@SillyGoose WHOOOO!!

along with silly goose i got two physics phd offers today !! best day of my life !!!

this means guaranteed hbar activity for 6 more years ;)

Congrats Jigglypuff!!!

thanks !!! im so excited ahhhhhh

@Relativisticcucumber WOOOOOOO

consider a group as a category $C \in \text{Group}$. is the set of covariant functors of $C$ the set of group homomorphisms of $C$ as a group?

LOL

My thinking is: $C$ has a single object $G$, so the to-be functor necessary is a map $\mathcal{F}: G \rightarrow G$. It must additionally satisfy $\mathcal{F}(g_1 g_2) = \mathcal{F}(g_1)\mathcal{F}(g_2)$ and $\mathcal{F}(\text{id}) = \text{id}$. But this is precisely the condition for $\mathcal{F}$ to be a group endomorphism (into itself) of $G$

H O N K

H S O C N H K W E G

oops in my first message i mean to say the set of group endomorphisms of $C$ as a group

@PM2Ring Try $$u=\frac13-\frac23\cosh\frac23\arg\!\cosh\frac{3\sqrt3}{2b}$$

oh god wait disregard my above messages.

@PM2Ring It even derives why the threshold is at 2.598

Let $G$ be a group as a category (we will abuse notation and also use $G$ to represent the object, which is a group, in $G$ the category). Consider the set of all functors $F: G \rightarrow G$. I believe that the of set of all such functors is precisely the set of group endomorphisms of $G$.

Because each such $F$ necessary sends the object $G$ to $G$ (there are no other objects in the category). Moreover, the functor axioms are precisely the group homomorphism axioms in this case because the morphisms are group elements. Hence, $F$ is a group endomorphism of $G$.

@PM2Ring when $b>2.598$ the solution that is smoothly connected to the above, is the one whereby you swap all the hyperbolic cosines for circular trigonometric cosines, no need for extra flair.

Now I'm confused as to why u<0...

@SillyGoose yes

do you think we should improve ourselves as a person everyday

@PM2Ring You cannot have b<2.598 because there is no way to prevent dropping beneath the Event Horizon when that is so and so your imaginary components appearing is unavoidable. The solutions were trying to tell you that you have been exploring an unphysical parameter regime. A correct expression for the physically sensible parameter space $b>2.598$ is thus $$u=\frac13+\frac23\sin\left(\frac\pi6-\frac23\arg\!\cos\frac{3\sqrt3}{2b}\right)$$

Congratulations cucumber and silly goose ! 🥳🥳

My hate for index notation has grown exponentially

Hate can only destroy the hater :P

@Mr.Feynman what happened now?

@naturallyInconsistent I've always hated the notation $\nabla_\mu T^{\alpha\beta}$ to denote $(\nabla_\mu T)^{\alpha\beta}$

But it's forgivable until you write it in semicolon notation

$T^{\alpha\beta}{}_{;\mu}$

Which really stands for $(T_{;\mu})^{\alpha\beta}$

I mean, you can easily see what's so bad about the notation :P

err... But the $_\mu$ is an index that is summable on the outside, so it should not be inside the brackets...

If you wanted to insist upon that, I would have had the ugliest $(T^{\alpha\beta})_{;\mu}$

semicolons and commas are too easy to misread, so I really prefer $\nabla_\mu T^{\alpha\beta}$

helps keep everything in order

Meow too

meowth

My point is that: say you have $T^{tr}=0$, then $T^{tr}{}_{;\mu}$ is not zero in general

Im going for dinner. It is freezing. Miserable. And I'm groggy

But the notation is misleading theee

@Mr.Feynman you unlocked a memory

@nickbros123 thanks !!

@Relativisticcucumber congrats :)

@Mr.Feynman thanks !!!

@nickbros123 when i hit jackson and zangwill i know who to call

@Relativisticcucumber congrats, just read your message above!

@ekardnam_ thanks !!!!!

@naturallyInconsistent om nom

How does it work? Will you start PhD as soon as you finish your undergrad?

Or have you finished it already?

RC and SillyGoose wishing u a happy phd

@Mr.Feynman why is this notation bad

@Mr.Feynman oh

@Relativisticcucumber lol I'm just an idiot, ACM NI and co would be more helpful

@Slereah do BH pertain your GR interests or are you more into less astrophysical stuff?

@Mr.Feynman They are hard to avoid but I don't focus that much on it

@Mr.Feynman in what sense?

@Relativisticcucumber om nom nom

@naturallyInconsistent i think Mr Feynman means u cant apply the covariant derivative component wise becuz the Christoffel symbol mixes the components

this is differen from partial derivative, $\partial _{\mu} v^{\nu}$. here u can separately differentiate the components $v^0(x)$...$v^3(x)$

for e.g. $\nabla _{mu} (T^{12})$ doesnt make sense, but $(\nabla _{\mu} T)^{12}$ makes sense

becuz u cant apply covariant derivative to $T^{12}(x)$

Apparently an Aristotelian kinematic structure is basically a Galilean and Carrolian structure together

Which makes some sense I guess

The only way to have a boost that behaves the same in both cases is to have the boost equal to zero

i thought an Aristotelian structure is a first order differential equation, so a 0 connection

Objects in math can be several things at once

@Slereah so here u hav Galilean symmetry and the Carrolian structure violates that symmetry?

Basically the Carrolian structure defines a prefered time direction since its "light cone" is the degenerate cone (a line) while the Galilean structure defines a preferred spatial plane as its light cone is the other degenerate cone (the plane)

Between those two, there is no boost that preserves both those structures at once

So objects at rest remain at rest, travelling along the Carrolian "cone"

oh

how are we characterising Aristotle's physics here?

is this equivalent to a 0 connection?

In the sense that there is a preferred time and space

so a preferred rest frame

If you attempt to define the spacetime "length" of your momenta it will be something like $g(v,v)$

hm

Actually

Not entirely sure what would be the "geodesic" here

a preferred rest frame is still kinda vague. for e.g. Newton's second order eqn with an external velocity dependent force gives u a preferred rest frame

Since you have 3 structures

It is a Klein geometry so presumably you can just define a Cartan geometry on it and get the appropriately defined generalized geodesics

But I couldn't tell you what they are for the Aristotelian group

Although I saw a paper use the Aristotelian connection, let's see

if u go one level below Aristotle's physics, u get a 0 order differential eqn

this is just a constraint

there isn't much lower you can go than Aristotelian

If you go below you stop being an isotropic kinematic at all

yeah. constraint physics is non sense

so u r exploring kinematics, which is the geometry. and dynamics are deviations from the geodesics? @Slereah

is kinematics defined to be the metric stuff

Kinematics is just the non-curved version

You just have a variety of Klein geometries

oh. i would say curved should still count as kinematics as long as it's just geodesics

@Slereah so kinematics is defined to be non curved Klein geometry physics?

I mean you can make the curved version by defining a Cartan geometry from those

But I am looking at the basic version for now

Klein geometry looks very general in its definition

doesnt even mention a metric

just a Lie group

You can derive your metric from Klein geometry

i always thought metric vs lorentz group was a chicken vs egg topic

idk what came first lol

but ur approach is taking the group as more fundamental

You can pick either

oh. theyre still equivalent. interesting

The Lorentz group is the group which leaves the metric tensor invariant, and the metric tensor is a structure which is invariant under the Lorentz group

There's equivalences like that for all the Klein geometries

starting from groups is pretty general because groups are a more well motivated concept than metrics

The action for the Aristotelian case is apparently $$S = \int \alpha(\dot{X}) + \delta(\dot{X}, \dot{X}) $$

With $\alpha$ the clock form and $\delta$ the spatial metric

Assuming a flat spatial metric that gives $\dot{X} = \alpha$, so the geodesic equation is that the velocity of a free particle is a specified timelike vector

ie. absolute rest

Errr wait

$$\ddot{X} = \dot{\alpha}$$

is this time metric different from the time metric that Galilean geometry uses?

it uses the time metric t_2-t_1

As I said it is the combination of a Galilean and Carrolian structure

yeah, but Galilean structure already has two metrics, and here u also hav two metrics. hav u thrown out the time metric of Galilean?

so is this Galilean space metric + Carroliean time metric?

The metrics are somewhat complicated because they are very similar looking

Aristotelian is technically four "metrics" : a timelike form, a timelike vector, a rank (0,2) tensor and a rank (2,0) tensor

although you only need 3 technically

oh

the action needs to be symmetric wrt all four?

Hard to say

Not a lot of papers on the topic

becuz as soon as u give the action Galilean symmetry, u no longer hav preferred rest frames

so it can have absolute time but not required

It does if the theory is torsion free

i feel Lorentz geometry might be the only one that makes perfect sense quantum mechanically

so in this sense, quantum mechanics constrains the geometries

but this is just a conjecture

ooh but there's Galilean QFT which also makes sense

i think the conjecture is false. but Galilean is a limit of Lorentz

Since any path has the same time interval, this means that the path of extremal spacetime length is just the path with the smallest distance between the initial and final location

ie. zero

it is at rest

yeah

but this is also a feature of Galilean action

and yet things move there

@Slereah ok but this is only zero when the initial and final locations are the same

if u choose initial and final locations to be the same in galilean action, u get rest

How would you identify the initial and final location though

They are different spacelike hypersurfaces

You can do it in the Aristotelian case presumably because the timelike vector field defines a foliation of the manifold by the integral curves

u can identify them in Galilean

Can you though

There's no timelike vector field, only a 1-form

@naturallyInconsistent I said it above: if $T^{12}=0$, in general $T^{12}{}_{;\mu}\neq0$

so how are we defining rest if we cant identify

rest means same location at different points of time

You can if you have an Aristotelian structure

oh

Two points of different spacelike hypersurfaces are the same if there exists a timelike curve integrating your absolute time vector field connecting them

Apparently this is equivalent to some condition on the timelike form

is ur diff eqn first order or second order? if the latter, what happens when the particle has an initial velocity? does it tend to come to rest?

If you have your Galilean timelike form $\omega$ such that $\omega \wedge d\omega = 0$, then you can identify things like that

@Slereah this action is the Galilean action with an extra 1 form term

i think this is effectively giving u a velocity dependent force

Oh I think I got it

Or do I

I think it's related to the properties of the clock form in the Aristotelian case

There is some properties that it obeys like $\omega = f dt$ and $\omega \wedge d\omega = 0$

theses.hal.science/tel-01283692v1/document

That paper seems to be the only one to go in detail into it

Why is the subspace orthogonal to two lightlike vectors spacelike?

Any subspace composed of two lightlike vectors is gonna be timelike

Since the sum of two null vectors is a timelike vector

That Aristotelian spacetim paper came out in 2016

pretty recent shit

@Slereah ah, but isn't 1 the maximum dimension of timelike vector subspaces?

"timelike" has a differen meaning for subspaces

Also if I have $n$ and $m$ null, $(n+m)^2=2n\cdot m$ so how do you conclude that?

I should say "in the same lightcone" since obviously $(n + -n) =0$

Ok, that can't be it $n-m$ would have opposite norm. How do I understand "timelike", then?

physicsforums.com/threads/…

@Slereah Ok, so the point is that is must contain a timelike vector (and thus a 1d subspace of timelike vectors), not that all vectors in the subspace are timelike

And if your subspace is orthogonal to two null vectors, you can pick them to be in the same light cone since $\langle v, n \rangle = 0$ is equivalent to $\langle v, -n \rangle = 0$

@Mr.Feynman Basically yes

Ok, check outs then

The proper definition is about the signature of the induced metric on the subspace

It will be $(+1,-1)$ for the span of the lightlike vectors

So the complement is $(1,1)$ aka spacelike

I wonder how I got to this point without ever using such an obvious property lol

Thanks

There's plenty of little properties like that that you either don't learn or forget

Wait I think I am falling for the old trap for that Aristotelian thing

I'm confusing vectors and dual vectors

The EoM isnt $\ddot{X} = \dot{\alpha}$, those are different objects

I'm not even sure the Lagrangian is like that tbh, tried to adapt a Lagrangian for field theories

There seems to be some condition that's the "proper time condition" $\alpha(\dot{X}) = 1$, which would basically constrain it to move like that, but no idea how important that is

Also wait I have a quadratic particle equation but no mass term

Maybe I need some version of Polyakov

Maybe the Polyakov constraint field gives you the appropriate constraint

$$\int \left[ \alpha(e^{-1} \dot{X}) + e^{-2} \delta(\dot{X}, \dot{X}) - m^2 \right] e$$

Doesn't seem to help much

arxiv.org/pdf/2311.04112.pdf

That paper proposes a different action

Oh wait I think it's just that by default you're gonna have that $\dot{X} = 0$ for a positive-mass particle

in the timelike direction

Any idea about where $(3.30)$ comes from? The Frobenius condition along with the Killing condition can be recast together as $$\ell_{\mu;\nu}\ell_\sigma+\ell_{\sigma;\mu}\ell_\nu+\ell_{\nu;\sigma}\ell_\mu=0$$

This is what happens with these one-liners "you may find elsewhere the proof" and half fucking course is has incomplete theorem

@naturallyInconsistent Thanks so much! I even understand why that equation works, and how to derive it. :) We can easily fix the u<0 problem of the first (cosine) version by negating the arccos argument.

@naturallyInconsistent I think you might have misunderstood what I was doing yesterday. I certainly was not exploring an unphysical regime! I totally understand that values of $b<b_0=\sqrt{27}/2$ give trajectories that plummet into the event horizon. However, using numerical methods near $b_0$ can take you into dangerous territory, due to the maximum of $k=u\sqrt{1-u}$ at $b_0$.

As I said in physics.stackexchange.com/a/680961/123208, which I linked earlier:

Here's a demo of the cosine version in Sage. When b is very large, it loses a few digits of precision, so I do one round of Newton's to fix that.

sagecell.sagemath.org/…

Hi, i have a question, there is someone to talk?

@JL14 you could ask the question and let it simmer in chat and someone will probably come along and answer it if they can :)

consider an elementary and textbook presentation of the standard model. if a particle reaction $A + B \rightarrow C + D$ obeys all conservation laws (energy, momentum, angular momentum, electric charge, baryon number, electron number, lepton type number, ...) is there necessarily a feynman diagram for this reaction?

From generic Casimir invariant arguments I think that Lagrangian for Aristotelian dynamics works

Literally the Lagrangian built from the two Casimir invariants of the theory, $E$ and $p^2$

But from what I can read, based on Souriau theory arguments, timelike particles are fixed if free

Hi silly goose, i am new at this web and i don't know how it works. I was studying the conservation of momentum and energy and a question came to my mind.Why is kinetic energy responsible for causing damage in an accident?I try to explain my question better. Both energy and linear and angular momentum are "numbers" that represent the symmetries in the interactions that our universe has (rotational, translational and temporal), so both momentum and energy are related to force.

My question is, what makes kinetic energy the cause of feeling the forces in a collision and not linear momentum? I have two contradictory examples, in one it says that a bull and a bullet have the same kinetic energy, but the bull has more momentum, and if they hit a target and come to rest the force exerted by the bull would be much greater than that of the bullet since F=dp/dt.

In another example we have a rifle firing a bullet, both the rifle and the bullet have the same momentum, but the bullet has much more kinetic energy, and I think we all know what happens if you get hit by the bullet or the rifle.

@Mr.Feynman seems like a chat fluke, didnt see it when I asked it. I think you mean that it can acquire a value different from zero because the Christoffel symbols sum other components too? I do not see why that is misleading?

@SillyGoose Pretty much by definition of it being able to happen under Standard Model, it will have a (lowest order) Feynman diagram.

@naturallyInconsistent I mean that the notation makes it look like you're differentiating something that is zero (a zero component), while the inner index denote the component after the differentiation

But really it's not worth even discussing :P

@Mr.Feynman I'm not following the "problem" here: If a tensor is identically zero $T=0$, then $\nabla T = 0$, too.

@Mr.Feynman I kinda think I have an idea what you are trying to express now. I'd have to point out that in the semicolon scheme, it is difficult to even express this.

@ACuriousMind He is saying that just one component is zero, but the Christoffel symbols sum over the other components, so it can end up with that component non-zero

ahhh

I think that's part of the reason why Wald and some others insist on interpreting indices as "abstract indices", i.e. you're not really allowed to just plug in some concrete choices for them

Yes, that's it

But it was just a rant so I didn't bother discussing much about it

@ACuriousMind I don't like that either. I just like to see is as omitting a couple of brackets $(\nabla_\mu V)^\nu$ is the way :P

I like how the modern non-relativistic kinematic stuff is all geometric structures named after ancient people

The zero mass Galilean particles are called Fermat particles

@PM2Ring ah, you are warning meow about numerical instabilities. I suppose it is difficult to be accurate over the entire range.

@Slereah it's all a ploy to make it look like they didn't just make up those concepts on the spot ;)

@ACuriousMind Some of them are pretty tenuous

Like one of them is an "Augustinian structure"

Saint August didn't do much imo

@imbAF it's a nice advertising slogan but there isn't really an answer to this

the conditions after the big bang were not the same as the conditions of two beams of particles colliding head-on :P

@imbAF Less than a picosecond. To find the Higgs boson, they needed to achieve collision energies similar to what occured during the electroweak epoch. See en.wikipedia.org/wiki/Electroweak_epoch & en.wikipedia.org/wiki/… Those article say the electroweak energy was ~169 GeV. The maximum LHC proton beam energy is ~7 TeV.

Also do we know on which day the big bang happened

Was it a monday

@Slereah so you are implying there was time prior to it?

hmmm

although presumably without the Earth as a reference point, that answer is observer dependent

The Big Bang happened on a Thursday. I never could get the hang of Thursdays..

what they really mean is that in collisions in the LHC, the energy available to the particles is comparable to those available to particles colliding in the very hot phases briefly after the big bang

I think by Christian mythos creation happened on a sunday

I mean they must have an idea how close they are to the event

Since God rested on the seventh day and that is the sabbath

@imbAF my point is that the two situations are so unalike that trying to pin down an exact value here is meaningless

I would say exact

Just go with an average I guess

I am asking for a proximity value

A range

we generally think of the universe after the big bang as a hot, but roughly equilibrated mess that had a well-defined temperature

Find the point of the universe where the average particle had a kinetic energy of a few TeV

two particle beams colliding are not an equilibrated system with a well-defined temperature

it's not clear how you are supposed to compare the "energy" of these systems, really, and depending on the method you choose your estimates are going to be orders of magnitude apart

So what, are we getting punished if we get it wrong

@ACuriousMind One question. Isn't the Big Bang theory and the First law of thermodynamics saying the opposite?

@imbAF I have no idea what you mean

Ok

It's difficult to do internet research on plutonium titanium nitride. When I type PuTiN into Google it mostly gives me stuff about some Russian guy.

::groan::

@PM2Ring Even harder if you're searching for copper nanotubes

I live in the state of New South Wales, which has the standard abbreviation NSW. So when I see NSFW my brain naturally reads it as New South F'ing Wales.

@Mr.Feynman stop stalking me via the transcript :P

schwank

do you guys drink coffee or tea

I like coffee but I had to stop drinking it regularly because I didn't like how I had no energy on days when I didn't have some

I'll have some tea if offered but I don't make any for myself

ah i feel similarly. during a busy semester i will end up relying on getting four shots of espresso at some point during the day to stay not dead :P. but i usually try to cease caffeine consumption over breaks or less busy times. i have now been trying to go back to just drinking green tea (which is like half or a third of the caffeine per serving of coffee).

tea is delish

I only occasionally drank coffee as a student, it just bloomed into addiction when I started working in an office with free coffee :P

@ACuriousMind you brought that up :P

... I guess? I can't find it

@SillyGoose I drink espresso every morning and occasionally after lunch

I mean, you're Italian :P

There are Italians not drinking coffe :(

My best friend is not a coffe drinker

Come on, it's like a German disliking beer

Exactly!

Or a Slereah disliking weird old papers about exotic geometric structures in GR

By the way, I was always able to drink coffe without sugar but this year I finally appreciate it

I...actually think I've never tried it with sugar

my parents just drank theirs with milk, and I didn't like that, so I drank it black

though oddly enough I do like cappuccino

I guess black coffe is more popular abroad? Most Italians drink it with sugar. Then again, those who are really fond of coffee (I mean real coffee nerds) would never put sugar but that's not representative of the population I guess

Cappuccino without sugar is not bad, actually

@ACuriousMind are you talking about espresso or american-style coffee?

@Mr.Feynman I'm not talking about espresso but I'm not sure about "american-style"

That's what I call a cup full of coffee

I think technically I'm talking about both drip coffee and caffee crema?

While this is what I mean by espresso

yeah, I know what espresso is, I just wasn't sure what you mean by american-style

afaik, "american-style" here would mean an espresso you'd stretched with water

which isn't what I'm talking about when I say "black coffee"

By "black" I mean "withour sugar" but I learned the phrase "black coffee" watching subbed anime...

I think the name originates because there is also coffee cream that makes it brighter...? So it's not even about sugar. Anyways, it's not something you usually say in Italy. "Caffè" alone can only mean "black coffee"

Although we don't use cream, there is milk, actually. Even then, coffee with milk changes name (e.g. cappuccino :P)

Or macchiato (which I guess is a word used abroad too? It means "stained")

i thought black coffee means no milk product and no sugar

hehe i just made some espresso imgur.com/a/bZwRDA3

So it also means no sugar! Anime subs worked fine :D

@SillyGoose I was about to type "coffee so late at night?!" Then I remembered :P

@Mr.Feynman Lol a few months the answer to that would be yes

I think what I really mean by "black coffee" is a) without sugar or milk and b) either drip coffee or caffee crema - often people here have drip coffee machines at home

If someone asked me "Do you want a coffee?" and I just said "Yes" I absolutely would not expect to get espresso

@SillyGoose do you mean that you'll move to Europe for your PhD or that you're planning to give up sleep (for your PhD)? :P

yeah i would expect a cup of drip coffee if i asked for "black coffee"

@Mr.Feynman oh i mean a few months ago* haha. i could drink coffee at night with no problem :P

i think now it would be a little problematic

@ACuriousMind that's the opposite of italian convention I guess :P

in the states and korea there is an "americano" which is watered down espresso

(maybe other places too)

yeah, that's what I first thought "american-style" was supposed to mean

i like espresso and honey lattes out of coffee drinks

There is also that in some cafes here

But the fact is that even what you call "drip coffee" translates to "caffè all'americana" according to google which as you can guess could be adapted as american-style

huh

@ACuriousMind Lol this happened to me last semester. the physics mentoring center at my uni has a nice espresso machine, so i would essentially drink a double shot espresso every time i went in

drip coffee is kind of fun hehe, but it is a lot more clean up than espresso :P

well maybe not if you have a machine i guess. I have a filter holder that you put atop a cup

with the machine the cleanup is literally just throwing out the filter :P

i realized hehe

Ah looks like the whole Aristotelian business is due to secondary constraints

@naturallyInconsistent hm is there a sort of proof of this statement?

@SillyGoose I think nI misunderstood your question :P

oh

also is there a braid theoretic formalization of feynman diagrams? or like a graph theoretic one

yes; Feynman diagrams are graphs (with some additional structure to mark "in" and "out" legs)

i am basically wondering if one can define feynman diagrams quite precisely other than (as is done in the textbook my course is following, which introduces primitive vertices as building blocks and leaves it at that)

@SillyGoose there's a pretty large mathematical literature on generally organizing perturbation series via graphs

oh

this is not really surprising - the Feynman diagrams are literally just mnemonic graphical ways to represent the terms in the perturbation series

this is a diagrammatical way of doing things that i find conceivable harder than working with the original maths :P, but maybe i will come to appreciate it after learning the scattering amplitude computation proper

Hm so by a perturbative theory, is what is meant like textbook QFT is perturbative because all our computations rely on expansions which conceivably only converge or are valid in some neighborhood of a variable? Or is this just different

the terrible truth is that QFT perturbation series do not converge at all :P

see e.g. physics.stackexchange.com/q/27687/50583, physics.stackexchange.com/a/368686/50583

"perturbation theory" just means that we approach "solving" the theory by expanding the quantities of interest in a formal power series, whose terms we determine order by order

Ah I found the bastard

so would a non-perturbative theory have exact, closed form solutions of quantities of interest?

The Aristotelian Lagrangian seems well-behaved enough, but that's its usual sneakiness

Obviously the Hessian is the spatial metric $\delta$, but that metric is degenerate

@SillyGoose it's not the theory that's perturbative, really, it's our approach to obtaining the solutions

oh

Ergo the extra constraint

so people saying perturbative QFT is like an abuse of language or something

I guess it makes sense to separate the methods of computation from the underlying frameowkr which is the theory

@SillyGoose what we mean by "perturbative QFT" is really both the theory of QFT and the perturbative approach to extracting quantities of interest from it

i see

"non-perturbative" does not imply exact or rigorous, e.g. lattice methods are non-perturbative but still approximate

but quantities like scattering amplitudes seem to be expanded in terms of a coupling constant, so do we expect these results to differ upon applying non-perturbative methods?

I mean, it's from the perturbative methods that you get these quantities in terms of power series in the coupling

non-perturbative methods will yield different results

is there an example of obtaining a different result using non-perturbative methods for a simplest case

it's pretty difficult to state exactly what it means for these methods to match (or differ), since as I said we know that the typical QFT perturbation series does not converge

right, okay I see

guys, regarding ang.momentum addition, I'm trying to add angular and spin momenta of a spin 1/2 particle (namely, $j_1 = l = 1, j_2 = s = 1/2$). the notation for the two basis is the following: $$\vert 1,\{-1,0,1\} \rangle \otimes \vert \frac{1}{2}, \pm\frac{1}{2}\rangle \longrightarrow \vert 1, \frac{1}{2}, J,M\rangle$$

one common example for a theory that doesn't behave as you might expect is triviality of $\phi^4$ in 4d from the lattice viewpoint

now I was able to obtain the combinations: $\vert 3/2, 3/2 \rangle, \vert 3/2, 1/2\rangle, |3/2, -1/2\rangle$

by applying $J_- = L_- + S_- $ and then equating the expressions