The h Bar

@Danu I'm primarily interested in functions of the form $F(x,y,z) = \det (z\mathbb I - M(x,y))$ where $M(x,y)$ is a square matrix with analytical entries

Cows

3:05 PM

@0celo7 The Pensyvania supreme court decided there was nothing. Even the prosecution refused to move on it.

@0celo7 the legal community is planning on coming up with creative ways of firing the judge involved

@0celo7 He left jail like the boss into a quarter billion dollar car, then into a hellicopter with a billionaire, then there were women all over lined up waiting for him at his crib. Total boss lol

Emilio Pisanty

as I understand it, in 1D there's plenty of results on that line, so you look at e.g. $S=\{(z,w)\in \mathbb C^2 : \det (z-M(x))=0\}$ where $M(x)$ is $n\times n$, which in a Bloch-function band-formalism perspective has $n$ bands and $n-1$ gaps, and in some circumstances one can show that $S$ is a manifold of genus $n-2$ or something

I'd like to take that stuff up one dimension

Cows

@0celo7 so the rumors are . . . . . the cop was dirty and conspired. Second the judge did not get the D from him at some point so . . . .

hehe

presumably the judge also wanted him to do a b2m remake and he refused. So she got angry

Presumably she decided to go after his producer, his manager then him

@0celo7 thegrapevine.theroot.com/…

0celo7

@Cows this sounds like an Alex Jones theory

Cows

@0celo7 it sounds so except it happens to be true which is crazy lol

@0celo7 It also helps that Meek's lawyer was en.wikipedia.org/wiki/Joe_Tacopina

Joe Tacopina

Joseph "Joe" Tacopina, is an American-Italian lawyer and chairman/owner of Venezia F.C.. He was born and raised in Manhattan to Italian immigrants. He and his family currently split time between living in Italy and Dyker Heights, Brooklyn. == Lawyer == Tacopina graduated from the University of Bridgeport in 1991, his law firm, "Tacopina & Seigel" is based in Manhattan. He has defended the former head of the New York Police Department, Bernard Kerik, actor Lillo Brancato, Italian sailor Chico Forti, and Joran van der Sloot, who was the suspect in the mysterious disappearance of the Natalee Holloway...

Danu

@EmilioPisanty One thing that might be good to know is that if you can get $F$ to be homogeneous (and with non-zero differential at 0) then you're in really good shape

Emilio Pisanty

3:16 PM

@Danu as in, a homogeneous polynomial?

nope

Danu

yes

Emilio Pisanty

reasonable examples are above, lemme find them

5 hours ago, by Emilio Pisanty

wait, simpler version. $$H(k) = \begin{pmatrix} 2 & -k & 0 \\ -k & -2 & k \\ 0 & k & 0 \end{pmatrix}$$ with characteristic polynomial $$ E^3 -2(2+k^2) E +2k^2=0$$

also

6 hours ago, by Emilio Pisanty

and $$ E^2 = 1+ k_x^2+k_y^2$$

Danu

what are teh complex variables here?

Emilio Pisanty

all of 'em

Danu

$E,k_x,k_y$?

Emilio Pisanty

3:18 PM

@Danu yes

Cows

cnn.com/2018/04/24/entertainment/meek-mill-free/index.html

Danu

and $k$ is the norm?

0celo7

@Cows he’d better make some good music now

Waiting for DC5

Cows

@0celo7 yeah I'd be looking out for his music.

Emilio Pisanty

@Danu also if you can swap out $k^2 = k_x^2 + k_y^2$ on that first one

Danu

3:19 PM

yeah, right

Emilio Pisanty

$$H(k) = \begin{pmatrix} 2 & -k_x\pm ik_y & 0 \\ -k_x\mp ik_y & -2 & k+ik_y \\ 0 & k-ik_y & 0 \end{pmatrix}$$ to be definite

Though to be clear these are local approximations at the avoided crossings in the band structure of a material. I want to understand those first but ultimately I want to be doing things like $$H(k_x) = \begin{pmatrix} 1 & 3\cos(k_x) \\ 3\cos(k_x) & -1\end{pmatrix}$$

Danu

OKay

Emilio Pisanty

or in 2D $$H(k_x,k_y) = \begin{pmatrix} 1 & 3(\cos(k_x)+i\sin(k_y)) \\ 3(\cos(k_x)-i\sin(k_y)) & -1\end{pmatrix}$$

Danu

So.. you're interested in what are called affine algebraic varieties (for the polynomial case, at least).

Emilio Pisanty

ah, shit

remind me

what's the difference between a variety and a manifold?

Danu

3:27 PM

zero set of polynomial (or multiple poly's) inside $\Bbb C^n$

varieties are not locally homeo/diffeo to $\Bbb R^n$

they may have singularities

Emilio Pisanty

ah

Danu

classic example is $\{(x,y)\in\Bbb C^2\mid xy=0\}$

that's the union of the coordinate axes

which is an affine variety but not a manifold

Emilio Pisanty

@Danu ?

wrong way round =P

Danu

lmao yes

Emilio Pisanty

@Danu so, like the branch points in $\{(x,y)\in \mathbb C^2 \mid y^2 = 1+x^2\}$?

Danu

3:28 PM

corrected

It's not like branch points; branch points on Riemann Surfaces are not singularities of the surface itself

Emilio Pisanty

@Danu hmmmm

Danu

if you just draw the set where $xy=0$ in $\Bbb R^2$ you see this is not a smooth manifold, cause it's a cross

same happens in the complex setting

Emilio Pisanty

@Danu ah

no

that probably classes as pathological in my setting

Danu

but there is one good, easy to check sufficient condition for smooothness

namely the "preimage of regular value theorem"

which tells you that if you have a smooth function $F$ (on a manifold such as $\Bbb C^n$) such that $\forall x\in F^{-1}(0)$, $F'(x)\neq 0$, then $\{F(x)=0\}$ will be a smooth manifold

Emilio Pisanty

@Danu but that fails at the 'branch points' when you project, no?

Danu

3:32 PM

in the example you gave to me $F^{-1}(0)=(0,0,0)$ and at that point the derivative is nonzero because it is $dF((0,0,0))=4d E$

@EmilioPisanty I'm not talking about Riemann surfaces here

I'm just talking about arbitrary smooth (real) manifolds

Emilio Pisanty

hmmm

Danu

I don't think that this whole branch point story is very useful in higher dimensions

In any case, so the zero set of that polynomial you wrote down is in fact a smooth manifold

so that's nice

(no need to think about varieties in this case)

Emilio Pisanty

@Danu ugh

Danu

You spent a lot of time getting into the whole branched covering story? :P

Emilio Pisanty

So for me it's relevant (at least in the 1D case) because 'standard' physics only cares about solutions where everything is real, and there you get two disjoint bands, but if you allow the variables to wander into complex values then they can loop around that branch point, at which point the bands swap places.

Danu

3:37 PM

sure

Emilio Pisanty

I.e. the two bands are just two different intersections with a single connected manifold.

but the nature of those connections (including e.g. the genus / multi-connectedness) feels relevant

Danu

sure

but you gotta realize

that in higher dimensions

there are MANY more possibilities than just one infinite family (classified up to diffeomorphism by genus)

Emilio Pisanty

@Danu wait, one infinite family of what?

Danu

Riemann surfaces

Emilio Pisanty

spheres with $n$ handles?

Danu

3:39 PM

sure

Emilio Pisanty

aha

great

Danu

in 2d everything is very very simple

but in 4d everything is IMMENSELY complicated

Emilio Pisanty

@Danu so yeah, I'd like to get a feeling for what those possibilities are

at least

Danu

So... even assuming your space is let's say a compact complex manifold

then the classification is already extremely complicated

Enriques–Kodaira classification

In mathematics, the Enriques–Kodaira classification is a classification of compact complex surfaces into ten classes. For each of these classes, the surfaces in the class can be parametrized by a moduli space. For most of the classes the moduli spaces are well understood, but for the class of surfaces of general type the moduli spaces seem too complicated to describe explicitly, though some components are known. Max Noether began the systematic study of algebraic surfaces, and Guido Castelnuovo proved important parts of the classification. Federigo Enriques (1914, 1949) described the classification...

But this is a classification up to isomorphism of complex manifolds; I don't know if you care about that. In the 2d case, did you jsut care about the genus?

Emilio Pisanty

@Danu I'm just lost in a sea of geometry =/

Danu

3:44 PM

So let's put it this way

given a manifold

Emilio Pisanty

so, generally, I'm integrating a function along $k$ and I want to deform that path into the complex plane

Danu

there are tons of different structures it can have

Emilio Pisanty

and in 1D there's that branch point at which the naive approach breaks down, if nothing else

Danu

(i) topological, (ii) smooth structure (iii) Riemannian structure (iv) complex structure (v) symplectic structure, etc

Emilio Pisanty

but in 2D that "point" becomes something much scarier

@Danu what exactly do you mean by complex structure?

Danu

3:45 PM

do you remember how manifolds are defined in terms of charts?

Emilio Pisanty

yes

Danu

which give local patches diffeomorphic to R^n

and the transition functions have to be smooth, right

Emilio Pisanty

Yes

Danu

So a complex structure is an atlas of charts to $\Bbb C^n$ with holomorphic transition functions

Emilio Pisanty

gotcha

Danu

3:46 PM

quite restrictive

anyways; let's try to get to the bottom of what your'e interested in

so let's recall the basic story for Riemann surfaces

essentially, if you're looking at the zero locus of $F$, what you do is something like the following

instead of looking at e.g. $F=y^2-x^2-1=0$ you "solve for one of the variables" and look at $y^2=x^2+1$ right

and then you sorta think of this as a graph

Emilio Pisanty

yes

Danu

where $x$ parametrizes the x-"axis" (=a complex plane) and $y$ the y-"axis"

Emilio Pisanty

yep. With multiple "branches " on $y$.

Danu

so this thing is not acutally globally a graph, of course, if the algebraic equation is of degree >1

however, if you look at things LOCALLY, it does look like a graph almost everywhere

except at the so-called branch points

where essentially the graph has a vertical bit

Emilio Pisanty

@Danu what do you mean exactly by a graph?

Danu

3:50 PM

In the real case, it's really jsut a graph with an x-axis and y-axis

the graph of a function $\Bbb R\to \Bbb R$

now you're in the complex case but you can still get intuition from the real situation

enumaris

@Slereah actually I think I figured it out, thanks :D

Emilio Pisanty

a subset where if $(x,y)$ and $(x,y')$ are in the graph then $y=y'$?

Sir Cumference

@enumaris Oh hey, you're back

Danu

Right, so that's the case LOCALLY almost everywhere @EmilioPisanty

enumaris

kinda lol

Danu

3:51 PM

(so what I mean is if you only allow x to vary a little bit then $y=y'$), except at the branch points

Emilio Pisanty

@Danu yes

Danu

OK. So that essentially means that you have 2 separate bands almost everywhere, except when you don't because they cross, and then that's a branch point

Emilio Pisanty

yes

though at that branch point you could choose some other projection and it would look perfectly regular

Danu

yes, of course

Semiclassical

this conversation looks fun

Danu

3:53 PM

that is to say, the surface itself is not actually singular

oh yes, @Semiclassical is really into this stuff too

Semiclassical

not sure I can do much right now beyond just listen, though

yeah, I'm the one who pointed Emilio in this direction

Danu

oh, I see

Sir Cumference

enumaris
enumaris
@Slereah actually I think I figured it out, thanks :D

Emilio Pisanty
Emilio Pisanty
11:50
a subset where if (x,y)
(
x
,
y
)
and (x,y′)
(
x
,
y
′
)
are in the graph then y=y′
y
=
y
′
?

Sir Cumference
Sir Cumference
@enumaris Oh hey, you're back

Danu
Danu
Right, so that's the case LOCALLY almost everywhere @EmilioPisanty

enumaris
enumaris
kinda lol

Danu
Danu
(so what I mean is if you only allow x to vary a little bit then y=y′
y
=
y
′
), except at the branch points

Emilio Pisanty

Danu

Anyways @EmilioPisanty if I understood correctly, you said that it was interesting to understand what the genus of the corresponding Riemann surface (which is a branched covering of $\Bbb C$ (or $\Bbb CP^1=\Bbb C\cup \{\infty\}$ if you will) is, right?

eulB

@SirCumference r u ok

Emilio Pisanty

3:56 PM

@Danu as a starting point, yes

Danu

OK, and you were hoping to do that in a higher-dimensional case, right?

Sir Cumference

@eulB ?

Oh what the hell

Semiclassical

stuff like $z^2=f(x,y)$ instead of $z^2=f(x)$, for instance

Emilio Pisanty

another way to put it is that in the 1D case the branch points will come in pairs, with the branch cuts going from one to the pair, and I can sort of see what that does to the geometry in terms of adding handles.

but if you add a second independent complex variable then all my intuition goes completely woozy

Semiclassical

(aside from othe case with $f(x,y)$ being homogeneous of degree 2 of course)

enumaris

3:57 PM

copy pasted a big chunk thar

Semiclassical

I'm not entirely sure what the terminology is in that case beyond just "complex manifold"

obv. it's still an (affine) algebraic variety

enumaris

my deep spell checker is not doing all that great...=/

Sir Cumference

@Semiclassical What's the difference between a variety and manifold

Semiclassical

A variety is the zero set of some polynomial.

0celo7

@SirCumference met JS’s student

Emilio Pisanty

3:58 PM

31 mins ago, by Emilio Pisanty

what's the difference between a variety and a manifold?

eulB

@0celo7 :o

Sir Cumference

@EmilioPisanty Oh wow lol

eulB

@SirCumference is a bot

nitsua60

What's up?

eulB

bad bot

0celo7

3:59 PM

@eulB why is that surprising

eulB

it's not

just putting the word out there

stay safe people

Sir Cumference

Haha that is very funny ~~fleshbags~~ fellow humans

Danu

@EmilioPisanty So in the case where you start with an equation on $\Bbb C^3$, under the condition on the derivative I told you about earlier (in other words, 0 is a regular value of $F$) the set $\{F=0\}$ will be a complex manifold of complex dimension 2

Semiclassical

A variety is a type of manifold, I think? (Only issue is re: singular points)

Danu

no

Sir Cumference

4:00 PM

@0celo7 How? Where?

I mean did he go to you or were you over here?

eulB

Jan 19 at 9:20, by Bernardo Meurer

Oh wow, lol

:o

Emilio Pisanty

@Danu gimme a few minutes, I need to check on something

Danu

the singularities are the problem; they're not manifolds in general. But like Isaid, if $0$ is a regular value of $F$ then it's fine and you get a manifold

Semiclassical

Right, that's what I was getting at

the story is simpler with projective varieties than with affine varieties

rob

2

Q: What is the fastest speed you can ride a bicycle on wet roads without getting wet?

Optimum speed without getting a line of muddy water splashing onto your back? no mud guards 26 inch wheels surface water on the road calculations to be based on my shirt and shorts remaining dry standard dimensions of an adult bike frame I am riding a bike and the water is being thrown upwar...

A flagger says,

> This question was closed by mistake by people who thought it was the same as the question: 'What is the angle at which water is splashed when a vehicle tyre rides in water on the road? '. It isn't. The latter is about sideways splash while the latter is about water following the bicycle wheel rim it is attached to, as it leaves the road, and then being thrown up behind the cyclist.

I think that's a sound argument but I don't want to reopen unilaterally. Thoughts from this group?

Sir Cumference

4:03 PM

I don't know what I'm saying at this point, I'm so tired

enumaris

dang itttt

Semiclassical

The Beatles | I'm so Tired

►

enumaris

I forgot to bring my cellphone to work...boring day ahead...:(

0celo7

@SirCumference I will probably meet JS this summer. I can recommend you to him

Sir Cumference

@0celo7 You don't know who I am

0celo7

4:05 PM

You’re the Jewish physicist. Good enough?

Sir Cumference

And I haven't been in his class for two years

Wait no, one year

enumaris

who JS

Sir Cumference

@0celo7 Well considering like half the physics majors here are Jewish (very small physics major here :P), that's not much

Well more like a third

enumaris

half, a third what's the difference to an astrophysicist?

Sir Cumference

@enumaris We're not engineers

enumaris

4:08 PM

if it ain't an order of magnitude, it ain't nothin

Sir Cumference

@0celo7 But please for the love of god back off a bit...

You're getting a bit too personal

0celo7

Trying to help you buddy

Sir Cumference

@0celo7 you're being an obe

Semiclassical

let's just grant each other our personal space, mmkay

enumaris

but we're all crammed in this room together

Sir Cumference

4:18 PM

The Parker Square - Numberphile

►

Best math video

Rithaniel

I've recently watched that one

rsadhvika

Anyone interested in helping with a tricky resistor circuit problem ? physics.stackexchange.com/questions/402242/…

Semiclassical

for context, here's the image:

to which my immediate response is... D=

b/c that's just horrifying

dmckee

Well, you can reduce part of the complexity with the usual series/parallel rules for sure, and then maybe more with the star-delta thing.

rsadhvika

Haha indeed it looks horrifying! All the resistors are 4 ohm and batteries are 10V. It does piss me off when the hint says it is trivial if you have a good eye

dmckee

4:24 PM

But sooner or later it is coming down to a fat system of linear equations. Just the sort of thing for students to do.

Semiclassical

If there's a trick to it, I'd say it's that you can establish some of the relative voltages by inspection.

So, for instance, take the middle-left junction between the two batteries to be at zero voltage

rsadhvika

Yeah I've already figured one neat thing : Looking at the outer most loop, current through the bottom most resistor is 0.

nice

@Semiclassical Yikes

can't see images...

4:26 PM

It's like 50 resistors in parallel

Semiclassical

And if there's no current through that resistor, you can basically remove it from said circuit without issue

Sir Cumference

Or you could just click the question

rsadhvika

Since the current through the bottom most resistor is 0, is it okay to remove that resistor from the circuit ?

Semiclassical

So if I had a pencil I'd write down the voltages along the edges and start eliminating resistors which have no current across them

Hmm

rsadhvika

Ahh nice looks you read my mind ;)

Semiclassical

4:27 PM

You can definitely replace the resistor by a wire in that case

Sir Cumference

@rsadhvika Do note that this isn't a homework help site

Semiclassical

But you may want to leave that wire in place because it tells you that two nodes connected by it are at the same voltage

enumaris

I can't see any images anywhere

my browser blocks like most of the servers

Semiclassical

the other nice thing is that you can get some of the voltages of nodes inside the circuit rather than just on the outside

so that may let you figure out a lot of them right off the bat

my guess is that you'll end up with a lot of the resistors having zero current across them etc

enumaris

someone draw the circuit using ascii characters in a comment, kthx

Semiclassical

4:29 PM

i'd be lying if I said I had the patience myself for that right now, but with a pencil it shouldn't be hard

@enumaris nothxbai

rsadhvika

Yeah looks some of the resistors in the mid section can also have 0 current. I'll give it a serious try once again..

Semiclassical

I think if you trace the voltage changes then you can find the voltage drop across the four-resistor subcircuit which includes the one you're interested in

rsadhvika

Just a quick question, when the current through a resistor is 0, should we remove the resistor, or replace it with a wire ?

Semiclassical

replace it with a wire, I'd say

rsadhvika

that essentially means joining both the nodes @enumaris I have a link to the image if it helps ?

Alex K Chen

4:36 PM

Do you know some good online sources which shows simulation of how angular velocity is added in 3d ?

Semiclassical

right. the point is that, if they're at the same voltage, then you can effectively merge them

without losing any info

enumaris

I can't access it...probably can't help you on this one, sorry D:

Semiclassical

I think, though, that doing this may be unnecessary. What I'd suggest is this

enumaris

if the current is 0, remove that whole circuit element...don't create a short circuit!

Semiclassical

Start from the bottom left corner, and take that to be zero voltage.

rsadhvika

4:37 PM

No worries @enumaris semiclassical gave some new hints which I'm going to try..

Semiclassical

Then trace the voltage throughout the rest of the circuit, just by looking at the orientation of the batteries you pass over. you should be able to get to the node just above $i$ in your diagram

once you've got that, you've effectively reduced it from a many-resistor problem to a four-resistor problem with a known potential difference

John Rennie

Is it just me or is the toolbar (at the top right of the page) broken on the main site?

ACuriousMind

@JohnRennie Broken in what way?

Alex K Chen

3 mins ago, by Alex K Chen

Do you know some good online sources which shows simulation of how angular velocity is added in 3d ?

John Rennie

@ACuriousMind The icons have changed to simple links instead of opening menus.

enumaris

4:40 PM

angular velocity adds like a vector...that's why it can be represented as a vector...

ACuriousMind

@JohnRennie They do that sometimes before the page has finished loading. Does this also occur for you on the fully loaded page?

John Rennie

@ACuriousMind So for example there is now no link to this chat room on the main site.

enumaris

@JohnRennie opens up menues for me

Semiclassical

works fine for me

John Rennie

@ACuriousMind yes, or at least the page claims to have completed loading. This is in Chrome. Let me try IE ...

ACuriousMind

4:42 PM

Works fine in FF for me

enumaris

I'm in chrome

works fine

John Rennie

Ah, it works in IE. So Chrome is playing silly buggers then

enumaris

maybe ur chrome...

rsadhvika

@JohnRennie yeah in Chrome chat link is not showing on the main page, it shows if you go to a question page..

John Rennie

Ah, just closing all instances of Chrome then relaunching it has fixed the problem. Oh well, one of those things.

Subho95

4:49 PM

@rsadhvika I know how to solve it. The answer is 4A upwards

Semiclassical

yeah, I think that's right as well

Subho95

@rsadhvika It follows really easily by choosing two inner loops that have all but one resistor

rsadhvika

@Subho95 What is your strategy ?

Semiclassical

eh. i'd say it's more that they have to deal with a lot of questions and that doesn't inspire patience

I wouldn't call it good by any means but i can see how it happens

Subho95

@rsadhvika Let me point out the two loops by using some good old paint...

ACuriousMind

4:54 PM

@rsadhvika What makes you feel that way?

John Rennie

→ 3 messages moved to trash

Please don't call people jerks in this chat room

Semiclassical

given that my message doesn't have context now, I'd ask it to be moved as well

rsadhvika

@Subho95 thank you, I'm waiting... this is not really my homework. So you don't have to feel guilty of helping

Subho95

John Rennie

→ 1 message moved to trash

@Semiclassical done!

Semiclassical

4:59 PM

"John Rennie has invited you to Trash"

rsadhvika

quick screenshot, before that solution is moved to trash

Semiclassical

I know why that happens, but its funny

John Rennie

:-)

Semiclassical

your solution isn't going to be trashed.

rsadhvika

do your thingy @JohnRennie

Semiclassical

5:00 PM

this isn't the main site. the policies here are to do with respectful communication not with the integrity of Q&A

John Rennie

@rsadhvika there's nothing wrong with discussing homework and/or worked examples in the chat room. In fact there is a chat room specifically set up for precisely that purpose.

Problem Solving Strategies

General chat for high school physics. For MathJax see [here](m...

2

47

Semiclassical

the point is not to cast shade on the mods

John Rennie

(or anyone else)

rsadhvika

Okay this is my first time chat so forgive my kindly

Semiclassical

well, doing so with the mods is a particularly perilous venture :P

rsadhvika

5:03 PM

Our mods are kinda superheroes

Semiclassical

eh, they're people

they make good decisions and bad decisions

ACuriousMind

@rsadhvika I'm happy to respond to criticism more concrete and constructive than calling us jerks.

Semiclassical

(I personally find the Be Nice policy to be silly and over the top, but that's what's enforced and I (usually) abide by it)

rsadhvika

Brilliant! @Subho95 Calling that center node 0 did the magic! May I ask how many hours you have spent before spotting that center loop with a single resistor ?

Emilio Pisanty

@Danu Well, that took longer than expected, but I'm back

Sid

5:18 PM

@Semiclassical nested parentheses are just... wrong

Emilio Pisanty

@Sid they're not. Nested parentheses are (normally, of course (but not always)) perfectly OK.

ACuriousMind

As long as you close all you open, nothing wrong with nesting them. If you don't, however...obligatory xkcd 1, obligatory xkcd 2

rsadhvika

@ACuriousMind I do like our moderators. The ones I've encountered are rly nice and decent ppl. Again this is my first time in chat here... I didn't use "jerk" to hurt you intentionally. I see it is not the right word to express sarcasm

Emilio Pisanty

@ACuriousMind is that all you're going to quote?

Sid

@ACuriousMind gah. You beat me to it

ACuriousMind

5:21 PM

@EmilioPisanty If I try to post every Lisp joke Randall ever made I'll be doing this for a while :P

Emilio Pisanty

'cause, I mean, this is also obligatory.

in this setting

ACuriousMind

Ah, right. I guess I don't relate to that one so much because the second alternative looks so obviously correct to me :P

Avnish Kabaj

Is xkcd a stackexchange user ?

I was crawling through chatrooms and saw this feed room full of his comics posted by an account "xkcd"

Emilio Pisanty

@ACuriousMind I guess as time goes on and you become more acutely aware of double-chin as something that happens to people you might change your mind

@AvnishKabaj no. It's just the fact that if you post a message that only consists of an xkcd url it will automatically onebox

look

The IAU ban came after the 'redefinition of 'planet' to include the IAU president's mom' incident.

ACuriousMind

@AvnishKabaj You can give names to automatic feeds, they just told the room to post things from the xkcd RSS feed and named the feed xkcd

Avnish Kabaj

5:24 PM

Aaah

It would have been cool if he was a user

Sid

I wonder if there ever was an xkcd comic on Stack Exchange...

ACuriousMind

@Sid The alt-text of this one refers to SO.

Loong

@Sid 979

Subho95

@rsadhvika One of the comments struck me to think along those lines.

rsadhvika

You

You've just figured it out ?

Subho95

5:28 PM

@rsadhvika Yeah. It took some 10 mins

Avnish Kabaj

https://imgs.xkcd.com/comics/wisdom_of_the_ancients.png
Selected Comics

2

Emilio Pisanty

@AvnishKabaj well, there's this if you want a poor substitute

it does include this remarkable screenshot

rsadhvika

Amaging! what must I do to have some of your quick reflexes ?

ACuriousMind

@EmilioPisanty Old topbar <3

Avnish Kabaj

@EmilioPisanty nice

Semiclassical

5:34 PM

dangit latex

what am I doing wrong

for reasons of time management I decided to go back to the thebibliography environment, since most of my references were from papers that had used that format

and since I was using \include to do the chapters, I figured i'd be able to \include the bibliography in a similar manner. but no joy

...I wrote \include{bibltemlist} when I meant \include{bibitemlist}

(first l should be i)

ffs that was annoying

Emilio Pisanty

5:53 PM

@ACuriousMind meh

enumaris

6:24 PM

meh

Semiclassical

6:48 PM