The h Bar

RIP

The Cranberries, famous singers of "Zombies"

and hypothetically other songs

vzn

← endorses antiwar msg also

skullpatrol

The Cranberries - Linger

►

vzn

7:27 PM

wow, rather monogamous for a sexy female rock star whosdatedwho.com/dating/dolores-o-riordan

skullpatrol

not as hypothetically as Milli Vanilli

Not everyone is a 70's hipster fuck

although the ones who are are the best

:3

@dmckee Isn’t that what the teacher is though?

7:46 PM

@Slereah do you want to do an awful GR calculation

Dunno

How awful is it

p. bad

ocelo asks the most inviting of questions.

gotta do a strange change of variables first

and then compute the ADM mass

I think it can be simplified but I'm not sure

I'm thinking some more about it

I bet going to polars right from the beginning helps

Show the integral I guess?

7:53 PM

the metric is the Euclidean metric in the coordinates $x^i$

but you introduce the variable $\rho$ defined by $r=\rho +c\rho^{1-\alpha}$, and then $y^i=\rho x^i/r$

and write $g_{ij}$ for the components of the metric in the $y$ coordinates

Anonymous

I'm having a bit of trouble justifying $\delta \dot{q} = \frac{d}{dt} (\delta q)$ ($q$ is suppose the position coordinate). And $\delta$ represents first order variation (in same sense as $\delta S$ as in this page). @BalarkaSen Any idea?

then compute $$\frac{1}{16\pi}\lim_{R\to\infty}\int_{S^2_R}(\partial_i g_{ij}-\partial_j g_{ii})\nu^i\, dS$$

where $\nu$ is the outer-facing normal vector

(I think just $x^i/r$)

Do you have $g$?

No, that's the obstacle

the change of variables formula is pretty horrific

Do you need 4 dimensions absolutely

7:56 PM

@Blue I don't know variational calculus.

it's a 3 dimensional problem

@Blue whom are you asking

oh, Balarka

Well that helps I suppose

too bad

Anonymous

@0celo7 I was going to ask you, but you seem busy

ask @0celo7

Hey, I am busy too!

Anonymous

7:57 PM

You can answer of course

What is this, 9gag?

So the starting metric is just $ds^2 = dr^2 + r^2 d\Omega$?

@Blue Cf. physics.stackexchange.com/q/93176/50583

@Blue so you have some function $f$, and you make a variation $f_\epsilon$ such that $f_0=f$

then $\delta f=df_\epsilon/d\epsilon|_{\epsilon=0}$, so that thing above is just commutativty of derivatives

Anonymous

@ACuriousMind Awesome. I'm reading

7:59 PM

that's what I just said!

Let's see, $$\frac{\partial r}{\partial \rho} = 1 + c(1 - \alpha) \rho^{- \alpha}$$

@Slereah well the issue is that the ADM mass has to be computed in rectangular coordinates

@Blue what's wrong with $$\frac{d}{dt} \delta q(t) = \frac{d}{dt}[q(t + \delta t) - q(t)] = [\frac{d}{dt}q(t + \delta t) - \frac{d}{dt}q(t)] = \delta \frac{dq(t)}{dt}$$

so $$ds^2 = \frac{1}{1 + c (1 - \alpha) \rho^{-\alpha}} d\rho^2 + (\rho + c \rho^{1 - \alpha}) d\Omega$$

@Slereah I think the best way to do this is to transform from x-polars to y-polars, and then to y-cartesian, no?

8:00 PM

@bolbteppa Everything, because the variation of the field/coordinate need not be induced by the variation in time :P

Anonymous

@0celo7 Ah. So commutativity is the thing I'm looking for. Josh says something similar. I'm reading through that answer now. Thanks for the help

and again everyone is ignoring my correct, concise answer

nvm

Maybe

The $y$ part does seem bad

@0celo7 I'm not ignoring it, I posted the link before you answered!

especially expressed in spherical coordinates

8:01 PM

@Blue be weary of physicists writing things like $\delta q=q(t+\delta t)-q(t)$

or is it wary?

whatever

just be aware of it

The weary should be very wary.

But of course it varies.

can u say that in another language

tired people should be cautious

or smth

same language tho innit

people should be tired and unconcious

I guess the $y$ coordinates would be like $$y^i = \rho(x) x^i / (x^2 + y^2 + z^2)$$

8:03 PM

10/10 translation skills

English to English translation

It's physicist shorthand for
$$\frac{d}{dt} \delta q(t,\varepsilon) = \frac{d}{dt}[q(t,\varepsilon + \delta \varepsilon) - q(t,\varepsilon)] = [\frac{d}{dt}q(t,\varepsilon + \delta \varepsilon) - \frac{d}{dt}q(t,\varepsilon)] = \delta \frac{dq(t)}{dt}$$
same thing :p

Does $r = \rho + c \rho^{1 - a}$ have an easy inverse

Balarka : English $\to$ English

Where $\varepsilon$ parametrizes the space of functions

that would help

8:04 PM

still wrong

Not wrong

@Slereah I don't know. Is it even invertible?

Don't look @0celo7, I'm gonna apply logarithms with little regard for rigor

I am immediately reminded of the "we're gonna Taylor expand the $\delta$" paper

$\ln r = \ln (\rho + c \rho \rho^{-a}) = \ln(\rho) + \ln(1 + c \rho^{-a})$

Hm

Not sure that help

8:05 PM

@ACuriousMind I showed that paper to my advisor and he asked me why I read trash

@ACuriousMind $$\delta(x) = \sum \infty \delta^{(n)}(0)$$

:(

wolframalpha.com/input/?i=Solve+y+%3D+x+%2B+c+x%5E(1+-+a)

not a good sign

Why can't everything be easy

Well we have the metric in $\rho$, and we know the expression of $y(\rho)$

8:10 PM

I said it was a hard calc

Hm, do we have $x(\rho)$

plus it's gonna involve the angular coordinates

yeah it wouldn't be pretty

the problem is that a priori the integral should diverge for $\alpha<1$

but it's in fact zero for $\alpha>1/2$, $c^2/8$ for $\alpha=1/2$, and does diverge for $\alpha<1/2$

$x^i = r \mathfrak{whatever}(\theta, \phi) = (\rho + c\rho^{1-\alpha})\mathfrak{whatever}(\theta, \phi)$

@ACuriousMind help me communist friendo

$y^i = \rho \mathfrak{whatever}(\theta, \phi)$

8:14 PM

this environmental ethics class is going to kill me

Well, it's not too bad

Same thing as the old metric except more angular stuff

@Slereah in theory the angular parts just get integrated out

Oh

also I bet the metric will be diagonal

in fact I'm sure of it...maybe

Do we only have that part : $$\frac{1}{1+c(1-\alpha)\rho^{-\alpha}} d\rho^2$$

Otherwise I'll have to remember the angular parts for the spherical coordinates

Wait, we're integrating on a sphere

So au contraire we only need the $d\Omega$ part

8:20 PM

remember that the metric has to be in the $y^i$ coordinates.

you cannot have it in polars

yeah, that would be...

\begin{eqnarray}
y^1 &=& \rho \sin \theta \cos \phi\\
y^2 &=& \rho \sin \theta \sin \phi\\
y^3 &=& \rho \cos \theta
\end{eqnarray}

Man I don't look forward to finding the jacobian

$$\frac{\partial y^i}{\partial \rho} = \mathfrak{whatever}(\theta, \phi)$$

$$\frac{\partial y^{1, 2, 3}}{\partial \theta} = \rho (\cos \theta \cos \phi, \cos \theta \sin \phi, -\sin \theta)$$

$$\frac{\partial y^{1, 2, 3}}{\partial \phi} = \rho (-\sin \theta \sin \phi, \cos \theta \cos\phi, 0)$$

I think we still have $\rho^2 = \sum (y^i)^2$?

So I guess from this you can get the components okay?

the relations between $y$ and $\rho$ are basically the proper spherical coordinates so you can just find what the angular parts are the usual way

right, that's why I was thinking do $(r,\theta,\phi)\to(\rho,\theta,\phi)\to(y^i)$

each of those seems easier than going directly

probably yeah

especially because going from rectangular to polar is Easy (maybe)

why does it have to be rectangular btw

Seems hard to integrate on a sphere

8:32 PM

it's a sphere of constant $r$ and the metric will be spherically symmetric

so the integral will just be Area of sphere * integrand evaluated at r

yeah but that won't really be apparent from the coordinates

the integral is easy once you have the $g_{ij}$'s

@Slereah come again?

Well why can't you just use the $\rho$ coordinates?

I meant sphere of constant $\rho$

what are the $y$ coordinates for, though?

8:35 PM

the point is that $\alpha$ and $c$ are parameters in the definition of the coordinates

the goal is to show that if you don't pick the right coordinates for the ADM mass, you can get literally any number in $[0,\infty]$

Is the ADM mass measurable

although you're doing it in Euclidian space so I'm guessing this isn't really an issue here

I don't think it is.

ADM mass is a very interesting thing

@ACuriousMind Jesus

@BernardoMeurer it was the most physics paper of all time

arxiv.org/abs/1507.04348

that's the one

What even

Why tho

8:45 PM

"[10] E.B. Davies, Heat Kernels and Spectral Theory, Cambridge University Press, Cambridge, U.K. (1990)"

how dare they reference one of the sacred texts

Oh my god look how they calculated $\int sin(x)/x$ in that paper, incredible

it doesn't seem to actually be referenced

Oh wait, it's not the taylor series of delta

It's a taylor series of delta functions

$$\sum a_i \delta(x - x_i)$$

Wait, no

It's actually $$u(x) = \sum_{i} a_i \delta^{(n)}(x)$$

Does it mean he can do exponentials of diracs

@Slereah don't physicists have rules for calculating $f(\delta(x))$?

or is that $\delta(f(x))$?

It's the latter

it's used for gauge fixing

8:55 PM

What does $\int_a^b f(x)dx = \lim_{\varepsilon \to 0} f(\partial_{\varepsilon}) \frac{e^{\varepsilon b} - e^{\varepsilon a}}{\varepsilon}$ mean

@bolbteppa I do not know

It's heavy in symbolism

notation is from arxiv.org/abs/1404.0747

I think

Those integrals are so horrible this might be worth learning

I'm sure it makes sense in a physics fast and loose way

how can physicists do that with a straight face

the trick is to pretend everything is analytic and pretend that anything can be replaced by anything

This is advanced physics wizardry

"g must be sufficiently well-behaved"

No kidding

9:06 PM

it's such a silly paper

Apparently the trick is that you obtain an "exact" representation of the dirac

Except of course that doesn't work

So it's divergent

So you only take an asymptotic series

I don't know what for

It keeps getting worse

I'll buy a function with a derivative as an argument, if it's analytic

But what the hell is it for a delta

it's clearly $\delta(\partial_x)=\frac{1}{2\pi}\int e^{ik\partial_x}dk$

He seems to be using it the usual way

$\delta(\partial_x) f(x) = \delta(f'(x))$

Holomorphic functional calculus

This is legit functional analysis

In mathematics, holomorphic functional calculus is functional calculus with holomorphic functions. That is to say, given a holomorphic function f of a complex argument z and an operator T, the aim is to construct an operator, f(T), which naturally extends the function f from complex argument to operator argument. More precisely, the functional calculus defines a continuous algebra homomorphism from the holomorphic functions on a neighbourhood of the spectrum of T to the bounded operators. This article will discuss the case where T is a bounded linear operator on some Banach space. In particular...

I think

I know it works for analytic functions and all

But on a distribution?

9:15 PM

why not sure

anything that is not forbidden is mandatory in holomorphic functional calculus too

@bolbteppa if $\delta$ were a holomorphic function, you might be right.

or a function

'holomorphic functional calculus is functional calculus with holomorphic functions', distributions are functionals, qed

and you wonder why people ignore you

is wikipedia intensifying again

Emilio Pisanty

9:21 PM

@JohnRennie Finally, my efforts at rebuking Marty Green have earned me a gold badge

physics.stackexchange.com/help/badges/49/populist?userid=8563

ahh that dude

I vaguely understand it, pretty crazy

@BalarkaSen he did link a wiki page

LMAO

9:24 PM

This is potentially a way to definite-integrate functions mainly by looking at the integrand, no joke

he always looks like he's gathering spit

he's going to throw some spit facts at you

@BalarkaSen I haven't played PUBG in hours

I feel like a new man

I don't get the PUBG fever

Seems like a boring game

9:26 PM

@BernardoMeurer it is

nothing happens

it's a very bad game

It's addictive though

but it's like heroin

Don't talk about heroin like you know what it's like

I noticed something in PUBG

I can guarantee you heroin is better than PUBG

9:29 PM

god, I can't upload the image

what garbage is this

Did someone say heroin

I’m an edgy teen

Count me in.

you can dab when skydiving

@BalarkaSen

it's the best thing ever

beautiful

My gf keeps on dabbing

I’m try a cut it out

Inglorious Basterds - Bear Jew

I'll dab you

relevant:

►

9:36 PM

2

Incredible
$$\int_a^b f(x) dx = \lim_{c \to 0} \int_a^b f(x) e^{-cx} dx = \lim_{c \to 0} \int_a^b f(-\partial_c) e^{-cx} dx = \lim_{c \to 0} f(-\partial_c) \int_a^b e^{-cx} dx = \lim_{c \to 0} f(-\partial_c) \frac{e^{-cb} - e^{-ca}}{-c} = \lim_{c \to 0} f(\partial_c) \frac{e^{cb} - e^{ca}}{c}$$

What's not to like about this craziness

It's great to annoy math people

"The challenge is to generalize the present methods to higher dimensions so that, for example, a new perspective into the algebraic workings of boundary and coboundary operators may be gained"

@Slereah Surreal

@0celo7 Tarantino movies always sucked

I'm going to get banned if I respond to that

9:44 PM

Get banned

how dare you say that

you're gone so communist that you're not even human any more

Tarantino's best movies are the schlocky exploitative ones

None of his high budget box office successes are worth watching

What do you have against shlock

Ocelo plz not more nazi films at me... ;(

@Slereah ? I'm literally saying his schlocks are the best

9:46 PM

His box office successes are all exploitative shlock

Death Proof is a great movie

Tarantino can only make 70's action movies

I don’t think I’ve seen that many Tarantino films

Django Unchained is exploitative schlock?

caman san

CNN is such absolute shit that I can't code while it's playing

9:49 PM

get banned

Well that’s a title

It's all blaxploitation

@BalarkaSen it's a great movie

only a white nationalist could dislike it

it's okay

@CooperCape Pulp fiction is a classic...

9:51 PM

sad

Werewolf Women of the SS (2007) Trailer

Speaking of campy shlock

He also made this trailer

oh god

fully torqued

Nicolas Cage as Fu Manchu

Oh God

9:57 PM

that's amazing

wonder how much cocaine produced that

Nicolas Cage have an odd liking for getting himself cast in B-roles

That's because he's a B movie actor

Have you seen him act

Did you see Vampire's Kiss

Yes

Nicolas Cage at his Nicolas Cagiest

Wasn’t there that weird thing where Nicolas cage was on the front of a Serbian textbook

10:01 PM

iwut?

Hell no

Yah

No weirder than him being in an American movie

I wanted to watch Wild at Heart at some point of time in my life

but then I saw the movie poster

Nick Cage standing with a chick in his arms

and Im like nope im done

I never watched it

it's probably one of his better movies because the director is David Lynch anyway

national treasure

only a commie could hate it

David Lynch went downhill after his first movie anyway

Erasorhead is his best movie of all time regardless of whatever he wants to feed his audiences

10:10 PM

Clearly this method will solve the Riemann hypothesis

Hollywood is cancerous. Nothing good comes out of it

the people who make marginally better movies from Hollywood are influenced by European filmmakers more

@ACuriousMind any simple reason why the automorphism group of a Lie algebra is a Lie group?

closed subgroup of a Lie group?

Why is $f'(\partial_c) = f(\partial_c)c - c f(\partial_c)$

::vomits::

@bolbteppa the notation is a bit hard to follow

Who knows

Not quite sure what that derivative even means

Is it $f'(x)$, $x=\partial_c$?

Or are you doing a variation around the derivative

10:20 PM

$f(\partial_c)e^{cb} = f(b) e^{cb}$

That one is fairly obvious

That's all you need :p

It's commonly used in QM

That relation is equation 52 in arxiv.org/pdf/1507.04348.pdf , the $f'(\partial_c) = f(\partial_c)c - c f(\partial_c)$

Since $p = -i\hbar \partial$

10:22 PM

> In folklore we find the following witty slogan: if there is a contradiction between physics and logic, you must change the logic.

:D

@0celo7 there used to be a big math program to rewrite logic

So it would fit QM better

Math can handle QM just fine

it's just that the required math is too hard

Yeah but they thought it might help

(It didn't)

us.metamath.org/qleuni/mmql.html

10:38 PM

I guess $f(\partial_c) (c e^{ac}) = (f(\partial_c) c) e^{ac} + c f(\partial_c) e^{ac} = [f'(\partial_c)]e^{ac} + c f(a)e^{ac}$ implies $f'(\partial_c) = f(\partial_c) c - c f(\partial_c)$