The h Bar - 2017-04-21 (page 5 of 5)

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6:03 PM

Good old Wolf Einstein

Greatest of wolf physicists

@Mostafa Not really, no. Stuff that is rude, offensive or just makes people uncomfortable doesn't become any less so if someone claims to be posting it "for research purposes".

@Slereah A werephysicist, perhaps? Half man, half physicist.

@ACuriousMind rude and offensive, yes, I agree. But not just makes people uncomfortable. Research is hard sometimes, and you need to get out of your comfort zone.

AccidentalFourierTransform

user image

@Mostafa I wasn't necessarily saying that everything that makes people uncomfortable is off-limits here, but if someone complains, I'd urge you engage with their concerns instead of retreating behind "research purposes"

@AccidentalFourierTransform Sigh...is that really necessary?

AccidentalFourierTransform

no, not really

feel free to delete it

6:17 PM

@Mostafa Eg I probably wouldn't post a snuff film here

@AccidentalFourierTransform Eh, I don't find it that upsetting. Just...pointless :P

Certain kind of research is just off-topic in a physics chat on the internet :P

AccidentalFourierTransform

like everything I do

You're right, this is a physics chat

AccidentalFourierTransform

have I ever posted something... not pointless?

6:18 PM

So how bout them apples

@AccidentalFourierTransform We've had some nice dicussions about QFT

And sometimes you're actually funny, I don't view humour as pointless, either ;P

hi

AccidentalFourierTransform

Jan 5 at 19:11, by AccidentalFourierTransform

user image

hi there

I have been thinking about solitons (really kinks) in scalar fields

mostly because I want to know about instantons and circles

::waits for AFT to make a joke about "thinking about kink"::

6:22 PM

:~)

AccidentalFourierTransform

what I am, private joker?

its not all jokes you know

theres more to life than that

@Cows circles are sets of points equidistant from the same point

circles are really cool. I swear, Atiyah told me so

@Slereah I learned that definition in primary school

hehe, still holds though

:D

It's amazing . Circles are amazing I swear!

I'm off to cook something. I will be back soon

Well no because circles don't have to be embedded in the plane

are you trying to say sphere is not homeomorphic to a subset of R?

anyways got to make some brunch hehe

:D

6:30 PM

@Cows solitons are awesome :) do you have interest in fluid dynamics? there is also an emerging new area called "pilot wave hydrodynamics"... are you a physics student? undergrad?

...

Trump’s War On Federal Science Will Stifle Innovation And Hurt The Economy / huffpost

@vzn I have been thinking about how if something is lorentz invariant, you can compute solitons and then do boosts. Fluids are awesome too. I sometimes fill buckets with water and try to physically manifest kdv related stuff. I am hoping to be enrolled as a student somewhere soon. I dropped out senior year in physics(money problems, about 5 or 6 years ago)

@Cows cool, what country are you in?

I am in the United States

@Cows state?

Los Angeles

6:48 PM

@AccidentalFourierTransform That sign can be used for research purposes.

e.g. in chiral or dichroic metasurfaces

user image

this paper ^

AccidentalFourierTransform

wow

how... how did you find that?

That's very close to my research field :)

or this:

user image

(top-right)

These are used in all chiral metamaterials....which are used when you want to manipulate waves' polarization

AccidentalFourierTransform

neat

its the symbol of peace ;-)

lol

AccidentalFourierTransform

7:14 PM

is vzn an acronym for something?

visual zygote newt

user image

user image

lol

Good night

@Slereah Do you know the RG book by Eisenhart?

7:28 PM

@0celouvsky can you tell me more about what slereah was trying to say about circles and planes. I had to grab brunch , but I am back

he said "Well no because circles don't have to be embedded in the plane" I said circles were amazing

A figure 8 is not embedded

But I'm it sure if that's what he meant

@JohnRennie I need another out of print book for my work

I see

AccidentalFourierTransform

physics.stackexchange.com/review/suggested-edits/172266

Bernardo Meurer

7:44 PM

@0celouvsky Halp

Why can't Daniel help?

Bernardo Meurer

He's busy

He helped earlier

When you weren't around. You two take turns

@0celouvsky plox

I

I'm in class

Bernardo Meurer

7:59 PM

Okay, ping me when you're free then

Bernardo Meurer

8:18 PM

@BenNiehoff HELP ME WITH ANALYSIS

RIGHT NOW

please

AccidentalFourierTransform

if its something simple I can try

lol, why did someone make microscopic swastikas?

AccidentalFourierTransform

I mean, if everybody else is busy

Bernardo Meurer

@AccidentalFourierTransform Okay, this is the exercise:

AccidentalFourierTransform

if its in Portuguese I'll log out

Bernardo Meurer

8:20 PM

Given a set $A = \{(x,y,z)\in \Bbb R^3\colon x+y < 1;y+z^2 < 1; x,y,z > 0\}$ write down the expression to compute it's volume using a triple integral of form $\int\int\int dydxdz$

@AccidentalFourierTransform Come on, I translate them to everyone but ACM

AccidentalFourierTransform

can you draw $A$?

it should be more or less easy

Bernardo Meurer

I have

user image

It's like a burrito laying down

Well

$y+z^2<1$ is

AccidentalFourierTransform

let me check

Bernardo Meurer

My problem now is finding the bounds for the integrals

I have $\int_0^1\int\int 1 dydxdz$

But now idk what the bounds on $x$ will be

AccidentalFourierTransform

well, the volume is 2/5

we already know that

Bernardo Meurer

8:26 PM

I don't

AccidentalFourierTransform

now you do

Bernardo Meurer

Okay

How do I write the bounds for $x$ now?

I have them for $z$

AccidentalFourierTransform

im thinking!

its been ages since the last time I did this :-P

like, three years?

Bernardo Meurer

Is it not just 1-y?

@AccidentalFourierTransform I thought we were about the same age?

AccidentalFourierTransform

no, unless you're 22

Bernardo Meurer

8:29 PM

Oh jee

You're old af

Granps

AccidentalFourierTransform

sad frog face

my drawing doesnt quite looks like yours

in the $x,z$ plane it should be a square, right?

or am I doing it wrong?

so, basically the region is $0<x<1-y$

and $0<z<\sqrt{1-y}$

right?

and $0<y<1$

so thats it

$$\int_0^1\mathrm dy\int_0^{1-y}\mathrm dx\int_0^{\sqrt{1-y}}\mathrm dz$$

yep, that evaluates to 2/5, as required

QED

pls answer

say something

Bernardo Meurer

Hey

Sorry

Got a call

Oh god why do you write it like that

AccidentalFourierTransform

8:45 PM

cause im a pro

Bernardo Meurer

Okay

I see the sense in that

So I start by "fixating" the innermost variable to find the bounds?

AccidentalFourierTransform

well in this case it is simple because the conditions are independent

one for $x,y$ and the other one for $y,z$

so you can solve them for the variables, in my case I solved for $x$ and $z$ in terms of $y$

Bernardo Meurer

Why do you write it like that though

AccidentalFourierTransform

in the general case, I dont think there is an algorithm you can follow

Bernardo Meurer

it's so bad

Just do triple integrals like normal peeps

AccidentalFourierTransform

8:48 PM

please stop

just acknowledge the superiority of my notation and move on

Bernardo Meurer

It's horrible

Please rewrite that using triple integrals normally concatenated

AccidentalFourierTransform

im so mad right now

ill burn everything that you know

Bernardo Meurer

Well, good luck because I DON'T KNOW ANYTHING

AccidentalFourierTransform

typical Bernardo

Bernardo Meurer

Why is that typical me?

AccidentalFourierTransform

8:53 PM

you always do that

the thing you did there, that thing

Bernardo Meurer

Must resist making your mom joke....

@AccidentalFourierTransform Okay, another one

AccidentalFourierTransform

said your mother

Bernardo Meurer

Using cylindrical coordinates compute the volume of $$B=\{(x,y,z)\in\Bbb R^3 \colon x^2+y^2 < 1; \sqrt{x^2+y^2} < z < 2; x>0; y>0\}$$

Why do I know that cylindrical coordinates are going to be handy here?

AccidentalFourierTransform

yes

Bernardo Meurer

Is it because z is dependent of (x and y)

AccidentalFourierTransform

9:00 PM

in fact, $B$ a quadrant of a cone

with a bit of geometrical visualisation/imagination, its volume should be clear

but anyways, draw $B$ first

Bernardo Meurer

I have no imagination at all

or notion of space

or any idea of volume or anything

My mind is just a lexer and a parser

Morning

Bernardo Meurer

@Mostafa morning

@AccidentalFourierTransform Let me try and draw this

AccidentalFourierTransform

spoiler: the volume is $\pi/3$

Bernardo Meurer

You're using matlab

AccidentalFourierTransform

9:03 PM

thats for peasants

I use Mathematica

Bernardo Meurer

Okay, so

$x^2+y^2 <1$ is just a circle of radius 1

AccidentalFourierTransform

in the $xy$ plane, yes

Bernardo Meurer

Ye

Now the second equation is a bit sketch

I know it will have height 2

@BernardoMeurer So you prefer algebraic viewpoint over geometric...

Bernardo Meurer

@Mostafa Yep, my brain does not support OpenGL

the OpenCL extensions work OK though

@AccidentalFourierTransform How do I figure out that it makes a cone?

I mean I see the base, and the tip

9:06 PM

Lots of CS guys I've met say they hate continuous math....only discrete.

Bernardo Meurer

but the side-surface I don't get

@Mostafa Agreed

AccidentalFourierTransform

well, $r<z<2$, right?

where $r=\sqrt{x^2+y^2}$

Bernardo Meurer

ye

@BenNiehoff Because that's a special sign.

AccidentalFourierTransform

its not exactly a cone

its kind of a cone

Bernardo Meurer

9:08 PM

Oh

It's like a cutoff cone?

Send me a pic dood

from mathematica

AccidentalFourierTransform

the figure in MMA makes no sense to me

user image

user image

user image

Bernardo Meurer

what the heck

AccidentalFourierTransform

whatev

$0<\phi<\pi/2$, right?

and $0<r<1$, yes?

and $r<z<2$, so the integral reads

$$
\int_0^1\mathrm dr\ \int_0^{\pi/2}\mathrm d\phi\ \int_r^2\mathrm dz\ r
$$

@BernardoMeurer Google it

google.com/#q=z%3Dsqrt(x^2%2By^2)

Result:

user image

Bernardo Meurer

@AccidentalFourierTransform Who the hell is phi there?

AccidentalFourierTransform

9:15 PM

@Mostafa much better

@BernardoMeurer a Klein-Gordon field

Bernardo Meurer

Oh come on

AccidentalFourierTransform

an angle

say, $\theta$

or $\varphi$

Bernardo Meurer

theta is nice yes

I'm still confused

But okay, let's not draw it

In cylindrical coordinates we have $(r,\theta,z)$ right?

AccidentalFourierTransform

fun story, in Spanish $\theta$ is pronounced like "tit", and we use the same word for "to grab" and "to take"

Algebra I, professor: "and if you take a big enough $\theta$..."

@BernardoMeurer yes

Bernardo Meurer

In Portuguese too!

Okay so we will have $$\int\int\int drd\theta dz$$

BUT

I am changing coordinates

and there's a hack

AccidentalFourierTransform

9:19 PM

recall that $\mathrm dx\,\mathrm dy\,\mathrm dz=r\mathrm dr\,\mathrm dz\,\mathrm d\theta$

Bernardo Meurer

Ah, yes, the added $r$

because of the coordinate change

Okay okay so we have

$$\int\int\int drd\theta dz r$$

But how do I find my boundaries?

This is what I don't know how to think of

AccidentalFourierTransform

the one for $\theta$ should be more or less clear

from the drawing :-P

Bernardo Meurer

0 to pi/2

Since it's a full circle

r goes from 0 to 1

AccidentalFourierTransform

more like the quarter of a full circle

but yes

Bernardo Meurer

z should go from 0 to 2, no?

AccidentalFourierTransform

9:26 PM

not quite

but close

Bernardo Meurer

Why?

It's lower bound is $\sqrt{x^2+y^2}$

and x,y are

Oh

They are > 0

So it's lower bound is $\sqrt{x^2+y^2}$

AccidentalFourierTransform

^.^

Bernardo Meurer

which is just r

Nice!

@AccidentalFourierTransform What other substitutions are usual? cylindrical, spherical, and?

AccidentalFourierTransform

Parabolic coordinates

Parabolic coordinates are a two-dimensional orthogonal coordinate system in which the coordinate lines are confocal parabolas. A three-dimensional version of parabolic coordinates is obtained by rotating the two-dimensional system about the symmetry axis of the parabolas. Parabolic coordinates have found many applications, e.g., the treatment of the Stark effect and the potential theory of the edges. == Two-dimensional parabolic coordinates == Two-dimensional parabolic coordinates ( σ , τ ) {\displaystyle (\sigma ,\tau )} ...

but they are very rarely used in practice

cylindrical and spherical are the only ones that are used systematically

I would even say that only spherical :-P

oh and en.wikipedia.org/wiki/Hyperbolic_coordinates

what's up?

@AccidentalFourierTransform parabolic is used in GR

AccidentalFourierTransform

9:35 PM

and hyperbolic

yeah there are some sinhs floating around

AccidentalFourierTransform

but they are not nearly as ubitiquous as spherical coordinates

@BernardoMeurer do you need help or not

Bernardo Meurer

@0celouvsky Yep :)

Teach me spherical coordinate shift

what specifically

Bernardo Meurer

9:39 PM

Say I want the volume of a sphere of radius 1

In cartesian coordinates that will suck

correct

you want to to that integral in polars?

Bernardo Meurer

I guess that's the smart choice, no?

So what's the issue with that?

Bernardo Meurer

What's the cartesian eq. for a sphere?

AccidentalFourierTransform

$x^2<1$...

9:41 PM

$$\int_{-1}^1 \int_0^{2\pi} \int_0^1 r^2 \,dr d\phi d\cos\theta$$

@AccidentalFourierTransform that's an open ball, not a sphere

Bernardo Meurer

Okay, what's the name of this factor we multiply at the end when we shift coordinates?

@BernardoMeurer $x^2+y^2+z^2=1$

@BernardoMeurer Jacobian?

AccidentalFourierTransform

Jacobian

Bernardo Meurer

Ah, right!

I had forgotten about that

Let me read my notes

In fact, I guess I should just know the jacobians for some coordinates

So, cylindrical it's r, spherical it's $\cos\theta$

AccidentalFourierTransform

nah, just learn how to calculate $\star$

Bernardo Meurer

9:44 PM

The hell is that star

no

Ignore him

@BernardoMeurer for spherical it's $r^2\sin\theta$

AccidentalFourierTransform

also, $r^2\sin\theta$

LPT: use dimensional analysis to remember the powers of $r$

you need $dxdydz$ to have units of $L^3$

in cyl. you have $d\theta$ which has no units

so you need one factor of $r$

I approve of this

AccidentalFourierTransform

in sph. you have $d\theta$ and $d\phi$, so you need two factors

and in $d$ dimensional spherical coordinates, $r^{d-1}$, etc.

Bernardo Meurer

In circular coordinates it will just be $\sin\theta$?

AccidentalFourierTransform

9:47 PM

what is circular coordinates?

polar?

thats $dxdy=rdrd\theta$

Bernardo Meurer

Yes

AccidentalFourierTransform

no $\sin$

Bernardo Meurer

So it's just $r$? Like cylindrical?

AccidentalFourierTransform

recall that $S^1$ has no curvature

@AccidentalFourierTransform ...

Bernardo Meurer

9:48 PM

The fuck is $S^1$

that's a horrible thing to say to someone in calculus

Bernardo Meurer

I make computers dawg

AccidentalFourierTransform

¯\_(ツ)_/¯

I was just messing with you

Bernardo Meurer

@AccidentalFourierTransform You're old

AccidentalFourierTransform

oh no

;_;

polar coordinates is pretty much the same thing as cylindrical coordinates

for the latter, you just add one independent (flat) coordinate

Bernardo Meurer

9:53 PM

I see

@Slereah I have to solve the EFE

I'm going to be reading chap 7 of HE soon.

2 hours later…

Bernardo Meurer

11:33 PM

@0celouvsky You around?

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