The h Bar

2:09 AM

Is this homework?

user218912

yes but I'm only asking if i'm right

user218912

i did the work myself

You've said math is boring af

why should I help you

user218912

I regret everything I said.

user218912

math is the most beautiful thing in existence.

user218912

2:12 AM

and 0celo7 is a pro mathematician.

I don't need your sass

user218912

xD

@IceLord I need three books. Hardcover Lee, Wolf, Dugundji.

Get me one and I will help w/ homework no questions asked.

user218912

If you were ACM I would. but you said you don't know physics anymore so it's not a good deal.

Ah. You're smarter than I thought :P

user218912

2:14 AM

wow

@IceLord Looks right.

user218912

thanks ^^

user218912

it works too so it must be right

@IceLord Hmm.

Stokes.

user218912

okay

user218912

2:16 AM

one problem though

I'm not sure how you want to use the E-L equations.

user218912

I forgot how to do it with stokes theorem xD

user218912

you showed me last year but bad memory

user218912

sigh... this will take like 5 hours to figure it out.

The derivative term is $\int_M \partial_\mu F^\mu\,d^dx$

user218912

2:17 AM

yes

$M$ being spacetime

user218912

okay so we assume that the fields vanish at infinity right?

Stokes theorem gives $\int_M\partial_\mu F^\mu \,d^dx=\int_{\partial M}F^\mu \,d\Sigma_\mu$

and you assume $F^\mu\restriction \partial M\to 0$

shit

user218912

what does that mean?

user218912

never seen it before

2:19 AM

Well, here's where physics meets math.

Technically $\int_M$ is undefined

it could blow up badly

Well...maybe it could have noncompact support but still be integrable

(btw lots of naive QM arguments fail here)

@IceLord basically just assume $\partial M$ is the "boundary at infinity" and $F\to 0$ as you approach it

so "on" the boundary you get zero

and the integral is zero

There's no rigorous way to prove this.

user218912

ah I see

user218912

thanks @0celo7 I love you

Well, there is. You just integrate over a precompact $D\subset M$ with empty boundary.

The usual physicist action isn't well-defined.

But you got a horribly divergent sum as a "correct" answer. So I give up.

user218912

@0celo7 actually apparently it's not divergent because you can choose $\psi_n(t)$ to be 0 for $n>1$

user218912

leaves only 1 term in the sum

2:23 AM

What the fuck...

That doesn't make it non-divergent.

Wait

user218912

that's what my prof said when I told him.

Lol

Ok, you didn't say $\psi$ was dependent on $n$

And even then...that's not making the series converging. That's just cutting it off at n=0. What the hell?

Of course the series converges if you set all but one term to zero. Sigh.

user218912

xD

@IceLord So...I'd like Dugundji first.

user218912

@0celo7 how much is it?

2:29 AM

Well, depends. A new one is $500 but that's insane.

user218912

._.

Used they go for $65?

user218912

even if I were to buy it for you it's not for this since I could figure it out myself in enough time, it's because I owe you a lot from before.

Wolf would not really be useful now

Lee can wait, honestly.

I'll get it when I write my thesis and I need it daily.

...well i already use it almost daily

user218912

still don't know what kind of engineer you're supposed to be.

2:31 AM

I'm probably not one.

user218912

why don't you just be a mathematician?

they're useless

user218912

wtf

user218912

you're full of contradictions for real.

$H\\M$

Yo what

What did I do

user218912

2:32 AM

@0celo7 for the integral where you used stokes theorem is that the integral for $\delta S$ for the lagrangian?

$a\\b\\c$

WHAT

$H\backslash M$

@IceLord What?

@SirCumference The problem is that a factor 1.0006 would render many pre-existing measurements incorrect and make tools from before the change incompatible with those from after.

user218912

@0celo7 what integral is that?

user218912

you just randomly integrate it?

All of which makes that a major no-no.

2:33 AM

what integral is what

user218912

or is it from somewhere else?

user218912

the integral where you used stokes theorem

oh, you're doing $\mathcal L\mapsto \mathcal L+\partial_\mu F^\mu$, right?

user218912

yes

It's the integral of that transformation.

So yeah it's $\delta S$.

2:34 AM

0celo

You like history, right?

user218912

kk thanks

Why wasn't the speed of light just defined as exactly 300,000 km/s?

@SirCumference maybe

Because it's not 300,000 km/s?

What kind of question is that

But we defined the meter in terms of it

And the meter changed by some crazy small factor, right?

2:35 AM

We could have made the meter a tad bit bigger and the speed of light would be 300,000 km/s

@SirCumference But not until after we'd been using meter for almost 200 years.

@0celo7 $300,000,000/299,792,458=1.00069229$

You can't retroactively change it's length without messing things up. Badly.

@SirCumference At work I measure things in terms of ten-thousandths of meters

So wouldn't the meter just be 1.00069229x bigger?

2:36 AM

And I don't do anything precise

Think about e.g. chip manifacturers

All right, but then again, we did redefine the meter so that the speed of light has no decimals

We'd end up with two standards: pre @SirCumference stupid decision and after

Because there'd be million-dollar equipment that would be useless

@SirCumference But the new definition was chosen to be compatible with existing precision measurements.

@SirCumference That's a lot.

All right, new plan

Define a new unit of distance

We'll call it the "SirCumference"

2:37 AM

@SirCumference You didn't understand what was wrong with the second varying by a few microseconds either

You don't understand something fundamental here and I don't know how to help

Nothing was broken by that move. It's that kind of thing that keeps metrologists busy.

It'll be the SI unit

Sigh...

Fortunately, the charging one has been solved now that we've all standardized on mini-USB. Or is it micro-USB? Shit.

@dmckee hah

Well, there's also the fact that pounds is both mass and force

If someone says "I have a 1,000lb object", is that weight or mass?

We are slaves to our history with all it's oddities (and sometimes inanities).

2:39 AM

I can't figure it out

Stupid Americans

@SirCumference Hey now. Look closely at how governments define the kilogram. For the purposes of commerce most of the (including Europe) define it as a weight.

@dmckee This world makes no sense

Interesting @dmckee

So you hear engineers talk about kilograms-weight and kilograms-mass in exact parallel to pounds-weight and pounds-mass.

@SirCumference Move to Somalia you commie

2:41 AM

@0celo7 It's not me! I swear!

Which makes me want to pound my head on something weighty.

It's the assholes outside in the morning

Or massy :P

Anyway, TIL Tycho Brahe had a pet moose that died after it drank tons of beer, got drunk and fell down some stairs

Our astronomy teacher went over that for like 10 mins

Kinda neat I guess

Now I'm waiting for the punch line...

2:42 AM

Nope, sorry

user218912

@0celo7 since I'm using $\delta S$ shouldn't the integral for the total derivative be something like $\int_m \delta \partial_\mu F^\mu d^4 x$?

no, cuz $\delta\mathcal L=\partial_\mu F^\mu$

user218912

can you explain?

No

user218912

how is that true?

user218912

2:48 AM

please :(

what definitions are you using

I don't see how that's not obvious

what could $\delta\mathcal L$ mean if not that

user218912

I know from the derivation of the regular E-L equations that $\delta L$ is not equal to that.

different $\delta\mathcal L$s.

user218912

but how

user218912

if you take $\delta \mathcal{L}$

user218912

2:51 AM

you expand it out

user218912

and get $\frac{\partial \mathcal{L}}{\phi}\delta\phi + \frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)}\delta\partial_\mu\phi$

user218912

am I being really dumb

Yes

search on PSE

user218912

what's not obvious to me?

there's probably 10 answers on this

user218912

2:54 AM

ok

user218912

can't find it. @0celo7

too bad?

user218912

sigh...

user218912

how am I supposed to know that it's true?

user218912

it's not in the notes or on wikipedia

2:59 AM

I don't know what your issue is, I can't help

user218912

prove that it's true please.

user218912

fine

user218912

don't do it

Prove that what is true?

I actually don't know what your issue is.

user218912

i'll figure it out myself gimme 5 hours.

user218912

3:00 AM

oh

user218912

@0celo7

user218912

15 mins ago, by 0celo7

no, cuz $\delta\mathcal L=\partial_\mu F^\mu$

user218912

why is that true

By definition.

user218912

where is that definition

3:03 AM

I don't understand what your issue is with that.

user218912

i've never seen it

user218912

how did you know that

I'm defining a variation of the lagrangian by $\delta\mathcal L=\partial_\mu F^\mu$

Bro, you asked me to prove that the action does not change when you add a divergence.

I added a divergence.

What's so confusing?

user218912

ooooh

user218912

i'm dumb, thanks for explicitly saying that though i understand now.

3:05 AM

ok

user218912

i'm just tired.

@ACuriousMind Lol, I think we proved the closed graph theorem in analysis today

""""intro"""" analysis

user218912

so is it right to say $\delta S' = \delta \int \mathcal{L}' d^4x = \int \delta\mathcal{L}' d^4x = \int \partial_\mu F^\mu d^4 x$ by definition that the action does not change when you add a divergence term?

user218912

@0celo7

Sure

user218912

3:10 AM

does it make sense?

user218912

full marks?

user218912

@MAFIA36790 hey!!! o/

Don't know about full marks

user116211

hey @IceLord

I would probably fail everyone in the class for not checking convergence, etc. at every step

user218912

3:10 AM

@0celo7 oh god.

user116211

@0celo7 GOOD

user116211

Anyways, Qmech left a comment yesterday:

user116211

Comment to the post (v1): Consider to double check for $x\leftrightarrow q$ typos, and $3N \leftrightarrow n$ summation limit typos. — Qmechanic ♦ 13 hours ago

user116211

Man, I'm still not getting where the mess up is ;/

user218912

@MAFIA36790 is that a good book btw?

user218912

3:14 AM

for mechanics

user218912

i'll need to do a refresher soon since i'm forgetting stuff.

user218912

i wouldn't mind using a new book

user116211

@IceLord At least it is better than Goldstein; I love it.

user218912

kk i'll check it out

user116211

@IceLord There is free pdf available online.

user116211

3:24 AM

And since now the book is published by the Dovers, you can easily comprehend the price ;)

user218912

@0celo7 can you link a reference for learning stokes theorem in index form?

no

user218912

well then...

user228700

3:35 AM

Hihi! Does anybody study Chem. here?

user218912

I did 3 years ago.

user228700

@IceLord If I asked u a small doubt, do you think you'd be able to answer?

user218912

nope but ask anyway.

How do you ask a doubt?

What does that mean

user228700

@0celo7 Huh? O.o

3:39 AM

I'm saying "ask a doubt" does not make sense.

user228700

@IceLord OK, I was wondering if you know what it means really, when a compound has some not charge like in the case of (ClO4)-

user228700

@0celo7 What then, is the correct way to say it?

user116211

The Periodic Table

Haikus are awesome / Chemistry's even better / So pull up a chair

user218912

@KaumudiHarikumar what?

@KaumudiHarikumar Say what? You seem to be asking a question.

user218912

3:42 AM

@KaumudiHarikumar do you mean net charge?

user228700

@IceLord Yes yes, I mean net charge.

user218912

it just means it has a surplus of electrons.

user228700

@0celo7 You just said that "ask a doubt" doesn't make sense. When you want to get your doubt cleared by asking someone a question, what's that called then?

user218912

damn I could have given you such a good answer 3 years ago.

@KaumudiHarikumar Asking a question, clearly.

user218912

3:44 AM

@0celo7 you're clearly a mathematician even when it comes to language.

user228700

@0celo7 I'll get back to you on this.

What is there to back to with?

user228700

@IceLord :-( OK. I've asked at the CSE chat, thanks.

user228700

@0celo7 I'm not entirely sure if u're right. I'll look it up another time and get back to you.

Of course I'm right.

user218912

3:45 AM

@KaumudiHarikumar he's right.

I'm a native speaker, why is this a debate?

user218912

you should have said "ask about a doubt"

user218912

not "ask a doubt"

user218912

it doesn't make sense.

Even that sounds awkward @IceLord

user228700

3:46 AM

@IceLord OK. In any case, I'll look this up to confirm. Thanks.

user218912

@KaumudiHarikumar so you don't believe us...

user218912

that's a waste of time to confirm it.

user218912

@0celo7 nah it's fine imo.

user218912

well

user218912

maybe it's kind of awkward.

3:47 AM

Yes it is.

user228700

@IceLord :P I believe you and all. Just want to make sure.

user218912

since i never heard anyone say that.

Would you ever say "hey prof can I ask about a doubt"

user218912

@KaumudiHarikumar then you don't believe us 0.o

Answer: no

user218912

3:48 AM

@0celo7 that's true @KaumudiHarikumar you should also have given context

user218912

for example

user218912

"hey prof can I ask about a doubt I have regarding ____"

user228700

@IceLord Omg, really? People use the phrase "Ask a doubt" a LOT, actually-even though it may not be correct.

I use "doubt" outside of "I doubt that" maybe once a year.

user218912

@KaumudiHarikumar show evidence for that since I'm a native speaker and I've never heard anyone say that ever.

3:49 AM

@KaumudiHarikumar Certainly not in English-speaking countries.

user228700

@IceLord How do you want me to prove it? People use the phrase here a lot in conversations.

@KaumudiHarikumar Because it's a direct translation of Hindi, which is false.

user218912

maybe in india, because I found this

The phrase "ask a doubt" does not exist in English.

What the hell is Indian English?

user218912

idk man

3:51 AM

My evidence: I Got a Question by YG & Lil Wayne is not I Got a Doubt.

user218912

lol...

user228700

Yeah, no, "Indian English" is not a thing.

user218912

@0celo7 I just don't know why in stokes theorem $d\Sigma_\mu$ has an index

@KaumudiHarikumar What is the Hindi for "ask a question" ?

@IceLord Outward normal vector.

user228700

@0celo7 Actually, I'm not a native speaker of Hindi so I wouldn't know about that.

3:53 AM

Oh you really, really don't want me to prove that. You really really don't want it.

user228700

@0celo7 What, you want me to type here, in Hindi?

user218912

@0celo7 okay :o

@IceLord It took me like a week and I think I have the only full proof of it.

user218912

@KaumudiHarikumar in indian english

No

What is your first language @KaumudiHarikumar

user228700

3:54 AM

@0celo7 I doubt you may have heard of it but it's a language called Malayalam.

Hmm, ok. please type "I have a question"

I want to try something

user218912

he wants to put it into google translate obviously

user218912

and get "I have a doubt"

എനിക്കൊന്നു ചോദിക്കാനുണ്ട്

...

lol that looks strange

user228700

@0celo7 Yeah, no, that's not "I have a doubt" lol.

user218912

3:56 AM

lol @0celo7

@KaumudiHarikumar that's what google gave me

user218912

it looks like georgian

what does it say?

നിങ്ങൾ ഈ വായിക്കുന്നുണ്ട് എങ്കിൽ വളരെ വൈകും

user228700

@0celo7 It's not a direct translation of "I have a doubt". A direct translation of that would be " I need to ask something".

user218912

@0celo7 $d\Sigma_\mu = d\mathbf{r}$?

3:57 AM

@KaumudiHarikumar how would you ask a question in your language?

that's what I want to know

@IceLord jesus, what?

I told you not to make me explain

We will be here all night

user228700

@0celo7 Dude, I should've told you this before getting into this, but I learned how to read and write Malayalam only 3 years ago.

user218912

pls

@KaumudiHarikumar You learned to read and write three years ago?

user218912

fixed^

user228700

@0celo7 xD No! Malayalam alone I learned 3 years ago.

user218912

3:59 AM

@KaumudiHarikumar but you said it's your native language

Then what is your first language??

user218912

so what did you know before that?