The h Bar

ACuriousMind

12:00 AM

@Relativisticcucumber what does the complex conjugate have to do with this? I'm just saying you can see a complex scalar field as the sum of two real scalar fields, namely its real and imaginary parts

Relativisticcucumber

okay so if we just have $\psi$ in our lagrangian, we still have 4 operators? $\psi^*$ is not necessary for this at all?

ACuriousMind

I'm not sure how you're going to make a real-valued Lagrangian without using the conjugate :P

Relativisticcucumber

okay poor question, but i think i see what you mean. the new DOF is purely from adding a complex part. it has nothing to do at all with treating $\psi$ and $\psi^*$ as independent?

ACuriousMind

yes

you can write everything in terms of $\phi_1$ and $\phi_2$ if you like

Relativisticcucumber

okay i was misunderstanding. that makes more sense.

and you said in the case of scalar field this notion of antimatter/matter does not make sense, so how should i interpret these different operators in scalar field / free field context then? @ACuriousMind

ACuriousMind

12:04 AM

you shouldn't

Relativisticcucumber

@ACuriousMind so do you mean bosons cant have antiparticles?

ACuriousMind

or, rather, you should just get used to calling them "particle" and "antiparticle" because that's what they'll turn out to be once you coupling this single complex scalar field to something else

@Relativisticcucumber where did I ever say that?

Relativisticcucumber

oh no

ACuriousMind

this has nothing to do with bosons, really - if you just have a single free field, I don't see any lot of meaning to saying there's a "particle" and an "antiparticle"

Relativisticcucumber

@ACuriousMind why not

ACuriousMind

12:06 AM

how would you define "anti-particle" :P

Relativisticcucumber

@ACuriousMind well cant any particle experience a free environment? so if i have 2 particle types why cant i act on vacuum state to make either one? and we still have free particle, right?

i think i might be missing smth

ACuriousMind

the two crucial aspects of anti-particles are that they a) have the opposite charges of their respective particle and b) can therefore very easily pair-annihilate with their partners

in a theory with a single free field you have no charges and there are no interactions that would allow annihilation

@Relativisticcucumber The field is free, not the particle

Relativisticcucumber

hm so what about a photon? it doesnt have an antiparticle?

ACuriousMind

what do photons have to do with what I said above?

Relativisticcucumber

you said antipartcles must have the opposite charge of their respective particle

but photons dont have charge and i thought no spinless particles have charge?

ACuriousMind

12:09 AM

I said charges, not charge

Relativisticcucumber

oh no what does that mean sorry

ACuriousMind

but indeed the photon is so chargeless that the idea of an "antiparticle" for it doesn't really make sense

it's either "its own antiparticle" or "has no antiparticle", depending on the context

Relativisticcucumber

oh is charge like all its characteristic features?

ACuriousMind

you will find out :P

you really don't need to worry about the exact meaning of "antiparticle" at this stage. Just note that the two sets of c/a operators create two distinct kinds of states and move on

Relativisticcucumber

XD

ACuriousMind

12:12 AM

it will become very clear later - once you look at a real theory like QED and not the toy model of free fields - why one might call one of these kinds of states the "antiparticles" for the other kind

Relativisticcucumber

i guess i should keep going since at this rate ill never even get to QED

2 months on free fields kms

ACuriousMind

I mean are you sure you need to learn QFT right now? :P

it's perfectly possible to realize that a topic is not the right fit at the moment and return to it later instead of suffering through it

Generally I find that the more classical mechanics one knows, the easier it becomes to learn QM, and the more QM and classical field theory one knows, the easier it becomes to learn QFT

This approach to QFT where we teach it to people who have just learned QM and often never even seen field theory or Hamiltonian mechanics proper is not the most pedagogical

and especially if you're trying to understand the details it gets very confusing which parts are "QFT", which ones are "quantum" and which ones are just field theory

Relativisticcucumber

12:36 AM

hm I think my issue might be that I dont feel I struggle in QM, but I could be wrong. I have worked through most of Sakurai and can do the problems and work through most of the sections without issues. some of the angular momentum stuff is a bit hand wavy to me though. i think what i need to transform is the way that i think about the things in quantum mechanics though. im not sure the most effective way to do that except approach it from different angles. @ACuriousMind

mechanics is just so boring to me but maybe i should return to it. i generally just want to learn qft tho and have spent years building up to it but maybe im still not ready :P

ACuriousMind

maybe you're also going in here with the wrong expectations - from your questions you seem to expect that the intro texts to QFT should build QFT slowly and methodically and carefully explain all the words they use or why exactly they're doing certain things

but that's not what they're designed to do - the typical intro to QFT is designed to get you as quickly as possible to a point where you could compute stuff in QED or the Standard Model

so in some sense they're actually building as little theory as possible, this is a very pragmatic kind of theory

naturallyInconsistent

Jigglypuff

B A H

Relativisticcucumber

i think i just feel like they skip so much stuff or that they dont come out and say it but i think im filling in the gaps slowly but surely XD i think this is the first theory that i have come to that doesnt really have a "way to think about it" and so im just acclimating to that.

BAH

naturallyInconsistent

Yes, ACM is most frustratingly correct here. So many QFT texts think that they are making it easy on students by making the path from first page to workable first QED computation supremely short.

ACuriousMind

12:53 AM

@Relativisticcucumber I mean if we're being honest hep-th QFT in practice doesn't need to be a giant theory that is applicable to all sorts of situations. We only have one universe, and it's described by the Standard Model, so what we need in practice is people who can figure out what the Standard Model tells us.

This isn't like QM where we go "here's a general framework that you can use with very different Hamiltonians and spaces to describe all sorts of situations" and then there's hundreds of examples, there's really only a single specific situation in hep-th QFT that matters in practice

the case where you have real-world systems described by different kinds of QFTs is really cond-mat, not the relativistic scattering hep-* world the typical QFT book is concerned with

doesn't mean I'm a big fan of this approach to teaching QFT, but I get where it's coming from

naturallyInconsistent

1:16 AM

@SillyGoose I read that you and ACM had a wonderful discussion while I wasn't here. But I have something slightly different from what you were talking about, to contribute: If you want to have a nice conception of Hamiltonians, I would suggest thinking of it as the generator of time translations. Firstly, you had Newtonian mechanics going F=ma, i.e. if you know the position and velocity/momentum at one time step, N2L gives you the next time step, Hamiltonians, as guaranteed by Noether's theorem,

is the appropriate generalisation of this concept in the case whereby you specialise to positions and momenta. In the cases we care about, this is a 1st order coupled set of equations, but it is not strictly necessarily the case. We might have started from energies and ended up with Hamiltonians, but I would point out that this is a pattern matching process: After we got Hamiltonian theory worked a bit out, it means that we can just write down more and more intricate Lagrangians,

Legendre transform into more and more Hamiltonian terms, and whenever the conditions are satisfied for those Hamiltonians to be identified with total energy, it means that we would be able to identify new terms in the total energy accounting for all kinds of systems. So, the relationship between energies and Hamiltonians is not severed, even though they are, strictly, not the same thing.

Now, you might think that, just because Hamiltonians do not have to be energies, that this complication would also arise in QM or QFT. The fact of the matter, however, is that QM and QFT are so difficult, that we have yet to need to consider Hamitonians that violate the conditions needed for them to be identified with energies in the classical setting. So, if you were wanting to generalise on this point, you were simply prematurely optimising.

Mind you, while in the classical setting, Hamiltonians might well not exist, as ACM already pointed out to you, the hodge podge of energy terms are still there, we can still speak of energies, etc. i.e. different generalisations have different potential crash zones and different interpretations. It is far better for you to think of the Hamiltonians in QFT context as the simplest interpretation. Or else, you will be drowning in the deep end of the pool for no benefit.

Note that even in the case whereby we are interested in modelling how a quantum system is interacting with the environment, the kind of situation whereby a classical system would just give up and assert that there is no Hamiltonian for it, we would simply just work with Hamiltonians in a larger (system+environment), and then trace away the environment part of the density operator. So, again, trying to go beyond the simplest conception of Hamiltonians is not really helpful.

Silly Goose

2:48 AM

@naturallyInconsistent this point certainly helps to contextualize things. I guess I have thought of quantum theory as being quite successful and consequently far reaching in its predictions, but this is not true really

Relativisticcucumber

4:02 AM

is it more common to use the heisenberg picture or schrodinger picture in qft? i assume schrodinger?

naturallyInconsistent

4:59 AM

@Relativisticcucumber Only ever interaction picture. But closer to Heisenberg picture. Schrödinger picture is for QM

@SillyGoose What do you mean? I was trying to say that quantum theory is exceedingly successful; I do not know of any prediction of QM or QFT that is in disagreement with experiment. It is just that we can use Hamiltonians that are identified with total energy (including energy in EM field et al) to get those predictions. We never needed to work with weird classical Hamiltonians non-equal-to energies for those predictions.

Mr. Feynman

6:09 AM

@Relativisticcucumber Schrödinger picture isn't used much (at least in standard treatment). You mostly work in Heisenberg picture, e.g. when you impose ETCR or ETAR (remember that the quantum fields are operators: they do not have a time dependence in Schrödinger picture) or when you define n-point GF. Interaction picture is used to set up the perturbation expansion

Silly Goose

6:54 AM

@naturallyInconsistent i think maybe i have an inflated perception of how many systems are out there in which $H$ is not total energy

separately, why do we care so much about the Poincaré group in special relativity?

naturallyInconsistent

@SillyGoose now you are on the train tracks straight into where the lorentzian vegetable was just on. If something obeys Poincaré group, then it at least has SR properties.

Silly Goose

I guess I don't get what the starting point should be taken to be

i used to think: 1) empirically, physical observations are invariant under Poincaré transformations. 2) okay, then perhaps a natural definition of a property inherent to an object or even of an object at all is for that property/definition of object to be invariant under Poincaré transformations. 3) then a natural way to tease out such properties and objects is by doing some rep theory

but talking with @Relativisticcucumber has made me confused

@naturallyInconsistent Maybe I am confused about what is meant by having SR properties

naturallyInconsistent

But it is manifestly not (1); Energy and momentum obviously transforms, iirc covariantly, if not, then contravariantly, under Lorentz transformations.

Because nobody else would put it as having SR properties. Literally just made it up. But you get the vague idea miao miao was trying to convey

(and miao miao is feeling particularly crazy today because of the repetitiveness at work. Have already automated most of the workflow, editing the input file rather than using the wizard interface to manually type in every run anew, but still, annoyingggggg.)

Silly Goose

@naturallyInconsistent but say we have a state $\lvert \psi \rangle \in \mathcal{H}$ and Hamiltonian $H$. Then, upon Lorentz transforming with $U(\Lambda)$ we get, by construction of the theory, $\langle \psi \lvert H \lvert \psi \rangle = \langle \psi \lvert U^\dagger U H U^\dagger U \lvert \psi \rangle$?

Relativisticcucumber

7:19 AM

@SillyGoose cf days of convo @ACuriousMind and i had ab this

painful but i finally understood so maybe get it directly from the source

Ryder Rude

@SillyGoose hello. probabilities like $\langle \psi |\phi\rangle$ are scalars under Poincaire transforms. $\langle \psi |P^{\mu}|\psi\rangle$ is not a Lorentz scalar.

the transformed expectation value is given by $\langle \psi U |P^{\mu}|U\psi\rangle$

where $|U\psi\rangle=U|\psi\rangle$

@SillyGoose u r not supposed to transform both the state and the operator to get the transformed expectation value. only one of them is transformed (according to Schrodinger or Heisenberg picture)

Ryder Rude

7:42 AM

to be clear, "by transformed four-momentum", we mean the expectation value of the energy-momentum operator of the other observer. this is not a scalar

but the expectation value of a fixed observer's energy momentum operator doesnt transform. it has the same value in all frames. this is what u r calculating when u do $\langle \psi U^{\dagger} U H U^{\dagger} U|psi\rangle$

different observers have different energy-momentum operators that they measure

Silly Goose

hm I guess I mean to talk about the case of expectation value of a fixed observer

because when I say "physical predictions are left invariant under trasnforamtions" I mean that the physical prediction is equal before transforming every object and after

Ryder Rude

ok u r keeping the observer fixed. but if u r boosting the experiment, then the same analysis applies. the measured energy of the boosted experiment is different from the original measured energy

however, the probabilities $\langle \psi |\phi\rangle$ are invariant, yes

different observer will agree on the expectation value $\langle \psi |H|\psi \rangle$, where $H$ is the energy operator of a fixed observer. then ur analysis is correct @SillyGoose

the agreement wont exist when either 1. the operator is fixed and u have boosted the experiment , or 2. the operators r of different observers and the experiment is fixed

Silly Goose

8:00 AM

Right okay i see

I am thinking in the context of a wigner symmetry is why i mean to transform every object

Ryder Rude

the Wigner symmetry PoV is useless in this context because, while the expectation value of a fixed energy operator is invariant in all frames, that energy cant actually be measured in every frame because each frame has its own energy observable

The Wigner PoV is good for invariance of $\langle \psi |\phi\rangle$ and the expectation values of scalar operators

this analysis is not specific to QM. in SR, u have a four-vector energy. each observer uses different basis vectors, so the components r not scalars. but the component in the direction of a fixed basis vector is a scalar

the invariance of $\langle \psi | \phi\rangle$ using Wigner symmetry is sufficient to establish invariance of predictions in all frames

ACuriousMind

8:23 AM

@SillyGoose But the Poincaré group isn't just a "Wigner symmetry" - like the rotation group in non-relativistic QM, it connects the states as perceived by different inertial observers

If you and another observer are related by a transformation $\Lambda$, then you assign the state $\lvert \psi\rangle$ to a system and they assign $U(\Lambda)\lvert \psi\rangle$

whether $\Lambda$ is from the rotation group, the Galilei group or the Poincaré group is irrelevant, there is nothing "relativistic" about this logic

Silly Goose

i believe i understand that beyond being identified as Wigner symmetries, that transformations exist and can be used otherwise. (i.e. if I want to mathematically model rotating my stern gerlach device $\pi/2$ radians).

but i think i am still confused about the privileged role Poincaré group has among physical transformations

and i brought up wigner because i thought the privilege comes from an empirical fact: that transition amplitudes (and expectation values and other physical predictions) are left invariant upon poincaré transforms (on the system; that is, every state; that is, when you have an observable $O$ represented in the original basis, you need to now conjugate it too by the operators to represent it in the transformed basis). And Wigner posits the mathematical statement that corresponds to this empirical

but it seems like this line of reasoning is not correct

ACuriousMind

that's just true for every (anti-)unitary operator

for any unitary operator, $\langle \psi \vert U^\dagger U\vert \phi\rangle = \langle \psi\vert\phi\rangle$

Silly Goose

well I guess I thought that it is only special transformations that get to be unitary

ACuriousMind

that's not special to the Poincaré group at all

@SillyGoose No, the point of Wigner's theorem is just that transformations that preserve transition probabilities have to be (anti-)unitary, not that unitary transformations are necessarily special physically

note that on an infinite-dimensional Hilbert space even in ordinary QM you have an infinite-dimensional group of unitary operators

they're certainly not all special :P

Silly Goose

hm well i still don't get why the Poincaré group is so privileged among all transformations (even among all transformations that are physically relevant)

what is special about rotations, translations, and boosts, that is not the same as say accelerating a frame

right but why should we care about frames related to each other by lorentz transforms

Ryder Rude

8:40 AM

QFT allows u to describe any frame, yes. this is a very good question. i think it ultimately comes down to the metric which is invariant under Poincaire transforms

the dynamical laws have this symmetry

ACuriousMind

@SillyGoose well, what is special about the Poincaré group in classical relativity?

Silly Goose

well in an intro. course on SR one is told that we postulate that the laws of physics take the same form in inertial frames (which are related by Lorentz transformations). So it seems like the task most approachable is to consider physics in inertial frames.

ACuriousMind

so what part of this logic do you think no longer applies to QFT?

Silly Goose

well i did present this as my current understanding of why put so much attention to Poincaré. but separately, I don't get why we should then define a one-particle state as an irrep of the Poincaré. this makes the Poincaré group seem fundamental when it is simply the "most approachable" set of transformations

Slereah

The one particle states are the ones that have a well-defined momentum

Surely that would imply some action on them by the Poincaré group

ACuriousMind

8:48 AM

@SillyGoose A particle is an irrep of the Poincaré group because where one Lorentz observer sees $n_i$ particles of kind $i$ another Lorentz observer should also see $n_i$ particles of kind $i$

i.e. Lorentz transformations should not create/destroy particles

Silly Goose

okay i see then I am making a wrong assumption

ACuriousMind

so states of definite particle number - and in particular one-particle states - are irreps of the Poincaré group

Silly Goose

why should one expect particle number to be preserved under lorentz transforms and lorentz transforms only?

ACuriousMind

@SillyGoose well, for one, because it would be extremely concerning if two inertial observers disagreed about how many particles there are in a box :P

Silly Goose

but why is not concerning that an observer and an observer accelerating relative to that observer do not see the same number of particles? (unless accelerating is part of Poiincaré somehow?)

ACuriousMind

8:52 AM

you really aren't willing to take anything for granted, are you :P

Let's try from another angle: Particles are described by mass and spin (among other things). Mass and spin are Poincaré invariants, so the space of objects of a fixed mass $m_0$ and spin $s_0$ is at least a subrepresentation.

Silly Goose

@ACuriousMind well if i am to believe this statement based on physical intuition (which seems to be the appeal) i would not expect accelerating to be any different :P. But more truthfully i wouldn’t be sure to expect particle number to be conserved or not at all under ang of these transforms

Ryder Rude

@ACuriousMind why do u say this? mass being $p^{\mu} p_{\mu}$ is a scalar in SR regardless of the co ordinate transforms

Silly Goose

Ive got to sleep now…but i will be back tomorrow

ACuriousMind

That subrepresentation $(m_0,s_0)$ decomposes into irreps. There has to be "particles of mass $m_0$ and $s_0$" somewhere in here. Assume it wasn't the irreps (but e.g. the sub of two irreps) - in that case you get that your particle consists of "two parts" that are separately Poincaré invariant and don't mix with each other under Poincaré transforms

what would you like to call indecomposable objects that carry a particle's properties like mass and spin if not "fundamental particles"?

ACuriousMind

9:10 AM

In the end, mass and spin are the Casimir operators of the Poincaré group, so it's really if you want to label something with fixed mass and spin, you need to put it into an irrep of the Poincaré group, but that would require you to first independently prove these statements about mass and spin and the Poincaré group and we usually don't even actually work out Wigner's classification in intros to QFT

bolbteppa

vice.com/en/article/epvkep/…

> In this latest paper, Oppenheim combines the two pillars, suggesting that spacetime is classical (not governed by quantum theory at all) and pairs it with a modified quantum theory. Under this reasoning, spacetime, even at the smallest level, is continuous and not chunked into little leaps.

Ryder Rude

r the Bogolibuov transformations just postulated in the Weinberg reps of Poincaire grp approach, or u can u derive them as the correct transforms for accelerated frames?? @bolbteppa

bolbteppa

> A new theory claims to solve one of the biggest problems in modern physics, but not everyone is convinced.

Another one in the bucket

Slereah

@bolbteppa Why should I trust Oppenheim and not someone who is Oppenheimer

bolbteppa

Bogoliubov transformation

In theoretical physics, the Bogoliubov transformation, also known as the Bogoliubov–Valatin transformation, was independently developed in 1958 by Nikolay Bogolyubov and John George Valatin for finding solutions of BCS theory in a homogeneous system. The Bogoliubov transformation is an isomorphism of either the canonical commutation relation algebra or canonical anticommutation relation algebra. This induces an autoequivalence on the respective representations. The Bogoliubov transformation is often used to diagonalize Hamiltonians, which yields the stationary solutions of the corresponding...

They are claimed to be isomorphisms, so if you want to prove something prove it's an isomorphism

Ryder Rude

9:20 AM

it says they diagonalise hamiltonians, which is correct. this part can be derived

bolbteppa

@Slereah er = ER = EPR = conspiracy!

Ryder Rude

but how do u know the hamiltonian of an accelerated frame using the Weinberg particle approach to QFT?

i think these transforms show that the field approach is more general

bolbteppa

Weinberg never mentions Bogoliubov

Ryder Rude

Weinberg's approach is good for simpler things like inertial frames only

@bolbteppa ooh

this is y the Weinberg particle approach is less general

u need field theory to know the Hamiltonian of an accelerated frame

the particle approach takes Poincaire transforms as the definition of the theory. so far, accelerated frames r just undefined

there is a long route to find the Hamiltonian of an accelerated frames in the particle approach

u start with reps of Poincaire groups, introduce fields, get the Heisenberg picture, and now define an acceleration transform of this time dependent field, and then fetch the Hamiltonian of the accelerated frame

Mr. Feynman

@ACuriousMind do you know a good paper explaining what group contraction is without getting too technical?

bolbteppa

9:29 AM

Accelerated frames is a completely different ballgame

Group contraction

In theoretical physics, Eugene Wigner and Erdal İnönü have discussed the possibility to obtain from a given Lie group a different (non-isomorphic) Lie group by a group contraction with respect to a continuous subgroup of it. That amounts to a limiting operation on a parameter of the Lie algebra, altering the structure constants of this Lie algebra in a nontrivial singular manner, under suitable circumstances.For example, the Lie algebra of the 3D rotation group SO(3), [X1, X2] = X3, etc., may be rewritten by a change of variables Y1 = εX1, Y2 = εX2, Y3 = X3, as [Y1, Y2] = ε2 Y3, [Y2, Y3]...

Slereah

@Mr.Feynman You can try the OG paper :

ncbi.nlm.nih.gov/pmc/articles/PMC1063815

ACuriousMind

@Mr.Feynman what do you need a non-technical explanation for? :P But you can always just read the original İnönü-Wigner paper - I usually find Wigner's papers very readable

bolbteppa

When you introduce a parameter into the commutators and take some limit, you get a whole new group

3

A: What is the relationship between the Galilean group and the Poincaré group?

It is assumed you have appreciated Inönü, E.; Wigner, E. P. (1953), "On the Contraction of Groups and Their Representations" Proc. Natl. Acad. Sci. 39 (6): 510–24, and the super-helpful Gilmore text in Group contraction. Very crudely, the Poincaré Lie algebra, $$ [J_m,P_n] = i \epsilon_{mnk} P_k ...

Slereah

There's a nice book on Lie groups that has a section on group contraction

Mr. Feynman

@ACuriousMind to get the gist of it and handwave my way though this cursed quantum phase transitions

Thanks all, I'll start reading

bolbteppa

9:34 AM

The wikipedia example of $SO(3)$ and the Poincare down to Galilei is good enough

Slereah

books.google.fr/books/about/…

p. 436

though those pages aren't on google books

bolbteppa

Both of his books discuss it briefly

PM 2Ring

10:06 AM

JCB just dropped a comment on the Kerr question. I wonder if he'll post an answer...

It's a mathematical claim so it can be resolved by thinking about the math. — John Baez 2 mins ago

Slereah

> And so, implying that the primary body is something else beyond earth, fire, air, and water, they gave the highest place a name of its own, aither, derived from the fact that it 'runs always' for an eternity of time. Anaxagoras, however, scandalously misuses this name, taking aither as equivalent to fire.

Ryder Rude

is Jon Baez a working physicist

wiki says he works for LQG

PM 2Ring

10:56 AM

@RyderRude He's a mathematical physicist. He mostly does mathematical / software stuff these days. See math.ucr.edu/home/baez

Here's some old stuff he did on GR. math.ucr.edu/home/baez/gr/gr.html

Ryder Rude

thanks

he loves algebraic geom

in what order shud i learn these topics : group theory, differential geometry, algebraic geometry, topology

i think differential geometry subsumes differential topology

dunno if algebraic geometry subsumes algebraic topology

what does Galois theory come under?

ACuriousMind

11:22 AM

There's a lot of search engines you could type these questions into instead of guessing

Ryder Rude

1:09 PM

im interested in physicist's math books

for topology and differential geom

can we say that group theory, diff geom, topology, algebraic geom is 99% of the math of physics

Slereah

Where's the analysis

and linear algebra

user 85795

^

Slereah

Where's the accounting for getting grant money

user 85795

Publish or perish.

Ryder Rude

1:34 PM

@Slereah these topics subsume those becuz u cant do diff geom without calculus and linear algebra

Slereah

Weird way to split the topics

You could argue you can't do most of the other without topology or group theory

Ryder Rude

oh

can u give me a good split ?

i think we just need two branches: diff geom and group theory. this, and the pre-requisites for this, covers 99% of physics

is there anything im missing

algebraic doesnt seem much relevant. is it?

what is a book for physicist diff geom which also subsumes topology

bolbteppa

1:54 PM

Logically: group theory, topology, diff geom, alg geom.

Ryder Rude

@bolbteppa thanks. is it better to learn from physicists or mathematicians

bolbteppa

Topology is mostly a waste of time, at most you need the most basic stuff in formal GR until you get to advanced specific cases which you can pick up as you go along

Definitely physicists no question

Ryder Rude

yes, i think mathematicians r too hung up on making things rigorous instead of the ideas

@bolbteppa what is algebraic geom relevant for

bolbteppa

Geometric Langlands correspondence

In mathematics, the geometric Langlands correspondence is a reformulation of the Langlands correspondence obtained by replacing the number fields appearing in the original number theoretic version by function fields and applying techniques from algebraic geometry. The geometric Langlands correspondence relates algebraic geometry and representation theory. The specific case of the geometric Langlands correspondence for general linear groups over function fields was proven by Laurent Lafforgue in 2002, where it follows as a consequence of Lafforgue's theorem. == History == In mathematics, t...

Good luck ever getting to it

Ryder Rude

thankss

pls suggest a diff geom book

and is measure theory much relevant? if yes, is it subsumed in these topics? @bolbteppa

bolbteppa

2:04 PM

Nakahara has probably all the formal stuff you'd ever need for physics but even this is simply too unintuitive and not worth it

Slereah

All you need to know for measure theory is that $\mu([a,b]) = b-a$

Ryder Rude

i dont like formal. something like Zee's book for grp theory

bolbteppa

The most basic measure theory might be useful if you want to waste your time on formal path integrals

Ryder Rude

Zee writes informally

@bolbteppa yeah. one of these measures called $e^{-iS[\phi]} d[\phi]$ shows up in QFT

idk how deep that idea is

@Slereah lol

Slereah

@bolbteppa It's not even a measure

Ryder Rude

2:08 PM

it's like they just took the path integral and called the whole integrand a measure

becuz the functional measure $d[\phi] $ is "too small"

but it's really badass to say that the entire theory is in a measure

so i like this measure $e^{-iS[\phi]} d[\phi]$

bolbteppa

The physics is hard enough without worrying about every last detail of the tools you're using to discuss the physics

ACuriousMind

you don't need "measure theory" to understand the path integral measure

the canonical mathematical text on the path integral measure is still Glimm & Jaffe, there's a lot of functional analysis in that book and almost no measure theory :P

Ryder Rude

so no to measure theory

Ryder Rude

2:30 PM

i wud never in a million years hav expected lattice field theory to actually give correct preditions but they literally just do a discrete path integral to compute non perturbative effects

i just couldnt put that much faith into the eqns we've discovered. but it all holds after such giant computations

really makes u think that the universe is a giant computer. laws of physics r extremely strict rules

one guy's reply to Wigner's unreasonable effectiveness of math was "physicists find math becuz they r looking for math. so all these discoveries have a bias"

such a ridiculous take

it is the response of Richard Hamming : en.m.wikipedia.org/wiki/…

math is def forced upon us by our mode of existence

naturallyInconsistent

3:24 PM

ACM, I answered a question before someone else pointed out that it is a duplicate; should I delete my answer and hope that the system will flag it as closed and delete it after some time?

ACuriousMind

@naturallyInconsistent If you agree it's a duplicate you should add your vote to closing it as one. I don't think deleting the answer is necessary, but if you think your answer adds something over the ones the dupe target already has, the best thing is to delete it and post it as an answer to the dupe target instead because ideally future duplicates will also be closed as duplicates of the same target, and so people will find your answer more easily there

naturallyInconsistent

Mine doesn't add much, just a lot shorter and to-the-point. I'll just mark it as duplicate and vote to close

@nickbros123 The abyss is gazing at yet another : physics.stackexchange.com/q/791515

Obliv

4:11 PM

what's a simple explanation of Liouville's theorem?

on wiki it mentions a phase space distribution function, what's that mean exactly?

a phase space represents the possible physical states of a system?

ACuriousMind

@Obliv why do you need an explanation of Liouville's theorem if you don't know what a phase space distribution is?

that's a bit like asking for an explanation of multiplication but not knowing what numbers are :P

Obliv

I think i get it, but what is the assumptions that lead to it

ACuriousMind

4:28 PM

@Obliv can you be a bit more specific?

Monty

4:50 PM

a phase space distribution is just a probability distribution on phase space which is usually the space of positions and momentums @Obliv

Monty

5:08 PM

Can anyone help me with the construction of the microcanonical ensemble

is the reason we cant average over the set $S(E)=\{(q,p) ~:~ H(q,p)=E\}$ that this set might have zero Lebesgue measure ?

ACuriousMind

@Monty Aren't you averaging over it with the measure $\delta(H-E)\mathrm{d}q\mathrm{d}p$?

Obliv

are*

ACuriousMind

@Obliv ?

Silly Goose

5:33 PM

@RyderRude it also seems relevant for classifying topological phases of matter

See eg Kitaev’s arxiv.org/abs/0901.2686

ACuriousMind

@SillyGoose K theory is not exclusive to algebraic geometry

Silly Goose

Hm actually i guess wiki says this sorr of stuff is alg topology

ACuriousMind

the classification you're referring to is via topological K-theory, which came historically after the algebraic version but you don't need to learn about schemes to use its topological incarnation

Silly Goose

Ohzies

Mr. Feynman

@naturallyInconsistent as a user, I don't think there could be any benefit deleting your answer; after all, the question duplicate or not won't be deleted and if anyone visits it, they might read your answer and find it helpful

naturallyInconsistent

5:39 PM

oh meowth

Mr. Feynman

Many times I find a useful answer under a duplicate rather than the "original" question

ACuriousMind

that's why I said to repost it as an answer to the original if you think it adds somethiing

Mr. Feynman

Oh my, there was a tie priced as much as a volume of Polchinski

Why are ties so gorgeous

Silly Goose

@RyderRude doesn’t wigner explicitly give examples in the article of instances that are not like this

ACuriousMind

@SillyGoose Yes, and Hamming explicitly says he find this (and the other explanations he proposes) unconvincing, i.e. doesn't actually think this objection is very strong. As usual Ryder is just skimming the Wikipedia summary badly instead of actually reading the source.

Silly Goose

5:48 PM

@RyderRude he has moved away from the category theory stuff that he used to work on (i emailed him a few months ago)

@ACuriousMind i still find it unconvincing having read the article hehe

i didn't even know who wigner was when i first read that article :P how things have changed. also, not the writing style i expected from wigner for some reason

@ACuriousMind Hm well I think what is confusing is the restriction to caring about Poincaré transformations. Like why should we not extend the Poincaré group to include "acceleration transformations"? What tells us that mass and spin are not invariant under these other transformations?

ACuriousMind

@SillyGoose No one claimed they necessarily aren't, but "acceleration transformations" (from a Minkowksi observer to a Rindler observer) are not symmetries of special-relativistic theories: Terms like $\eta_{\mu\nu}\phi^\mu\phi^\nu$ are invariant under transformations $\phi^\mu \mapsto \Lambda^\mu_\nu\phi^\nu$ iff $\Lambda$ is Lorentz

decomposing your space of states into representations of something that's not a symmetry is not all that useful

Slereah

If you had to have a theory invariant under all possible transformations I think that would just be the trivial theory

Silly Goose

okay so now we're getting closer to maybe the heart of the question. let's say we are dealing with a special-relativistic theory. why should I care about its "symmetries" which seems to be synonymous here with "isometries (?)"

ACuriousMind

and my claim is that mass and spin are the Casimirs of the Poincaré group - it doesn't matter what else they might be invariant under

@SillyGoose it's not synonymous to "isometries"

the notion of symmetry is the same as always: Transformations that leave the action invariant

Silly Goose

@ACuriousMind is there a unique identification of a group with a (set of) casimir element(s)?

ACuriousMind

6:03 PM

since $\eta$ is not a dynamical variable in SR, transformations only act on the fields $\phi^\mu$, not on $\eta$. You derive that the only spacetime transformations that can be symmetries are the isometries of $\eta$ by the argument about $\eta_{\mu\nu}\phi^\mu\phi^\nu$ above

@SillyGoose I'm not claiming that

Silly Goose

@ACuriousMind hm i guess i am missing still why we should leave the action invariant (other than it is nice to have some conserved quantities probably) and what the metric has to do with the action. is the action written always and unavoidably in terms of the metric?

ACuriousMind

How are you going to construct scalars out of stuff like $\partial^\mu\phi$ without using the metric?

and the reason we care about symmetries is because they're useful, yes, because of conservation laws etc

Silly Goose

well what if our objects are all already scalars (if I am understanding correctly that you are saying that generally we are dealing with vectors and covectors and etc. which need to be mapped in some canonical way into scalars to write an action which is scalar valued)

ACuriousMind

the quantum manifestation of a "conservation law" is for example that for a true symmetry in the above sense, it will commute with the Hamiltonian (ignore the question of time-dependent symmetries for the moment), and so the time evolution generated by the Hamiltonian can only map states inside an irrep of a symmetry to states inside the same irrep

the same way the classical phase space decomposes into the level sets of the moment map of a symmetry, the quantum space of states decomposes into irreps

Silly Goose

@ACuriousMind i can buy this sort of stuff

ACuriousMind

6:10 PM

think about it in direct physical terms: when angular momentum is conserved, time evolution can only map a state with a given angular momentum to states with the same angular momentum

Silly Goose

and i suppose if we are taking as our starting point the action and we would like to bias our physical theories to utilize noether's theorem as much as possible, then maybe this would make conceivable paying attention to Poincaré

ACuriousMind

@SillyGoose you want a kinetic term in the action, which even for scalar fields is made out of derivatives $\partial_\mu \phi \partial^\mu\phi$

you're again worrying about extremely pathological ideas that don't have anything to do with how the theories we actually want to apply this formalism to look like

Silly Goose

@ACuriousMind i definitely get this (especially in a quantum context). e.g. the spin-1/2 hilbert space is the 2D irrep space with the appropriate operators and what not. so you are "making the proper restriction of Hilbert space to spin-1/2 states"

ACuriousMind

@SillyGoose yes, and it's the same with Poincaré symmetry: The irreps are classified by mass and spin, and we're restricting to the spaces of fixed mass and spin

that's why I keep saying this is just like the rotation group

Silly Goose

well i can understand this whole story better if it is truly just a clinging for labels to start to make sense of some physics :P. where the machinery of noether's theorem paired with representation theory is our label maker

but it seems quite hard to understand for me if we are claiming that the labels we print out for these objects are something fundamental (just from the considerations we made today. maybe together with empirical evidence and whatever that could change).

ACuriousMind

6:15 PM

everything in physics is just labels that impose order on the chaos of reality!

"fundamental" is a meaningless word

Silly Goose

@ACuriousMind i definitely see this connection and it was helpful for me to have learned a bit about angular momentum (in rep theory language) before approaching Wigner classification since it is broadly mathematically the same thing (at least to my understanding)

@ACuriousMind well i shall change my views to reflect this :P

is there an example of a theory that is made to be Poincaré and "acceleration transformation" invariant?

ACuriousMind

you're trying to invent general relativity now :P

Silly Goose

also to tie this up...so the choice of metric determines the transformations that leave the action invariant?

I would think the form of the action itself also has a say (since it also seems that when someone says "a physical theory" they mean to invoke a particular langrangian)

ACuriousMind

and closer to QFT, considering what Rindler transformations do to a QFT leads us to the Unruh effect - this is as far from "invariant" as it gets - the accelerated observer sees the Poincaré invariant vacuum as filled with particles

Silly Goose

similar to what happens in classical mechanics when we consider accelerated frames: like we see centrigugal and coroilis effects?

naturallyInconsistent

6:25 PM

@SillyGoose This, yet again, goes back to ACM's point: It is an observed fact of our universe that our universe obeys SR, not Galilean relativity, and SR comes with its own symmetry group that, thanks to the SR equivalent of N1L, privileges position-time-s, via homogenity of spacetime, and constant velocities. Because of that, we are only looking for Lagrangians also obeying this symmetry group. Your logic is backwards

ACuriousMind

@SillyGoose you have a tendency to assume I'm making some sort of all-encompassing statement when I'm not - I merely pointed out that actions usually contain terms like $\partial_\mu \phi \partial^\mu\phi$, and for these to be invariant under a spacetime transformation, that transformation has to be Lorentz. I made no claim about all transformations that leave the action invariant being determined by the metric, just spacetime transformations

Silly Goose

@ACuriousMind if considering a QFT which is special relativistic isn't it possible that this result appears because our theory is not being formulated in a nice way with respect to acceleration transformations (as it is with respect to Poincaré)?

Obliv

@ACuriousMind ah sorry my phone was bugging out and not updating the room. I get what a phase space distribution is now, but my prof said he'd upload the derivation so I'll take a look at that later

Slereah

more general QFT theories are not gonna have different interpretations I fear

As I said if you tried to have a theory invariant under every transformation you'd just end up with a trivial theory

The Lagrangian would be a constant number

And as far as we can tell physics is not invariant under acceleration

Silly Goose

@ACuriousMind hm but then i am back to being confused about why Poincaré is important. if we previously motivated Poincaré to be the privileged set of symmetry (in your sense) transformations, but there are others! So why not include these others (on a action by action basis) in our labeling

naturallyInconsistent

6:28 PM

@SillyGoose note that ACM already asserted, within this convo, that if you manage to extend the Poincaré group of SR to include accelerations of GR, then the resulting QFT will have GR too, i.e. be a quantum gravity candidate. I hope you understand how difficult it would be for anybody to find such a generalisation.

ACuriousMind

@SillyGoose because all the theories we care about are on spacetime and hence will have Poincaré symmetry; what other symmetries they have is particular and nothing we can talk about for all theories at once

of course you will care about the additional symmetry groups for specific theories when they arise, when you get to non-Abelian gauge theories the theory of representations of the gauge group will play a crucial part

naturallyInconsistent

@SillyGoose It comes back down to experiment and observations! The as-yet-maximal symmetry group of spacetime that we have come to realise and to trust, is the SR one. The Poincaré group, and no more. If someone later finds a bigger group that generalises the Poincaré group, then we will switch to that

Mr. Feynman

Damn, ACM. You got 3 starred messages in 20 minutes

Slereah

> Another idea is that geodesic incompleteness by itself is indicative of a singular spacetime. Certainly the ghost of an observer whose world line is an incomplete timelike geodesic would have grounds for complaining that the spacetime is pathological.

naturallyInconsistent

Oh, yeah, since this topic is also one whereby @Relativisticcucumber was also confused upon, we should summon. Just in case there is still interest on it.

Silly Goose

6:50 PM

Hm so Poincaré is just omnipresent in modern theories is why we learn it. As in it is the common denominator of symmetries of all (or almost all) modern theories of physics

Slereah

7:01 PM

You can have larger spatial symmetry groups for theories

they just happen to not be the ones of our universe

Silly Goose

Is anti de sitter space of our universe:P

Slereah

I mean also locally

I think if a theory was invariant under acceleration it would fuck up with the principle of inertia

user 85795

honk

ACuriousMind

@SillyGoose I mean, that's not really true? Lots of condensed matter etc. has no trace of Lorentz invariance

it's just that you're currently trying to learn relativistic QFT - the Poincaré invariance here is part of the definition

user 85795

7:17 PM

Quantum Chemistry doesn't even use relativistic QFT.

naturallyInconsistent

7:37 PM

@SillyGoose One thing to always keep in mind, especially over all the craze in the AdS/CFT correspondence, is that our universe, by having dark energy and thus expanding, is dS.

user 85795

Why did you choose the username naturally in consistent?

naturallyInconsistent

@user85795 your emphasis is on the wrong thing. Quantum chemistry uses a lot of SR QM. They might rarely use SR QFT, but they definitely need the SR corrections. They might start with finding a NR QM approximation, then upgrade that to scalar SR QM, or use NR Pauli spinors, before flattened SR Dirac spinors and then actually SR Dirac spinors.

@user85795 Because in life, as in physics, we should strive to be as consistent as possible, and only be inconsistent when it is natural to be so.

user 85795

I don't get it, please elaborate.

Consistent with what?

naturallyInconsistent

7:57 PM

Consistent use of conventions and mathematical notation, consistent physical interpretations and premises,...

Im going to bed

good night yall

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