The h Bar

12:01 AM

lol

i would describe my short experience in reading Infinite Jest as "reading the dictionary"

because every other word had to be looked up

even though English isn't my native language, that actually wasn't the problem (I have been called a walking dictionary at times :P)

mostly I think I just didn't find any reason why I should care about the (at first glance) disconnected miserable stories the book wanted to tell me

Silly Goose

If I am recalling correctly is the main plot of the book about some sort of perfect entertainment movie or something

ACuriousMind

I never got far enough to be able to identify that as the main plot :P

there's one vignette early on where that movie plays a role but the rest seems completely disconnected. Which isn't bad as such - I don't dislike non-linear plots in general - but generally I didn't feel interested in any of the stories I was reading so I always stopped a few chapters in

Silly Goose

perhaps i should read a synopsis before judging but i am not very partial to these sorts of counter-consumerism books (like White Noise by DeLillo) they take 300 (or perhaps in the case of Wallace 1000+) pages to tell you that consumerism is bad without a personally valuable reading experience

i would like to find more books like mrs. dalloway and the waves by woolf :P some nice peaceful abstraction

ACuriousMind

to be fair I'm probably not the target audience for postmodern literature since my favourite author is Terry Pratchett

Silly Goose

12:15 AM

i have never read any of his work 0:. i have still not yet made any advancement beyond classical literature

ACuriousMind

books are not temporally ordered, you can read anything you want anytime ;)

Ryder Rude

1:08 AM

@Mr.Feynman omg same. In the last dream, I cud fly instead of jumping. And it was in third person. This enabled me to shoot them down

But the dream ending had some cliffhanger about a bigger villain. I guess my brain got inspired by movies

@Mr.Feynman u should wake up like 1 hour before your normal wake-up time. U will extremely sleeply and annoyed and go back to sleep within minutes. This will probably give lucid dreams

Or at the very least, this gives extremely HD graphics dreams

Silly Goose

2:48 AM

lol "and annoyed" made me laugh

Silly Goose

6:50 AM

oh god it has all led back to string theory

Mr. Feynman

7:16 AM

@RyderRude lol

@RyderRude Oh, I tried that too

The HD dreams part worked

DIRAC1930

8:34 AM

@ACuriousMind Have you looked into the Faddeev-Kulish reasoning behind asymptotic states?

Mad

9:40 AM

Hello

Why is it in the theoretical physics one demands that the semi linearity of a scalar product to occur in the second argument not in the first (as per the mathematical canonical definition)?

Prähilbertraum

In der linearen Algebra und in der Funktionalanalysis wird ein reeller oder komplexer Vektorraum, auf dem ein inneres Produkt (Skalarprodukt) definiert ist, als Prähilbertraum (auch prähilbertscher Raum) oder Skalarproduktraum (auch Vektorraum mit innerem Produkt, vereinzelt auch Innenproduktraum) bezeichnet. Man unterscheidet dabei zwischen euklidischen (Vektor-)Räumen im reellen und unitären (Vektor-)Räumen im komplexen Fall. Die endlichdimensionalen (n-dimensionalen) euklidischen Vektorräume sind Modelle für den n-dimensionalen euklidischen Raum. Die Nomenklatur ist aber nicht einheitlich. Manche…

ACuriousMind

10:26 AM

@Mad historical accident

Mad

Okay so no atual physical meaning?

ACuriousMind

there isn't really a lot of reasoning behind it, it's just that that matches with how Dirac notation works

Mad

Alright thanks!! :)

ACuriousMind

@DIRAC1930 No, because I am completely satisfied with the Haag-Ruelle construction.

ACuriousMind

10:52 AM

@SillyGoose if you're saying that because you're trapped reading nLab articles: you have to keep in mind that the nLab POV is very openly not neutral and so they always try to bring things around to higher category theory and string theory. If you don't know the topics you're reading about there all that well it can be a bit hard to separate the core mathematical facts from the editorialising

DIRAC1930

11:23 AM

Can I just pick which one is easiest to learn (i.e. do they all the same result done a different way) or are the others (other than Haag-Ruelle) doing something else?

ACuriousMind

what do you mean?

you don't have to learn anything

I mean, plenty of people clearly get by without ever questioning the usual handwave construction of asymptotic states in the standard texts

Ryder Rude

11:59 AM

@ACuriousMind but I don't understand how the handwave construction, in e.g. Schwartz, doesn't end up with the bare mass for the asymptotic free Hamiltonian

ACuriousMind

that's questioning the construction :P

I said plenty of people get by without questioning it!

the reason why you can get by is because renormalization in the handwavy approach blows up one of the masses anyway, so the distinction doesn't really matter

Ryder Rude

How can one even handwave that the asymptotic particles have the physical mass instead of the bare mass? Is there any handwave for this?

I'm looking for any simple explanation

ACuriousMind

@RyderRude what is "the bare mass" and "the physical mass"

once you realize you need to normalize there is no finite bare mass

Ryder Rude

The physical mass is what u observe at ur energy level.

ACuriousMind

so if you're asking why the asymptotic states don't have infinite mass, the answer is because that would be silly :P

Ryder Rude

12:02 PM

Yes, but that's not mathematical handwaving. That's just a meta reason

Usually, I would expect the asymptotic hamiltonian to be the usual hmailtonina minus the interaction part

This is the kind of handwave i would expect

And this is the stuff that books seem to follow. But they never give any reason for y we changed the mass in the free hamiltonian

ACuriousMind

don't think about it - once you start thinking too hard about this construction, it falls apart

Weinberg explicitly discusses the point about the mass, for instance, and is careful to define the "free" part of the Hamiltonian as containing the corrections to the mass

so the handwave is "just absorb the mass corrections into the free part"

Ryder Rude

Oh so defines it as bare mass plus dressing. So he is not completely getting rid of interactions becuz there r dressings

How is this physically justified?

ACuriousMind

how is what justified?

Ryder Rude

Like, how we get rid of the interaction term but keep the correction due to the dressing due to the interaction

ACuriousMind

what's there to justify? We're just separating terms in the Hamiltonian into two different packages

Ryder Rude

12:06 PM

Oh

Yeah. We can just split it

But what justifies that the particles would actually evolve due to this split hamiltonian in the asymptotic limit? Why not some other split (like just ignoring the interaction hamiltonian)?

Like, we also need a physical justification for y particles wud physically evolve due to this hamiltonian we have split out

ACuriousMind

now that's the part where the handwaving doesn't work

Ryder Rude

Lol

Ok but this problem of handwaving cant be related to Haag's theorem? Becuz the same problem would arise in lattice qft too

ACuriousMind

but also, this doesn't matter at all for getting to Feynman diagrams, so that's why the texts don't bother

you don't need to think about the actual time evolution in the asymptotic limit, or what exactly these particle states are to start computing scattering amplitudes

Ryder Rude

Yeah. This only refers to the afterwards evolution

ACuriousMind

it's not so dissimilar from how scattering is treated in normal QM courses (if they treat it at all) - there you usually also don't start by questioning whether any particular potential really has states that look like plane waves far from the potential "source"

even though you would need to ask the same question there

@RyderRude Haag's theorem plays a role here in that the "normal" QM scattering theory uses the interaction picture

so you can't directly use the resolution of the same question you know from QM (if you know it)

Ryder Rude

12:13 PM

But i think this same problem would arise in lattice qft too. There too, we've got two choices of splits of Hamiltonina between bare and physical mass splits. It's just that the bare one isn't infinite

So i think this is independent of Haag's theorem

But it needs some handwave

I'm not sure what the handwave could be for this. Usual QFT books just don't discuss it becuz they r only interested in the scattering amplitudes instead of the afterwards evolution

Ryder Rude

12:28 PM

@ACuriousMind Weinberg defines asymptotic particles as eigenstates of the free hamiltonian with physical mass. But is this the same free Hamiltonian Weinberg uses to evolve the fields in the interaction picture?

Or does he choose the bare mass free Hamiltoniant to evolve the fields? (i think this is necessary for Feynman rules to be derived)

Slereah

1:43 PM

Trying to remember the plot of Torment Tides of Numenera and I am not remembering much

Mr. Feynman

has anyone here ever used a paper (e-ink) tablet?

Ryder Rude

1:58 PM

@Mr.Feynman I have not but they r good for the eyes I guess

ACuriousMind

@Slereah that's because it was extremely forgettable :P

Mr. Feynman

@RyderRude I hope they feel like paper. I need friction when I write

I took a leap of faith and got one

Ryder Rude

Just return it if it doesnt feel like paper

Mr. Feynman

Yeah, I have 15 days to do so

Ryder Rude

If a physics model uses pansychism, would it be considered crackpot?

ACuriousMind

2:05 PM

what does that even mean

Slereah

@ACuriousMind Hard to remember what part of it differentiates it from Planescape :p

Ryder Rude

@ACuriousMind i mean suppose a model says everything is conscious and makes predictions with that assumption

ACuriousMind

@Slereah Hm? I remember a lot of Planescape's plot (and side plots) even though I haven't played it in years; I don't remember a lot about Tides

Slereah

I mean that a lot of Tides' plot parallels Planescape

ACuriousMind

@RyderRude yes, what does that even mean? I know of no physical theory where it is relevant whether something is conscious or not

Slereah

2:07 PM

being a mysterious immortal being with an unknown past

in a land of many worlds crossing over

ACuriousMind

yes, it tried to replicate Planescape without understanding anything about why people liked Planescape :P

Slereah

was it the skull

ACuriousMind

absolutely

imbAF

In the Born-Oppenheimer approx. the effective potential is the potential felt by the electrons of each atom?

ACuriousMind

Morte is a brilliant first sidekick

Slereah

2:08 PM

having an evil skull is always a good point in a game

Ryder Rude

@ACuriousMind what about a hypothetical model which says everything is conscious? The consciousness spread over the universe can objectively collapse the wavefunction, so this will be a testable prediction

ACuriousMind

@RyderRude how is that "testable"? The whole reason we can argue forever in circles about interpretations and whether or not there is "collapse" is because it makes no difference in practice!

Ryder Rude

In relational QM, the collapse is subjective. So we can never test if other things can collapse it

But if we make that objective in some model...

That would be make it testable

ACuriousMind

no, it wouldn't

you can't get around the no-communication theorem by waving your hands and muttering about objective collapse

Ryder Rude

I read objective collapse implies FTL information transfer and p=np

ACuriousMind

2:11 PM

unless you actually want to change how QM works on the technical level, in which case you need to actually write down that physical theory that is not QM

@RyderRude You probably read about specific objective collapse theories that actually modify the Schrödinger equation

Ryder Rude

@ACuriousMind yes, this is what i mean! Objective collapse theories r not QM. Usual QM is just relational QM which must be subjective collapse

@ACuriousMind yes

ACuriousMind

so what does adding specific terms to the Schrödinger equation have to do with "panpsychism" or consciousness?

@RyderRude no, relational QM is just one of many interpretations

"usual" QM does not even necessarily have collapse

Ryder Rude

In usual QM, u can only test that ur own consciousness can collapse it. But relational QM says all objects can collapse it, but it's untestable becuz of its subjective nature. But if we make it testable by transitioning to objective collapse, those tests will show that everything is conscious, wouldn't they? @ACuriousMind

Slereah

why must we be eternally cursed with talking about the intepretations of QM

3

can't we do the interpretation of another theory

I know some soviet paper that argues against structural chemistry, maybe we could argue about that

Down with structural chemistry

ACuriousMind

@RyderRude I have no idea what you're saying. Quantum interpretations are subtle, and you are carelessly throwing terms like "collapse" around that are relative to your interpretation. E.g. MWI is an interpretation of standard QM and does not have collapse.

Ryder Rude

2:16 PM

I read that usual QM is Solipsistic

@ACuriousMind yes, but MWI has tons of problems. The wavefunction alone cannot derive collapse

It can only derive branching. That alone doesnt derive Born rule

At best, MWI can derive 50-50 collapse. The branching implies that

But it cant derive non 50-50 probability collapse becuz u would need to add the Born rule by hand for that

ACuriousMind

why would anyone derive the Born rule

what you're saying is almost fractally wrong, but I'm not really in the mood for an extended discussion about the fundamentals of quantum interpretations

Ryder Rude

Ok I have a very nice thread about this

physics.stackexchange.com/questions/536522/… this one

OP's comments on the first answer are very good. He puts it very well. It's what I'm trying to say

It's saying that even after you add the Born rule by hand in MWI, non 50-50 probabilities do not have any physical meaning, except for sentient beings

Slereah

Let's get mad at structural chemistry

Mr. Feynman

2:31 PM

I lost my soul to understand how the torsion appears in the commutator of covariant derivatives and now I hate indices more than ever

Why do Physicists apply covariant derivatives in such a non standard way as to make torsion appear aaaaaargh

Loong

@Slereah Well, indeed, benzene doesn't have any double bonds. What's the problem?

Slereah

@Loong It is bourgeois to claim that it does, even

Secret

pi systems, are shorter than single bond of rings

neither double nor single

Ryder Rude

@Mr.Feynman i stopped learning GR when proofs involving indices came up

Slereah

@RyderRude you're not gonna like QFT either then

Mr. Feynman

2:40 PM

@RyderRude So after the EP? :P

Ryder Rude

QFT is easier imo. I at least made it to the dynamics. With GR, I never left the mathematical machinery @Slereah

Mr. Feynman

@Slereah man why is your comedic timing so good

Ryder Rude

They were just proving stuff about curvature when I left GR. The actual dynamics was one chapter away

I shall try it again tho

Mr. Feynman

Covariant derivatives do not affect the tensor rank i.e. they take vectors to vectors, scalars to scalars and so on. Why in the world would GR people decide to intepret them as if they did creating a mess in stuff like $[\nabla_\mu, \nabla_\nu]X$?!

ACuriousMind

@Mr.Feynman Given any $n$-tensor $T$, the map $(X_1,\dots,X_{n+1})\mapsto \nabla_{X_{n+1}}T(X_1,\dots,X_n)$ defines a $n+1$-tensor

and it is that tensor that has the components $\nabla_{\mu_{n+1}} T_{\mu_{1}\dots\mu_{n}}$

DIRAC1930

2:46 PM

@ACuriousMind Yes but the potential $V$ drops off to $0$ in the asymptotic limit for most potentials you study in QM. It's simple compared to never being able to switch off the interaction operator in QFT

Mr. Feynman

@ACuriousMind I mean, yes that is no more than saying that $\nabla_\mu X$ is the $\mu$ component of a vector valued 1-form

Ryder Rude

I think I just hav a fear of looking at so many indices

ACuriousMind

@Mr.Feynman so why do you claim that "the covariant derivative" does not affect the tensor rank?

you can use it to construct n+1-tensors from n-tensors as I just did

Mr. Feynman

Because you can also see it as $\mathcal{T}^1\ni Y\mapsto\nabla_X Y\in\mathcal{T}^1$

ACuriousMind

so?

that doesn't mean there's anything wrong with the $\nabla_\mu$ notation or viewing it in terms of the rank-increasing operation

Mr. Feynman

2:49 PM

It's correct sure, only messy :P

for a scalar $f$, $[\nabla_\mu, \nabla_\nu]f=0$ in one interpretation of the notation

ACuriousMind

well, the messy part are the indices, not the interpretation in terms of a rank-increasing operation

Mr. Feynman

In the other it yields the torsion

ACuriousMind

it is actually very natural to think about the rank-increasing operation because that's the way you generalize the covariant derivative to gauge covariant derivatives

Ryder Rude

I think u mean directional derivatives vs the covariant derivative @Mr.Feynman The former doesnt increase rank. The latter is like taking the Jacobian matrix

ACuriousMind

for any gauge field $A$, you have the covariant exterior derivative $\mathrm{d} + A\wedge$

Ryder Rude

2:51 PM

The former can b derived from the latter using dot product

ACuriousMind

for the ordinary covariant derivative, the "gauge field" are the Christoffel symbols, and you have that the rank-increasing operation is $\mathrm{d} + \Gamma\wedge$

Mr. Feynman

Sure it's legitimate, here I'm only considering covariant derivatives without resorting to differential forms where what you mention is quite more natural

In fact I said above that it makes sense precisely because in that case we have the components of a vector valued form

ACuriousMind

@RyderRude that's not how the terminology works, the covariant derivative is the generalization of the directional derivative in $\mathbb{R}^n$ to manifolds

Ryder Rude

@ACuriousMind but directional derivative is the contraction of the direction vector and the jacobian (which is like Covariant derivative)

The contraction can reduce the rank which was initially increased by the taking the Jacobian

ACuriousMind

@Mr.Feynman well, the reason we have to resort to differential forms in the general case is precisely because without a connection on the tangent bundle we need to use the exterior $\mathrm{d}$ to have some sort of derivative that increases rank and is well-defined

Mr. Feynman

2:55 PM

@RyderRude Directional derivative are defined for scalar functions. In this case the covariant derivative is defined to be exactly the directional derivative

Then, we generalize that to tensor fields

ACuriousMind

but in the LC case, $\nabla_\mu$ is constructed exactly so that we don't need to anti-symmetrize because the Christoffels cancel the terms that otherwise spoil the derivative being tensorial

Ryder Rude

Maybe I'm remembering something different as called directional derivative

Mr. Feynman

I mean, it's not like I didn't understand what was going on (after a good hour of thought) but just saying I find indices a bit nasty :P

ACuriousMind

but I still claim it is very natural to consider the derivative as a rank-increasing operation because that's what $\partial_\mu$ does in flat space, too

@Mr.Feynman oh, I agree with you about the indices, I'm just not agreeing that the root cause of the problem is somehow considering the derivative as a rank-increasing operation

the root cause is that physicists are in love with coordinates :P

Ryder Rude

I'm remembering it as $u^{a} \partial _{a} f$ (for Scalar functions). Is this not directional derivative?

DIRAC1930

2:58 PM

Where does this come from $$e^{-i H t} \int d\alpha f(\alpha) \Psi_\alpha^\pm \to e^{- i H_0 t} \int d\alpha f(\alpha) \Phi_\alpha$$ as $t\rightarrow \pm \infty$?

It is the first comment here physics.stackexchange.com/questions/717995/…

Mr. Feynman

I once read on PhysicsForum something like "I can believe that only geniuses understood GR back then. Index notation is a complete mess"

@RyderRude Yes and it is also the covariant derivative of $f$ wrt the field $X=v^a\partial_a$, that is $\nabla_X f:=X(f)$

ACuriousMind

@DIRAC1930 ..Prahar in that comment explicitly says it's an assumption, and once again this type of asymptotic state is exactly what Haag-Ruelle theory constructs

I really don't know why you keep asking the same question if you're always getting the same answer :P

Ryder Rude

@Mr.Feynman oh. Then I'm defining covariant derivative differently. I'm defining it to be $\partial _a f$ itself. So this increases the rank in my definition.

Mr. Feynman

@RyderRude That is the covariant derivative wrt the vector field $Y=\partial_a$

ACuriousMind

again, the two definitions are equivalent via the construction I mentioned above

Ryder Rude

3:02 PM

@Mr.Feynman oh u can just call this to be another vector field :P. This is a good viewpoint

Mr. Feynman

For the sake of completeness, I'll mention why the index mess above originally appeared. I was dealing with the coordinate free form of the Riemann tensor $R(X,Y)Z=\nabla_X\nabla_Y Z-\nabla_Y\nabla_X-\nabla_{[X,Y]}Z$ and then I considered $R(\partial_\mu,\partial_\nu)Z=\nabla_\mu\nabla_\nu Z-\nabla_\nu\nabla_\mu Z$

And I wrote that as $R(\partial_\mu,\partial_\nu)Z=[\nabla_\mu,\nabla_\nu]Z$ and the mess started

@RyderRude That's a way to define covariant derivatives. The one ACM mentions above with bundle-valued forms is much more general and shares many similarity with the standard exterior differential

DIRAC1930

What is a good source on Haag-Ruelle theory

Is it in here? cdsweb.cern.ch/record/212242/files/p1.pdf

I can't tell

Slereah

Like all fancy tensor operations the covariant derivative is neither mapping two vectors to one vector nor one vector to a rank two tensor

It's both

That's the tensor-hom adjunction

Ryder Rude

3:20 PM

I think any tensor can b interpreted in multiple ways like that

Slereah

$$\mathrm{Hom}(Y \otimes X, Z) \sim \mathrm{Hom}(Y, \mathrm{Hom}(X, Z))$$

Ryder Rude

Omg is this category theory

Slereah

I mean in the sense that everything is category theory

it is categorier than most

Ryder Rude

What is its simple translation?

Slereah

If you have a map from a tensor product of two vector spaces to another vector space, it is equivalent to two chained maps between those three vector spaces

ACuriousMind

3:25 PM

@DIRAC1930 vol. 3 of Reed and Simon

Ryder Rude

Oh

DIRAC1930

Thanks

Ryder Rude

Chained product means function composition

I read category theory focuses on functions as foundations

Like u have a bunch of sets and u hav functions which r transitive

This is a category

Mr. Feynman

@DIRAC1930 In a recent question I asked, ACM mentioned the third volume of Simon&Reed

@Slereah Is that $\sim$ a $\cong$?

Ryder Rude

I always feel like this is just additional (restrictive) structure defined on sets. Set theory allows u any relation, even non-transitive

Mr. Feynman

3:29 PM

@ACuriousMind You are assuming the authors commute :P

Ryder Rude

I think the real benefit of category theory is not as foundations, but as a study of a specific functional structure defined on sets

There r restrictions placed on the morphism properties

They r required to be reflexive and transitive iirc

DIRAC1930

4:24 PM

So what it seems like is that the asymptotic assumption arose from a fudging everything but it turned out to be correct

I suppose you can somewhat motivate it by the fact that QM and RQM give the correct predictions for the dynamics of particles just sitting there not interacting with other particles

Amit

5:22 PM

Did category theory contribute to physics?

Slereah

5:39 PM

What does contributing to physics mean

you can use it for physics certainly

Amit

I mean even indirectly, if someone used it to develop math that is applied to physics

Slereah

they certainly do

Amit

I feel it's like the functional programming language of math... everyone knows it's great but it isn't used enough

But I don't trust feelings, lol

Silly Goose

6:55 PM

@ACuriousMind do you know where i can read about opinions that category theory is not all that. I.e. like a group with the opposite partiality as the nLab people. Or even better just an accurate review of such work :P

Amit

@Mr.Feynman Torsion is usually the first thing we set to zero in GR no? :) Why are you concerned about that?

Silly Goose

These category theorist written papers seem to promise the world while these 10+ year old answers on stack seem to relegate category theory to being a nifty toy with no fruit to bear

My naive hope is that people just don’t like category theory and that it actually is or can be a powerful tool :P

Mr. Feynman

7:12 PM

@Amit I mean, that's like forgetting a piece that would eventually go to zero in a calculation and say "who cares, it is zero" afterwards :P

@SillyGoose how dare they say that

Silly Goose

Lol i was surprised to read such responses on math overflow too @Mr.Feynman

Amit

@Mr.Feynman I rather thought it's the assumption that partial derivatives are commutative in that case

Slereah

I'm never safe from stupid GR statements

Mr. Feynman

@SillyGoose How did you fall in the category hole btw? :P

@Amit See the first part of this

And this together with the comments

Silly Goose

One of the questions that has most interested me is how to formally talk about a physical theory as a mathematical object and so i asked something like this to my linear alg professor and he mentioned category theory. I just happened to finally have some time to look at it hehe

And then i found the doring papers which purport to do this but of course i cant understand them:P not yet at least

Mr. Feynman

7:21 PM

I like the fact that eventually we all converge to the same things

Although my resolution to learn some category theory was soon killed by other stuff and it's in my to-do-list

DIRAC1930

Doesn't there have to be an asymptotic limit otherwise RQM/QM would not work

Mr. Feynman

::*angry ACM spawns*

Silly Goose

My interest in math has increasef relative to my interest of physics after this category theory business but i don’t thinki have the hardware to do mathematics :P

at least not analysis

perhaps more algebraic stuff i have some chance at

Mr. Feynman

I don't think I have the hardware to do neither seriously. These days I've been thinking about the future and if a PhD is really what I should go for

After my master's degree, so I have ~1.5 years to understand it

Silly Goose

indeed it is quite hard to make this judgement

Amit

7:28 PM

@Mr.Feynman Thanks, I did forget that the definition of torsion is more involved than that... when it's zero it implies what I said, but what I said (commutativity of partial derivatives) doesn't imply, I think, zero torsion

Well you know that Cartan probably was the first to try and make a non zero torsion theory from GR: en.wikipedia.org/wiki/Einstein%E2%80%93Cartan_theory

Mr. Feynman

Partial derivatives always commute (necessary condition for a coordinate frame). Covariant derivatives don't. Torsion is $T(X,Y)=\nabla_X Y-\nabla_Y X-[X,Y]$

Silly Goose

it is a bit hard to imagine that even after a phd one would know enough to do work in physics and category theory :P

Mr. Feynman

So it is related to a kind of "commutation" (the first two terms) but that's not what I'm talking about, that is $\nabla_X\nabla_Y-\nabla_Y\nabla_X$

acting on something

Anyways, the zero torsion condition is that the Christoffel symbols are symmetric in the lower indices

@SillyGoose you never know enough, you'll have to learn for the rest of your life

DIRAC1930

I think I'm going to go back to this $sin^2$ thing

And see what happens in QM

Mr. Feynman

The real question is whether resolution and dedication are really enough to succeed in Physics

DIRAC1930

7:35 PM

And then just accept the asymptotic condition in QFT

Silly Goose

i would think it depends on the area of research

lol the hbar chat is something of a collection of vignettes itself. a mini odyssey for everyone

DIRAC1930

The $\sin^2$ seems to suggest that if we accept on faith that the initial state is an eigenstate of the free Hamiltonian with energy $E$, then the final state at $t=+\infty$ will have the same energy $E$

Mr. Feynman

@SillyGoose you always find two people discussing, someone making a comment on their discussion and also someone speaking to themselves

Silly Goose

LOL

Mr. Feynman

And ACM handling three simultaneous conversations: one about rep theory, the other about QFT/QM foundations and the last one about the animal of the moment (e.g. pigeons)

DIRAC1930

7:42 PM

I am always speaking to myself :'(

Mr. Feynman

@DIRAC1930 I do that a lot here too :P

You need Bolbteppa back

DIRAC1930

Everytime I look at the chat history, I realise that Bolbteppa has asked the exact same question 3 years ago

Silly Goose

XD

Amit

@Mr.Feynman Ah I see, it comes down to the assumption that, if you'll excuse the lame notation, $\nabla_{\partial{j}}i = \nabla_{\partial{i}}j$ right?

DIRAC1930

The thing happened with spinors and also the whole rel. measurement thing

Mr. Feynman

7:43 PM

@ACuriousMind @SillyGoose @Slereah what's your take on capybaras?

Slereah

I am not opposed to them

Amit

Which is just a fancy way to write the christoffels

Silly Goose

they are large and they are cute

Amit

I'm not gonna answer 'cause you didn't ask me, but they are cool, lol

I worked with a guy that noticed I have a slight obsession with them, so he bought me a mouse pad that has a capybara printed on it. No kidding.

Mr. Feynman

@Amit this is what happens for a null torsion tensor. The "problem" was with taking two covariant derivatives one after another. The answers I linked explains it

@Slereah that sounds too neutral

@Amit lmao

DIRAC1930

7:48 PM

I'm convinced something fishy must be going on because it feels like the most important and crucial part of QFT is buried in some obscure papers and is barely mentioned properly in QFT books

Mr. Feynman

Accept QFT as a set of rules to write and compute the perturbation expansion (when perturbation theory works)

I decided to throw away any form of understanding about it

Amit

@Mr.Feynman I kind of lost the thread when they started talking about taking a covariant derivative "before" the indices, but I still think I understand/recall what is torsion better now, thanks

@DIRAC1930 Rubber duck physics ^_^

DIRAC1930

I suppose maybe it's not too much of a bad thing to just accept the asymptotic condition

And just thinking of them as the single particle states of QM/RQM

Amit

8:08 PM

Chat history search --
"asymptotic": 430 messages found
"capybara": 1 message found

C'mon! :D

Mr. Feynman

8:41 PM

Look up pigeons @Amit

Amit

impressive

ACuriousMind

9:16 PM

@Mr.Feynman pretty cute

DIRAC1930

Capybaras are an asymptotic state

Going to try and figure out how to do a capybara scattering experiment

2 capybara's in - 3 capybara's out

You need this experimental data to calculate the above scattering amplitude google.com/…

That gives you the lifetime of the unstable state

Okay noone found my joke funny lol

Silly Goose

11:09 PM

Capybara moment

Amit

11:43 PM

when two slits are open

ACuriousMind

@SillyGoose uh, that's like, every place but the nLab :P

I've repeatedly said in here that I personally haven't experienced category theory as useful in learning new things, but that it is exceptionally useful in recontextualizing stuff I already knew

for instance, the first time I saw a definition by universal properties I found it weird, the fifth time I saw the same pattern of proof of existence and uniqueness I started thinking "there has to be something behind this"; very basically I don't think you can appreciate something like the general "tensor-hom adjunction" or the Yoneda lemma without having seen it in examples without all the categorial language

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