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meep

M I A O ~

h o n k

who let these animals in the bar

who, who, who who

Conjecture: Imagine I have a spherical jar of balls. I shake the jar (non-randomly) such that the balls come back to its original configuration in time $t$. If my shaking algorithm takes $1$ unit time and $1$ unit volume to complete $1$ cycle. Then if I scale the volume by $v$, time by $k$ then the number of cycles my system completes is $C = v^{1/3}/k^3$. Assume simple $F = ma$ for equations of motion.

Is this better on computer science stack exchange or PSE?

Hello Everyone...

Just before $(1.1.2)$, are GSW assuming integer spin? Since $\sigma$ apparently carries Lorentz indices

Also, I'm not sure what contraction would give rise to that $s^J$ factor (check the footnote)

Reading on cone theory for GR reason and what we may call a lightcone is referred to here as an ice cream cone

Hyouka

@Slereah hello u have been gone lately

I have been around

just not said much

@ACuriousMind Is symplectic reduction in classical symplectic mechanics related to this "modding out by gauge transformations" business in field theories?

@SillyGoose you should not have tagged ACM on this. Like, Slereah probably could answer this too

i shall refrain in the future ~~

S C H W E G G ~

@naturallyInconsistent The text box should be filled with his tag by default :P

lol

More realistically, imagine a system based on levels which only lets you tag ACM if your rank is high enough

uugh, two ppl on the main site are asking meow for old history. Feynman calculating QED corrections for constant background E field. Can't remember where such tales are...

hi

i hav a charge density $\rho(x,t=0)$, its time derivative $\rho '(x,t=0)$ and the electromagnetic field $F^{\mu \nu} (x, t=0)$. this is insufficient to determine the full time evolution, right?

in this section, it says that we can drop the k label which specifies the state, but it also mentions particles occupying the ground state here. im not sure how these two things relate? k specifies the state in terms of momentum, right?

@SillyGoose tsk tsk ignoring the site policies

wait it seems they use k to refer to energy eigenstates bleh i thought they are momentum eigenstates

gr wait this is like harmonic oscillator.

@SillyGoose yes, see this answer of mine

what is something about learning physics that physics students shud know

@ACuriousMind I request your assistance above :P

@Mr.Feynman yes

@Mr.Feynman you have two vertices, and $J$ momenta coming from each (due to the Fourier transform of the $J$ derivatives in the coupling). Each pair of momenta contracts to an $s$, so you get $s^J$

@ACuriousMind But aren't the momenta contracted with $\sigma^{\mu_1\mu_2...}$?

Or does it happen something like QED where $\gamma^\mu p_\mu\gamma^\nu p_\nu=p^2$?

@Mr.Feynman Hm? Remember the Feynman rules - the $\sigma^\mu$ etc. turns into equivalent indices on the propagator for that field, which are the contracted with something suitable with indices on the vertices

and that something here are the momenta from the derivatives in the coupling

do the derivation explicity if you don't see it

I find it interesting that my answer on this question has 10 upvotes even though this question isn't eligible for HNQ

Let's hope it isn't illegitimate voting

nah, I think it's just a popular question even if not HNQ and your answer is just very nice since it's so succinct

I see; this has never happened to me before

Well thanks I guess

@ACuriousMind oooh you're right, that's the propagator so you'd get metric tensors contracting vertices

@Relativisticcucumber what text is this?

@Relativisticcucumber i would guess that $k$ and $x$ do refer to "states of definite" momentum and position, respectively. this then allows you to write an arbitrary state (even if your Hamiltonian isn't free) in terms of vacuum $\lvert 0 \rangle$ and creation/annihilation operators

@Relativisticcucumber oh i see who the text is by lol

oh nevermind it does indeed seem like $k$ is used to label the energies and that this discussion is specific to the harmonic oscillator

i feel like philosophers use "Science" like physicists use "Math" lol

i am wondering what the "new technology" is in high energy theory. for instance, it seems like putting field theories into bundle theory language was popular (and continues today) in say the mid-late 1900s (maybe this is inaccurate). what new sort of mathematical structures are being used today to "mathematicize" physical theories?

why would you expect there to be "new sorts of mathematical structures"?

well not new in the sense of actually newly developed mathematical structures, but more like newly picked up by physicists

or rather: What question exactly are you trying to answer here? Bundles etc. are the answer to "what is the proper mathematical formulation of what fields are?", what's the question you expect more structures to appear while answering it?

hm well maybe a question could be like: what sort of mathematical structures are involved in a given supposed theory of quantum gravity (since quantum gravity is what I would guess is a more modern and in development physical theory)

ah, but we don't know the correct theory of quantum gravity :P

right right that's alright. but even what people are trying out

unless that's just everything :P

we don't even know the "proper" mathematical formulation of ordinary QFT - the Yang-Mills millenium problem is still, as of today, unsolved

so demanding to know the proper mathematical structure of QG strikes me as one step too far - in all other cases, we have first figured out the correct theory at a physical level of rigor, then formalized it in mathematical terms

and we haven't even quite caught up with "physical" QFT yet

hm i see

is the Yang-Mills millenium problem just proving the existence of a structure (subject to some carefully defined axioms) that is supposed to be a physically relevant QFT? or is the problem more than that

i mean to more ask: do we have a set of axioms for a "QFT" we think is reasonable, but the hard part right now is proving that a structure even exists satisfying them

yes; the challenge is establishing the mathematically rigorous existence of a realistic QFT - namely non-Abelian Yang-Mills theory in 4d, i.e. essentially the Standard Model - that satisfies any equivalent of the Wightman axioms

are there alternative formulations of classical mechanics that do not use manifold structures? more precisely, assuming we are dealing with a "nice" system: is there a formulation of classical mechanics that does not define a configuration space and then use its tangent or cotangent bundle as the "space of states"?

none that would be useful :P

you can do the parlor trick of Koopman-von Neumann mechanics and do "quantum mechanics" with commuting $x$ and $p$, I guess

are our current formulations of algebra and analysis nice languages for creating physical models? Maybe measuring niceness by being able to compute and prove things

do you have any other languages to compare them to? :P

nice or not, the math we have is the one we use

Okay so it’s just all we have

Well i guess i am wondering if anyone does work in reformulating (in a loose sense) mathematics or if that is even a thing one could do

you sound like you're about to become a follower of the nLab :P

also reformulations happen all the time; most of modern mathematics, after all, isn't even a century old in the way we now talk about it

@ACuriousMind xD

@ACuriousMind maybe if i could understand anything they write lol…

we like to pretend there is some unbroken continuity from Newton to us, but it's much more that we can't understand the principia without a lot of effort and context and Newton wouldn't understand any of our math, either

there's something continuous there, sure, but it certainly isn't the way we talk about math

Hm i see your point

mayn always hard to decide what to learn :P

is there a resource which talks about electromagnetism historically? In particular to make sense of the following observations. It sort of seems like the theory of electromagnetism is quite special relative (lol pun not intended) to any other physical theory. (1) it is automatically relativistic, (2) it was the first field theory to be quantized successfully, to my understanding. Or maybe it is not so special really?

I don't know a specific resource but really the core of what distinguishes EM is that it is the only part of the Standard Model that is better approximated in terms of waves and fields in the classical regime than particles due to photons being the only massless unconfined objects

and then if you look at the EM waves you inevitably will notice that their speed in vacuum is fixed by the form of Maxwell's equations, whether you wrote them in a manifestly relativistic formalism or not

what is an unconfined object?

I mean a particle not subject to confinement (since gluons are the other thing that remains massless, but confined)

Anyone know what the square here is supposed to represent?

I hate how this book doesn't define its damn notation or what a "conformal transformation" is

it's literally a square

oh i see

$\Omega(x)$ is a scalar function, $\Omega^2(x)$ is its square

Wait, so the box is just $\Omega^2(x)$?

are they not referring to the literal symbolic square $\square$?

why wouldn't they just write that?

@SirCumference what do you mean by "the box"?

the second term in the parentheses

with the coefficient of 6

oh, you didn't mean the superscript 2 by "the square"

I am having a hard time seeing how the 2-form defined in (13) is a 2-form. In local coordinates it looks like it takes in a tangent vector from $TM$ and a tangent vector from $T(T^*M)$

nope, didn't know what to call it tho

the $\square$ is the d'Alembertian

kind of looks like the wave operator

oh, so it actually was

it's perfectly ordinary notation

thanks. wasn't sure if it was something else since I don't know how that operator would be relevant in this context

but it's not like the book explains anything so eh

but presumably the (will be called symplectic) 2-form should take in two tangent vectors from the tangent space of phase space $T(T^*M)$ (being loose with global v. local notation here)

@SillyGoose no, it's just a 2-form on $T^\ast M$. The coordinates of $T^\ast M$ are the $(q^i,p_i)$, so $\mathrm{d}q^i \wedge \mathrm{d}p_i$ is a perfectly ordinary 2-form on it

hm okay so we have our manifold $T^*M$. what does a tangent vector in local coordinates look like?

sure, the basis of tangent vectors is all the $\partial_{q^i}$ and all the $\partial_{p_i}$

i.e. any tangent vector $v$ is $v = \sum_i v_i^{(q)} \partial_{q^i} + v_i^{(p)} \partial_{p_i}$

i am wondering if i have written (9) then in a confusing way. since i wrote the tangent vector as a tuple of the partials

yes, no one writes $\partial$s with vectors

well, perhaps not no one as you clearly demonstrate, but this is not standard notation and it's not very clear what it is supposed to mean

so should i rewrite it as $D\pi(v_i^{(q)}\partial_{q^i} + v_i^{(p)}\partial_{p_i}) = v_i^{(q)}$

that's one option

because the partials themselves are the "basis" so actually i don't even understand what it would mean to write the basis itself as if it were a vector in a basis as (9) seems to do

the other is just to write it on a basis: $\mathrm{D}\pi(\partial_{q^i}) = \partial_{q^i}$ and $\mathrm{D}\pi(\partial_{p_i}) = 0$

oh okay

i shall make this change :P

i wonder why the notes i was working off of wrote it like that

so can i think of $D\pi$ like a linear map from linear algebra? so its action is specified by its action on the basis vectors. so us specifying its action as you wrote is all the data we need to define the map

the differential is a linear map on the fibers, yes

in practical terms it's just the Jacobian matrix of the function

and matrices are linear maps as you know

I see

Also so is the differential a particular example of a pushforward? Or are these distinct concepts

like can it just be called the pushforward of the projection $\pi$

Or the pushforward of whatever your are differentialing

it is the pushforward

those are just different names

Oh

also it seems like this wedge product is additional structure than the cotangent bundle itself

is that true?

or i more mean to ask: where does the structure allowing us to now talk about differential $k$-forms come from if we only start with a cotangent bundle of some manifold

there is no additional structure

differential forms exist on all smooth manifolds, and the cotangent bundle of a smooth manifold is a smooth manifold

(so is the tangent bundle, for that matter)

are they defined just by this?

I mean that's not a definition of differential forms

that's a definition of the deRham complex given a definition of differential forms

is $\Omega^0(M)$ as the space of smooth functions not a sufficient definition? And then you obtain the $+1$ forms in terms of $\Omega^0(M)$

no

how do you think you get $\Omega^1$ from $\Omega^0$?

if you want a succinct definition, differential k-forms are sections of $\Lambda^k (T^\ast M)$

something roughly like "the space of functionals over $\Omega^0$"

@SillyGoose no, where do you get that from?

a differential k-form is not a function on $\Omega^{k-1}$

i think maybe i am conflating what a $1$-form does with what a general $k$-form does

oh wait

i see i am misunderstanding entirely

if anything functional, it's a multilinear antisymmetric functional on $k$ tangent vectors

yes i was conflating (in the linear algebra picture of things) vectors with covectors and etc.

@ACuriousMind what is $\Lambda^k$?

the k-th exterior power

oh

just the antisymmetric part of the k-fold tensor product

wait but i don't get where is this exterior product coming from

@SillyGoose it just an equivalent way of saying "multilinear antisymmetric functional on $k$ tangent vectors"

if i have a vector bundle it is specified by $(E, M, F, \pi)$ a total space, base manifold, fiber, and projection map.

linear functional on tangent vectors at every point = element of $T^\ast_p M$ at every point $p\in M$ = section of $T^\ast M$

linear functional on $k$ tangent vectors = element of $\bigotimes^k T_p^\ast M$ at every point = section of $\bigotimes T^\ast M$

antisymmetric functional = lies in $\Lambda^k$ instead of $\bigotimes^k$

hm wait isn't that just the definition of a tensor

of course

differential forms are just antisymmetric tensors (with all indices lower)

i thought tensors are automatically antisymmetric

@SillyGoose the metric tensor would like to disagree

(oops yes i should have said tensors with arguments only of tangent vectors)

the metric tensor is a 2-tensor on tangent vectors

it takes two tangent vectors and spits out a number (their inner product)

hm okay maybe the definition of tensor i got from linear algebra is not general t hen

the general definition of tensor from linear algebra is just multilinear functional

there's nothing about antisymmetry there

ohhh i think i keep conflating the properties of a tensor with the properties of the determinant as a particular type of tensor... and the determinant is antisymmetric or "alternating"

okay i see

okay so this idea of a tensor is inherited from the vector space structure of the tangent spaces or cotangent spaces

(or that is how i am thinking of it, perhaps there are other ways)

then of course we can restrict to particular types of tensors, the antisymmetric ones, and we give them the name differential forms

sure, a tensor field (often just called "a tensor" when the context is clear) is just a tensor in the sense of linear algebra on the (co)tangent spaces of a manifold at every point

okay so at each point $x \in M$ we have a tangent space $T_xM$. from this tangent space we construct a sequence of tensor products of this space $\{ \otimes_{i=1}^k T_xM\}$. We can construct the dual sequence as $\{\otimes_{i=1}^k T^*_kM \}$. These constructions are just fine since this is usual linear algebra. This is the data we were after to talk about differential forms

@SillyGoose Yes

and so then we define $\Omega^k(M)$ to be the space of (global?) sections of $\otimes^k_{i=1} T^*M$ such that at every point the tensor (living in the tensor product of cotangent spaces) is antisymmetric.

@SillyGoose You need to construct the kth exterior power first

it's basically the antisymmetric subset of the kth tensor power

okay so the tensor product structure induces a natural exterior product

then we can define the exterior derivative $d: \Omega^{k} \to \Omega^{k+1}$. this part here still seems like additional data

@SillyGoose The exterior product is really just the antisymmetrized tensor product

it's not additional data, it works on any smooth manifold

You take the tensor product and then take its projection onto the alternating subspace

okay well i can accept that

the trick is that the terms that spoil proper tensorial behaviour of the derivatives of arbitrary tensors (the ones the Christoffels of a connection cancel!) are symmetric

so antisymmetrization kills them, and the exterior derivative is well-defined without a connection

so in a physical theory what would the de Rham cohomology theory over our differential forms as defined above tell us

it tells you whether there are closed forms that are not exact :P

for instance, whether there are divergence-free vector fields that are not the curl of a vector potential

e.g. there are none on $\mathbb{R}^n$, but in the case of the idealized solenoid that removes the origin from the space under consideration, you have a magnetic field that's not the curl of a single vector potential

@ACuriousMind Aharonov-Bohm?

yes, that's one place where you run into this

I see

@ACuriousMind hm that seems like very course grain data. in a physical situation would the mere existence of a non-exact closed form really matter? since if you are working with particular potentials, it seems you would have to check anyways whether the potential you are working with is non-exact and closed.

who said cohomology gives you anything but coarse-grained data? :P

well i guess i thought it would have more use in this physical context

im not sure where that impression came from though :P

is de rham cohomology’s only purpose in physics to give such course grained data?

it's just cohomology

but it shows that the topology of the space is related to the existence of certain kinds of differential forms

Hm well maybe i should ask: what is a useful physical result that cohomology theory gives you

in some contexts, e.g. dimensional reduction, you will be interested in harmonic forms $\Delta \omega = 0$, i.e. zero modes of the Laplacian. There is exactly one harmonic form in each cohomology class

but like with much math it's more that it's an organizing principle rather than the only way to arrive at some fact

for instance for the solenoid and its magnetic field: you can very explicitly show that there is no vector potential on all of $\mathbb{R}^3 \setminus 0$ for it without using much machinery at all

but then it remains somewhat mysterious why it can sometimes happen that such magnetic fields exist and sometimes not

the Poincaré lemma and deRham cohomology are the answer - this happens if and only if your space has certain non-trivial cohomologies

@SillyGoose XD

cant ask the author bc everytime i do he says smth incomprehensible

ah, the "eldritch being" approach to professorship

@Relativisticcucumber lol classic behavior

@ACuriousMind ah okay i see