The h Bar

00:18

@DannyuNDos The only case whereby that would not be toooooo horrible, is when the operator is merely one single state's projector. The moment you have more, it would be too confusing to use such a way of writing things down. Operators are not vectors; why would you have thought so in the first place?

A: ???
F: _replies_
A: I know
_loop_
F: _breaks into many numbered questions_
A: _finally replies giving some context_

Dannyu NDos

00:35

@naturallyInconsistent Operators are vectors. They admit addition, negation, and scalar multiplication.

naturallyInconsistent

00:47

@DannyuNDos Look, yes, of course they satisfy the mathematical definition of vectors if you are going to put it that way. But they are not vectors in the Hilbert space in the same way that bra and ket vectors are. It is just not how people use the terms colloquially

Dannyu NDos

Fine;

Slereah

@DannyuNDos they don't typically form a Hilbert space rho

Dannyu NDos

Yeah..

Slereah

It's a pretty specific property to have bras and kets be isomorphic

Silly Goose

the operators live in banach land

naturallyInconsistent

01:04

bananaaaa

qwerty

b-a-n-a-n-a-s

Mr. Feynman

01:28

Banana.

So this is what the chat looks like at night

naturallyInconsistent

it is bright here

blindingly so

Mr. Feynman

I'm in the darkness

naturallyInconsistent

bite someone

Silly Goose

03:21

do yall have charms

naturallyInconsistent

@SillyGoose which kind

qwerty

@SillyGoose no only strangeness

Silly Goose

@naturallyInconsistent the key chain like ones

@qwerty only honkzies here

have you all ever heard of thévenin's theorem?

i am hearing about this for the first time while TAing for intro EM :P.

naturallyInconsistent

03:37

@SillyGoose There is also Norton's theorem. Both are due to the excessive simplicity of linearity and a bit more

qwerty

04:22

I have not

naturallyInconsistent

07:20

For those who have suggested Zangwill, thank you. The little bit miao miao hath read, have been enjoyable

qwerty

07:39

all the contentious stars from yesterday were removed

John Rennie

@qwerty You're welcome to ping me if we get another round of random starring and I'll remove the stars.

Mr. Feynman

Great. The starboard was... uneasy

qwerty

@JohnRennie i dont have opinions either way, just an observation.

qwerty

08:01

> One can take any family of formulas and turn it into a category where each formula is an object and there is an arrow A→B when there is a proof that A implies B. The identity arrows are trivial (empty) proofs and two proofs can be composed by concatenating them.

fun

Mr. Feynman

@qwerty beware, my friend. You start with "fun" and you find yourself in a hole deeper than you expected. Remember: you don't feel anything when you cross the event horizon

Madder

Why would the Energy of a some ice molecules depend on their orientation or on the way which they connect?
Oh oh, wont let me draw it here

Meaning, at vertex one and two, we have different energies?

I do not understand the difference, it is the same connections, and some of them are actually just rotational symmetric to each other, wouldnt they all have the same energy?

Atleast locally if i zoom in around the oxygen in vertex one and two, and look in all four directions, whatever lattice point i choose, most of the configurations are equivallent up to a rotation. Unless then the energy calculation exceeds the local neighbourhood.

qwerty

08:33

@Mr.Feynman i will gladly be slowly consumed by mathematics rather than be confined to this mortal coil

naturallyInconsistent

@qwerty Those "set of 5 equivalent expressions" kind of statements, provable from each other, make loops

fqq

@naturallyInconsistent they make cliques

naturallyInconsistent

@Madder how big is the difference? Is it within numerical convergence error? Also, it looks like there are closer/farther connections for the only H atom different in orientation, so there is at least some goodness there

Madder

@naturallyInconsistent oh it is not clear. they are building a model, and assigning values to different configurations. Where in this context, a configuration is just the way the connections are made around a vertex.

naturallyInconsistent

Well, we cannot be expected to be helpful when the initial problem statement is so vague...

@fqq sure? miehehe

qwerty

08:43

ugh, im sleep-deprived from having too much fun with morphisms too late into last night and barely functioned at work. still #worthit

fqq

@qwerty that's two of us! Or it'll be once I get to work

qwerty

@fqq what were you reading/working on?

Mr. Feynman

@qwerty Thy words exude hubris, mortal

I mean, who in this chat is not sleep-deprived?

qwerty

@Mr.Feynman nI always seems bright eyed and bushy tailed

Mr. Feynman

You mentioned the only guy who clearly sleeps less than any of us. Try to check the time on his messages :P

qwerty

08:49

...having "fun" and partying half his work nights

I dont see the contradiction with being bright eyed and bushy tailed :P

besides, cats are nocturnal, right?

Mr. Feynman

He sent a message at around 1:18am, which should be 8:18am for him; His last message was at 3:54pm yesterday, which should be 10:54pm for him (in any case, if the timezone is wrong, we'll just make a subtraction). He was absent for about 9h24m. DAAAAMN, I was wrong

Why did I think he slept like 5 hours at night :P

qwerty

@Mr.Feynman well i somehow don't think that's abnormal for him either lol

Mr. Feynman

yeah, that must be right

naturallyInconsistent

09:51

@qwerty sneeppuuuu

qwerty

@naturallyInconsistent WOOF

naturallyInconsistent

M I A O !!!

@qwerty thy spaketh le truthe

qwerty

10:42

a dream is a morphism on the space of hallucinations

Slereah

13:16

@ACuriousMind How does one do the grim business of variational theory if the configuration space isn't a vector space

Do I just slap some coordinates on a local patch and call it a day

Vector bundles have a Frechet space structure for their sections sayeth the Michor, but what do you do if they are not vectoresque

Are you left with a Frechet manifold or some other such horror

ACuriousMind

13:46

@Slereah Infinitesimally, what does it matter? Calculus of variations considers small variations $f +\delta f$, if $f$ is valued in a manifold instead of a vector space, you just do the calculus in a coordinate patch where $f+\delta f$ makes sense again

It's just like you can talk about the derivative of maps into smooth manifolds - we can pretentiously talk about "smooth structures" but all that you operationally do at the end of the day is resolve everything by coordinate charts and do the usual derivative of maps $\mathbb{R}^n \to \mathbb{R}^m$

I think you're right that in the infinite-dimensional setting this implies you need to demand the target spaces of the $\sigma$-models to be Frechet manifolds

but I'd need to check that when I'm not at work :P

Slereah

A lot of things in the global picture seem to assume that the configuration space is a vector space

For instance the additivity condition on the action

ACuriousMind

14:01

@Slereah Well, if you do that, you're excluding all $\sigma$-models, I find that hard to believe

Slereah

I'm guessing they are typically more interested in field theories

ACuriousMind

and certainly I know that formal Hamiltonian mechanics does not assume that $Q$ is a vector space when it constructs the theory on $T^\ast Q$

really, since you only need the calculus of variations locally you can always do it in coordinate patches

Slereah

Likely true

but I'm still curious

ACuriousMind

We're usually only interested in the statement that something is a critical point of the action, which is a local statement

Mr. Feynman

Oh people, I might as well go crazy before it's time to

ACuriousMind

14:32

@Mr.Feynman keep calm and read about Haag's theorem :)

Mr. Feynman

This is a moment in which I would benefit from that!

Exams in Italy really suck for me. I have so many courses that for a good half are a list of applications that are barely hinted at, like a list of results that to understand properly you would have to take a course of its own

Imagine a course that is half "you may calculate that with such a hamiltonian you get this qualitative/quantitative behaviour"

So, yes, I could just make all calculations but wasting so much time for an unimportant exam is overkill D:

How can I better explain with an example? These are the kind of courses in which you define something and then explore a handful of applications which are not *per se* the subject of the course, so you have two possible paths:
1. You only know what a Green function is and what is the perturbation expansion;
2. For a single exam you get an expertise in superconductivity, transport theory... To be able to understand half a page of notes

And one may argue: who care about "half a page" of notes? I wouldn't if half my notes weren't made of this half-pages :P

Sorry for venting

Slereah

15:16

Why is it called hag's theorem anyway, was it made by witches

ACuriousMind

15:46

@Mr.Feynman In my experience, that's just the nature of some more advanced courses - they're more pointers to the vast range of things one can do with some idea than the neat self-contained bundle of knowledge the intro courses are designed to be.

Slereah

just sit the student in front of Bourbaki and go through every page

LEARN

Mr. Feynman

@ACuriousMind which would be fine if the exams were structured compatibly and not with any question that could in principle be asked :P

Then some professors remember every example they did, others don't even know lol

ACuriousMind

well, that sucks :P

Madder

i am stuck here and i dont understand the argument, maybe a more experianced eye can help me out

In this model, the vertices (ie the points in the middle of the arrows) represent an Oxygen atom, and the arrows point the direction of the hydrogen bond. like so:
(it is lattice representing ice so we have many molecules)

Now there is only six possible ways to have a junction

The issue is this: in order to build a statistical model, we assign weights for these junctions. figure 1 and 2 get the weight a, 2,4 get weight b, and 5,6 get weight c. The argument is : "Under no external field, the Boltzmann weights must be invariant when reversing all the polarizations simultaneously"

But i dont understand this rule, if we are talking symmetry. the first figures are literally the same just rotated 90 degree. They even comment on this

"Let us consider π/2 rotation of the square lattice. If we rotate vertex configuration (1) of Fig. 2.2
by the angle π/2 in the counterclockwise direction, then, it becomes vertex configuration (4). Under
the π/2 rotation, the weight a is exchanged with the weight b, while the weight c does not change."

But what? why would you not assign the same weight for 1 2 3 4? why this inversion stuff? it makes no sense to me. They are all the same difference by a rotation. i dont get their symmetry law.

Mr. Feynman

16:13

@ACuriousMind yes, it does

ACuriousMind

@Madder I have no concrete expertise here, but my guess would be that you have no guarantee that the overall crystal structure is invariant under rotations, I. e. rotating the entire lattice is not the same as rotating each junction around it's center, while reversing all polarizations in the lattice is the same as reversing the polarizations at each vertex

Madder

I am not sure i understand you?Do you mean to say that, it is not possible to rotate a vertex, or reverse it spin, without affecting the whole lattice?

ACuriousMind

16:28

Yes, that's part of it. But the way to extend the "reverse" operation to the whole lattice is by just reversing at every vertex, while a rotation has a single center - the operation that rotates every vertex around its center is not the same as rotating the whole lattice

I'm not saying I understand the context well enough to explain why that's relevant, but it's the difference between the two operations that immediately strikes me

Madder

So i just did an experiment, and you are right in that regards.

But i still do not understand the physical explanation behind the symmetry argument:
How can we follow from :
" reversing all polarizations in the lattice is the same as reversing the polarizations at each vertex"
To:
""If there is no ambient electric field, then the total energy of a state should remain unchanged under a charge reversal, i.e. under flipping all arrows. Thus one may assume without loss of generality that energies E_1 = E_2 .. E_3 =E_4, E_5 = E_6"

Its from wikipedia btw. en.wikipedia.org/wiki/Ice-type_model#The_zero_field_assumption

@ACuriousMind Sidenote: Wow! really impressive you immediatly caught that!

ACuriousMind

@Madder ah, the argument goes in reverse: the zero-field assumption implies that reversing the whole lattice should be a symmetry. Then because reversing the lattice is the same as reversing each vertex, for this to hold for arbitrary lattices, it must already be a symmetry for each vertex

Madder

@ACuriousMind Oh wow.

I am really amazed how you can catch these things.
But i see the point now.

ACuriousMind

16:44

It's mostly training - I've seen enough arguments, so I start seeing the same few patterns, such as this kind of global/local argument, everywhere

Madder

Thank you so much! you just unstucked me

Mr. Feynman

18:20

Quantum finance theory

QFT

Mr. Feynman

18:35

Money is the charge conserved under transaction

ACuriousMind

oh no he broke

Mr. Feynman

18:52

he broker

The Empty String Photographer

Time to measure temperature in nm/ms

Actually maybe nm/cs would be better

Slereah

19:17

Worst part is

Quantum finance is a thing

Although like most things in economics and finance, it is an attempt at a veneer of respectability riding the tailcoat of real science

Slereah

19:39

If you wish to know more about the use of mathematics in the Humanities, see

Sociology Song

►

Silly Goose

21:15

this is from jackson's section on macroscopic maxwells

do we usually choose $\textbf{x}_n(t)$ to be the center of mass of the molecule/atom so that the dynamics of the molecule/atom are more tractable? It seems this prescription collapses the information of the atom into a single point located at the atom's center of mass. This center of mass then obeys newton's 2nd law and so on

if we were to use a different prescription for $\textbf{x}_n(t)$ seemingly the equations of motion to track the particle dynamics would be more complicated

ACuriousMind

21:37

@SillyGoose I'm not sure what you mean - for the molecules, you have that they are composed of several point charges, which is what 6.73 tells you - the $\eta_n(x)$ is the sum of the point charge densities of the point charges that constitute the molecule

there's no center of mass anywhere here

oh, you mean the $x_n(t)$ at the end

I'm not sure what the question is, then

if you want to describe some composite object by a single position, it seems extremely natural to me to choose its center of mass

Silly Goose

i am wondering why we would want to use the center of mass for $x_n(t)$

right it is natural, but i am wondering about a more quantitative reason

ACuriousMind

what else would you choose?

the question "why would we want to use this" implies you have some other candidates lined up :P

qwerty

@SillyGoose I think you're probably correct

Silly Goose

@ACuriousMind well in class we talked about the idea of using the center of charge for $x_n(t)$ so that the polarization density vanishes

but i think the catch is that it should make something else harder to compute, that thing being what i described above

ACuriousMind

I mean as of the passage you posted we don't know what you use the $x_n(t)$ for yet

did you finish the chapter before asking this :P

because what you're going to use the $x_n(t)$ for would probably be a pretty good hint for why you'd choose the center of mass

Mr. Feynman

22:16

Self-energy is one of the coolest names in physics

Oh god, that sounds RRish

2

Slereah

It sounds like a self-help term

Mr. Feynman

Self-energy according to psychology:

Slereah

a monoid apparently?

qwerty

22:31

I think reading the very basics of category theory (i.e. just wiki and SEP) was the right move xD

the terminology is making more sense

@Slereah where is that from?

I love it it's so bizarre

Slereah

bartoszmilewski.com/2018/05/16/…

Mr. Feynman

@qwerty Don't EVER ask that question to Slereah

qwerty

lol

Silly Goose

23:50

@ACuriousMind hm but i only see that they are used in defining the polarization, quadrupolarization, and etc.

Transcript for