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1:24 AM
@Obliv The issue is that the B field is already determined by the original flat surface loop, so the same B field has to be found from deforming the flat surface into a balloon surface that doesn't pass through the wire, and instead passes through the gap between the capacitor plates.
@SillyGoose yes, this is covered in basically every treatment of EM that has this level of maths. Not that they would say a lot about it, but they would note that two of them are mathematical requirements of consistency, and two are not. And often they would also cover the magnetic monopoles, in which case the "always zero" pair is also possibly circumvented.
 
1:59 AM
the more i study physics the more i get confused about things i once thought i understood
onto my 50th revisit of the pauli exclusion principle
i understand that on physical ground, it should be that probability amplitudes dont change upon particle exchange. this would mean that if we do exchange particles, we should just pick up a phase. is there a reason that we only have the case of symmetry and antisymmetry as opposed to a range of phases?
 
2:14 AM
what do u guys think
"from the ground" means
nvm I see it now
 
 
1 hour later…
3:28 AM
@Relativisticcucumber That goes all the way up to spin statistics theorem, but at your level maybe you could just see that, if you apply the same exchange operator twice, it needs to go back to itself. i.e. the eigenvalue is only $\pm1$ because only $(\pm1)^2=1$. That there are no other possible phase relationships is due to spin statistics theorem.
 
how does your whole state/system picture look for e&m? @SillyGoose
 
@Obliv SR requires specification of which frame you are measuring things by. The question is measuring everything in terms of the ground, i.e. takes the ground as stationary, and then asks you what the car will think the train's length would be.
 
@naturallyInconsistent I get that if I have a ground state and excited state, for example. Then if I have two fermions. If they were both in the ground state, I would need to swap them and pick up a negative sign. But I cannot do that because they would be indistinguishable, so I should really just get the same state back when I swap. This means that fermions cannot be in the same state. [...]
[...] But this is quite confusing to me because how do they just distribute in the proper way — I asked this on the main site awhile back and it seems fermions naturally experience repulsion, but I dont understand how that happens physically
 
@Relativisticcucumber no, when you swap, including indistinguishability, you still get a negative sign.
 
and iiuc, this q is purely about the stat mech side of things. i dont care so much about relating this to spin. but when i ask this to many people, they always answer w SST. im not sure why though because iiuc, all SST tells us is the spin value of the particles we classify in stat mech
@naturallyInconsistent right so isnt that why the fermions cant be in the same state -- because we cannot have this happen?
or am i misunderstanding
it seems a fermion must know somehow that another fermion is already in that state but this seems absurd
 
3:32 AM
@Relativisticcucumber The "repulsion" from wavefunction needing to obey Fermi-Dirac statistics, which is the core underlying principle that enforces Pauli Exclusion Principle, is not a repulsion at all. There is no force involved. It is purely that the wavefunction has a certain antisymmetry behaviour in there.
 
i dont understand how to develop intuition for this. i worked through the calculations for the BE stats and FD stats but i am left just puzzled
i feel it just tells me the answer but gives no mechanism
 
@Relativisticcucumber It is not absurd. If you think of it as fermions running around, then you will feel it is absurd for one to know another. But if you think of it as that there is one single underlying field, and what we call as particles are merely excitations on top of that one field, then it makes much more sense that all excitations on the field will know that there are other excitations on the field.
 
@naturallyInconsistent really? :o
hm so the information is somehow engrained in the field?
 
@Relativisticcucumber Yes. And so there is a need, say, for the spin-half portion to be antisymmetric, so that when the orbital part is symmetric "due to being in the same orbital state", the fermion system is still acceptable.
@Relativisticcucumber only in the form of "these states have already been occupied", which is not possible to communicate faster than light. The proofs surrounding these are quite intricate and Feynman famously asserted that he could not explain this to a child.
@Relativisticcucumber the spin statistics theorem is the only thing that enforces which set of particles obey BE stats and which set of particles obey FD. Trying to separate the behaviour (of FD v.s. BE stats) away from spin is thus somewhat futile. But yes, you can easily consider some spinless yet FD stats obeying nonsense, if you want to consider them theoretically.
Anyway, this is also why a popular modern thinking is that all these quantum weirdness are due to quantum fields. It is not waves and/or particles. It is that the quantum fields underlying them are the real thing, and excitations on these quantum fields can look like waves or look like particles. No need to keep the outdated dichotomy of classical waves v.s classical particles
 
3:50 AM
@Relativisticcucumber now it is in bundle language >:D (sort of). at a high level, it is still there exists a state space $S$ of all fields. On this state space, we place a differential equation (the Maxwell equations) and so solutions are a subset of the state space.
mathematically, we have a $U(1)$-principal bundle. we consider the space of all connection $1$-forms on this bundle modulo "gauge transformations". This is the state space $S$. Then, we define the Maxwell equations over $S$. Solutions to the Maxwell equation are then a subset of $S$.
 
this is the language of classical e&m? @SillyGoose
@SillyGoose what does "still there exists a state space of all fields" mean
@SillyGoose and what is the characteristic feature of this state space? i dont really understand this at all
@naturallyInconsistent i see. i will look into SST maybe bleh thanks
 
@Relativisticcucumber bundle theory is the language of any field theory
 
@Relativisticcucumber beware, it is mostly technical and does nothing to explain anything. Purely mathematical relationship
 
classical or quantum
@Relativisticcucumber hm i think it means, formally, the set of all connection $1$-forms over a principal bundle with gauge group $U(1)$. Intuitively, this is the set of “all objects that could be an electromagnetic potential”
then we impose gauge equivalence and a differential equation on all these could-be electromagnetic potential objects, singling out what we call solutions
@Relativisticcucumber what do you mean by characteristic feature?
 
4:13 AM
i think the field picture would clear this up and another issue i have been having
the above message was a delayed response to nI not you @SillyGoose hehe
starbucks wifi had a glitch XD
 
glitchy B A H ~
 
@SillyGoose I think the bundle is in general part of the solution
for example let us take Yang-Mills, instantons are different solutions of the YM equations which correspond to different bundles
you dont fix the bundle before solving the YM equations is finding a bundle with a connection on it which satisfy the YM equations
just like in GR you dont fix spacetime and then find the solution to Einstein equations. The solution to Einstein equations is a spacetime $(M, g)$ whose metric satisfies Einstein equations
for now im speaking about vacuum stuff so there is no sources
of course with $\mathbb{R}^4$ you cant do much with bundles since they all are trivial
so in the end you dont realize that finding a bundle is part of the problem there
 
4:45 AM
ah i see
 
 
1 hour later…
5:48 AM
@lucabtz is life better once you graduate?
 
pay is better...
 
6:13 AM
assuming this isnt your situation youtube.com/watch?v=VjgmcZwaY1Y
 
6:36 AM
@Mr.Feynman kinda the same tbh
 
 
1 hour later…
7:43 AM
@naturallyInconsistent I mean, you don't get paid as a student so it can't be worse
 
@Mr.Feynman The stipend is supposed to be minimum wage, before minimum wage was a thing.
 
What I can say for sure is that it's not the case here
 
So, who do we stab to improve things there?
 
8:14 AM
i am aware of someone who launched a lawsuit against the government in my country over phd stipends
as far as i can tell it probably wont amount to a lot but it's been going on for a few years now
 
fqq
@naturallyInconsistent he's not talking about PhD
 
Oh, right there is also that problem with the word "graduate"
 
fqq
Also there's no min wage in Italy
 
@fqq is that a big problem?
 
fqq
And I think with the increase in min wage in the UK, the standard PhD stipend went below it there too
 
8:20 AM
confusion
 
123
Hello Everyone...
 
in the states the phd stipend seems to in general be a livable wage (and certainly at or above minimum wage). but at the less wealthy schools in places like boston you might be barely living lol
 
@SillyGoose is the funding not set at standard rate from the govt?
 
i think it is set by some committee at each school depending on cost of living of where the school is located
but that doesn't mean that the set amount makes sense hehe
where i am going the stipend to cost of living is quite nice
 
so in a sense it's privately funded?
 
8:32 AM
but it is more a problem with places like southern california or boston
or new york city
 
i kinda wish that phds could be treated like apprenticeships, with all the trappings and benefits of employee rights
 
9:29 AM
hi
academia is predatory
 
 
5 hours later…
2:05 PM
@qwerty they did that here and now it's basically impossible to get funding for a theory thesis
They're not paying real money for that
 
 
1 hour later…
3:29 PM
@Slereah wait really, what's the country? and the funding is (private?) allocated at the discretion of universities?
 
The France
They have to actually pay you a normal work amount for a PhD
which sounds good but it does mean they are very picky on thesis topics
 
3:52 PM
Oh interesting
@qwerty to my understanding a lot of ppl’s funding in the states for physics comes from the government (national science foundation)
i think maybe some universities give new professors hires funding to start their labs and stuff though
(For phd) i think masters students only pay to go to school here, though im not entirely certain
 
4:20 PM
This is a real downside and something I personally encountered (although I live in a different country). Even in areas such as condensed matter theory, projects became insanely applied to where they were borderline projects you would essentially expect in industry.

Even some of the top cond matter theorists in my old departments [who are less applied but nonetheless work in standard (real world scenario) areas e.g. optics, many body theory etc.) have barely managed to obtain funding for a PhD student for the past couple of years (unless it came directly from the department)
 
not sure what is the problem with that
 
I might be misunderstanding but does this paper seems to have calculated quark masses from string compactifications!
This was also featured on Quantamagazine!
 
@Sanjana Great to hear.
The work was based on Ruehle's package, who is at NEU where from I got a Ph.D. offer :) Thanks to all PSE members who helped me a lot during my highschool and UG times.
 
@ManasDogra Congrats
 
@Sanjana It's "quark masses" only in a very broad sense - as the paper itself says, "the purpose of this paper is to provide proof of concept that a calculation of fermion masses can be carried out, by using machine learning techniques. Obtaining a realistic (up-quark) mass spectrum was not expected (and is indeed not achieved)."
 
4:36 PM
@ManasDogra congratulations
 
5:34 PM
@ManasDogra congrats :D
@ManasDogra is NEU northeastern uni?
Can gauge theory be thought at all as some generlization of cohomology? Or is it just that quotients are ubiquitous and are a part of gauge theory in a way in general distinct as in cohomology
Well i mean of de Rham cohomology whose primary objects of interest are sections of forms over a manifold
 
why would it be a generalization of cohomology?
also depends on what you mean by "gauge theory" :P
the Chern and Pontryagin classes that form the "topological" terms in gauge theory are cohomological in that they classify the bundles they live on via cohomology classes of the base
but I see nothing remotely cohomological about the ordinary YM action
 
5:50 PM
i see
is there a mathematical structure whose elements are the on-shell and gauge equivalence classes in non-abelian chern-simons?
e.g. the first de Rham cohomology class is what i am describing (to my understanding) for abelian Chern-Simons
 
it's the space of flat connections modulo gauge transformations, no?
more generally the space of "on-shell" fields of any (gauge) theory is the zero locus of the EL operator/equations of motion inside the space of fields modulo gauge transformations; that's a perfectly fine mathematical structure
 
so the de Rham identification in the abelian case is just a "fact"?
 
not sure what that means - it's because $F=0$ is the condition for $A$ to be a closed form and "modulo gauge transformations" is the same as "modulo exact forms" in that case, which is exactly what de Rham cohomology is
 
well i mean to say "is a coincidence" but in previous discussions that is equivalent to "is a fact"
and by is a coincidence i mean not a general theme of chern-simons theory
so what is abelian chern-simons telling us? (1) we have distinct classes of gauge potentials $A$ iff we have a nontrivial first de Rham cohomology class. (2) nonetheless, on shell gauge potentials are all "physically" the same field due to the "trivial" eom $F = 0$.
 
6:13 PM
@SillyGoose I mean the curvature is the obstruction to $\mathrm{d}_A^2 = 0$, since $\mathrm{d}_A^2 = F\wedge$, and so since the Cherm-Simons e.o.m. means the connection is flat with $F=0$ you indeed have a complex of forms with the covariant exterior derivative $\mathrm{d}_A$ instead of the usual $\mathrm{d}$ whose cohomology you can take
 
so many structures at once :0
 
there's a bunch of high-level applications/connections of this cohomology modified by a flat connection in this MO post
 
oh tak
also per what lucabtz had mentioned, what exactly is the input data for a classical field theory?
 
6:32 PM
@SillyGoose what do you mean by "input data"?
 
the (reasonably, but not necessarily certainly) minimal data to specify a "classical field theory"
 
@SillyGoose second try: Note that this isn't quite like "the space of solutions of Chern-Simons theory is a cohomology", but to each of the solutions $A$, there is the associated cohomology of $\mathrm{d}_A$. In the Abelian case, these $\mathrm{d}_A$ end up all being the same, so you only have the single ordinary cohomology
 
e.g. i thought it was a base manifold $M$, a bundle structure on it, a specification of the mathematical object that your dynamical variables are on this structure, and an action
 
you could say that
though I find "a specification of the mathematical object that your dynamical variables are on this structure" a strange phrase
 
hm what would you say
 
6:41 PM
you've given a bundle, fields are sections of bundles, what's the point of the bundle if your dynamical variables are not sections of it?
and if they aren't, you need to give the action anyway as a map from the space of whatever your fields are to the reals
 
can't i have a section of my bundle that is not a (local pullback of a) $\mathfrak{g}$-valued $1$-form
 
what's $\mathfrak{g}$?
 
$\text{Lie}(G)$ where $G$ is the structure group
 
you said "classical field theory", not "gauge theory"
 
i see
hm but even if i have the classical dirac field theory (governed by dirac eqn) is a section of a bundle really all the necessary structure to call a thing a field?
 
6:44 PM
you have a spinor bundle with connection, the dirac fields are sections of the spinor bundle
i.e. the configuration space is the space of all sections of the bundle, the space of solutions is those sections that fulfill the Dirac equation; what more do you think you need?
 
hm i see
so in a relativistic theory, is relativistic "invariance" or whatever solely baked into the governing differential equation?
 
well, in that case you have a metric on the base, and your e.o.m. differential operator is invariant/covariant under its isometries, sure
 
that seems sort of neat
 
but that's not really special to "relativistic" theories
we just usually don't talk that much about the fixed Euclidean metric of non-relativistic space :P
 
i guess i am wondering where the representation theory is then induced or naturally comes from. we start with a bundle with a metric on the base and sections of the bundle are fields. then we define an action, which induces a differential equations over the space of sections of the bundle. this differential equation respects the isometries or what not.
 
6:51 PM
what "the representation theory"?
 
for instance when i first was learning some of the introduction to qft, some resources start by supposing that a classical or quantum field's target space transforms in a finite-dimensional representation of the Poincaré group
and this is how they talk about a "field"
 
that's just the (local) way to state that tensor fields - in the general sense of sections of some tensorial powers of the (co)tangent bundles - carry representations of the frame bundle, and since having a metric that's invariant under some $G$ (in this case the Lorentz group) means you have a reduction of the structure group from the $\mathrm{GL}(n)$ of the frame bundle to the Lorentz group, and so all the tensor fields inherit a representation of the Lorentz group
 
hm i haven't read about frame bundles i think i'll have to look at that to understand what you said
 
7:13 PM
separately, can passing from tensor to direct sum representations for spin systems be viewed as a sort of supersymmetry? since we at times turning fermionic degrees of freedom into bosonic ones without changing any of the physics
 
you're not "turning" fermionic stuff into bosonic stuff
you just observe that e.g. the combination of two fermions is a boson
there's no (super-)linear operator hear where the source would be "fermionic" and the target "bosonic"
 
 
3 hours later…
10:39 PM
@Slereah oh i see. because i guess there are less phd slots then
 

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