Who?

Lagranian

Click the "he's back?" link

Got it.

I'm on mobile @Phase

It has no links to click :-/

RIP

Im gonna play Max payne for the first time

Good luck

Fun fact: it was the first game to use photographs to texture character faces, and the protagonists face was made from the Boss of the company that made the game

Sounds like the photo shop of vanity type of game :P

wow, I got almost as much rep from one answer today in cross validated than I got from all my answers and questions in stack overflow combined...sure makes my stack overflow participation seems sad...lol

Link please

to the question in cross validated?

sure

oh, fancy hyperlinking

@0celo7 I thought he graduated?

gosh dang it, I can't comment in the physics stack exchange...limits me a lot...cus I usually don't wanna provide answers...just some comments lol

Ask a good question :-)

@enumaris If you get 200 rep on any one site, it'll add 100 rep for every site you go on.

So you'll automatically have the ability to comment

@enumaris Please don't use comments for anything but improving or clarifying the question, or linking related questions.

I'm at like 66 on cross validated...

oh..hmm, what about partial answers?

@enumaris Usually looked down upon

@SirCumference who?

@0celo7 ACM

I see

@enumaris If it's an answer, it belongs as an answer.

That's a pain...have to do so much math to answer one question...lol

ACM might be a physicist on paper, but they want him less than we do

._.

what kind of physics did ACM study?

I'm right here

@enumaris Particle physics

Oh, forgot he was here...

@SirCumference Ehhhhh

ACM does not study particle physics

Honestly, do you pay attention to the people in here?

@0celo7 'Course I do. Erm...if it's not particle physics, then GR?

No, that's me.

oh

I studied astroparticle physics

@0celo7 I thought you were nuclear engineering

This is all spinning my head

so, kinda related

@enumaris high fives

@SirCumference Obvs you don't stalk people's profiles enough ;p

Wow, you're really behind @SirCumference

@enumaris We got an astronomy SE...just saying...

@SirCumference dmckee does nuclear & particle physics - is that what you were thinking of?

@SirCumference astrophysics $\neq$ astronomy

@SirCumference No. Quantum field theory, especially gauge theories, and string theory. QFT is technically particle physics, I guess, but "particle physics" usually denotes the more experimentally oriented QFT people.

@Mithrandir24601 I thought dmckee was a stellar physicist??

Now you're thinking of either Chris or Kyle :P

whenever I read that I just think of stellar as in good...

hmmm

@SirCumference Go to a memory specialist. Please.

@Mithrandir24601 Ugh no...no matter what branch of astronomy you go into, you're work is dominated by physics

It's a meaningless distinction at this point

QFT never clicked with me that well, could use it for applications, but could never feel like I understand it lol

@SirCumference Sure, but an astrophysicist still does different things than an astronomer

Now astrology people - that's what we're missing out on.

@ACuriousMind TBH I've asked profs about whether they consider them to be the same, and I always get the same answer: "Eh, I don't really care"

I'm sure you can get away with calling yourself either if you do work in that field

I'm gonna call myself an astrologer

Try calling a theoretical astrophysicist an astronomer and let me know how it goes :P

just to piss off the astronomers

and astrophysicists

@ACuriousMind Is astronomy not just the study of the Universe beyond the Earth? :P

We've had this pointless discussion before :P

I wouldn't mind if u called me an astronomer

so survey of 1

don't mind

@enumaris Yep, although to be real it's more prestigious and cool to say "astrophysicist" :P

@ACuriousMind I was thinking the same thing.

@SirCumference What's your derogatory term for me?

Inequality specialist?

@0celo7 ::stores in brain::

@0celo7 "Applied theoretician". You enjoy the theory but only seem to respect those working in applied...

@0celo7 I'll use that

I feel like something along the lines of 'hand-wavey manifold muncher improper notation' would annoy you the most...

@SirCumference Because we theoreticians are a lazy bunch

Let's not try to annoy anybody :P

If anyone can name an "astronomer" who does not do physics, please do. But so far, the definition for astrophysics as "the application of physics to astronomy" seems redundant.

It literally encompasses all of astronomy.

Sagan?

@skullpatrol Ugh...a real astronomer

@SirCumference Hans Ringstöm might object.

lol

He's Ivy League @SirCumference

All right, let's make sure I know everyone. 0celo is an undergrad. ACM is a prof. rob is a prof. dmckee is a prof. John Rennie is a physical chemist.

@SirCumference It's not about abstract categories (who even made that weird definition?) but about the fact that the community of people calling themselves astrophysicists and the community of people calling themselves astronomers may overlap, but are not the same.

@SirCumference 3/5, not that bad :P

@0celo7 Careful there.

@ACuriousMind Who'd I get wrong?

I actually don't know a community that calls themselves astronomers

@SirCumference ACM and rob.

don't they all just say they are astrophysicists :P

@SirCumference Me and rob, I think. JR is also questionable, depending on whether you mean training or current occupation.

John's a laptop buyer.

@ACuriousMind I agree.

@ACuriousMind Ok come on, what rule did I violate

@0celo7 I recall rob mentioning he graded papers

@SirCumference He's a lecturer

Not everyone who grades papers is a prof

Be nice @0celo7

@0celo7 I...was close...

@skullpatrol I am nice!

It's @SirCumference who apparently ignores all of us

Sigh, I'm an idiot...

I graded papers as a TA

"Idiot" not a nice term @0celo7

Ocelo's just here to collect flags...

crap, was I flagged?

dont think so

Jesus people, you don't need to take everything as a literal hate message...

flags flags flags flags flags flags flags flags flags

Nah I'ma listen to the words of our glorious leader.

Chill out, people, no flags were raised.

Ahhh the words of JR.

(removed)

@0celo7 I mean, there probably more than 30 active people on the chat. You can't expect me to look at everyone's profiles and remember them perfectly, right?

JR... huh that's close to... GR... suspicious.

@SirCumference so now you read my profile and know what I do?

@0celo7 nuclear engineering...

@SirCumference I dunno, I have a pretty good idea of what everyone does and I'm stupid

(that's a joke.. okay)

@CooperCape Jeneral Relativity.

@ACuriousMind Rennieral Relativity

@ACuriousMind Jeeneral relativity?

JEEneral relativity?

@SirCumference To be fair, I still don't really know the proper term for exactly what I do. I mean, physicist, sure, but any more than that starts to get shrugs

The plot thickens...

lol

@0celo7 I'll leave JEEneral relativity to the Indians

@Mithrandir24601 Quantum engineer

that's way cooler than anything a physicist could be called

see ocelos so good he tells others what their careers are...

@0celo7 I suppose that's the closest...

Yeah, once I actually get round to looking at things like chip design, that'll be an appropriate term :)

I stole that from the neighborhood octopus

(though the past half-year or so has been unusually theoretical for me)

Theory is where it's at

@0celo7 ... who's definitely much more of a quantum engineer than me :P I suppose my PhD broad area is 'quantum engineering' though :)

@enumaris ^

ooooof

guys

shit's getting to me

Although I have trouble convincing that to people

I could use some random stranger on the internet saying that this is indeed a nice-looking avoided crossing

associative law is (AB)C=A(BC), right?

a hyperbola

Nice

@0celo7 secondary

is that a mass splitting graph for neutrinos

avoided crossings are hyperbolas

Two lines. They don't cross. Looks like a nice avoided crossing. (I have no idea what it's supposed to look like :P)

@ACuriousMind two states with an avoided crossing as you spin them up ;-)

@EmilioPisanty Looks like what I'm going to term a shifted avoided crossing to my inexperienced eyes :P

that reminds me of the effective mass splitting graphs for neutrinos

like for explaining the MSW effect

I'm not sure what it's supposed to look like either

that's why I'm sciencing it out =P

@Mithrandir24601 why shifted?

the graph does violate the holy rules of graph making though

needs a title, needs labeled axes and units

@enumaris indeed it does

It's not symmetric?

and if you show me someone who does all of that with all their informal notes, I'll show you someone who doesn't work as fast as they could ;-)

@ACuriousMind On the topic of garbage rigor, I wrote my advisor about an issue in a paper and a BS reference, and he sent me "I think the spirit of Rick's reference is "any analytical fact can either be found in Bartnik's thesis, or derived quickly from it, or is already found somewhere""

@skullpatrol no, doesn't need to be

okay

@ACuriousMind could you give me a hint as how to use the notion of "extending the basis" to prove rank nullity?

Because I really don't get what the point of extending a basis is. It just seems trivial

@skullpatrol generically avoided crossings will only be symmetric the uncoupled states have the same slopes

That's probably my uncultured mind

here one is stationary and the other is getting detuned

@Phase Given some vectors you can add others to get a basis of the whole space

starting with some linearly independent guys, of course

Yeah I get that but I dont get how I'm supposed to prove anything with that

I already wrote out something going along the lines of vanishing basis vectors, and equivalence classes, but I've been advised that extending the basis would be a quicker and less convoluted way

you're trying to prove rank-nullity?

Use the first isomorphism theorem.

@Phase work backwards from what you need to prove

@EmilioPisanty To me, it looks like if you continued both graphs without 'avoiding the crossing' (there must be a technical term for that) they'd cross at around 0.035-ish on the x-axis, as opposed to the turning points. Alternatively, looking at either side of either line (i.e. say the bit from 0-0.02-ish) and extrapolating it, it doesn't look like it would meet the other half of the other line. Don't know if that makes any difference to the definition or not though :/

@Mithrandir24601 indeed they would

@0celo7 is that the thing where you say that there exists a bijection from $V/E$ to [field of numbers]^(dim(V) - dim(E))?

I already did that if so

they're the eigenvalues of $$\begin{pmatrix} E_1 & \varepsilon \\ \varepsilon & E_2 -\Omega\end{pmatrix}$$

oh god

It says that for $f$ a group hom, $\im f\cong G/\ker f$

latex without latex

if you set $\varepsilon=0$ you'd get a true crossing

my eyes are bleeding

@enumaris see the info on top right for LaTeX in chat

Top right corner

the MathJax stuff?

A group.. what? Hom? @0celo7

Yup

@enumaris yes

@Phase homomorphism

Idk what that is

@EmilioPisanty It's probably just a case of the amount of zoom deceiving my eyes then :P

"preserves algebraic structures", so like, an inner product space is changed to another inner product space?

@Phase Sigh

@Mithrandir24601 ah, I see what you meant

no, that's just because you're extrapolating from bits of the hyperbola that aren't far enough into the asymptotic region

@0celo7 nevermind :|

a map $f:G\to H$ is a group hom if $f(ab)=f(a)f(b)$ for all $a,b\in G$

if you did the extension from the asymptote, they'd meet smack in the middle of the plot

test: $f$

nope

What does $f(a)f(b)$ mean though, say you take vectors?

Does it mean the operation defined on the space, so $f(a)$ [some operation defined in the original space] $f(b)$?

@EmilioPisanty Makes sense - as in, if the graph was plotted from -arbitrarily large no. to arbitrarily large no. it would look like a regular avoided crossing :P It's definitely the right Hamiltonian for an avoided crossing though :)

he's omitting the symbol for group element multiplication

@Mithrandir24601 yeah, exactly

@EmilioPisanty ::relaxes with the realisation that optics wasn't wasted on me::

a group homeomorphism, roughly, is a mapping of one group to another that preserves the group operation.

@Phase Exercise: show that for $f:G\to H$ a group hom, $\operatorname{im}f\cong G/\ker f$.

why isn't the latex rendering for me...I got the chrome extension...

I can give the proof if you need.

Fuck that, im still not even sure if I understand homomorphisms. Does that mean, applied to a vector space, that all the possible operations in the original space are preserved? Like say you have a homomorphism of a euclidean space, you'd have a defined inner product and vector addition in the new space?

it's enabled...

A vector space has more structure than a group