The h Bar

Slereah

13:00

And there is a total of $4^4$ states total

yuggib

@Slereah I always am

Slereah

All particles in the same area will have a probability of $4 / 4^4 = 1/64$

But all particles in different areas will be ~ 1/10

hence it's a much more likely scenario

0celo7

@ACuriousMind dude

I must suck at that game, I have no money

I'm trying to get the Wolven Armor but it broke my bank

I literally cannot buy enough Dimeritium

Shing

@Slereah thanks! isn't it enough for a "physicist's proof"?

0celo7

I've done basically all of the side quests and contracts

Slereah

13:03

@Shing well you can prove it properly afterwards

0celo7

What is @Shing trying to prove?

Las Vegas Raiders

16

Q: How do you prove the second law of thermodynamics from statistical mechanics?

How do you prove the second law of thermodynamics from statistical mechanics? To prove entropy will only increase with time? How to prove? Please guide.

ACuriousMind

@0celo7 Money is a bit scarce, but you should be able to sell surplus crafting ingredients to get enough for most things

You don't need 100 drowner brains, for instance ;)

Shing

@Slereah so ppl are trying the proof in a more rigorous way?

0celo7

@ACuriousMind I need like 3000 more gold to complete my stuff

I already sold a shitload of swords and armor

I don't think I have 100 Drowner brains.

Maybe I can find some more quests. There are some remote parts of Skellige I haven't explored.

Shing

13:06

or Ben Crowell just saying the 2nd law is just a consequence of " all states are equally probable", and that is a postulation ?

ACuriousMind

@0celo7 There are some ingredients that are surprisingly valuable. But, well, breaking the economy is not possible here as it is in other games, you'll never have ALL THE MONEY

yuggib

@0celo7 The collection of all possible spacetimes (smooth 4-dim Hausd paracomp connected manifolds with lorentzian metric) is not a set, right?

Slereah

@yuggib why not

yuggib

@Slereah it seems too large

Slereah

I think all spacetimes can be built from a simplicial manifold

Which do form a set?

not sure though

yuggib

13:11

symplectic manifolds? I doubt they form a set, but maybe since they are of finite dimension...

Slereah

Simplicial

different thing

yuggib

anyways

I doubt they are a set

Slereah

if they weren't Hausdorff or second countable then definately not a set

yuggib

e.g. the collection of hilbert spaces is not a set (this I know for sure)

and hilbert spaces are hausd and second countable, albeit there is no restriction on the dimension

Slereah

Well do you impose a restriction on the dimension of spacetimes

Balarka Sen

13:13

It's a proper class, but I don't think it's a set

Slereah

It's kinda never really specified (except in my book, which is great), but I don't think anybody considers spacetimes built on $\Bbb R^\omega$

yuggib

@BalarkaSen if it's a proper class it is not a set ;-P

Balarka Sen

Ok, fine, delete "proper"

yuggib

don't get mad, but that is a very trivial statement...

Balarka Sen

I realized, which is why I wrote down my next statement.

yuggib

13:15

anyways, I will simply specify I am considering a small subcategory of the category of spacetimes

Slereah

Hm

Let's see

Do the spacetimes manifold form a set

Balarka Sen

I actually don't know efficiently how to prove a class is not a set

Slereah

They're all embeddable in $\Bbb R^{2n}$, so they should be of a cardinality inferior or equal to... $$\sum_n 2^{\text{Card} (\Bbb R^{2n})}$$

yuggib

@BalarkaSen mmmh I have seen the proof for Hilbert spaces somewhere

Slereah

Is that a proper set?

yuggib

13:19

but I don't remember neither where nor the proof T__T

Slereah

I'm not sure

Also you have to put all the metrics on it afterwards

which is another hornet nest

Since all $\text{Card} R^n = \mathfrak c$, we have $\sum_n 2^\mathfrak{c}$

yuggib

$2^{2^{\aleph_0}}$ is an upper bound for the cardinality of a single manifold

Slereah

So embeddable manifolds do form a set

How many metrics tho

Hm

Wait, metrics are manifolds too

yuggib

no no

a manifold has at most the cardinality above

Slereah

well yes, but they're spacetimes

They all have cardinality $\mathfrak c$

since they're second countable

0celo7

13:23

@yuggib what are you considering spacetimes for?

yuggib

this does not tell you much on the eventual cardinality of the collection of all possible manifolds.

@0celo7 semiclassical analysis in AQFT

Slereah

I think spacetimes should form a set

0celo7

ugh, you're doing AQFT now?

yuggib

@0celo7 just an application of semiclassical analysis to AQFTs

Slereah

You can just associate every manifold + metric as an embedding in $\Bbb R^{4n}$

Wait, what's the dimension of the metric bundle?

yuggib

13:24

it should (supposedly) be "easy"

Slereah

I think the set of all metric sections is just gonna be $\mathfrak c$, no?

0celo7

@yuggib It's known that one cannot classify 4-manifolds because there are too many fundamental groups. Not sure exactly what that means, but if they formed a set I would expect one to be able to write down the groups

Slereah

@0celo7 Plenty of sets aren't constructible

What's a set and what's a solvable problem are two different things

$\Bbb R$ contains plenty of uncomputable numbers

Balarka Sen

@0celo7 That just says every finitely generated group appears as fundamental group of 4-manifolds.

Do we know if f.g groups form a set?

0celo7

I don't even know what a set is, so I'm going to make breakfast.

@BalarkaSen For what he wants you need the restriction $\chi(M)=0$ in the compact case or simply noncompact.

Balarka Sen

13:28

I dunno what restrictions $\chi(M) = 0$ pose on $\pi_1(M)$

Slereah

@BalarkaSen : A spacetime is gonna be a manifold of dimension $n$, with an embedding in $\Bbb R^n$, and a metric that's a function $\Bbb R^n \times \Bbb R^n \to \Bbb R$, roughly

0celo7

@BalarkaSen @yuggib People use this terminology loosely anyway. Gromov calls the class of all metric spaces a set, and it's certainly not a set.

Slereah

I think that can be indexed by $\Bbb N \times 2^\Bbb R \times 2^{2^{\Bbb R}}$

Balarka Sen

@0celo7 lol

@Slereah Is the embedding specified?

Also, er, most manifolds of dimension $n$ do not embed in $\Bbb R^n$

Slereah

Well all spacetimes can be embedded in $\Bbb R^{2n}$

Most manifolds, no

But every spacetime can

0celo7

13:29

All manifolds can be embedded in $\Bbb R^{2n}$.

Why would you be certain about spacetimes but not in general?

Slereah

@0celo7 not all

0celo7

@BalarkaSen No. I actually have no clue what he is on about.

Slereah

Long line and whatnot

0celo7

Manifolds are paracompact, always.

Slereah

Anyway

That would mean that we can put manifolds as a surjection from the set of functions on $\Bbb R^{2n}$

Which does form a set

Balarka Sen

13:32

Jun 20 at 15:38, by 0celouvskyopoulo7

It should be noted that Kobayashi and Nomizu do not assume manifolds are paracompact.

NANDA?! - GET REKT (OFFICIAL MUSIC VIDEO)

►

Slereah

they're not even required to be Hausdorff

Secret

(Please refer to starboard)

And, the most common weird idea is mistaken nonlocal correlations as nonlocal signalling (Looking at you, quantum mysticists)

Emilio Pisanty

@Secret you've seen the top of the star board, right?

Secret

ooooooops

Anyway, the reason I dug that xkcd out is because recently people in HK are talking about a recent wave of pseudoscience concerning something called animal telepathy, and a news programme managed to debunk it by telling those telepaths to talk to a toy turtle without them knowing

those telepaths then say their telepathy is done by entanglement (classical case of how quantum mechanics is misrepresented)

0celo7

@BalarkaSen Yeah well those damn geometers

Slereah

13:39

I can do animal telepathy with a can of tuna

Secret

Btw, compared to western countries, east asian countries tend to get more affected by quantum mysticism and new age stuff. If you ever travelled to these countries, you will noticed a lot of shops talking about a whole lot of quantum mysticism to the point of eyebleed (o, and there are expensive courses on that)

Slereah

Western hemisphere best hemisphere

as always

Secret

These scammers means that even if there is a very infintesimal chance that true esoterists do exists, they reputations are completely ****ed and swamped by these scammers and money makers to the point that nobody will believe them

I think it is very important to get the message across to the public that entanglement is NOT nonlocal signalling, though how to get that concept across as easy as possible is not that easy

4 hours ago, by Slereah

"On the other hand, most physicists presume that feminist critique is incapable of generating ideas that will make a superfluid colder, a plasma hotter, or a particle beam more intense. This is correlated with the fact that feminism has done essentially nothing to transform what one might call “orthodox physics”

WTF have you been reading...

Sid

... What?

Yashas

@Sid hi, where are you joining?

Sid

13:46

Hey @Yashas ! Well, in JOSAA, I have got NIT Rourkela Electrical Engineering. Might join that.

Yashas

@Sid HS quota?

Sid

Yes.

yuggib

So if manifolds are subsets of some set with cardinality $\aleph_{\alpha}$ (probably $2^{2^{\aleph_0}}$), then their collection is a set with cardinality at most $2^{\aleph_\alpha}$

Slereah

@yuggib If they're embeddable, yes, I think that's fine

If they're not second countable then all bets are off

I know that non-Hausdorff manifolds don't form a set

But I don't know about non-paracompact ones

There's $\text{Card}(V)$ non-Hausdorff 1D manifolds for $V$ the category of all sets, but only two non-paracompact ones

I don't know if that generalizes well for more dimensions

yuggib

@Slereah that you can't write

the cardinality of a class :-D

Slereah

13:56

And yet here I did!

You'll never stop me

yuggib

You just "sound" funny

you can write it, still it's weird

Slereah

"In extensions of set theory where classes are allowed (not just formally as in ZFC, but as actual objects as in MK or GB), sometimes it is suggested to add an axiom (due to Von Neumann, I believe) stating that any two classes are in bijection with one another. Under this axiom, the "cardinality" of a proper class would be ORD, the class of all ordinals.
(By the way, by class forcing, given any proper class, one can add a bijection between the class and ORD without adding sets, so this assumption bears no implications for set theory proper.)"

yuggib

@Slereah I want definitions to be convinced ;-P

Slereah

From what I can read, basically the cardinality of a class is either the cardinality of a set if it's a set, or its own thing otherwise, since all proper classes have a bijection to $V$

So there's only set cardinalities and a really big class cardinality

yuggib

but you need an axiom to have the bijection

Slereah

14:03

It's apparently in there : gdz.sub.uni-goettingen.de/dms/load/img/…

5

Q: When do we have a bijection between a proper class A and its power set class P(A)?

We work in the set theory NBG with the axiom of (local choice but without global (class) choice. For every class A P(A) is the class of all sets x included in the class A. We know that P(A) is a set iff A is a set and a proper class iff P(A) is a proper class. We also know that if A is a set th...

set-theory

yuggib

anyways, cardinality of sets make sense because the cardinality is a set as well

Slereah

Well cardinality of a class is a class as well!

Since it's just $V$

yuggib

yeah, but that's like saying an apple is an apple is an apple

Slereah

Well no, because not all proper classes are $V$

yuggib

cardinals are defined up to bijection

Slereah

14:06

Same here

7

Q: Bijective-equivalent collections of proper classes in set theory

In ZFC set theory (or better in NBG set theory, where the language is more flexible with proper classes), we have that every unbounded class of ordinal numbers is a proper subclass of the class On of all ordinals and that every such proper class has a unique bijection (by the enumeration function...

set-theory

the horror

yuggib

well, it seems a bit moot to me but if you're happy...

Slereah

@yuggib How very pissy :p

yuggib

I am essentially just quoting the famous sentence "a good definition should be the hypothesis of a theorem" ;-P

Slereah

Hypothesis : if a class has the cardinality of $V$ it is not a set

Corrolary : Manifolds do not form a set

Balarka Sen

14:34

what are we fighting about

Slereah

Skub

Sid

@BalarkaSen Good question. That can have both a literal answer as well as a philosophical answer

Balarka Sen

@Sid Well, you are thinking of literalism as different from philosophism, which is a much controversial matter, but it is a metaphysical convention that whatever that is philosophical also has literal truth value equivalent to the philosophy as proposed before.

#comebacktroll

Mostafa

Hi. I had a (dumb) QM question

In Shankar's QM (page 210), it is said that (squares) of all the matrix elements of an operator, $| \langle \psi_1 |\Omega | \psi_2 \rangle | $ are physically relevant and can be measured.

Slereah

that is true

Mostafa

14:38

The diagonal elements are clearly the expected values of $\Omega$.

But I don't understand the physical significance of the off-diagonal elemetns.

Slereah

The off-diagonal elements are transition probabilities

The probability of passing from a state to another

Emilio Pisanty

@Slereah in the vaguest of senses

Mostafa

@Slereah How?

There were some questions about this, but couldn't help much. like this one physics.stackexchange.com/questions/209350/…

Slereah

If the state is initially in state $E_2$, what's the probability of being $E_2$ after measurement by $X$

Emilio Pisanty

@Slereah what?

measurement by $X$, as in, projective measurement on an eigenstate of $X$?

which eigenstate?

ACuriousMind

14:42

@Slereah I hope you mean that in the sense that $X$ is supposed to be a projector, and not the observable being measured

Emilio Pisanty

^ that. Or rather: that's the only way to keep the dimensions consistent with $|\langle E_1|X|E_2\rangle|$ being a probability.

Slereah

Well after the collapse of the wavefunction by measurement of $X$

Emilio Pisanty

@Slereah ... onto which eigenstate?

Slereah

An unknown one

such is the quantum mystery

Emilio Pisanty

@Slereah sorry, mate, but no

Slereah

14:43

itisamystery.com

nvm then

Emilio Pisanty

if it's an unknown one, then you're just doing a decoherence channel on the $X$ eigenbasis

Slereah

been a while since I've done QM proper

Emilio Pisanty

@Mostafa The quantity $|\langle a_1|B|a_2\rangle|$ is interpreting by initiating the system in the eigenstate corresponding to $A=a_2$, then evolving for a time $\tau$ under the hamiltonian $H=g B$ with a weak coupling $g$ such that it can be treated as a perturbation, i.e. the unitary can be decomposed as $$U=e^{-i\tau g B/\hbar} \approx 1 -\frac{g\tau}{\hbar}B,$$ measuring $A$ again, and looking for the eigenvalue $A=a_1$.

(apologies for repeated edits.)

In that sense, it's a transition probability

but really, the name "transition probability" is a moniker that stuck from when first-order perturbation theory was much of what QM did, and it's a poor description for the full range of roles it plays.

Mostafa

@EmilioPisanty Thanks a lot. It makes (some) sense now.

But, does that inner product have any other physical meaning by itself (without considering time evolution)?

Because the time evolution only approximately gives that inner product

@EmilioPisanty What do you mean the full range of roles it plays?

Sorry for the delays...I've just started QM (~2 weeks) and my mind is a little sluggish processing QM

0celo7

15:05

@yuggib @Slereah Might have a point. You can identify manifolds with their images under Whitney embedding. So there are at most $2^{\aleph_0}$ of them for a given dimension. Maybe?

Slereah

Damn right I'm right

Well, not identifies, since two embeddings can be the same manifold

But there will be less of them

Hajicek actually made two papers on non Hausdorff spacetimes

aip.scitation.org/doi/10.1063/1.1665474

Secret

Meanwhile in maths chat:

in Mathematics, 13 mins ago, by Semiclassical

Oh man, today's xkcd:

(see starred message)

and followed by a short discussion about entanglement and quantum interpretations

0celo7

@Slereah Sure. That's why I said at most $2^{\aleph_0}$.

Emilio Pisanty

@Mostafa That inner product has a huge range of roles ─ it's the meat and bones of QM, really

Semiclassical

Hence my deleted message above :P (I posted the link, then noticed Slereah's starred comment and felt careless)

Emilio Pisanty

15:12

it's hard to say "what it represents" because it can be used in a huge number of ways

@Mostafa regarding transitions, keep in mind that for 50+ years, first-order perturbation theory was pretty much exact

That doesn't mean it is exact

but it informs a lot of the terminology

Mostafa

Yeah!
yesterday I Googled this question for like 1 hour and couldn't find a satisfying answer clear as this :)

Emilio Pisanty

@Mostafa the key term you should be googling is Fermi's Golden Rule

so, again, a first-order perturbation-theory framework

but it's more grounded in the technical parts of the formalism and it's easier to connect with treatments that explain its domain and limitations

0celo7

@ACuriousMind help

Anonymous

@Yashas sup? All admission formalities over? :D

Yashas

@Blue I think so

Anonymous

15:22

When are your classes starting ?

Sid

@Yashas IIIT-Hyderabad is your place, right?

Yashas

@Sid yea

@Blue idk :D I think it is on the last Saturday of July

Anonymous

@Yashas Same here...the seniors will be holding a welcome fest for us (something like that) and tell us the schedule there XD

Yashas

......

Anonymous

Probably last week of July or first week of August

Sid

15:25

Oh, Ragging..

Yashas

do you know the date of the welcome fest?

Anonymous

lol...they are thin chaps

Anonymous

They can't rag

Yashas

IIITH is 100% ragging free it seems :d

Anonymous

We have strict anti-ragging rules here though...so nothing to worry :P

Anonymous

15:26

@Yashas Obviously :'D

Sid

"Strict anti-ragging rules" are everywhere..

Balarka Sen

"IIT"s and "ragging-free" together in a sentence forms an oxymoron

Anonymous

@Sid Dude, I met those guys today. They were so shy to even talk to us. I can't even imagine them ragging us. :P

Balarka Sen

@Blue are you going to J

Anonymous

@BalarkaSen yea

Balarka Sen

15:28

They are rather notorious for their political conflict with the authority.

Anonymous

@BalarkaSen That's in the arts department. The engineering campus is very calm.

Balarka Sen

Ah

Yashas

IIITH has proper students who are interested in serious academics

Anonymous

I need to sign this and submit it tomorrow...lol

Yashas

IITs have majority of the people who think life is set after making it to IIT

Sid

15:31

^ That

Yashas

and majority of the students at IITs are studying in a branch which they don't like

Anonymous

Okay...okay....enough of talking against IITs. :P Not everyone is like that there. I know very good students too.

Yashas

do you seriously think all the top 60 students are interested in CS?

the olympiad medalists make up the top ranks and are they really really really interested in CS? ;)

Sid

Nope, they get paid better..

Slereah

If you've done physics CS is basically a snooze

Anonymous

15:33

@Yashas No, that's a different issue altogether. Mostly they are forced by their parents to study CS. But many change branches after first year if they don't like it. For example Chitraang who was AIR 1 in 2014 left CS and went to engineering physics department. I can feel the amount of pressure the society puts on them. It's not fair to blame them for choosing CS.

Slereah

Modern CS classes are basically kindergarten

Yashas

@Blue Nope. Chitraang left IITB and joined MIT

Semiclassical

"Tedious but straightforward", to borrow a phrase from Jackson

Yashas

and took up Physics @ MIT

Anonymous

@Yashas Before that he went to engineering physics in IITB.

Yashas

15:34

LOL

Anonymous

Search quora

Yashas

IITB does not have pure Physics, right?

good for him

Anonymous

wait a second...i'm linking it

Yashas

that's what he gets for choosing CS :D

Slereah

Society decided that we needed a lot of CS people so they dumbed it down like hell

Yashas

15:34

eng physics isn't as good as pure physics

Slereah

So that all the numb brains can do it

Sid

Oh, there was a thing that Allen bought him or something during that time..

Yashas

I have heard of some two digit AIRs joining IIT Kanpur Physics

instead of IISc

not sure why they would do that

Anonymous

@Yashas quora.com/How-did-Chitrang-Murdia-get-into-MIT-by-IIT-JEE-rank/…

Anonymous

@Yashas Kanpur Physics is very very good with professors like HC Verma. It's definitely not a bad choice if location is a concern.

Yashas

15:36

HC Verma resigned :P

Anonymous

Yeah, few days back

Anonymous

I heard

Anonymous

Great guy

Sid

Yeah, there was some thing going on that Allen paid some money to Chitraang to advertise his name or something like that

Yashas

.

Anonymous

15:37

They pay money to all top 100 guys to put them in ads.....

Anonymous

Most of them don't accept but some unfortunately do

Yashas

The education system in India is nonsense. The coaching centers are businesses who want to abuse the system to make as much money as possible. The end result is that graduates leave India or the best students (who have the potential to make a Google or Microsoft in India) never make it to good colleges.

Anonymous

Grow up and change the education system then :3

Anonymous

I mostly agree though

ACuriousMind

@0celo7 ?

Yashas

15:40

Not to mention that a 3 time IOI medalist did not even qualify for JEE but was accepted into MIT :P You know how crappy this system is when that happens :P

Anonymous

@Yashas XD

Physics Meta

1

Q: Did I award the bounty?

I recently put a bounty on this question: Prove isometry preserving excision is Killing-like? and Valter Moretti answered it just before it expired (I think). I took the weekend to consider his answer and accepted yesterday (Monday, July 10) and don't recall seeing an award bounty button. Val...

support specific-question bounty

Sid

Wow... people don't even know if they awarded a bounty or not..

Shing

I am refreshing my knowledge on classical coupled oscillator by working on a problem. but the solution seems a bit odd to me, would anyone give me a helping hand?

Slereah

"Jack Little lamented the tendency of manufacturers to design languages “for use by some sub-human species in order to get around training and having good programmers.”"

Heh

Shing

15:47

I think $y_1(t)$ in the picture is wrong?

omega 1 and 2 are the two normal mode freq

Anonymous

@Shing What are d1,d2,x1,x2 ? Can you upload the full question?

Shing

@Blue sorry for the confusion

here is the full question:

David Z

@Shing BTW we do have a chat session coming up in... 8 minutes now, so if you don't solve it before then you can take the discussion to the backup room until after the chat session is over

Shing

@DavidZ okay, no problem

Sid

what happens in the chat session?

We chat, of course but on what?

David Z

15:53

@Sid Depends... sometimes we have particular topics to discuss, but I don't know of any today

Does anyone have something to put on the agenda?

In general, if there is something about site policy we need to talk about, we'll talk about that, or if there is some big event in the physics community we could talk about that, or whatever

Sid

This site's tag-wikis are not at all updated..

David Z

@Sid very true

Emilio Pisanty

and in a related observation, this site has a perennial drought of volunteers that are up for maintaining the tag wikis

::discrete cough::

;-)

Sid

editing tag-wikis privileges are what? 2.5k rep?

Emilio Pisanty

@Sid I'm pretty sure you can edit from any rep at all? It'll go through peer review until 2k rep

Sid

15:58

I thought for tag-wikis there is a threshold

ACuriousMind

@EmilioPisanty I think it goes through review for much higher rep than 2k

Emilio Pisanty

@Sid doesn't seem to be

ACuriousMind

Ah, yes, you need 20k to edit a tag wiki without review, but everyone who can suggest normal edits should be able to suggest tag wiki edits

Emilio Pisanty

@ACuriousMind ~~you're thinking of reviewing tag wiki edits~~

physics.stackexchange.com/help/privileges/…

Transcript for