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12:00 AM
@0celo7 Because you kept telling people facts about your real-life life and person.
 
^
Yes, Ryan
 
Yeah, so it's more complicated than simply not using a real name
 
@ACuriousMind he hasn't. But as for the question you linked (which I posed), I see what you mean. I figured that perhaps this final part was an easy addition to the actual answer, which is why I included it. It would make sense to make it a follow up question though, once the first one is answered.
 
Stop telling people your real name, Ryan
 
If you want to be anonymoous you have to be committed to that in your on-site interactions.
 
12:00 AM
@Danu wtf how do you know who I am
 
I heard it from your mom
(jk)
 
@dmckee pls ban Danu he's a stalker
 
He might prefer Mr Unger :P
 
I actually believe I know his family name too.
^ That
 
ahhh ban all of them
 
12:01 AM
welp there we go
 
Anyway, the question is already a difficult one requiring a nontrivial amount of time to answer well. I can some times psych myself up for those, and indeed I really enjoy answering questions like that, but there's no way I'm going to write a tutorial on quantum noise spectral density and answer another question.
 
@user129412 just split the posts
 
That sounds fair. Shall I edit out the final part?
 
Please do.
 
12:02 AM
Daniel will fulfill all your wishes and desires
 
I will?
 
You might want to examine my desires first before committing to that
 
Wait, I'm fulfilling @ACuriousMind's desires?
 
BDSM confirmed
 
I am confuse.
 
12:03 AM
I knew the latex duck suit was a sign
 
Also @user129412, as soon as you type "This leads to my first, short question." you should know the question is too broad.
 
@DanielSank Danu didn't specify whose desires
 
@ACuriousMind ~Sigh~
very well.
 
@0celo7 Your imagination is so...shallow
 
@user129412 edit the question and ping me.
 
12:04 AM
@ACuriousMind what does that mean
 
@ACuriousMind Yeah I'm confused. I figured I'd get out that bottle of mead and we'd play a board game or something.
 
I strongly associated with you BDSM, I blame one of your former avatars.
 
@DanielSank Now that does sound nice!
@0celo7 ???
 
@ACuriousMind See, I know what you like, baby.
 
Which one of my avatars carried that connotation for you?
@DanielSank Careful, you're married
 
12:06 AM
@ACuriousMind I can't remember all of them.
Do you have a list?
 
@ACuriousMind We're only exclusive about sex, not board games and/or sharing a bottle of hooch.
 
@DanielSank What about affectionate names like "baby"?
 
@ACuriousMind I don't call her that.
 
@0celo7 ehhhhh, not really
 
@DanielSank I have reduced it to a single question.
 
12:07 AM
@ACuriousMind We call each other "homey" for whatever it's worth.
 
@DanielSank Hey... homey
 
Is pronouncing "honey" that hard? ;)
 
must make for great pillow talk :3
 
ok this is getting weird...
 
Yes, time for me to bow out
cyas
 
12:09 AM
oO
Never seen Danu flee weirdness before :P
 
I'm getting out before I'm sucked in too deep
Don't wanna go too deep, homey ;)
 
I think you just duck out because you want to slightly better your sleep schedule
 
@ACuriousMind Very different meaning.
 
you and your logic... GET OUTTA HERE
 
@Danu Interesting how it sounds so sweet from Anna but so amazingly horrific coming from you.
 
12:11 AM
Ok, let's switch topics.
 
Must have been the winky face.
 
@Danu They'll have to carry me out of here feet first!
 
@DanielSank You'll get used to it
Anyways, I should stop freaking y'all out
Bye!
 
@ACuriousMind Let $V_p$ be the vertical space of a principal bundle $(P,M,G)$ and let $\bar{ \mathfrak g}$ be the fundamental vector fields evaluated at $p$. I can see that $V_p\cong \bar{ \mathfrak g}$ by arguing that the map $\bar{ \mathfrak g}\to V_p$ is injective, hence surjective since they are of the same dimension. But can one show directly that it's surjective? i.e. given $v\in V_p$ there is a fundamental vector field $\bar X$ such that $\bar X(p)=v$?
 
12:24 AM
So, $p$ injects $G$ into $P$ along the fiber over $\pi(p)$.
Well actually, I think that's a diffeomorphism onto its image
So the differential is an isomorphism onto its image (maybe)
 
12:48 AM
@ACuriousMind Ahhh, I can't prove it's surjective! I have another proof that it's injective
Bishop-Crittenden say that for $t\in V_p$, there exists $\bar X\in\bar{\mathfrak g}$ such that $\bar X(p)=t$, since $p$ maps $G$ onto $\pi^{-1}(\pi(p))$.
Now, I'm pretty sure that map $G\to \pi^{-1}(\pi(p))$ is a difeomorphism, so the dimension of the image of $dp$ must be at least $\dim G$
Since each fundamental vector is vertical, this shows that any vertical vector is a fundamental vector.
Hmm, but we don't know that the vertical space is $\dim G$-dimensional
Eh, work in a trivialization to show that it is.
 
@kaylagilmour Sup
 
user218912
@BernardMeurer they don't have enough rep to chat...
 
@IceLord Shame, she was in like 1000 rooms lols
 
1:11 AM
@knzhou You're my idol
 
@BernardMeurer I thought I was.
:(
 
what did he do?
 
@DanielSank You're more of a god
@0celo7 He helped me with computation theory once
 
I've helped you with math way more than once
 
@0celo7 Well yeah but you're my friend and you wear hideous shoes
@knzhou Is some dude
meh I mmeeed sleep
 
1:16 AM
hideous shoes my ass
 
user218912
I still didn't figure out how to take the d'alembert of $e^{-i k \cdot x}$.
 
lol
 
user218912
I tried things.
 
user218912
just not the right things.
 
user218912
can somebody help please :(
 
1:18 AM
what is the definition of the dalembert?
 
Where's @ChrisWhite? I really miss the guy
 
he deleted his account
 
@BernardMeurer He requested that his account be deleted a few weeks ago.
 
@dmckee What?
Why?
 
@BernardMeurer If he explained it wasn't to me.
 
1:19 AM
@dma
 
No one knows
 
@dmckee That sucks man, I really liked him
 
We can speculate
 
Nah, not interested in that
 
The most positive spin I could put on it was that he thought it would distract from a job or a job hunt.
 
1:19 AM
I genuinely appreciated him
 
A less positive spin involves a conflict with another user.
 
@dmckee Yeah right...why delete for that?
Just don't use the site
@dmckee This.
 
user218912
@0celo7 $\square$
 
@IceLord I can't help you then
I don't know what box means
 
@0celo7 Some people find it very hard to self-regulate their internet usage.
Not that I think there is a high probability of that explanation.
 
user218912
1:21 AM
@0celo7 sorry it's $\partial_\mu \partial^\mu$
 
user218912
do you want it expanded form?
 
@dmckee He has a PhD in astrophysics. Somehow I think he has the self-control to not use the internet for a while.
@IceLord I don't want anything
 
@dmckee Both of those seem rather unlikely
@dmckee Man this really sucks
 
user218912
@0celo7 I do though.
 
user218912
$\frac{\partial^2}{\partial t^2} - \nabla^2$
 
1:22 AM
@0celo7 No good. I've known too many <s>Spaniards</s> Ph.Ds.
 
Spaniards?
 
Hell. Strikeouts don't work. Meh.
 
@dmckee use --- on either side.
 
@dmckee they do
 
user218912
@0celo7 are you helping me?
 
@IceLord you can help yourself
 
user218912
help me help myself.
 
@dmckee Good movie.
 
@IceLord He's mine
 
But the wrong clip. I'll try again.
 
user218912
1:24 AM
@BernardMeurer nah I've known him for longer.
 
@IceLord Just compute the derivatives
I don't see the issue
 
user218912
@0celo7 I know that, but in the answer there is like a $k_\nu$ term.
 
user218912
I have no idea how that comes from the derivative.
 
@0celo7 Hmmm now I just have to remember that.
 
1:25 AM
@dmckee I still don't get it.
@IceLord is the answer $k^2 e^{-i kx}$
modulo a sign
 
@0celo7 I've known a lot of Ph.Ds. They all have something special going for them, but self-control in the face of an internet interest isn't uniformly one of them.
 
user218912
@0celo7 no.
 
@IceLord what is the answer then
 
user218912
@0celo7 the answer is $k^2 \epsilon_\nu$ where the epsilon is the coefficient for the exponential.
 
user218912
no exponential though
 
user218912
1:27 AM
i don't get it.
 
is it evaluated at $x=0$ or something
@dmckee Oh, I see.
 
user218912
maybe... it doesn't say.
 
user218912
I guess that's the only way.
 
pic
 
user218912
@0celo7 pic
 
1:28 AM
Are you telling me you can't divide?
 
@IceLord I've met both his sisters & stayed at one of their houses
 
Because that's sure what it looks like.
 
@0celo7 M. invited me for thanksgiving lol
But I dunno if I could possibly make it
 
what the fuck, I didn't get invited
 
user218912
oh fuck
 
user218912
1:29 AM
they just divided it out
 
I get to drive 9 hours back to virginia
 
@BernardMeurer wat
 
user218912
@0celo7 so this whole time I thought I was doing it wrong because they divided it out.
 
@IceLord Yes.
How long did you spend on that?
 
@0celo7 I got invited lol, I wish I could go, I want to meet Kevin
 
user218912
1:31 AM
like 30 minutes total.
 
@knzhou You're my idol
 
@BernardMeurer What? Kevin is going to my parents'.
 
user218912
trying new derivative techniques i invented myself to get the exponential to disappear.
 
@0celo7 Isn't he?
@0celo7 Also K. I really want to meet her
 
Yes, Kevin and I are going to Virginia.
 
1:32 AM
@0celo7 Yeah, I want to meet Kevin
 
user218912
@BernardMeurer how?
 
You're not coming to Virginia...
 
user218912
that's really weird.
 
@0celo7 Probably not lol, I',m just saying it'd be nice though lol
@IceLord I'm weird and I really love his sister & her husband
Ron is awesome lol
 
user218912
@BernardMeurer how did you find them?
 
user218912
1:33 AM
coincidence?
 
I agree @IceLord it's weird.
 
@IceLord When I was in LA earlier this year @0celo7 got me in touch with his sister who lives there because she works at GitHub
 
user218912
oh, makes sense.
 
@IceLord And idk I thought she was super cool so we became friends lol
 
user218912
I will definitely meet 0celo7 irl someday.
 
user218912
1:35 AM
idk how but I will
 
I've also had extended conversations with his father
 
that's creepy af
both of you are creepy af
 
@IceLord He has the shittiest shoes
 
user218912
@0celo7 no, isn't it fair to see the person you probably messaged 5k times irl.
 
no, it's not
@BernardMeurer You and R should just form the "hate on Ryan's shoes club"
 
1:37 AM
@0celo7 Come on, I've lived for a few days with your sister lol
 
user218912
@BernardMeurer okay... that is creepy af if anything.
 
yes it's very creepy
 
@0celo7 Me and R made talked about you for a long time back in LA lol
 
user218912
@0celo7 you got yourself into it xD
 
what does that sentence mean
@IceLord regretting it
 
1:38 AM
@0celo7 K and M have adopted me basically lol
 
ok wait
 
@0celo7 It means we mocked you
 
my R not M's R
@BernardMeurer I don't see what there is to mock.
wait who is K?
I have two Ks in my family
 
@0celo7 Your sister
 
also there's 3 Rs
well, 2 Rs and one B
 
1:39 AM
@0celo7 R is M's R
 
did you congratulate him today?
 
@0celo7 Why?
 
Michigan is 5-0
 
Sweet, I'll congratulate him lol
 
do you even know wha I mean
 
1:41 AM
@0celo7 That's wrong man lol
@0celo7 yes, Michigan Wolverines
 
user218912
@0celo7 how about the second term?
 
user218912
why is there a $k_\nu$ and $k$ dotted with $\epsilon$
 
user218912
how does that come out of a derivative?
 
not telling you, sorry
 
user218912
why? :(
 
1:51 AM
cuz I can do it in my head
it's e-z
 
user218912
idk how to do it though, if I see it once I'll be able to do it again easily.
 
just take the derivative
what more do you want?
 
user218912
please show me the steps.
 
user218912
please
 
user218912
this isn't homework.
 
1:53 AM
@BernardMeurer what do you think?
 
user218912
@0celo7 you big meanie
 
user218912
how am I supposed to learn
 
@0celo7 About what?
 
should I help him?
 
Hmm
@IceLord What have you tried?
 
user218912
1:58 AM
@BernardMeurer I took the derivative of the exponential and used chainrule.
 
user218912
I don't get that answer.
 
@IceLord let me see what you got
 
user218912
I got the same thing as the first term.
 
user218912
means I'm using indices wrong.
 
SHOW ME
 
2:00 AM
^ that
 
I can't read your mind
 
user218912
I got $k^2\epsilon_\nu$
 
user218912
again
 
user218912
idk how indices work in this situation
 
@IceLord Dude how did you get that
Show your development
 
2:01 AM
@BernardMeurer Good!
 
@IceLord It's pointless to help you if we can't point what you're doing wrong
We need to see your mistake to understand what you're not getting
 
Exactly.
 
user218912
I did $\partial_\nu \partial_\mu \epsilon_\nu e^{-i k\cdot x}$
 
@IceLord I will no longer help until I see every single step
 
@dmckee Do you have a contact email? I want to ask you some site-related things
 
user218912
2:04 AM
from that I first took $\partial_\mu$ of it
 
user218912
and I got $\epsilon_\nu ke^{-ik\cdot x}$
 
user218912
but idk how to take the $\partial_\nu$
 
user218912
does it contract with the epsilon?
 
hmm
what's the problem statement
 
user218912
to do $\partial_\nu \partial_\mu \epsilon_\nu e^{-i k\cdot x}$
 
2:06 AM
I want to check my answer
I will tend to help if I cannot get the right answer
 
user218912
fuck I'm retarded.
 
I could flag that and get you suspended
 
user218912
it's $\partial_\nu \partial_\mu \epsilon^\nu e^{-i k\cdot x}$
 
lol
 
user218912
@0celo7 wanna bet.
 
user218912
2:07 AM
that won't get me suspended.
 
remove it, I'm being triggered
 
user218912
even worse now.
 
the r word offends me
 
user218912
@0celo7 lol?
 
user218912
it's directed at ME.
 
user218912
2:08 AM
it doesn't offend anyone.
 
I won't flag because we have too many flags already and we don't want Shog to show up and ban everyone
but I've gotten suspended for saying I like a little pain with my women
anything can get you suspended here
 
user218912
no because that's offensive.
 
user218912
mine is not offensive to anyone.
 
to whom?
 
user218912
idk what you mean by that so idk.
 
user218912
2:12 AM
it just sounds offensive.
 
user218912
probably to feminists.
 
ok so are you done now?
 
Hey guys, anybody know what an equation of the form u_xx-u_tt=1 is called? Its almost the famous diffusion equation, but that would be u_xx-x_t =1...
 
=1?
 
user218912
@0celo7 $\partial_\nu \epsilon^\nu$ is 0 right?
 
user218912
2:14 AM
because epsilon is constant.
 
I don't know.
 
user218912
but you said you could do it in your head.
 
@0celo7 I just choose some kind of constant
 
@Mikhail I'm unfamiliar with the equation you are calling a diffusion equation
Usually diffusion is u_xx=u_t
@IceLord no this is PhD calc 3
 
user218912
@0celo7 ._.
 
user218912
2:18 AM
I just don't get how a $k_\nu$ appears.
 
user218912
how does it get an index?
 
user218912
oh shit
 
what bothers me is that no one had to explain this to me. I'm not trying to be an asshole, but I am really confused how you're confused
 
user218912
I figured it out.
 
user218912
well part of it.
 
user218912
2:29 AM
now I don't know how the dot product term appears.
 
I give you...2 minutes to show all your work
 
user218912
$\partial_\nu \partial_\mu \epsilon^\mu e^{-ik\cdot x} = \partial_\mu k_\nu \epsilon^\mu e^{-ik\cdot x}$
 
user218912
then idk how to proceed.
 
user218912
oh I know
 
user218912
are you taking $\partial_\mu(\epsilon^\mu e^{-i k\cdot x})$
 
user218912
2:36 AM
because then it becomes $k\cdot \epsilon$ i think.
 
user218912
which is what I wanted.
 
user218912
am I right or wrong? @0celo7
 
dude show me the answer
let me make sure what I'm doing is right
 
user218912
the answer is $k_\nu k\cdot \epsilon$
 
the FULL answer
ok as I thought
you're doing the chain rule wrong
 
user218912
2:39 AM
:(
 
that's not your issue here, but you are doing it wrong
 
user218912
where did I make a mistake?
 
you need some $i$s
 
user218912
but don't you remove them in the end?
 
no
they go away my themselves
and you need to prove that
 
user218912
2:50 AM
k I give up
 
user218912
@0celo7 i know
 
what did you try?
god dammit that's the last time I'm asking that
I just won't respond any more
 
user218912
21 mins ago, by IceLord
$\partial_\nu \partial_\mu \epsilon^\mu e^{-ik\cdot x} = \partial_\mu k_\nu \epsilon^\mu e^{-ik\cdot x}$
 
user218912
is this right so far?
 
no
 
user218912
2:55 AM
which part is wrong?
 
Last chance. Let $f:\Bbb R\to\Bbb C, x\mapsto \exp(-\mathrm ixk), k\in\Bbb R$. Find $f'$.
 
user218912
$-ike^{-ikx}$
 
Ok, so you do know the chain rule.
 
user218912
yes...
 
Now figure out the rest, no more help from me.
 
user218912
2:57 AM
wat...
 
user218912
but there are indices.
 
user218912
just tell me how to treat them
 
user218912
use 1 example
 
See you tomorrow.
 
user218912
:|
 

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