@KyleKanos I'm talking in general. Forcing users not to talk about politics even though it is allowed by SE is blatant misuse of power. If we have a poll where majority of users agree with the message in the pinned post then maybe it will be fair. But at present it seems like an order agreed upon only by a few mods/room owners.

@anonymous Ever seen a political discussion in Stack Exchange chat? It's rarely full of smiley faces and unicorns.

@ACuriousMind If I have a uniform magnetic field $B$, then is $A=\frac{1}{2}B\times r$ the vector potential? I don't want to have to calculate, is this well known?

@anonymous You're free to complain on meta or to the SE staff. But several users (the author of the pinned message, the ones who starred it, and Emilio by reposting it) have expressed a desire for this room to be free from politics and by pinning the message the moderators and room owners have decided that this is a reasonable request to make in times like these where political discussions have a tendency to blow up (and have done so, repeatedly, across the SE network, both in chat and on metas).

@0celo7 Since we have gauge freedom, "the" vector potential doesn't exist :)

@ACuriousMind Fine. Is it a function such that $\nabla\times A=B$?

You know what I meant

@ACuriousMind Okay, I'll write a complaint on Meta SE. Thanks for your reply.

@0celo7 I think not, the identity for the curl of two cross products is pretty ugly

@anonymous Yeah, not sure I'd agree with that. I'd warrant that room topics are decided by room owners; in this case the owners feel that politics introduces a certain negative atmosphere they'd rather not appear

@0celo7 yes, $A=\frac12 B\times r$ is a well known expression for a constant magnetic field

(perhaps, up to a sign, I dont remember)

How does one derive it? "By hand" is not the correct answer

one doesnt

Why isn't "by hand" not the correct answer?

one checks that it works

@AccidentalFourierTransform How?

well, calculate its curl

Does Stokes theorem or whatever work?

use index notation

I don't know what index notation is

Ah: Rotate your axes so that $B$ is aligned with the $z$-axis, then observe that $(1/2By, 1/2Bx,0)$ gives the correct $B$-field. (possibly up to a sign)

$A_i=\frac12\epsilon_{ijk}B_j r_k$

$$\text{curl } \left(\textbf{F}\times \textbf{G}\right) = \textbf{F}\text{ div}\textbf{ G}- \textbf{G}\text{ div}\textbf{ F}+ \left(\textbf{G}\cdot \nabla \right)\textbf{F}- \left(\textbf{F}\cdot \nabla \right)\textbf{G}$$

Good lord

$(\nabla\times A)_i=\epsilon_{ijk}\partial_j A_k$

What are $j$ and $k$ doing there?

They're hanging out

They need to get off my lawn

my milkshakes bring all the indices to the yard

^not very funny

@0celo7 ew

@KyleKanos I didn't get it

I am in my usual state of confusion

@0celo7 That's because you're young

So, $\nabla\cdot r=3$?

yes

@0celo7 why's that a problem? half the derivatives are zero

I get a random 3 in there

@0celo7 yeah, but that'll go away with the final term

it's easier just taking an explicit componentwise curl of $A$

Looks like it

but you can also have $\mathrm{curl}(\mathbf B\times\mathbf r)$, with a term in $(\mathbf B\cdot \nabla)\mathbf r = B\partial_z \mathbf r = B\hat{\mathbf z} = \mathbf B$.

Yes, I see that

@0celo7 That'll take the 3 down to a 2, which then cancels with the $1/2$.

Arithmetic is hard sweating emoji

@KyleKanos Yes, I find that argument reasonable :).

So I didn't need the gauge for this at all, using the $A=\frac{1}{2}B\times r$ thingie

@0celo7 well, your $A$ does satisfy $\nabla\cdot A=0$

no it doesnt

wait

no, its an axial gauge, $r\cdot A=0$.

@AccidentalFourierTransform what? $\mathbf B\times\mathbf r$ has zero divergence.

(assuming uniform $B$ of course)

How do you even compute that

@0celo7 compute what?

lol right

The divergence of $B\times r$

We sent an ICBM to the sun :D

@0celo7 You move to a frame where $B$ is along $z$, in which case $\mathbf B\times \mathbf r = (-By,Bx,0)$. That has zero divergence by inspection.

Frames, whatever happened to coordinate free

@0celo7 oh, you can do that if you want to. But if it's true in one coordinate frame, it's true in all.

And incurring a half-hour of pain to do a calculation that's by inspection in the right frame is just incurring pain for no good reason.

I've forgotten what the curl is in terms of differential forms

Let's see

@0celo7 oh wow, that's really asking for pain

Oh, not curl. Divergence

Its just the codifferential

In case anyone want to see the eclipse in August, that band is where you need to be

So $\nabla\cdot(B\times r)=d^*\star(B\wedge r)$

@KyleKanos is that actually what the picture shows?

Now $d^*\sim \star d\star$, and $\star^2\sim 1$

@EmilioPisanty That is what APOD is telling me

They just added it to some cloud cover map

So we get $\sim \star d(B\wedge r)= \star[(dB\wedge r)-(B\wedge dr)]$?

first term is zero

so what is $B\wedge dr$

@KyleKanos huh?

Well I thnk $dr=dx+dy+dz$, so this doesn't work

@EmilioPisanty Uh, yeah helps to see what you're referring to

Sorry

@0celo7 how'd you figure that?

That was the ISS meeting up with the sun, Mars & Venus.

@KyleKanos ah

apod.nasa.gov/apod/ap170204.html

mmm, its actually a 2-form

so that's wrong

it sounded pretty implausible

Well then, $dr=0$, no?

Since $r=xdx+ydy+zdz$

@EmilioPisanty Yeah, seems that way

@0celo7 it seems to me that you're mixing notations and it's leading you badly astray

But apparently someone is trying to turn earth into the Starkiller Base:

@EmilioPisanty I am identifying vectors with one-forms, sure

but the metric is Euclidean so it's fine

$dr=0$ just means that the dyad $\nabla r$ is symmetric.

It's actually the identity, so all is well

@KyleKanos why are we waging war on the sun?

@0celo7 So that vampires finally can go out regardless of day time

@ACuriousMind this isn't Skyrim!

Bethesda should sue whoever did this

clearly an IP violation

love the shadow on the clouds

Why are we posting huge cloud dildos?

ACM works in a BDSM club

@BernardoMeurer Kyle started it

@0celo7 I've missed you

we talk every day

Yes, I mean I've missed you on the h-bar

Did Reb finish her CS homework?

↑ old favourite

idk

just asked

shuttle reentry seen from the ISS

Shuttle reentry seen from ISIS

whats up with the stars people?

Eh, I don't know, space?

Galaxies?

Other planets?

HAHAHAHAHHAHAHA

:-/

@AccidentalFourierTransform :(

i.imgur.com/N2gdXlZ.jpg?1

She finished it

@0celo7 she did? tell me more ( ͡° ͜ʖ ͡°)

@heather are you here?

@0celo7, yep.

@AccidentalFourierTransform it was everywhere

Let us rejoice on this moment :P

lol

#NeverForget

@heather That literally made my week <3

@BernardoMeurer Let's...not. Stay classy, people.

@0celo7 Have you seen the sun? It is a glowing ball of death

@ACuriousMind Really? Classy?

@BernardoMeurer ::sighs::

@EmilioPisanty That's pretty sweet

sassy

@heather Come on, it was one of the great moments, we all have our shameful h-bar quotes :P

@BernardoMeurer I can't believe I made that mistake.

@heather Many more will come, don't worry about it

Ask ACM to quote @0celo7 and you'll see :P

I have no shameful quotes

@BernardoMeurer I don't.

you don't?

@KyleKanos Your existence is a shameful quote

@BalarkaSen No, why would I?

If I thought something was shameful I wouldn't say it

@0celo7 Tell me more

@BernardoMeurer What is that?

@KyleKanos What's what?

@BernardoMeurer That box doesn't look like anything I've seen on SE

@KyleKanos Ah, it's a desktop notification

@heather while I'm not old enough that I made those sorts of comment in unrecordable offline conversations, I am old enough that they only made it into MSN chat logs that are thankfully completely gone.

This right here is public and eternal

well, eternal-ish

@BernardoMeurer Oh, okay

He swiftly deleted the message, the notification stays on for a while though

ah, good old MSN

@BernardoMeurer Huh? Geodesic balls are awesome

Nice exit

@AccidentalFourierTransform yup, back in the good old days of beeping, booping, khzzzzping dial-up modems

hey! I remember that!

@AccidentalFourierTransform It's pretty hard to explain to e.g. a 14-year-old just how frustrating it is to browse at 56kbit

well, how frustrating it was, I imagine it'd be even worse if you tried it today

I loved dial up

I turned off the AOL dial-up sounds

I used to pick up the phone while browsing to hear the music

@heather Quite common at 14 :'D I remember at that age I would go around asking people their sex instead of their gender (on the internet). When I understood my mistake I felt like digging a hole and hiding underground. XD

I've never used dial up

PRRRRR-PIIIIIIIIIIIII-POOOOOON-BEEEEEEEEN

yikes

Dial Up was the OG Dubstep

@0celo7 That's because you're young

@0celo7 Huh?

@BernardoMeurer bee-boop, bee-booop-beep, khhhhzzzzzzzzzzz rrrzzzzzz

@AccidentalFourierTransform ↑

"talk"

@EmilioPisanty I miss that

YES

Speaking of youth, I got a little shocked/scared yesterday when I went to the Total Wine store and it said you had to be born before 1996 to buy alcohol

@KyleKanos No I'm not

@BernardoMeurer beat me to the punch there

Higher quality one

@EmilioPisanty It's music to my ears

good god

why does it make that noise

who thought that was a good idea?

@0celo7 uhhhh

@0celo7 Becuase it used a phone to connect to the internet

hard to explain

theatlantic.com/technology/archive/2012/06/…

Go read

it brings back so many memories

@KyleKanos but someone had to program that noise

why did someone do that?

@0celo7 how old are you btw?

@0celo7 it's all functional

I'm 18 and I had dial up

@AccidentalFourierTransform 15

@0celo7 The sound is a feature, not a bug

SMRPG Remix album came out