The h Bar

Emilio Pisanty

7:28 PM

oh yes i am really sorry i didn`t see the question thoroughly — Avnish Singh 1 min ago

::facepalm::

Lozansky

What is meant by "classical point of return" for a harmonic oscillator? :O

Semiclassical

the turning point

Lozansky

Isn't that the amplitude?

Semiclassical

it's also that, yes.

point is that, if you know the total energy of the harmonic oscillator, then it'll oscillate back and forth in [-A,A]

so x=A is where the classical trajectory stops and turns around

Lozansky

Well, in this case I only know $\omega$ and $m$ for a Weber rod and I am supposed to find the turning point in the ground state.

And iirc $\psi_0 = (\dfrac{m \omega}{\pi \hbar})^{1/4} e^{-\dfrac{m\omega}{2\hbar}x^2}$

Wait, I'm also given the length of the rod

So normalizing and finding the amplitude is probably a good idea

This is so wrong

Sir Cumference

7:49 PM

Sigh

At some point I just need a break

Lozansky

@Semiclassical If the oscillating rod is of length $L$, would solving $1=|A|^2 \int_0^{L} \psi_0 ^2 dx$ for $A$ give me the turning point?

Semiclassical

No. You get the turning point from you knowledge of the total energy of the system

Which you definitely know for the ground state

Lozansky

Yeah $1/2 \hbar \omega$

Semiclassical

The other thing that’s useful is that the probability density of finding the particle at a given spot should be inversely proportional to its velocity there

(the longer the particle stays in a given interval, the more likely you are to find it there)

One important point of comparison here is that the wavefunction is nonzero everywhere in space. In particular, it’s nonzero for x beyond the classical turning point

Lozansky

How can it be nonzero outside the rod?

Semiclassical

8:00 PM

So there is a small but nonzero probability of finding it outside the range you’d expect for a classical harmonic oscillator

Because the harmonic oscillator you solve in QM doesn’t know about that length

if you wanted to enforce that, you’d need to set V=infinity for large enough x

But for the harmonic oscillator you only have V=infinity at x=infinity rather than any finite distance

As such, the ground state wavefunction is nonzero everywhere

It becomes exponentially small as you move away from the origin, so the probability gets small very fast

But there’s no sharp cutoff

bolbteppa

Is there some online program where you can draw crappy pics and it will output the latex code, e.g. like you make your own inclined plane drawing with paint and it turns into code

Lozansky

Okay so should I try to find the expected value for the velocity? @Semiclassical

The revival time is $T=4mL^2/(\pi \hbar)$ so maybe if I multiply $T$ with $\dfrac{1}{m} \langle p \rangle$ I get something useful lol

bolbteppa

oh my god there is

mathcha.io

2

A: Online LaTeX diagram graphical editor

A wonderful option is mathcha: www.mathcha.io just open the editor, you log in and you make your drawings and equations. this online program will create different export formats, including Likz, which is compatible with Latex.

can convert it to tikz

Semiclassical

8:21 PM

@Lozansky no. You’re trying to work out the probability distribution for the positions of a classical harmonic oscillator, yes?

Lozansky

@Semiclassical I just want the magnitude of the classical turning point in the ground state :>

Semiclassical

Ah. Again, though, that’s just the amplitude

Main thing is that the classical turning point doesn’t know about the wavefunction

Classically, what you’ve got is stuff like energy conservation

Lozansky

Yeah, but isn't the amplitude $|\psi_0|^2 = (\dfrac{m\omega}{\pi \hbar})^{1/2}e^{-m\omega/\hbar x^2}$?

Semiclassical

No. That’s a probability density, for one thing

For another, when I say amplitude in reference to a classical harmonic oscillator I mean precisely “how far does it move from the origin before it stops and begins moving the other way”

Lozansky

Yeah, if the oscillations are given by $\psi_0 e^{-i\omega t/2}$ then $|\psi_0|^2$ would give me the amplitude, no?

Semiclassical

8:29 PM

again, no

Lozansky

Hmm okay

Semiclassical

A classical turning point is where v=0

Lozansky

Sure

Semiclassical

All it knows about is the potential energy

It knows nothing about the wavefunction

Moreover, the units of what you’re saying don’t make sense either

A classical turning point is a location

DanielSank

Hi, everybody.

Semiclassical

8:32 PM

Psi^2 is a probability density and has units of 1/length (in 1D)

Lozansky

Yeah

DanielSank

More accurately, Psi(x)^2 has dimensions of 1/length.

You can write wave functions of things other than position.

Important to keep in mind.

@EmilioPisanty Confirmed: It's now illegal to post answers on Stack Exchange unless they get a high score.

Lozansky

But I don't know anything about the potential, do I?

Semiclassical

What, exactly, are you given then?

Lozansky

I mean $1/2 \hbar \omega$ is the energy of the system?

Semiclassical

8:35 PM

The potential energy of a harmonic oscillator is 1/2 m omega^2 x^2

So if you know the mass of your particle and you know omega, you know the potential

Lozansky

Hm OK, and so we solve for $x$ in $1/2 \hbar \omega = 1/2 m \omega^2 x^2$ (all energy is potential)?

Semiclassical

At the turning point, yes

Lozansky

@Semiclassical So the length of the rod doesn't affect the turning point?

So in effect, I could have a rod smaller than the turning point...

Semiclassical

To the extent that you take the length of the rod as being independent of the frequency, yes

On the other hand, you could take the ends of the rod to be the classical turning points, and infer the frequency from that

So in that setup they’d coincide.

eulB

8:55 PM

@SirCumference play the drums

Emilio Pisanty

@DanielSank how so?

Ah. I missed the pointer.

@DanielSank you mean good answers?

Lozansky

Okay so the rod absorbs some energy

And jumps from energy state $E_n$ to $E_{n+1}$

Then the change in amplitude should be $\Delta x = \sqrt{\dfrac{2\hbar}{m\omega}}(\sqrt{n+3/2}-\sqrt{n+1/2})$

How can I rewrite this in terms of $\Delta x = C /\sqrt{n}$? :O

Semiclassical

Take n to be large probably

Lozansky

Yeah, that'd do it I guess =)

G. Bergeron

@bolbteppa Me too and big time!

Semiclassical

9:03 PM

Which is really just the correspondence principle in action

0celo7

9:14 PM

@BalarkaSen ok revisiting the issue

are you here

Balarka Sen

no

i am over there

0celo7

@BalarkaSen Let $M$ be oriented, and let $D$ be the PD map $H^k\to H_{n-k}$, ok

Balarka Sen

OK

0celo7

Is PD nice in the sense that $D(\alpha)\frown\beta =D(\alpha\smile\beta)$?

Semiclassical

PD = Poincare dual ? (Just curious)

0celo7

9:17 PM

yeah

Semiclassical

Kk

Emilio Pisanty

@0celo7 what's \frown?

cap product

what

In algebraic topology the cap product is a method of adjoining a chain of degree p with a cochain of degree q, such that q ≤ p, to form a composite chain of degree p − q. It was introduced by Eduard Čech in 1936, and independently by Hassler Whitney in 1938. == Definition == Let X be a topological space and R a coefficient ring. The cap product is a bilinear map on singular homology and cohomology ⌢ : H p ( X ; R ) × H ...

Emilio Pisanty

9:18 PM

@0celo7 oh god that's horrifying

Balarka Sen

I think this follows from $D(\alpha) = \alpha \frown [M]$ and how cap product interacts with cup product.

Semiclassical

Does that imply a certain asymmetry between A and B, or am I reading too much into that

0celo7

@BalarkaSen Right I don't know 100% how they interact

hence the question

Balarka Sen

I know that at the chain/cochain level, $\psi(\varphi \frown \sigma) = (\varphi \smile \psi)(\sigma)$

Hm.

Semiclassical

If D(a)-cap-b = a-cap-D(b) then my remark is superfluous

0celo7

9:19 PM

@BalarkaSen that's the definition, no?

@Semiclassical I'm being sloppy with the order

Balarka Sen

There are many definitions

Semiclassical

Ah

0celo7

what I said is probably only true mod 2 as written

i think I have to switch some things around to get the right sign

Semiclassical

Confusing

Emilio Pisanty

so

is this an HNQ in the making?

1

Q: Why are the $\mathbf E$ and $\mathbf B$ fields of an electromagnetic wave mutually perpendicular?

Why are the wave number $\mathbf k$ and the electric and magnetic fields $\mathbf E$ and $\mathbf B$ are perpendicular to each other? I know it but I haven't thought about it deeply. How can I prove this conclusion mathematically?

Semiclassical

9:24 PM

Would circularly polarized light be a counterexample? I’m forgetting

G. Bergeron

@EmilioPisanty ?

@Semiclassical no

Emilio Pisanty

@G.Bergeron Hot Network Question

Semiclassical

Fair enough

Emilio Pisanty

the ones that get pushed on the sidebar on the right across the whole of SE

sometimes known as the Stack-Exchange-Wide Advertisement for General Enlightenment list

Semiclassical

Aka “protect question incoming” :p

G. Bergeron

9:25 PM

@EmilioPisanty Seems a bit over the top for the question, no?

Emilio Pisanty

@G.Bergeron it's more of a systemic, software issue

Balarka Sen

@0celo7 The right order should be $\beta \frown D(\alpha) = D(\alpha \smile \beta)$, by checking for $\alpha = pt \in H^0$.

Semiclassical

Well, 5 answer in 40 minutes

Emilio Pisanty

the most cogently I've managed to articulate my views on the matter is here.

G. Bergeron

@BalarkaSen I hate that I can't run mathjax properly at the office

It's seems like a very ''dramatic'' part of math from that non-mathjax vantage point... :p

Emilio Pisanty

9:27 PM

@G.Bergeron what?

Balarka Sen

its annoying to not being able to render mathjax

Emilio Pisanty

are they blocking the mathjax CDN?

0celo7

@BalarkaSen So you think it's true in general?

Emilio Pisanty

you should probably be able to provide the script locally

G. Bergeron

@EmilioPisanty I don't have admin privileges, it's an old browser on centos...

0celo7

9:28 PM

If not, I can say that $\alpha$ is a cup product of 1-chains

Emilio Pisanty

@G.Bergeron yeah, figures

0celo7

and $\beta$ another 1-chain

Emilio Pisanty

on chat you may be able to change where the bookmarklet pulls the script

on main that's probably harder

Balarka Sen

@0celo7 Give me a minute to do the computation. I think $\alpha \frown (\beta \frown \gamma) = (\alpha \smile \beta) \frown \gamma$.

Let me see if I can do this.

G. Bergeron

@EmilioPisanty similarly, the tikz package for commutative diagram (tikz-cd) is not available. Trying to simply put in the same directory led to a several hours long dependancy chase only to realize that the version of pgf on our tex build is not compatible with tikz-cd

-_-

@EmilioPisanty Could probably make it work, there is a fair amount of lazyness at work here...

Emilio Pisanty

9:31 PM

@G.Bergeron yeah, if it's a system you control and a known fix then that's one thing

but figuring out how to get an old browser do things it wasn't really built to do on a machine with no sudo, definitely another order of magnitude

G. Bergeron

@EmilioPisanty Yes, but more simply, I could probably just run the script on the javascript console

Emilio Pisanty

I'm at a similar sudo-less desktop at work and that can indeed be a pain

G. Bergeron

@EmilioPisanty Yes, and wait for it.... WE ONLY HAVE 500 MB 8(

0celo7

@BalarkaSen Oh, Hatcher has the formula way back in the back of the chapter

Emilio Pisanty

thankfully it's a mostly fully-kitted-out texlive (I think bundled with ubuntu 16.04?) but at some point IT just utterly failed at installing a font I ~~needed~~ wanted because texlive just doesn't have it.

G. Bergeron

9:34 PM

That, coupled with the fact the every damn unix applications wants to put their stuff on home is a major PIA

Like my dropbox!!!

Emilio Pisanty

oh, and also

booom! rep-cap

cc @ACuriousMind

G. Bergeron

Then you have to mess around with symlinks and stuff which dropbox definitively does not like

@EmilioPisanty rep-cap? Reputation is capped?

0celo7

@BalarkaSen I have another question for you

Emilio Pisanty

... unless somebody unupvotes-then-upvotes a post of mine a minute before and a minute after GMT midnight

@G.Bergeron daily rep is capped

0celo7

@BalarkaSen actually 2 more

Balarka Sen

9:36 PM

@0celo7 Where in Hatcher

What are the questions

Emilio Pisanty

200 max, with accepts and bounties exempt

you get a badge if you hit it

0celo7

259

Emilio Pisanty

and then a gold badge if you hit it 150 times

G. Bergeron

@Semiclassical btw, I loved the greek quote you posted! Especially in the context it was posted! I was becoming quite exasperated before I just quit! :P

Semiclassical

lol

I wondered who all would get that :P

Balarka Sen

9:38 PM

i dont have the hard copy w/ me rip

Semiclassical

(I'm personally fonder of the bit of Kant I quoted here, though it's not nearly as pithy)

0celo7

@BalarkaSen Suppose I have $\alpha_1,\dotsc,\alpha_{n-1}$ classes in H^1

denote the total cup by $A$

Balarka Sen

OK

0celo7

what does it mean for D(A) to be nonzero

G. Bergeron

@Semiclassical I already rage-quit by that time. Just saw it 5 minutes ago.

0celo7

9:38 PM

geometrically

Emilio Pisanty

@Semiclassical there's this thing called "google"

it's really quite good

Semiclassical

lol, true

0celo7

and why does, $M\# T^n$ always have this nonzero

Semiclassical

I guess I should more say: I wondered who all would bother to look that up :P

0celo7

or does it?

G. Bergeron

9:39 PM

@EmilioPisanty Yeah, but had I not starred it at that point, would you have noticed it?

Emilio Pisanty

@G.Bergeron nope

Semiclassical

lol

0celo7

or maybe can you find classes for which that's true

I'm not sure what the right phrasing is

Emilio Pisanty

@Semiclassical c'mon, Greek on the star board?

G. Bergeron

@EmilioPisanty Let's just say it was really on topic.

Semiclassical

9:40 PM

I mean, if one wants to poke fun at certain parts of modern physics with regards to how speculative and out there it can be, I can't say I entirely object

e.g. the struggle of string theory / SUSY to find something that can actually be empirically validated

G. Bergeron

@Semiclassical What pissed me off was the dismissive and quite condescending attitude AFTER I took the time to seriously read and analyze the paper

Semiclassical

yeah

why I like the Kant quote is that it captures how important it is to have stuff to push against when it comes to scientific questions

Balarka Sen

@0celo7 For the second question, I think the PD of the generating classes of the meridians of $T^n$ have that.

$H_*(T^n)$ is an exterior algebra on those

Semiclassical

without that, you're more likely to convince yourself that you're making progress than to actually do so

G. Bergeron

@0celo7 I need to get hands-on practice with those cup and cap products... They're pointing their noses left and right in a few talks I attend.

@Semiclassical Indeed

Semiclassical

9:45 PM

I'm fine with being poetic. But one shouldn't conflate that with the actual labor of mathematical/scientific investigation

2

0celo7

@BalarkaSen but even when you connect sum with M?

Balarka Sen

@0celo7 The ring structure doesn't change. There's a map $M \# T^n \to M \vee T^n$ given by punching the sphere along which you took the connected sum. This gives a map $H^*(M \vee T^n) \to H^*(M \# T^n)$

G. Bergeron

Should I put my face in that pink square beside my name?

0celo7

what is * in M*T^n

Balarka Sen

Wedge sum

0celo7

9:47 PM

why not \vee

whatever

is that map an isomorphism?

Balarka Sen

cuz I forgot notation whoops

0celo7

@BalarkaSen I don't get the point of this comment

G. Bergeron

Anybody have a clue on how to generalize the coordinate ring (seen as a formal dual) of Lie group to the Weyl group?

From what I tried, it seems to lead to something infinitely generated, which in that form, is not that useful...

0celo7

@BalarkaSen I think the (co)homology in this paper is done really shittily and not correctly

Emilio Pisanty

@G.Bergeron that's unfortunately par for the course with that user

Semiclassical

10:00 PM

relevant: chat.stackexchange.com/transcript/message/43967407#43967407

G. Bergeron

@EmilioPisanty Yeah well I learn my lesson and will never again engage seriously... -_-

learned?

Emilio Pisanty

So, you know, engage at your peril. Can be worth it if the paper is good, but unfortunately you can't count on a proper engagement in return if you do.

@Semiclassical there was an excellent bit on this subject in A serious man

G. Bergeron

@Semiclassical Yeah, but poor debate is different than calling every physicists clueless and telling me I have low self-esteem. Like, ok, I was just trying to help.

Balarka Sen

@0celo7 Sorry I ran away for a while. That map is an isomorphism in grade less than $\dim M + n$, me thinks

Semiclassical

I'm fine with people making portentous statements tbh so long as they acknowledge that they're not a demonstration

Emilio Pisanty

10:02 PM

Roughly saying "the math is what counts, the stories are just stories"

Balarka Sen

I don't have a quick argument though

0celo7

@BalarkaSen I'm going to restart this conversation when I get through the rest of the paper

G. Bergeron

@EmilioPisanty :)

Emilio Pisanty

Unlikely to be youtubeable though

0celo7

just looking at the rest of the proof, it seems very shady

no need spending time on that now

Balarka Sen

10:03 PM

Mmmk

Emilio Pisanty

It's a great film and well worth watching in any case

G. Bergeron

@EmilioPisanty about?

Emilio Pisanty

@G.Bergeron hard to define

Life?

I guess

Semiclassical

It's not a movie about a specific plot from what I know about it

I mean, there are certainly events and themes

Emilio Pisanty

Yes

Semiclassical

10:05 PM

It's a Coen Brothers movie, for reference

G. Bergeron

I'll go check the trailer, anyway I've slept too little to be really efficient in my work now. Damn those early dentist appointment...

Emilio Pisanty

@G.Bergeron I don't know if the trailer will do it justice

Balarka Sen

Was the greek quote conversation with le epic troller and paradigm shifter

Semiclassical

The chalkboard might

G. Bergeron

@BalarkaSen probably?

Balarka Sen

10:07 PM

That's one annoying fellow

0celo7

who is le epic troller

G. Bergeron

@EmilioPisanty Well, there was a dentist in it! :p But, yeah, it's not one of those crazy post-modern movie like that one ''Pi''?

Balarka Sen

@0celo7 the Very Ziggy Nun

0celo7

Milo?

I don't get it

Semiclassical

I wouldn't say it's crazy post-modern from what I know of it

Balarka Sen

10:09 PM

you're thinking too hard

vzn

Semiclassical

though I think there is a certain magical realism to it

0celo7

I don't think he's a troll...

Semiclassical

Eh. I find him more frustrating than trolly

Balarka Sen

I think he's a crank but that's not Nice

G. Bergeron

@0celo7 Doubt it

Semiclassical

10:10 PM

(mr last letter of the alphabet over in the math chat, by contrast, is one of the few people I've actually Ignored in a while)

0celo7

I think Balarka is projecting tbh o_O

G. Bergeron

@Semiclassical Ok ok

Balarka Sen

Accurate impression

@Semiclassical I kinda like that guy

Semiclassical

I very much don't.

0celo7

please delete that

David Z

10:11 PM

@BalarkaSen that's also not nice

Balarka Sen

@DavidZ What if I haven't specified who is an idiot

G. Bergeron

@BalarkaSen Are you the kind of person going through youtubes comments to find the worse? :P

Balarka Sen

That guy over there is an idiot

Is that not nice?

@G.Bergeron Lol that's kind of spot on

David Z

@BalarkaSen it doesn't really matter. Though of course none of this is as bad as saying "you're an idiot" to someone in the chat.

G. Bergeron

@BalarkaSen I do that sometimes. I call it ''ethnological field trip''

Balarka Sen

10:13 PM

@DavidZ I was joking around a little. I deleted it.

Semiclassical

I think that's a pretty silly standard, tbh. But it's one which is consistent with what's been said before.

G. Bergeron

@DavidZ Are there written rules to this chat somewhere?

Balarka Sen

LOL

Semiclassical

I don't mind people who I disagree with, but I do mind people who (intentionally or otherwise) invalidate my or other people's experiences

Balarka Sen

Well I think it's mostly a sign of shallow thinking skills to have radical views about most things

David Z

10:14 PM

@BalarkaSen Cool. This one wasn't such a big deal though, because of its abstractness... I think deleting it is good choice, but not strictly necessary.

Semiclassical

The person in question doing so was what moved him from "tolerate" to the "ignore and forget about" category

David Z

@G.Bergeron good question :P

Balarka Sen

Which is what mr last letter does

0celo7

@BalarkaSen Given $\alpha\in H^1(M,Z)$, can one always find a smooth hypersurface in the class of $D(\alpha)$? I was under the impression this is not possible with integer coefficients.

G. Bergeron

@DavidZ I am sometime not sure where is the line, not necessarily in cases as above, but also, say, about acceptable topics, etc.

Balarka Sen

10:16 PM

@DavidZ Yup, I understood :) I intended to delete it in a minute or so actually

David Z

In general the expectations for people's conduct are pretty similar to what is in the help center, e.g. don't be rude, do be accommodating of other people's views. But it's not as clearly specified as on the main and meta sites.

G. Bergeron

@BalarkaSen From what I'm currently listening: ''Dude, your mixes are out of this world, I feel like I'm travelling through a robots intestines.''

David Z

@G.Bergeron As far as topics, in general anything is fine to discuss here, though it's best to stay away from certain topics that are widely considered not suitable for public conversation (e.g. sex, that's come up in the past I think).

G. Bergeron

@DavidZ So ''PG13''?

David Z

Something like that

Semiclassical

10:20 PM

@BalarkaSen My only regret from my last interaction with them was not linking this: penny-arcade.com/comic/2006/04/10/i-hope-you-like-text

0celo7

@BalarkaSen please, the suspense is killing me

Balarka Sen

@0celo7 I am multi-talking!!!

0celo7

I need to go eat, but have to know the answer to this first

Balarka Sen

Ok so $D(\alpha)$ is an element of $H_{n-1}(M; \Bbb Z)$

0celo7

yes

Balarka Sen

10:22 PM

I believe it's true that every integral homology class in codimension 1 can be represented by a submanifold

0celo7

oh really

amazing

oh yeah

G. Bergeron

I don't like homology in Z...

Balarka Sen

36

Q: When is a Homology Class Represented by a Submanifold?

Possible Duplicate: Cohomology and fundamental classes Given an oriented manifold $M$ and an oriented submanifold $\phi:N\to M$ we can obtain a homology class $\phi_*[N]\in H_*(M)$ where $[N]$ is the fundamental class of $N$. In general, it is not true that every hom...

at.algebraic-topology gn.general-topology

0celo7

the canonical MO answer says that

Balarka Sen

yup

0celo7

10:24 PM

not that I understand the proof

Yau lives on for another day

Balarka Sen

Lol

It takes more work to unwrap the proof given

I tried once

0celo7

I need to understand it

it must go in the thesis

chapter 6 of the thesis is alg.top $\cap$ GMT

the horror

Balarka Sen

I can recall and shoot the proof at you some time after

@G.Bergeron Amazing

G. Bergeron

@BalarkaSen Not that it is particularly surprising, it's a psytrance mix.

Balarka Sen

Wanna link me?

ACuriousMind

10:26 PM

@EmilioPisanty It sure sounds like it, but I don't think it is an actual German word.

Sir Cumference

1667

Astronomy – astronomy.stackexchange.com

Beta Q&A site for astronomers and astrophysicists.

Currently in public beta.

Jesus

Just 19 visits a day?

0celo7

@SirCumference most people dislike astronomy, no?

G. Bergeron

@BalarkaSen Doubt you'll like... but ok youtube.com/watch?v=1cyVgOdXVPc

0celo7

not bashing it, most people dislike my field too

not that I blame them

Sir Cumference

@0celo7 If you mean the general public, I'd imagine they are far more interested with the stars than mathematics

I mean no one writes pop-math books like they do pop-science

G. Bergeron

10:28 PM

@0celo7 algebraic topology? If yes, then I don't see people disliking it!

0celo7

I think the average person is concerned with the waste of money that telescopes are

@G.Bergeron I'm no algebraic topologist

I am just dealing with homological variational theory right now

Sir Cumference

C'mon, astronomy is probably the most interesting branch of physics to the general public. The vast majority of pop science books are based on it

Semiclassical

I think the average person is more interested in astrology than astronomy unfortunately

2

G. Bergeron

@SirCumference Probably, though it wasn't the case for me.

@Semiclassical LOL

Balarka Sen

@Semiclassical Hah I saw that comic from earlier

It's 100/10

0celo7

10:29 PM

I am going to have a 2-page definition. Ugh.

Sir Cumference

@Semiclassical "There's a difference?"

Semiclassical

@SirCumference .…

Sir Cumference

@Semiclassical Note the quotation marks :P

Semiclassical

Ow

Yeah, fair

I think astronomy has a better shot at public consciousness than a lot of other disciplines though

Sir Cumference

Tbh when I mention studying astronomy, the most common questions are "can you read my fortune?", "so you must be a fan of Neil deGrasse Tyson, right?" and "what's the seventh planet in the solar system? durr durr durr"

Balarka Sen

10:32 PM

@G.Bergeron This sounds like some cyberpunko-futuristic shit

G. Bergeron

@BalarkaSen Is it simpler if one considers $H_k(M;\mathbb{R})$?

@BalarkaSen It's called psytrance, short for psychedelic trance. You know dirty hippies on a beach in Goa... ore rather what evolved from that.

Balarka Sen

Lol

G. Bergeron

@SirCumference What about Planet X or nibiru stuff?

Sir Cumference

@G.Bergeron Most people have never heard of those hypotheses

G. Bergeron

@SirCumference Or where was the moon landing film after all?

Balarka Sen

10:34 PM

I have heard some trance music of various sorts, so I could catch that this falls into the genre

G. Bergeron

@SirCumference I though it was all the rage with that 21 December 2012 end of mayan calendar.

Sir Cumference

@G.Bergeron Well you get people like this

G. Bergeron

@SirCumference It's a new language base around ''not'' ''very'' and ''good''. It turned out to be not very good.

2

Sir Cumference

Yep sounds right

Balarka Sen

@G.Bergeron This has some of my favorite comments of all time

"When I grow up, I want to be a stuffed capsicum"

G. Bergeron

10:39 PM

@SirCumference On topic: When we were little, my father showed us a documentary about the fake moon landing that was well made and all. The goal of the entire exercise was to show us what bullshit looks like with a nice coating of paint.

vzn

@0celo7 lol thx for the ringing endorsement :P

Balarka Sen

"I am a zygote and i love this kind of music. I don't understand why my fellow pre-embryonic eukariote colleagues only listen to chart music but i've been told i'm an old soul"

G. Bergeron

@BalarkaSen Poetic fail to my ears...

Semiclassical

@G.Bergeron to clarify: the moon landing was BS, or a documentary trying to debunk it was BS?

DanielSank

@EmilioPisanty Yes.

G. Bergeron

10:42 PM

@Semiclassical Don't worry, the documentary was. :p

Semiclassical

Oh good

vzn

@G.Bergeron many thx for reading the paper & the analysis/ discussion (on the other hand, its not mine...). fyi the snarky comments were performance art, sorry you took them seriously/ personally

Semiclassical

“We are what we pretend to be, so we must be careful about what we pretend to be.”

vzn

@Semiclassical yeah and theres also this thing called projection identification in psychology... o_O

G. Bergeron

@vzn Look, I don't hold a grudge. It's just the entire exercise seemed like a waste of time in the end.

In any case, I have to leave for home

See you people

vzn

10:49 PM

@G.Bergeron am sorry to hear/ disappointed you think it was a "waste". not sure what you expected. sorry the discussion and/ or paper did not meet your expectations. enjoyed it myself at time but will have to reevaluate that feeling now.

Balarka Sen

theres also this thing called nAmEdRopPing in conversations OoOooo_oooOOo

vzn

← yeah lots of "droppings" on my name in here lately :(

0celo7

11:07 PM

@vzn you're the trump of the hbar

Emilio Pisanty

@vzn maybe you should evaluate whether that's because you treat interactions with others as "performance art" (?) instead of actual scientific discussions

0celo7

@BalarkaSen Is this MO thing really that hard to make precise?

I guess it's not clear why the preimage represents the homology class?

Emilio Pisanty

170

A: I believe I have solved a famous open problem. How do I convince people in the field that I am not a crank?

First, make sure you are not really a crank before trying to convince others. Read these common characteristics of cranks. If they apply to you then get professional help. For the rest of the answer I will assume that you have really solved a famous open problem. In the following "he" refers...

Cf. in particular point 4 in this answer

> He does not understand it is not a puzzle.

We do care about physics.

0celo7

11:22 PM

@BalarkaSen this is Theorem VI.11.16 in Bredon

good enough for me

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