speed of light is $c$. Thus, any transformation between coordinates, must preserve $\eta_{\mu\nu} v^\mu = 0$. This means that for a transforamtion $T$ to be consistent with the postulates (as taken here) of relativity, $T\eta T^{-1} = \eta$

You can view it as a diffeomorphism that applies to the spacetime, and then you can lift that diffeomorphism to a variety of fields

@SillyGoose $\eta$ has either both indices up or downstairs :)

Or you can view it as a transformation on the tangent bundle, and then there are maps that you can apply to other bundles from there

oh oopsies thank you

@SillyGoose there is the third route using homogeneity stuff and group theory assumptions

hm I think i saw that in the MITOC lecture on SR though maybe im confusing it

it starts with assuming symmetries of spacetime?

No, that wud already b assuming the Poincaire algebra

oh okay yeah i was abt to say that sure takes all the hard work out of it

But it does use principle of relativity. That's a symmetry between frames

It's here en.m.wikipedia.org/wiki/…

In the "using group theory" section

It also derives Galilean transformations all at once

But tbf that only requires $c\rightarrow \infty$

@SillyGoose also I think you mean to say that $\eta_{\mu\nu}v^{\mu}v^{\nu}=0$ but then you're referring to a light like vector. the transformations however more generally preserve the norm of any $v$ , then we say it is a "true" 4-vector (be it space-like, light-like or time-like)

oh gosh yeah i mean to write the inner product between $v^\mu$ and itself...lol index notation

yeah, it takes getting used to

@SillyGoose Your step 3 is imprecise (but in a way physics ways of talking about this often are): A transformation between coordinates is a priori just a diffeomorphism $f : \mathbb{R}^{1,3}\to \mathbb{R}^{1,3}$ - the thing that acts on the metric like the "$T$" you wrote down is its Jacobian. What we have to show is not that $T$ is linear (it always is), but that $f$ is, and that $T$ is constant, so that $f = Tx+a$

this follows mathematically indeed purely from assuming $f$ is an isometry, but the proof is somewhat lengthy (see the earlier math.SE thread I linked)

@ACuriousMind Oh, sure. Josh's answer has nothing to do with the Physics of it. Once you start with the requirement on Minkowski metric, as you say everything follows. I meant that once you impose that on the metric, you could see that from it homogeneity follows from the constancy of $\eta$ and you could proceed in a way similar to josh's. I'm not saying that the requirement on the metric alone is not enough but that I think that it being enough can be shown more easily

@RyderRude they're requiring that $c$ is invariant though, it is a statement about the spacetime metric

@Mr.Feynman c only shows up as a degree of freedom in the transformation iirc. Its interpretation as an invariant speed is derived later. In the derivation, the role of c is only as a parameter of the matrix

hm i see i must be misunderstanding. so how do we get that the coordinate transforms themselves are isometries of the metric from ad1?

Aftr u hav finish that derivation, u can later interpret c as an invariant speed

@SillyGoose Before you waste time looking for this proof in Weinberg: Note that Weinberg makes exactly the argument I make here in the QFT book, and then has a sneaky (3a) note on the claim that indeed the isometries are just Poincaré transformations, referring to his GR book - he's outsourced the proof to himself

@Mr.Feynman In fact, there derivation also derives Euclidean geometry where there is a c but it's not an invariant speed

Like u can hav an euclidean metric like $x^2+c^2y^2$

ACM, I'm not asserting my previous message is correct. Maybe the passage I'm glossing over regarding the constancy of $\eta$ is the critical part

lol i saw a question and posted it earlier here which provided the proof that lorentz transf are linear out of weinberg's GR book and i thought man am i going to have to look through that :P

@SillyGoose What do you mean? "The metric is the same in all frames" is equivalent to "the transformation from one frame into another is an isometry of the metric".

err I guess I am having trouble the difference between frame transform and coordinate transform

and translating the words into math

My reasoning was in this message along with the next ones

there is no difference

I mean the frame transform is like a lift of the coordinate transform

It just happens to be about the same for minkowski space and lorentz transform

I don't think they were using "frame" in that technical sense

Hm well I am confused about how to edit my step 3. Because I came short and only came to the Lorentz condition, not that $f$ is an isometry

also now i am confused how an affine transformation (lorentz transform) can act on the metric; in particular, how to deal with the translation

The affine transformation acts on coordinates

@SillyGoose What, exactly, do you think the $T$ there is? The metric lives in the (co)tangent spaces of the manifold, not on the manifold itself. A coordinate transformation is a map on the manifold, and it acts on the (co)tangent spaces (and hence the metric) via its Jacobian.

Any transformation on coordinates induces a transformation on tensors

we need not induce this transformation. We just choose to pick the co-ordinate basis for the vectors.

I think the $T$ is the jacobian between the coordinates you are transforming between, written as $\Lambda$ in weinberg, i think

Iirc there is some study field which uses the diagonal basis for each tangent space, irrespective of the co-ordinates

@SillyGoose In Weinberg's notation the isometry itself is $x'$ and the Jacobian is $\frac{\partial x'^\mu}{\partial x^\nu}$

wait so the right most object is not identified with $\Lambda$?

What is $\Lambda$ then :0

what "right-most object"?

the jacobian

well, after you have figured out that $x' = \Lambda x + a$, then you have that the Jacobian is just $\Lambda$, so yes, they're are "identified"

Which Weinberg book are you reading? iirc his books are a not exactly beginner friendly

QFT I

I feel like there are better ways to introduce yourself to QFT :P

@ACuriousMind could you tell me what fails with the message I have linked about (and the next ones)? Hope this ping doesn't bother you

U can also try Schwartz's book. He is more level-headed about particles vs fields being fundamental

@Charlie yes, the rest of this chat agrees :P (see chat.stackexchange.com/transcript/message/63614249#63614249 and after)

i thin kafter solving this confusion i will either try to learn GR or from another QFT book, but many resources begin with field theory which i know naught

@Charlie I think the best way is to avoid QFT for a happier life

suppose its in the name though :P

Schwartz does mention that it is not known whether every theory with cluster decomposition is a QFT. Weinberg instead tries to reduce fields into a calculational trick

But Weimberg's view is not proven yet.

You need to learn a bit of field theory before you do QFT, I'm not sure trying to avoid that is a good strategy lol

And while Weinberg is ofc a very respected scientist and writer his books are generally not on the "recommended background reading" of any QFT course for a reason

@Mr.Feynman which one? I don't really have anything more to say about "homogeneity" or whatever, I just find the approach via the Minkowski metric the cleanest

@Charlie GR is already giving me all I need in my life: freaking wormholes, no Haag's theorem, timelike loops...

@Mr.Feynman but imagine time loops combined with quantum stuff!

And the one after

@Charlie My QFT course explicitly recommended Weinberg with the description "oftentimes takes an approach different to the one of this course"

@RyderRude that's why I'm waiting for my love for QFT to be revived. The renormalization group kinda killed my hype rn

@ACuriousMind almost all my courses recommend Weinberg and eventually not much of it is used :P

how much of qft is group theory

@ACuriousMind lol

@Mr.Feynman lol. I also just want to do quantum gravity. Quantum gives u non realism. GR gives u time loops. Their marriage wud b very pretty

There's more Lie group theory in QFT

@SillyGoose 11.92%

They r both very cooool theories. But yeah, QFT suffers from. haag's theorem while GR's math is well defined

darn

Also yes that percentage is the accepted number

well now what is the accepted percentage for group theory (lie theory) in usual quantum mechanics, so i can have a point of comparison :P

@SillyGoose quoting my QFT I course "...during this course I'll show you that group theory is Physics"

angry ACM noises

Imagine time-loops but the time is quantum, like the time flow is not well defined. It wud b a heavenly theory

so where is the QFT book which is all group theory with physical interpretation thrown in here and there ;)

@Charlie that's always what we mean in Physics :P

i just peaked at tong's lecture notes and the section on lorentz transforms seems to hand them out with no explaination :P but maybe i skimmed too fast

@SillyGoose I mean, chapter 2 of Weinberg is all about group theory just that the unusual notation hides it :P

I don't honestly like Tong's notes

@RyderRude I want too but I feel like I'll never bring myself to understand QFT. GR is not easy at all but with enough time and effort it is easier to understand...

@Mr.Feynman yes becuz no one in the world understands qft. We just know how to do scattering using the infinite time formalism. Bound state qft is still being worked out

I feel dizzy whenever i try to do reductionism with qft

Like, trying to reduce my everyday experience to qft, as quantum fields r all that exist

It's like having a field day

Lol

that is why i enjoy it though as opposed to other resources i have had a cursory glance at ! @Mr.Feynman

also okay I think i see how ad1 works now. so 1) blah blah SR postulates. 2) frame transforms leave $\eta_{\mu\nu}x^\mu x^\nu$ invariant. Let $f$ be a frame transform. Then ${f_\mu}^\alpha {f_\nu}^\beta \eta_{\mu\nu} x^\mu x^\nu = \eta_{\mu\nu}x^\mu x^\nu$, yielding the isometricality of $f$

where really $f$ is supposed to act on coordinates, but index notation builds in how $f$ can also act on $\eta$?

@RyderRude I don't think that's the case, maybe no one understands certain weird aspects of QFT but I don't think the common opinion is that QFT is sorcery :P

Which incidentally would be mine

@Mr.Feynman but u r troubled by Haag's theorem. Most physicists indeed dont care about it. For them, qft works inspite for that becuz of sorcery :P

And no one in the world has resolved Haag's theorem for the Standard Model. So people use SM without understanding how it works by magic @Mr.Feynman

GR has changed me into a different person, at this point Slereah lies in my chronological future

lol

ahgod another index mistake. the lhs metric should be $\eta_{\alpha\beta}$

Friendship ended with ACuriousMind. Now Slereah is ur best friend :P

@SillyGoose again @SillyGoose... you need to balance them free vs. dummy indices :)

@RyderRude Oh no @ACuriousMind is that true? Are we officially enemies now? :P

I still like QFT becuz it gives some spirituality vibes. Particles popping of fields. It is a mathematical mess but an incredibly cool theory

@SillyGoose the $f^{\alpha}_{\beta}$ you're talking about is simply $\frac{\partial{x}^{\alpha}}{\partial{x}^{\beta}}$ and I like to write it concisely as $x^{\alpha}_{,\beta}$ with comma indicating partial derivative :)

@Mr.Feynman yes. Slereah has seduced u to the dark side

@SillyGoose If $f$ acts on coordinates, then you shouldn't just put indices on $f$ - the thing in that equation is the Jacobian of $f$, not $f$ itself

@Mr.Feynman nah we're good

Nooo