Reading the discussions involving JD is fun (at times, for short timespans); but more than that it is a perfect argument supporting mathematics and strict logical systems, where it is crystal clear what truth is.

the mathematicians are invading again

throws some infinitesimals to drive them away

no mathematician is scared by infinitesimals

smbc-comics.com/index.php?db=comics&id=2675#comic

you should try harder

$(\mathrm{d}x)^2+\mathrm{d}y=\int \otimes \mathrm{d}z$

solve for $z$ please

@yuggib @Slereah I see someone linked metamath. Any thoughts on computer automated proof systems? I kind of wanted to go through parts of euckids elements w/ hilberts axioms for geometry but it seems like a waste to learn metamath

in my master thesis, there is the formula $dx^2 = \hbar dt$

dear god

link

beware, it's in French : dl.dropboxusercontent.com/u/19940612/Science/…

wtf translate that into something sensible

like dutch?

no English or German

also

>stochastic calculus

I know nothing

$dx^2\propto dt$ is a common dirty trick in stochastic processes

-sigh- I promised myself I'd close all chat clients when working. I want to read bad arguments though :(

Basicallt it is somewhat equivalent to $\Delta x \Delta y \geq \hbar/2$

is appendix B self contained

what if you do $\Delta\to \mathrm{d}$

it is

@NeuroFuzzy A paper using an automated proof system has been accepted on annals of mathematics, so this type of proofs are starting to be more accepted by the mathematical community

then $\mathrm{d}x\wedge\mathrm{d}p\ge \hbar/2$

is that correct

IIRC the "good" demonstration is in...

"The Dimension of a Quantum-Mechanical Path"

please translate appendix B

neurotheory.columbia.edu/~larry/AbbottAmJPhys81.pdf

appendix C is autistic

so many indices

@ACuriousMind

Appendix C is because I spent 2 weeks on that integral :p

I didn't want to write "the reader can show"

the reader can, but does not want to, show

what on Earth is this even about

@NeuroFuzzy However I am old fashioned, and still prefer a human made proof ;-)

Well, when you do path integrals

Like, by hand

No cheating with fancy formulas or perturbations

You have to do stochastic integrals

@yuggib is the Riemann hypothesis true

But those are only defined up to a parameter lambda

dude I need to learn analysis already

this is getting silly

the point of that integral was showing that $\lambda = 0$ is the best choice

(This is equivalent to operator ordering in the Heisenberg formalism)

I formalism'd your mom's Heisenberg

Like $\lambda = 0$ is equivalent to $\hat{x}\hat{p}$

$\lambda = 1$ is equivalent to $\hat{p}\hat{x}$

@0celo7 A lot of people would like to know...

and $\lambda = 0.5$ is equivalent to $(\hat{x}\hat{p} + \hat{p}\hat{x})/2$

@yuggib why not just solve $\zeta(z)=0$...seems pretty trivial to me

you math people get lost in your fancy stuff

@0celo7 i prefer to see it as a "quantum mechanical problem" arxiv.org/abs/math/9811068

just calculate the inverse zeta function and do $\zeta^{-1}(0)$...

or calculate zeta - 1 and add 1

I don't even know German calculus, how do you expect me to know that

it's like

Ein, zwei, drei

Wrote a mathematica script to solve the riemann hypothesis. Solve[Zeta[x+I/2]==0, x]

Wow this program is useful

exactly

QED

go get your $1,000,000

I will give you like $10 since it was your idea.

Mathematica is pretty shit at solving equations once the functions become weird

uh I get at least 10%

I will sue your pants off, Cali boy

Over my dead body :)

you can joke as you wish, but coming from a discussion that contained the brilliant phrase "Spacetime is an abstract mathematical space which models space at all times" you do not seem so prone to interesting stuff

you can't defend yourself

that's perfectly alright with me

:-P

@yuggib if that's directed at me, I have no clue what that means

@yuggib that was @NeuroFuzzy

@0celo7 nope...neither it was you

but it was a silly argument indeed

@Danu Bombay Sapphire is not bad.

@ACuriousMind I'm meeting with Dr. Lang Research Group Gmbh. later

I need some Heidelberg stories

much sillier than Riemann hypothesis anyways

who was it @

As far as dranks go I prefer some Sobieski vodka

I still have no idea what you meany by it

@ACuriousMind no love for Tanqueray?

@0celo7 Never had it, can't tell

@Slereah vodka is shit-tier

why even drink vodka if you're not an alcoholic

We usually stick to the cheap gin, has a 1984 feeling ;)

Potatoes are magical

are you referencing the book there

or am I just dumb as usual

@0celo7 Yep, that's a book reference

@Slereah you're not Latvia

@ACuriousMind the book or some other book

@0celo7 The book. (To prevent further confusion, Orwell's 1984)

dude

your fucking memory

how

how on Earth do you remember they drank gin in 1984

Oh, come on, the phrase where Winston's tears smell like the gin they're always drinking is quite memorable.

another reason why I'm not smart I guess

welp

@Slereah why in (3.4) do you do $\langle\psi,\phi\rangle=\int \psi\phi^*$ and not $\int \psi^*\phi$

@Slereah also why is $\operatorname{det}$ not upright

@0celo7 Ah....this is mathematicans' convention :-P

you did not even upright $\ln$

shameful notation

@yuggib my ass is as crystal clear as mathematical notation

@0celo7 well...either your ass is pretty clear, or I do not see that much of a difference, given the evident graphical symmetry of the notation

please explain

is $c=1$ or something

@0celo7 you can write either :) or (: but I do not see that much of a semantic difference

(: is wrong

this is clear from 0celo7's theorem

I agree...but everyone is entitled to his opinion

no, it's a fact

why are there like 5 integrals in this paper

this is unreadable, it's in some foreign language

@Slereah No hyperref? Shame on you!

I hate unclickable tables of content.

I hate the way you do them :/

those red boxes are ugly

@0celo7 I have switched to doing them without boxes, and just tint the text blue

@ACuriousMind yay

But back when I wrote what you're referring to, I just didn't know one could adjust that :D

you used to be so naive

back when your QM innocence was intact

Lol, that's a long time ago

@0celo7 It describes how a fractal curve looks infinitesimally, basically

you probably took QM in 11th grade

@Slereah please explain

It's an infinitesimal version of a stochastic process where, for every step forward you take in t, you have a standard deviation change in x

Except this is taken to the continuous limit

I need equations

Well $dx^2 = dt$ isn't much of an equation, unfortunately

It's more of a cheap trick

@0celo7 I think the QM course that finally taught me actual quantum mechanics instead of handwaving everything with "wavenature" and "double slit" was two and a half years ago.

if you want the serious version, try en.wikipedia.org/wiki/It%C5%8D_calculus

But since in path integrals you usually take a limit, it's not too problematic

:-)

@Slereah Stochastic calculus and Ito integral have physical significance only in euclidean time. And it is not always possible to rotate back...

@yuggib : Tell me about it.

That is why path integrals in QM tend to be poorly defined

@Slereah You could say undefined (at least in a rigorous fashion)

but well

It works

And you can annoy mathematicians with it

@0celo7 Why don't you tell him something interesting about yourself

@Danu I did

we talked for a solid two hours on Friday

in Denglish

@Slereah :-D sure it works, and it is not so annoying for a mathematician

since it is mostly used as an elegant notation, but a limiting procedure is always intended

tell me what annoys mathematicians, then

I must know

What abuse of notation or sloppy use of math annoys them

@ACuriousMind Despiable

@ACuriousMind n00b ;D

I'm way into texing my stuff nowadays

Produce an infinite collection of sets A1,A2,A3,... with the property that every Ai has an infinite number of elements, Ai ∩ Aj = ∅ for all i̸=j,and􏰈 union∞i=1Ai =N.

wat

I had a professor who did $\pi \approx 2$ and $\pi^2 \approx 10$

The downside to text-color is that it messes up black-white printing.

@Slereah :-D I am very annoyed by he fact that in QM physics papers the Hilbert space is never explicitly stated

@Slereah That's pretty dumb. I hope you mean $\pi\approx 3$, at least :P

@Slereah ah, a fellow engineer

@Danu Nope, 2!

@Danu not with dark blue

@Slereah That's pretty dumb :D

Because 2 you don't need to use a calculator to do :p

@yuggib I doubt that

I can't do 2^34 in my head

@yuggib : Usually it's $L^2(R^3)$!

I need a calc for that

@Slereah If you consider spin it isn't; if you consider QFT it isn't; if you consider 5 particles it isn't

True

Hm

What is it for spin

Like just some dumb Pauli spinors

@Slereah it depends, a particle with one-half spin is usually represented in $L^2(\mathbb{R}^3,\mathbb{C}^2)$

Makes sense

from what I recall things get a little weird once you do gauge fields

another thing that annoys mathematicians are sentences like "complete orthonormal basis of eigenvectors" applied to operators with continuous spectrum

where is the problem located

complete?

Oh wait

Is it orthonormal

Because it's a dirac delta

and not a kronecker

no, there are no eigenvectors corresponding to continuous spectrum

whaaaat

details plz

@yuggib They're all isomorphic anyway ;) [I know the isomorphism is rather useless, I guess the abuse of that statement also annoys you]

@ACuriousMind ahahahahah exactly

@Slereah The "eigenvectors" do not belong to the Hilbert space, you need the rigged space to talk about them.

@Slereah $(H-\lambda)$ is not surjective but injective on continuous spectrum

Oh right, that thing

@ACuriousMind It is ok for deltas of position/momentum, but it is also not true in general

Wait, coherent states form an orthonormal basis and they belong to the Hilbert space

They are overcomplete, but still

rigged hilbert spaces also are quite annoying to me

@Slereah they are eigenvectors of a non-self-adjoint operators anyways

well you didn't specify that :p

@Slereah true...

and actually, they are also not orthogonal

$\langle\beta|\alpha\rangle=e^{-{1\over2}(|\beta|^2+|\alpha|^2-2\beta^*\alpha)} \neq \delta(\alpha-\beta) $

true

damn

I admit defeat

:-D

@0celo7 Hint: $2^{10}\approx 1000$

Hence $2^{34}\approx 16,000,000,000$

And the error estimation is pretty simple too, based on the intuitive picture of a square

(the good ol' $(x+dx)^2\approx x^2+2xdx$)

So we have a chat session coming up...

half an hour. I'll be back

@DavidZ Woohoo! Finally one that I won't miss

yay

GR chat session I'm assuming

:p

Preemptive just-for-fun possible topic, SR not GR: Does anyone else find the Mobius Transformation & Riemann sphere with its striking similarity to quantum formulation delightfully unexpected? It's just not intuitive me in any way that such an approach would capture so much of SR so neatly. mathpages.com/rr/s2-06/2-06.htm

@TerryBollinger correct me if I'm wrong but is that covered in chapter 1 of Penrose and Rindler "Spinors in Curved Spacetime"?

Wikipedia en.wikipedia.org/wiki/M%C3%B6bius_transformation

oh wow that's a detailed page

thanks!

@FenderLesPaul Very likely, but I don't have that one. I've been poking at the Dirac Equation lately and wow, has my profound respect (already high) for Dirac increased. Those whacky little spinors, who would have thought...

@FenderLesPaul Looks like it

A lot of fascinating differential geometry is attached to it too

@Danu sweet thanks

@Danu cool link, thanks!

I'm too mechanical, I keep thinking of spinors in terms of 1/2 gear ratios... :)

(be back in 10)

@TerryBollinger lolwat?

let's get in a good mood! :D

@Danu Eh??

@TerryBollinger You no likey?

@Danu I likey, what a wonderful world!

On that note,

"Don't know what a quaterion is for..."

is anyone here familiar with affine collineations in GR i.e. vector fields $\xi^a$ such that $\nabla_c \mathcal{L}_{\xi}g_{ab} = 0$?

I'm reading a paper that says these vector fields preserve the affine structure in the sense that $\nabla_a$ and $\mathcal{L}_{\xi}$ commute on any tensor field

@TerryBollinger But I do know that $SU(2)$/will always fascinate me & you

Where's John Rennie? I see a faded image...

but for the life of me I've tried for hours and hours and can't prove that

so if anyone knows how to show that I'd be much appreciative :)

What a wonderful world, this must be!

(hey, that actually came out quite nicely!)

@Danu True!

@yuggib : if you'd like to discuss something with me, feel free.

@TerryBollinger It's not unexpected, that the proper orthochronous Lorentz transformations are the Möbius trafos is just not as well known as it should be. It's actually a more general phenomenon - the algebra of conformal transformations in $d$ dimensions at signature $p,q$ is $\mathfrak{so}(p+1,q+1)$ (the Lorentz algebra of signature $p+1,q+1$, and the Möbius trafos are the conformal trafos of $2,0$

@FenderLesPaul Did the papers have no refs to earlier work? Sounds like they considered it a given, if I read you rightly.

(Slight cheat because the conformal trafos in 2D are actually larger, but the Möbius trafos are still in some sense the "right" conformal trafos because they are globally, not locally defined)

@ACuriousMind It's more just a beauty thing... I love the connectedness of it...

it was just stated in passing

@FenderLesPaul Hate it when that happens, it can be a bear to dig it back out... sometimes associated papers by the same authors... dig dig...

@TerryBollinger or keep trying it myself until I am forced to run around in circles and pass out

:)

@ACuriousMind Also, that is just the sort of thing that to me should be better known and more widely taught. Sometimes the randomness of what gets emphasized most is a bit distressing.

I've actually wondered whether there is a deeper reason that the Lorentz algebra of $3,1$ is one of the few Lorentz algebras that is not the full conformal algebra of two dimensions lower, but I didn't find a way to make that into a sensible question.

@JohnDuffield No thanks, but I have indeed fun in seeing you arguing with the others.

I like to see when strong opinions clash :-D