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12:00 AM
There's gonna be a Hodge star in there somewhere
 
@ACuriousMind uh, what is that equation
 
user54412
@0celo7 what is there to know about it?
 
And it "changes its sign" from the Riemannian to the Lorentzian case, which essentially inverts the sign of hte area element vector $\mathrm{d}S.
 
@ChrisWhite how to pick the direction of the normal vector
 
@0celo7 lol, I fucked that one up
 
12:01 AM
Wald, Carroll and HE don't bother explaining
 
It's supposed to be the divergence theorem :(
 
lol
 
But apparently I'm a bit more drunk than I thought :D
 
@ACuriousMind it follows from what I did in that post
 
But...isn't the question about the "orientation of the normal" about what $\mathrm{d}\vec S$ is?
 
12:03 AM
@ACuriousMind hmm
 
Or the $\vec n$ if you insist in writing $\int\int \vec F \cdot \vec n \mathrm{d}S$
 
I don't think I understand why it has to be outward directed in that case either...
 
user54412
I'm not following at all -- last time I applied the divergence theorem, all my vectors pointed outward, as they always do.
 
@ChrisWhite Why?
According to Wald, HE and Carroll the normal points inwards in some cases
 
user54412
Because the boundary inherits an orientation from the thing it is a boundary of?
 
user54412
12:08 AM
@0celo7 o.O
 
@ChrisWhite exactly
@ChrisWhite exactly
 
@0celo7 usually they don't, but in some case it may be useful
it depends on the approach, and on the problem at hand
 
Dammit, I found a thread on MO last week that led me to a handout that wonderfully explained the passage between the diff. form version and the vector calc version, and from the appearence of the Hodge star in there it would be obvious why the direction of the normal changes in the Lorentzian case
 
@ChrisWhite do you care enough to want the page no.s?
@ACuriousMind :(
 
user54412
@0celo7 sure
 
12:11 AM
I can't find it. I mean, I am on the pages I know where the link should be, but I can't find it...
 
It has to be obvious, it's just nonobvious enough to confuse the hell out of me
 
@0celo7 actually, it probably is really just triangle inequality (provided that the function $f$ is sufficiently smooth)
 
@ChrisWhite Wald: 434, HE: 50, Carroll: 455
Now I wouldn't trust anything Wald says about diff geo.
But HE...that's convincing.
 
Ha!
Found it
Anyway, I think the main point is in the divergence.
The divergence operator is $\mathrm{d}{\star}$
In the Lorentzian case, the Hodge in there will introduce a minus relative to the Riemannian case for either timelike or spacelike things
 
either?
 
12:17 AM
The fact that the normal points inward in one of the cases has to come from this because there's nothing else in Stokes theorem that even uses the metric
 
do you mean one or the other?
or both
 
@0celo7 Yes, I mean one or the other
It depends whether your Lorentzian signature is mostly plus or mostly minus which one it is
 
@ACuriousMind so you tell me about this wonderful thing and then don't link it
nice
 
Not quite as useful as I hoped
But nevertheless, it shows where the Hodge enters
 
@ACuriousMind yeah, that's what I need
what page?
 
12:20 AM
4
 
what
 
It's just in the definition of the divergence
 
wth is $dV_X|_U$
 
@0celo7 The canonical volume element on $X$ restricted to $U$
 
@ACuriousMind I could have figured that out
umm
now what
 
12:22 AM
0
Q: How many downvotes does this answer have?

user1717828I posted this answer, which I suspect is getting downvotes because I have a bunch of karma from it but it only has a score of 3. A commenter leads me to believe people find the answer unsatisfactory because they are mixing up the result of Bell's Inequality with the concept of measurement resolu...

 
@ACuriousMind sorry, I don't see how this helps
what is $\omega_F$
$F^\flat$?
 
@0celo7 Reading your post, I think it doesn't help you at all because your approach to Stokes is completely different
you should think about $\langle X,N\rangle = X\wedge{\star}N$ to get the Hodge, I think
 
@ACuriousMind I used a very elegant method
but I can change that
@ACuriousMind you mean $\langle X,N\rangle\mu$?
 
user54412
> Now recall (by definition!) that the outward unit normal field $\hat{N}$ along $\partial X$ satisfies the property $\langle \hat{N}, \partial/\partial x_n \rangle > 0$ at each point of $\partial U$.
 
@0celo7 sigh...yes. I should probably stop typing, the formulas aren't coming out right.
 
12:25 AM
@ACuriousMind ok, so we get $i_X\mu=X\wedge\star N$?
and...what about this?
wait that's wrong
 
$i_X\mu=i_N(X\wedge\star N)$
better?
@ACuriousMind that does not help
once again, the result is independent of the sign of $N$
so I'm just confused all to hell now
@ACuriousMind Oh, it relies on the metric at more than just $\star$
that $X$ should really be $X^\flat$, and $N$ itself is defined via the metric
so Stokes theorem does not depend on the metric, just everything else does :P
 
Stokes' theorem doesn't
The expression in terms of normal vectors does (by the sign)
 
@ACuriousMind I just said that
@ACuriousMind where?
the sign of the normal vector falls out
in Riem geom one fixes the sign of $N$ by demanding a positive measure on the boundary
 
Wait a moment
How do you define "inward" and "outward"?
 
12:34 AM
@ACuriousMind um
no data
 
I mean, the statement "you have to use the inward normal" is completely meaningless if no one tells you what "inward" means
 
I could give a contrived definition
 
Well, if you want to prove that "you have to use the inward normal", you have to define what that means first, no? ;)
 
@ACuriousMind you know, I kinda saw this earlier
 
Well, I guess, physicists don't
 
12:37 AM
@ACuriousMind dunno if I said this already, but I talked to my lit prof, and he said that moral lessons for young women are a characteristic of early English/American novels
he didn't know if German ones had the same thing
 
@0celo7 No, you did not say that, but it is interesting
The German enlightenment novels/plays don't have that at all
 
@ACuriousMind not just enlightenment
all early novels
 
At least not in any way that would have been obvious to me, and we never discussed that
@0celo7 Hmmm
 
the English novel only really existed starting in 1750s or something like that
it came to America in the 1780s-ish
 
anybody here knows programming
 
12:38 AM
Well, there was Effi Briest where I don't remember what its moral was because it was the most fucking boring thing I have ever read
 
@ACuriousMind Wow, and you've read Schutz.
 
@0celo7 I devour fictional books. Effi Briest is the only fictional book I have ever actually stopped reading because watching paint dry would have been more interesting
I have to this day no idea what it actually was about, or what its moral was, if any
The only thing I remember is a three-page description of a flower pot
 
must have been one fucking awesome flower pot.
 
any programmer here 1 simple question i want to ask
 
@sharafzaman ping @ChrisWhite
 
12:41 AM
@0celo7 It was a fucking ordinary flower pot. The author has the infuriating propensity to describe the most mundane things in the most mundane way
 
@ACuriousMind should I give up on this thing
 
In the most lenghty way possible
 
@ChrisWhite are you here?
 
I got nerd sniped but now I realized I wasted so much time on it
and I don't really care
 
@0celo7 I think it would be very useful to have an answer explan wtf all those books mean when they say "you have to choose the inward normal"
 
12:42 AM
@ACuriousMind hmm
@ACuriousMind well, I tried
 
But I have no idea how to answer that because I have no idea how one defines "inward" or "outward" in the Lorentzian case
 
@BernardMeurer you around
@ACuriousMind Hmm, maybe you pick $N$ as having the "natural" direction
and that's outward
but wtf is natural
 
@0celo7 that would have been my follow-up question
 
@ACuriousMind how about this
let $\theta:\Sigma\to M$
 
In the Riemannian case, I can see how to weasel out of this with "naturality", in the Lorentzian, I don't
 
then there exists an atlas with a chart in which $\theta(\Sigma) $ is $x^0=0$ locally
 
I like that people are trying at perpetual motion via electrical circuits :)
 
not necessarily time, just some coordinate
then pick $N^\flat=\alpha\mathrm{d}x^0$
$\alpha>0$
then $(x^1,...)$ inherit the orientation
 
@DanielSank In my experience, that's one of the users not worth interacting with. They have a not-so-hidden agenda, and they aren't here to learn something.
 
wait, that makes no sense either
 
12:48 AM
@ACuriousMind The one who posted the bounty?
 
@DanielSank Yep
 
@ACuriousMind That sucks.
 
@ACuriousMind And the two books specifically devoted to Lorentzian geometry don't cover integration, wtf?
Gorgoulhon also just claims it, lol
 
@DanielSank It does. Not so much as those who use answers to push their agenda, though.
Where "agenda" can be anything from personal pet theory to "quantum mechanics is wrong and everything is explained by classical physics if you just think hard enough"
 
@ACuriousMind Hmm, I like that second one.
 
12:55 AM
@0celo7 Everyone does, that's why all the questions about determinism get so many upvotes.
 
GR sucks.
 
@ACuriousMind That really pisses me off, actually.
 
You can't even integrate in GR without issues.
 
That's why I was mentioning the idea of a close reason which says "This is just totally wrong."
 
Who the hell decided to make spacetime Lorentzian, anyway?
Probably the biggest design flaw ever.
 
12:56 AM
@DanielSank Well, that's what unclear what you're asking is for, imo
 
@0celo7 Oh, that was Steve.
@ACuriousMind I meant for answers :)
 
Hah, well, we don't "close" answers :D
 
@DanielSank Steve can go jump off a cliff.
@DanielSank That's what your downvote is for.
 
@0celo7 Without an upper speed limit, the world would be hella weird. Think about a world where more energetic photons moved faster.
 
@ACuriousMind So?
We could at least integrate properly :P
 
12:57 AM
@0celo7 Yes.
 
@0celo7 I'm saying a Lorentzian world is conducive to life ;)
(And no, I can't back that up, and that's not really a meaningful statement)
 
@ACuriousMind If we weren't here we wouldn't have to worry about integrating on some contrived space.
 
@0celo7 A single downvote only does so much. It doesn't stop other people from believing that crap ten other people upvoted because they want to believe it's true.
It even gives the poster the feeling of being "oppressed" by actual physicists who just dislike that he disproved their wonderful theory :P
 
@ACuriousMind We've managed to downvote certain Einstein advocates into oblivion.
 
Okay, I won't say anything more because I don't want to defend myself from the accusation of coordinating downvotes when I next sign on
 
1:02 AM
@ACuriousMind Skype?
 
I also should go to bed
@BernardMeurer No, I have 5.5 hours of sleep left to catch, and I'll catch them now
 
@ACuriousMind you break my heart
 
@ACuriousMind Since when do you need sleep?
 
@0celo7 I'm only one here so long if I don't have to get up in the morning
I have to, this time
Lecture on non-perturbative Yang-Mills thermodynamics
 
eww
 
1:04 AM
It's awesome :D
 
in the five minutes that have passed you could have taught me to pronounce your name
that's all we want
 
Try again next week or the weekend, I don't have to get up early there (at least not that I'd know now)
G'night
 
@ACuriousMind just know well love you anyway
 
^ Except for me.
 
If you look at:
$$S = - m \int ds = - m \int \sqrt{-g_{\mu \nu} \dot{X}^{\mu} \dot{X}^{\nu}} d \tau = - m \int \sqrt{- \dot{X}^2}d \tau = \frac{-m}{2} \int \frac{-2\dot{X}^2}{\sqrt{-\dot{X}^2}}d \tau \\ = \frac{1}{2} \int \frac{\dot{X}^2 + \dot{X}^2}{\sqrt{-\dot{X}^2/m^2}}d \tau = \frac{1}{2} \int \frac{\dot{X}^2 - m^2(-\dot{X}^2/m^2)}{\sqrt{-\dot{X}^2/m^2}}d \tau = \frac{1}{2} \int (e^{-1} \dot{X}^2 - e m^2)d \tau$$
Everything up to the last equality makes sense, but then in the last equality it's like they just define something complicated by $e$ out of nowhere, I know you can get this from Lagrange multipliers and equations of motion, but if you just had this equaity, how would you ever think to define that thing by $e$?
Seems to me like the point is to just play with the action until you have $\frac{1}{m}$ so that you are dividing by mass so that in the case of massless particles that $e$ term becomes infinity so we can call it a new particle, but that's f*cked up and can't be right, right?
 
1:18 AM
@ACuriousMind I can define inward/outward.
 
user54412
1:47 AM
@sharafzaman in this room, you should just ask your question -- people will respond to it if interested
 
user54412
no one wants to commit to answering a question before hearing it
2
 
2:07 AM
What's the haps?
@sharafzaman Oh yeah seriously never do that on the internet. Just ask the question. Asking "can I ask a question" is nothing beyond annoying :)
 
user54412
2:18 AM
you know what's the worst? debugging code that only develops problems when using 30,000 threads
 
@ChrisWhite I was gonna say integrating on Lorentzian manifolds
 
2:29 AM
@ACuriousMind oh my god the ending of this book
the villain has some incredible power that explains everything, but then the author doesn't explain everything
what the hell
@ChrisWhite please help me with this integral
 
user54412
what would a programmer know about integrals?
 
@ChrisWhite you're a GR-ist
@ACuriousMind I can define inward!
@ACuriousMind At least, I can do it at a point.
 
2:44 AM
 
@Secret your diagrams are getting better
 
user54412
plot twist: that diagram was meant to define "inward" on Lorentzian manifolds
 
@Secret crappy diagram
 
@0celo7 @ChrisWhite
Rotational mechanic confusion again:
I knew from experience that the left end of the rope will not have tension pulling upwards. But I don't know how to rationalise why, as all I can see is the pulley basically rolling without slipping on the rope and the weight is directed downwards
therefore how is the left end of the rope remain slack by the rolling motion of the pulley?
 
PS plot twist is wrong this time (there is no plot twist)
 
user54412
2:49 AM
uhh, pretty sure there is tension?
 
^
@Secret wait you're in new zealand
or Australia
or something
mb gravity pull up down there or something
 
Because I kinda remember when I pull just the right end, the left end does felt a bit loose (and the pulley being lifted up by the pulling of the rope), except since I am bad at wrapping my mind about rotating objects, I don't know how to rationalise it
@0celo7 Yes, I live in Australia
 
@Secret ok pulleys behave differently down there than up here
we can't help you
@Secret wait, how large is $F_A$?
 
3:13 AM
Can you guys check this, any comment or answer
1
Q: Before and During the sunrise

CuriousLast year we went to a remote place near forest to a site with dark and clear skies. I captured two photos of the East horizon before and during the sunrise. I have questions regarding the effects you can see here. In the above image you can see the sky seems Dark above the horizon above it is ...

 
user54412
@ACuriousMind I was tempted to approve this -- might get you a few downvotes to stem the tide :p
 
@Secret If the pully has non-trivial mass then it has non-trivial moment of inertia. If it also is accelearting (and thus has an angular acceleration) then something must be exerting a torque on it.
That something is the rope, and that explains why the tension in the two sides is different.
@ChrisWhite I suppose there must be an existing data-explorer query for fraction of edits approved broken down by anonymous, very low rep, and non-trival rep but still too little to edit without review.
 
@dmckee dood Dr. dmckee how do you integrate on a Lorentzian manifold
 
I don't know. How do you integrate on a Lorentzian manifold?
 
I don't know!
 
3:28 AM
(I assume that was a set up and I was meant to be the straight man, right?)
 
@dmckee No, I really don't know
ACM doesn't think it's well-defined and @ChrisWhite refuses to do anything other than programming
 
Neither do I. I don't understand where you get this idea that I know the esoteric math of GR. I'm a complete dunce about this stuff.
 
so you know it's GR
 
I would desperately like to have the freedom to go back and takes some classes about it, but that is not in the cards for the time being.
 
that's more than 99% of people
 
user54412
3:30 AM
@dmckee the real problem is I don't think we can tell the other half -- what's the breakdown of edits that are rejected
 
^ Bummer. Maybe the team has a way to extract that kind of data.
 
user54412
I've voiced my opinion before on the matter -- that anonymous edits are pretty useless -- but I've gotten lots of backlash to the tune of "show us the data or we won't believe you"
 
Guys anything about the question ?
 
@ChrisWhite those people would make poor string theorists
@dmckee did I ask you something else about GR lately?
I feel like I have
 
You asked me about Hodge duals just last night. I don't know them from Adam.
 
3:33 AM
@dmckee oh, right
 
user54412
@Curious patience -- the bulk of the user base is asleep right now
 
user54412
also, I'm not sure the answer is anything more exciting than "clouds and haze"
 
@ChrisWhite I think I've seen a couple of gems that came in that way, but it mostly cabbage leaves and banana peals with the occasional used diaper thrown in for good measure.
 
@dmckee Oh, the Hodge dual is the unique map $\star:\Omega^p(M)\to\Omega^{n-p}(M)$ that satisfies $\alpha\wedge\star\beta=\langle\alpha,\beta\rangle \mu$.
 
@ChrisWhite we were out for astronomical observation the sky was clear ... I have too observed it at different places too .. There are few timelapses of sunrise where you can observe similar thing which I asked for
 
user54412
3:37 AM
I've never seen such a thing myself. I've also never seen a sunrise except over the mountains.
 
user54412
Hmm, it's nearly 70 with a raging thunderstorm outside. In February.
 
@ChrisWhite gg NJ
@ChrisWhite Did you know that $\nabla\omega=\frac{1}{2}(-\mathrm{d}\omega+L_{\omega^\sharp}g),\omega\in\Omega^‌​1 (M)$
what does that exponent not work :V
 
user54412
@0celo7 you have \Omega^<200c><200b>1 where those are the "zero width non-joiner" and "zero width space" codes
 
what
$\Omega^1$
it works there
 
user54412
yeah cuz there's no hidden unicode in that one
 
3:48 AM
how can there be hidden unicode
are you hacking me
 
user54412
if anyone here can hack you, it's @ManishEarth
 
@ManishEarth stop hacking me
 
lol
 
@0celo7 Roughly > weight of the pulley such that the pulley accelerates upward
 
@ManishEarth you think hacking me is a laughing matter?
 
3:57 AM
I kinda understood how if there's an angular acceleration it means a torque was exerted on the rolling object and that torque has to be from the rope. But I am not sure how to resolve the tension of the right end of the rope further so that I can essentially draw a free body diagram for the point mass at the point of contact where the pulley touches the left end of the rope to show that its tension is a lot smaller than the right end

(NB assuming the rope's mass is negligible because this is some steel pulley)
 
The torque exerted is equal to the difference in tension between the two ends of the rope times the radius at which the rope winds around the pulley.
You put that in with the usual Newton's Laws stuff you get from the freebody diagrams and then algebra the #*&^ out of it.
 
ok I see
 
Speaking of hacking I have this following case resolved
Feb 5 at 13:29, by Secret
`````````
That was not me being hacked by someone else. That was mum cleaning up my keyboard and accidentally key in this message and with the cloth hit enter
 
@ACuriousMind she just sent me a snapchat in which her socks did not match
sigh...
@ChrisWhite where's the answer where you destroy the misbelief that GPS needs GR
either you or someone like you
 
@ChrisWhite Vielen thanks
@ChrisWhite how do you define the outward and inward normal
@ChrisWhite Hmm, doesn't it fail in dimensions greater than $2$ that the "inside" and "outside" of a hypersurface are topologically disconnected?
@ACuriousMind Let $X$ be some subset of $M$ and $N\bot T_p\partial X$. Let $\gamma:(-\epsilon,\epsilon)\to M$ be a curve with $\dot\gamma(0)=N$.
 
user54412
4:38 AM
pretty sure a closed (n-1)-submanifold disconnects an n-manifold?
 
If $\exists \alpha>0$ s.t. $\gamma|_{(0,\alpha)}\cap X=\emptyset$, then $N$ points outwards.
 
What should be the report reason for a question that is metaphysics (philosophy)?
 
I'm 99% sure that works, but it's useless :P
 
@GPhys opinion based
 
@ChrisWhite do you agree with my definition
 
user54412
4:45 AM
sure
 
The Spivak pronouns are a proposed set of gender-neutral pronouns in English promulgated on LambdaMOO based on pronouns used by Michael Spivak. Though not in widespread use, they have been employed in writing for gender-neutral language by those who dislike the standard terms "he/she" or singular they. Three variants of the Spivak pronouns are in use, highlighted in the declension table below. The original ey has been argued to be preferable to e, because the latter would be pronounced the same as he in those contexts where he, him, his loses its h sound. == History == The precise history of the...
this is apparently THE Spivak
@ChrisWhite ok, now how to we prove the theorem
 
hey guys
Abhas Mitra (born 18 June 1955) is an Indian astrophysicist best known for his distinct views on several front-line astrophysics concepts, particularly black holes and Big Bang Cosmology. His research has received widespread attention, especially in India, which is reflected from the fact that he is one of the most frequently mentioned Indian physicists on the web. He has regularly questioned the mainstream cosmological concepts of the "big bang" and "black holes". He claims to have offered exact proofs that: (i) The so-called black holes cannot be true black holes even within the context o...
any thoughts?
heard of him?
He has regularly questioned the mainstream cosmological concepts of the "big bang" and "black holes".[21][22][23][24] He claims to have offered exact proofs that:
(i) The so-called black holes cannot be true black holes even within the context of classical general relativity and
(ii) The Big Bang solution is illusory and actual universe must be fundamentally different from the big bang paradigm.[25][26]
Consequently, dark energy, his research claims, is an illusion caused by the departure of the complex universe from the simple big bang model.
 
user54412
sounds pretty crackpot
 
Though Mitra stresses that the `black hole’ solutions are correct, his contention is that black hole masses, arising from relevant integration constants, are actually zero._ His peer reviewed paper published in Journal of Mathematical Physics of the American Institute of Physics supports this contention by showing that Schwarzschild black holes have M = 0._
@ChrisWhite sup?
-1 seriously?
come on...
 
 
2 hours later…
7:28 AM
45
Q: Has anyone ever hacked Stack Overflow?

PraveenWhile chatting with friends about hacking, a guy said that "if a site is getting popular/ or any competitive, it will somehow face severe hacks. Some used to expose it, others hide it". Out of curiosity I wanted to see if any such thing happened to Stack Overflow. Since Stack Exchange (being 5 ...

beware of the "zombie apocalypse"
the horror, the horror
 
8:01 AM
Hello
 
Bonjour mon ami
 
 
4 hours later…
12:05 PM
OMG so that's what a question off topic to philosophy look like:
0
Q: If magic is "fake," why is science real?

Tom HartatkkConsider this: I watch a Harry Potter movie when I am a kid (I did) and, as early as kids, of course, older people tend to teach young kids at some point that motion pictures are "fiction" and not to be taken seriously; however, in this same sentiment, we are expected to experience emotions from ...

 
12:39 PM
I've found philosophy SE to be pretty much useless
I answered a question once about whether Mao Zedong was a philosopher. The answer, regardless of your moral judgement of the man, is yes -en.m.wikipedia.org/wiki/Mao_Zedong#Writings_and_calligraphy
But that was closed as primarily opinion based. After that, I couldn't be bothered to contribute.
 
1:22 PM
Don't have time to read through the article itself
https://www.sciencedaily.com/releases/2016/02/160224070647.htm
I will look up non-intuitive sequence later
 
@0celo7 : Oh so you've managed to downvote certain Einstein advocates into oblivion have you?
@ACuriousMind : what's all this about coordinating downvotes? Have you been coordinating downvotes? Tut tut.
 
1:41 PM
@JohnDuffield Yes.
I downvote incorrect answers.
 
Let's all downvote Einstein advocates!
2
 
@JohnDuffield No, do you have evidence that we've coordinated?
 
Einstein would be nothing without THE evidence.
 
@skullpetrol Without evidence we'd have lots of murderers running around
So he probably would have been killed or his parents would have been killed
 
@0celo7 : we have your confession. That's the evidence.
@0celo7 : you don't downvote incorrect answers. You downvote Einstein.
 
1:53 PM
Feb 16 at 23:02, by 0celo7
Must...not...engage
;-)
 
Guys let's coordinate to downvote Duffield
Let's downvote all his things without reading them
Also mb we can ring his doorbell and run
 
Ding dong... ::runs::
 
@JohnDuffield Huh?
@JohnDuffield Einstein has a habit of being wrong.
@Slereah What an inspired idea!
 
Also Einstein isn't even on stack exchange
Our most famous guy is probably Lubos
so I tried to do a copy paste from a pdf
Here's the result
"➘❹➶♣➟✲➇❨➢♣➞✟➭➯➥➧➠♣➟❭➲❪➨➼➩✓➶♣➟✟❻❅➈➝➆➼➦➤❻★❻❪➄❂➴
➩✙➲❏➥➯➨♣➟✟➠☞➥➧➦✝❸♣➷✚➬❝➮❞➱✾➥➧➨➜✃❪➲❏➈➤➥❐➲❏➨❞➩❹➲❪➨♣➠✩➷➹❸✇➂❶❻❅➦➤➥❒➩✓➥❒✃❅➟❪❼"
3
Is that a bug or an anti-piracy thing
 
2:09 PM
Whatever it is it looks cool on the star board!
 
2:26 PM
@0celo7 : er, no. Einstein had a habit of being right. You have a habit of being wrong. Like you were wrong about being smarter than Einstein.
 
2:37 PM
See this question for an example of where you should pay more attention to what Einstein said. Dismissing Einstein in favour of something else from somebody else without any kind of explanation is not a good idea. That's not to say Einstein was right about everything of course. He wasn't. But when he was, you ought to be able to give some kind of explanation as to how or why.
Like lose the dust. OK, I gotta go.
 
user116211
@Slereah I also did face this.
 
user116211
Check this pdf:
 
@JohnDuffield Hmm, looks like a perfectly good mathematical proof.
 
user116211
No matter what thing you copy, if you paste you would get some thing of that sort:
 
user116211
2:49 PM
❚❤❡r♠♦❞②♥❛♠✐❝s ❞❡s❝r✐❜❡s t❤❡ ♠❛❝r♦s❝♦♣✐❝ ♣r♦♣❡rt✐❡s ♦❢ ♠❛tt❡r ♣❤❡♥♦♠❡♥♦❧♦❣✐❝❛❧❧② ✇✐t❤
t❤❡ ❛✐❞ ♦❢ ❡q✉❛t✐♦♥s ♦❢ st❛t❡ ✇❤✐❝❤ ❛r❡ ❞❡r✐✈❡❞ ❡♠♣✐r✐❝❛❧❧②✳ ❋♦r t❤❡r♠♦❞②♥❛♠✐❝s ✐t ✐s ♦❢ ♥♦
✐♠♣♦rt❛♥❝❡✱ ❤♦✇ ❛ ❝❡rt❛✐♥ ❡q✉❛t✐♦♥ ♦❢ st❛t❡ ❝♦♠❡s ❛❜♦✉t✳ ❚❤✐s✱ ♦❢ ❝♦✉rs❡✱ ✐s ❝♦♥s✐st❡♥t
✇✐t❤ t❤❡ ❧❛r❣❡ ✉♥✐✈❡rs❛❧✐t② ♦❢ t❤❡r♠♦❞②♥❛♠✐❝s✱ ♥❛♠❡❧② t❤❛t s❛♠❡ ❧❛✇s ♦❢ t❤❡r♠♦❞②♥❛♠✐❝s
❤♦❧❞ ❢♦r ❞✐✛❡r❡♥t ♠❛t❡r✐❛❧s✳ ❍♦✇❡✈❡r✱ t❤✐s ❞♦❡s♥✬t t❡❧❧ ✉s ✇❤❛t ♠❛❦❡s✱ s❛②✱ t❤❡ s♣❡❝✐✜❝
❤❡❛t ♦❢ ♦♥❡ ♠❛t❡r✐❛❧ ❞✐✛❡r❡♥t ❢r♦♠ t❤❡ ♦t❤❡r✳ ■t ✐s ✐♥t✉✐t✐✈❡❧② ♦❜✈✐♦✉s t❤❛t s♣❡❝✐✜❝ ❤❡❛t ♦❢
 
Interesting.
 
Odd
 
r ❛t♦♠ ♦❢ t❤❡ ❣❛s ✐s ❞♦✐♥❣ ❛t ❡✈❡r② ✐♥st❛♥❝❡✳ ■♥ ♦t❤❡r ✇♦r❞s✱ ✇❡ ♦♥❧② ♥❡❡❞ s♦♠❡
❛✈❡r❛❣❡ ♠❛❝r♦s❝♦♣✐❝ q✉❛♥t✐t✐❡s✱ ❛♥❞ ♥♦t ❡✈❡r② ♠✐❝r♦s❝♦♣✐❝ ❞❡t❛✐❧✳ ❙♦✱ s♦❧✈✐♥❣ t❤❡s❡ ❝♦✉♣❧❡❞
❞✐✛❡r❡♥t✐❛❧ ❡q✉❛t✐♦♥s ✇♦✉❧❞ ❛♥②✇❛② ❜❡
That's a BS PDF
 
user116211
❖♥ t❤❡ ♦t❤❡r ❡♥❞✱ t❤❡ ♠✐❝r♦s❝♦♣✐❝ ❧❛✇s ♦❢ ♣❤②s✐❝s ❞❡s❝r✐❜❡ t❤❡ ❜❡❤❛✈✐♦✉r ♦❢ ✐♥❞✐✈✐❞✉❛❧ ♣❛r✲
t✐❝❧❡s ✇✐t❤ t❤❡✐r ✐♥t❡r❛❝t✐♦♥✱ ✈❡r② ✇❡❧❧✳ ❋♦r ❝❧❛ss✐❝❛❧ ♣❛rt✐❝❧❡s ✇❡ ✉s❡ ◆❡✇t♦♥✬s ❡q✉❛t✐♦♥s ♦❢
♠♦t✐♦♥✱ ❛♥❞ ❢♦r q✉❛♥t✉♠ ♣❛rt✐❝❧❡s ✇❡ ✉s❡ t❤❡ ❙❝❤rö❞✐♥❣❡r ❡q✉❛t✐♦♥✳ ❆❧t❤♦✉❣❤ ♠✐❝r♦s❝♦♣✐❝s
❧❛✇s ❞❡s❝r✐❜❡ t❤❡ ❜❡❤❛✈✐♦r ♦❢ ♣❛rt✐❝❧❡s ❛❝❝✉r❛t❡❧②✱ t❤❡② ❞♦ ♥♦t t❡❧❧ ✉s ❤♦✇ ❛ ❤✉❣❡ ❝♦❧❧❡❝t✐♦♥
♦❢ ♣❛rt✐❝❧❡s✱ ♦❢ t❤❡ ♦r❞❡r ♦❢ 1023✱ ✇♦✉❧❞ ❜❡❤❛✈❡ ♦♥ t❤❡ ❛✈❡r❛❣❡✳
 
that's getting annoying
 
user116211
2:50 PM
more and more and more
 
Proof. We consider the map f1(y) = f(y) and the constant map f2(y) = f(x), for all y ∈ A; the former has center of mass q, the latter center of mass f(x). We apply (5.8.20).
The next result will be applied in §8.6 below only:
Corollary 5.8.7. Let f1 : (A1,ν1) → Y,f2 : (A2,ν2) → Y be measurable maps from probability measure spaces into Y .
Let q1,q2 be the corresponding centers of mass.
Let φ : (A1,ν1) → (A2,ν2) be measurable, with f2 = f1 ◦ φ. Then
􏰚􏰚
d(q1 , q2 ) ≤ d(f1 (x), f2 (φ(x))) dν1 (x) + d(f2 (x), q2 ) |dν2 − φ∗ dν1 |(x). (5.8.22) Proof. Let q2′ be the center of mass for f2 ◦ φ w.
 
user116211
@0celo7 BS pdf?
 
@user36790 bullshit portable document format
 
user116211
@0celo7 hahahaha
 
@yuggib not unlike yourself :)
 
2:51 PM
@0celo7 I take pride on it
 
what
I would not call anyone jihadi
if I were you
 
sounds like a huge mods-attractor
 
user116211
@0celo7 try to understand the emotion ;D
 
what
@yuggib Let $V$ be a finite dimensional vector space and $\alpha\in V^*$. What is $\operatorname{dim}\operatorname{ker}\alpha$?
Seriously. Someone called the mods?
 
user116211
2:59 PM
@0celo7 whom?
 

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