I want to ask a secret question

crap

the physics package does not work here :P

$\newcommand{\ket}[1]{\lvert #1 \rangle}$ $\newcommand{\ii}{\mathrm{i}}$

There. Although I'm not sure what ip and pdv are supposed to do

ip is the inner product

still broken on my screen @ACuriousMind

and pdv is the partial derivative operator

You might have to restart ChatJaX to see the effect

I just have to solve the SE for a 2D system and I want someone to check my calculus/algebra

@ACuriousMind halp ;-;

@BernardMeurer He won't skype with you

@BernardMeurer With what, exactly?

give it up

he never Skypes

@ACuriousMind Skype plox

I need to understand chemistry

hmm

@BernardMeurer No, Skype/having to carry a spoken conversation interferes too much with my multitasking.

called it

@0celo7 You win this time

he's an AI, that's why he doesn't want us to hear his voice

You've heard my voice already

@ACuriousMind Aha, my solution is correct.

Yeah, we have

It's consistent with my solution using the propagator.

@0celo7 He is not only German, but also an A.I. Interfering with his efficiency would be suicidal.

8 page QM homework...delicious.

@0celo7 i'll do it for you if you do my lab report

I just did it.

Oh fuck I have a lab report due tomorrow

yeah maybe you should worry about that instead of reading topology books

I have a topology midterm on Wednesday

midterm/exam

hence why I'm reading a book on it

@ACuriousMind Is $c\in\mathrm i\Bbb R$ standard for "imaginary"?

@0celo7 I'd say so

in a vector function, is retrieving the tangent line at a point the same as getting its derivative?

no

The general definition of a vector function's derivative makes no sense to me. $\lim_{h\to 0} \frac{r(t+h) - r(t)}{h}$ what is the point of dividing by $h$? Derivatives define the rate of change of a curve at a point. In this situation, $r(t+h)-r(t)$ gives the secant line between $r(t)$ and $r(t+h)$. As $h$ approaches 0, the secant line gets smaller and smaller. The direction is already determined by the secant line. Why can't we just use this as the derivative?

@BernardMeurer lol at your perplexity. plz recall the greatest physicists in the field share your perplexity wrt QM foundations for nearly the entire 20th century & continuing. think of the challenge now as "learning enough to solve the problem/ pass the test" etc... then when feeling less overwhelmed consider going deeper... take some solace that you dont get the "full harsh dose" in chemistry...

@Obliv Because without the $h$ that limit is always 0 for a continuous function.

The definition by components makes 100% sense to me. $\lim_{h\to 0}\langle \frac{x(t+h)-x(t)}{h}, y(t+h)...\rangle$

I would ask for a proof but I actually understand @ACuriousMind

but why not any number, then? Isn't it always going to be a scalar multiplication of $r(t+h)-r(t)$? @acuriousmind

@vzn I'm not angry because this stuff is hard. I find analysis hard and I love it. I'm angry because my class was horrible and no one understood a thing. QM is the one thing I want to get, and this was a bitter taste that maybe I'll fail in that

wait..

@Obliv I don't understand what you mean by "any number". If you divide by any constant, the limit is still 0.

@BernardMeurer At my university, (theoretical) QM is a fourth-semester course. It's highly unusual to throw it at freshman that non-chalantly in my experience.

@BernardMeurer how do you know you dont get it? there is a "deeper challenge" (ie ongoing cognitive dissonane even by top experts in the field!) and then there is just the challenge of solving problems/ passing tests. think of how many young ppl had a similar feeling as you & managed to pull it off... you dont have to understand QM to pass a chemistry class.... think of it as a foreign language where you (roughly) learn a few words at a time, right? rome wasnt built in a day :)

@ACuriousMind Noob. It was 3rd for Marco.

Oh, semester.

Jesus, QM as a second year student?

Wtf is wrong with Europe

@ACuriousMind It's bloody stupid. He wrote the wave equation, and then was like "Y'all know the partial derivative right?" To which the class responded in unison "No" And he just kept going anyway

I think if you have to make sure the class knows what a partial derivative is, you're doing QM wrong.

Well, not make sure, cuz he clearly didn't do that.

But if he asked, he's a dick.

@0celo7 The progression in the theoretical physics courses here is: Newtonian mechanics - Lagrangian/Hamiltonian mechanics - Electrodynamics - Quantum mechanics - Statistical physics. Although I know that many universities start that progession only in the second year, and may not have two mechanics lectures.

my freshman chemistry book also ref'd QM but its done in a sort of "imprecise/ brief aside"... there are some simple math eqns associated with it eg debroglie wavelength etc where one can manipulate them without understanding the "full meaning" (and arguably not even experts do...)

@ACuriousMind Well, we have quantum as a 4th semester course too, but it's probably nothing like yours.

@BernardMeurer have you taken any calculus?

@ACuriousMind What's the wave equation, how do you get to it?

@0celo7 Judge for yourself, this is the script from the QM course I took.

@vzn I just started my analysis course. We haven't gotten anywhere yet

@ACuriousMind That's literally my class.

Your prof took the material from Sakurai.

@BernardMeurer the wave eqn is ubiquitous and shows up even with simple waves eg sound etc & then was adapted to QM, so if you want to learn more about it thats another direction to pursue, the math is quite related, alas the compartmentalized way that edu works, this (historical) insight gets lost...

Same notation. Shame on him :P

@BernardMeurer If you mean the Schrödinger equation, it's basically a postulate. There are various ways to motivate it (there should be aplenty in various questions on this site), but in the end, it's not derived, just like Newton's laws are not derived.

@0celo7 Well, he took it from another prof, who probably took it from Sakurai, yes

@ACuriousMind I meant the wave function, sorry :p

But I'm taking a graduate level class. Although the prof says he's movin way slower than he thought we would because the homework is terrible

@BernardMeurer Ehhh, well, the answer is basically the same: It's a postulate that a quantum state is a vector in Hilbert space, and one of the most convenient ways to represent such a vector is by a square-integrable function. That's it's a "wave" then comes from the Schrödinger equation being very much like an ordinary wave equation in the free case.

@ACuriousMind Hm. I get none of that

I'll go run, this is pissing me off

thanks for the help guys, see y'all

The experimental evidence that this wave function represents something meaningful is traditionally the double-slit with electrons, where you can see "wave-like" interference pattern even without doing any precise math.

@BernardMeurer Well, I'd argue you're not supposed to get it instantly! If I could explain QM in a few pithy chat messages we wouldn't teach so many courses about it.

@ACuriousMind Did you see the unbounded L2 function I pinged you with?

Unbounded at infinity.

@0celo7 Yes

Neat, huh? Or did you know about it already?

It's a neat example, yes. If I knew it before I at least didn't recall it.

@ACuriousMind I'm glad I was able to enrich your life, then.

Hmm how does this work.. $s(t) = \int_{a}^{t} |r'(u)|~du \to \frac{ds}{dt} = |r'(t)|$? Shouldn't it be $|r'(t)-r'(a)|$?

What on Earth are you doing

@DanielSank : because I took it step by step and gave supporting references. But hey, if you don't want to pay attention to what I said, that's up to you.

@0celo7 I don't know.. looking at definitions in my book

$s(t)$ is arc length.

this is confusing.

@Obliv Why should it? That's not what the fundamental theorem of calculus says.

$\int_0^a f'(x) dx = f(a) - f(0)$ so I thought $\int_a^t |r'(u)|~du = |r(t) - r(a)|$ then take the derivative of that with respect to $t$

unless I'm missing something

$\lvert r'(x)\rvert \neq \left(\lvert r(x)\rvert\right)'$.

Trying to read this on a phone is hard

aren't you at the library? @0celo7 just get on a computer

@acuriousmind so $|r'(t)| = |r(t) - r(a)|'$? Still doesn't make sense to me

@Obliv No

did I at least evaluate the integral correctly

No.

I don't get what your problem is. You have a function $f(x) := \lvert r'(x)\rvert$. By the fundamental theorem the derivative of $\int_a^t f(x)\mathrm{d}x$ w.r.t. $t$ is $f(t)$ itself.

but it's a definite integral so I thought..

What?

@Obliv I was in class

topology

@0celo7 are you finished, then?

I have a laptop, you know

Finished with?

I did my QM homework

topology

yes

@Obliv No.

OH

What?

That was in reply to that..

I thought you were trolling me.

I hope that flag made you happy, @0celo7 :P

@ACuriousMind Very.

@ACuriousMind Other people just assumed J, K, Q, A didn't count instead of making them worth 10

fml

I'm going to fail this class

What?

Ah

although that seems dumber than what I did

because the deck is not 52 cards if you throw that out

and the problem said explicitly that it's a 52 card deck

@Obliv Sorry, you shouldn't have said such a hurtful thing. I'm literally in tears.

@0celo7 That's alright.

I have a throwaway 8^)

Flagging for IP ban.

NO

You shouldn't have done that.

Erm...circumventing bans with sockpuppets is not exactly welcome here.

It was a joke. I am two different people hehe

Too late.

Wonder if they ban the whole university if you're on campus wifi.

Oh well.

@0celo7 I don't think they impose IP bans for a simple sock precisely because of that possibility

A simple sock?

@acuriousmind even though the block resulted from miscommunication? That's irrational. Anyway, I'll get off if it if it matters that much.

As opposed to a nontrivial sock?

@colour Miscommunication?

Bans are sacred and infallible.

You were banned for a reason.

I wonder why no mods showed up.

@colour The whole chat flag system is irrational, but that doesn't make willfully circumventing it okay.

It doesn't make willfully abusing it okay, either, but that's harder to prove :P

Someone is abusing it?

Shame on that person!

@acuriousmind why would you side with the rules in a situation where you know the circumstances?

@0celo7 I don't believe for one nanosecond that you were actually offended by that message.

Because he's a German.

@0celo7 : who was?

Rules exist in general not to be broken. If it is evident that the rules were broken for a good/not bad reason then punishing it is silly IMO.

Sigh.

@colour Well, what circumstance do I know here? I think I personally know both of you well enough that neither did you intend offense nor was actual offense taken. But that's not evidence, it's opinion. The only evidence is that someone flagged your comment as offensive and the flag was validated. I'm not for following rules for their own sake, but the reasons to break a rule must be objective.

@Obliv Why did you do that?

Uh huh. This user... cannot chat for 17 minutes.

And This user... cannot chat for 26 minutes.

@ACuriousMind Right. But offense is not objective. Thus we should flag when offense is possible in order to have some form of objectivity.

We have two possible objective avenues: either never flag or always flag.

@0celo7 No, we should flag when we are actually offended. Precisely because offense is subjective there is no such thing as "possible offense", although one can certainly say that some statements are more likely to cause offense than others.

@ACuriousMind There is no way for anyone to know if I am actually offended or not.

For the record, I'm not convinced "offended" is a valid concept.

@0celo7 Correct. It's assumed all users act in good faith - the flag is supposed to be precisely your means to say that you're offended (apart from saying that out loud, of course).

If you don't like the word "offended", you might as well call the "offending" message "inappropriate for respectful discourse" or something like that.

@ACuriousMind Does it? It seems like the flag is more asking the 10k users if a message is possibly offensive.

Nowhere do I give my input, and I can certainly flag things that are not directly offensive to me.

If someone calls you X bad thing I will flag it.

That doesn't offend me.

@0celo7 The way the flag is actually handled is broken. No argument there. But it's not the intended purpose of the flag to be subject to the whims of random 10k users.

The issue is that SE doesn't care enough about chat to fix it

But I don't see what you think you achieve by using the broken review system to get others banned.

Me neither.

@ACuriousMind What we need is to get Shog9 chat banned for something.

How do we lure him here?

@0celo7 I don't think mods are subjects to autobans, and even if they are, they can easily unban themselves

@0celo7 You don't want that.

Why don't I?

I've been banned for a month, didn't bother me.

You haven't seen the things I've seen.

Jeez, why do I get banned so much, anyway?

@Obliv Welcome back.

Gee, thanks @0celo7

:D

Hello

@Slereah are you getting your PhD?

Am I?

I would be surprised since I'm not doing anything for it

Wasn't there that Vietnamese guy who you were gonna talk to?

imgur.com/gallery/sjQ2C2c

I just moved in m8

I don't even have proper internet or hot water yet

Where'd you move to?

First world?

Where else would I go

@Slereah Don't know. Somewhere without an extradition treaty? Just what have you done, anyway?

Well I have been looking into explosively pumped flux compression generators tonight

Which probably puts you on some lists

Other than that not much

I just moved because Paris is awful

@Slereah Ah. With that kind of list it's not entirely clear that the lack of an extradition treaty will help.

Eh, I don't think people really care about EMPs these days

It seems to have gone out of fashion

Like flared trousers and communism

But just don't accept the aid of anyone who says he can get the parts you're missing. That would be a Three Letter Agency triclk.

Or so it seems from the news.

Psh who needs FBI parts

It's basically gunpowder and electric components

The guys they catch are all losers who would never have made any progress is the cops didn't tell them they could get the [hard to get stuff].

It's pretty rare that terrorists are competent, yeah

At least in the west

I hear that they can get pretty crafty in the middle east

And thank FSM for that, too.

They get more experience, I suppose

Terrorist attacks are almost universally dumb blunt force, though

It's always bombs and guns

and trucks

I'm guessing that EMP bombs aren't terribly interesting for small terrorists, though

Because if you have the material to make them, you can make regular bombs

And you can kill people with regular bombs

I'd hate to be the guy in charge of human resources for some band of 'freedom fighters'.

^thusly

I mean, competent people and volunteers for suicide vest duty must have almost no intersection in paces that aren't already hopeless hellholes.

we invented political terror, baby

I need to get Robinson's book on hyperreal numbers

Can't seem to find a good ressource for them elsewhere

Fugg it's like 70 dollaridoos

Oh well

Let's buy it

Hm, let's see

What else is in my list

It is ordered

It better be good

Anyone knows if there's an easy link between non-archimedean valuations and the usual ordering relations of archimedean fields?

@Slereah I'm not sure what you mean by "link".

Well, in $\Bbb R$, we have $1 < 2$ [Citation needed]

And the distance between them is $|2 - 1| = 1$

But in let's say $^*\Bbb R$, the distance is $d(1,2) = max(2,1) = 2$

Or something like that

Even though it is an extension of it

Is there any relation between those two metrics

Or is it not worth considering them like describing the same thing

As two numbers become closer in $\Bbb R$, they don't really seem to move in $^*\Bbb R$

Because of the max

wait wait wait

What structure exactly is your $d$ on ${}^\ast\mathbb{R}$ supposed to be

An ultrametric

Even an ultrametric must have $d(x,x) = 0$, yours doesn't.

Hm

Lemme think

@DanielSank : Edward Farhi I recognize.

I might be thinking of the absolute value

$|0| = 0$, $|x + y| = max(|x|, |y|)$ if $x \neq y$

@Slereah The second equation doesn't make much sense.

does it not

It's from Toka Diagana

How am I supposed to use it? We have $5 = 5+0 = 4+1 = 3+2$, how do I compute $\lvert 5\rvert$?

A very good question

Hm

Lemme see specific examples

I think he does p-adic norms

My point was that as written, I get the contradictory $\lvert 5 + 0\rvert = \max(5,0) = 5$ and $\lvert 4+1\rvert = \max(4,1) = 4$.