The h Bar

Albert Einstein

16:00

@JohnRennie, could you tell me for which level is this stuff appropriate?

Avantgarde

Very nice to see you here, Albert. Finally got to meet you

Slereah

@0celouvskyopoulo7 I do not

0celouvskyopoulo7

@AlbertEinstein Do you know multivariable calculus and liner algebra?

Slereah

though vanishing Weyl tensor means that there's no gravity "propagating"

0celouvskyopoulo7

@Slereah that's physicist nonsense

Slereah

16:02

No Weyl tensor means that if it's Ricci flat, it's flat

it is, but it's true

John Rennie

@AlbertEinstein you need to have mastered special relativity, and when I say mastered I mean really understand it not just use the Lorentz transformations.

0celouvskyopoulo7

"$S^3\times R$ is the universal cover of the compactified Minkowski spacetime"

Ok that's just nonsense.

Or maybe she means the Penrose compactification. Do you remember how that goes?

Yashas

@AlbertEinstein hi

Secret

> I sense a combo breaker

Slereah

I don't think Penrose compactification is $S^3$?

Is there even a flat metric on that?

0celouvskyopoulo7

16:09

on $S^3$?

Slereah

yes

0celouvskyopoulo7

I don't know off the top of my head. None of the even-dimensional ones do.

You would have to read Wolf for that

He has a chapter on flat 3-manifolds.

but remember the Penrose compactification is a confomorphism, not an isometry.

Balarka Sen

aren't parallelizable manifolds flat

feels like it should be but i dunno

0celouvskyopoulo7

@BalarkaSen If $S^3$ were flat, it would be isometric to a quotient of $\Bbb R^3$, something about that feels wrong.

Balarka Sen

flat n-manifolds are isometric to quotients of R^n?

0celouvskyopoulo7

16:18

If they're complete, yeah.

Balarka Sen

Huh.

0celouvskyopoulo7

Quotient of $\Bbb R^n$ by a discrete subgroup of the isometry group

Balarka Sen

"intersante"

ya i got that

0celouvskyopoulo7

@BalarkaSen It's in do Carmo

page 163

Balarka Sen

I dunno curvature yet

hopefully we'd get to that

0celouvskyopoulo7

16:19

we?

Balarka Sen

me and my ego

whistle whistle

0celouvskyopoulo7

lol

$A=1,\dotsc, 4$

It's less characters to write them out

Slereah

The slightly odd thing about the wavefront set is that you can define products but they may not be differentiable

0celouvskyopoulo7

@Slereah Do you want me to read Hormander and get the real deal on the wavefront set?

Slereah

The $(H * H)$ business only works because you can't multiply $H$ by $\delta$

@0celouvskyopoulo7 Do as u wish

0celouvskyopoulo7

16:30

@Slereah Why isn't AdS globally hyperbolic?

Slereah

Depends what you mean by AdS

0celouvskyopoulo7

Anti de Sitter?

Slereah

Therer's a version that has time identified

And another version that's the universal cover of it

0celouvskyopoulo7

I want the universal cover

Slereah

IIRC if you take the pseudosphere, it has CTCs

Or something

0celouvskyopoulo7

16:32

Yvonne says it's $R^3\times R$

Slereah

that one is alrightI think?

Wait, what

no I think that one should be free of CTCs?

0celouvskyopoulo7

check page 121 of her book

Slereah

"see chapter 12"

Oh wait is that just the chapter on causality

0celouvskyopoulo7

do you wanna check chapter 12?

It will be a while before I get there

Slereah

it is, yes

Let's see if there's ads in it

Oh wait

I think it's related to like

Horizons?

"No light ray emitted at time $t_0$ will go beyond the space slice $t = t_0 + T$"

So obviously no Cauchy surfaces

0celouvskyopoulo7

16:37

> the isometry group $G_r$

is that a pun?

Slereah

It is not

I never thought about the fact that a spacetime could have no singularities and still not admit Cauchy surfaces due to horizons

0celouvskyopoulo7

what do you mean?

draw a picture

Slereah

the picture is literally on the page you said

0celouvskyopoulo7

I actually have no clue what that picture is supposed to represent.

Slereah

Oh wait

On the other hand

This seems to only apply to null rays

Causal curves can totally go beyond that point

0celouvskyopoulo7

16:40

Oh.

Well

Cauchy surfaces can be defined in terms of null curves

Slereah

I don't know what she's on about

0celouvskyopoulo7

That was how I "proved" maximal Schwarzschild is GH

@Slereah can we just write a proper GR book please

> called the Lie bracket, defined by $[X,Y]=\mathcal L_XY-\mathcal L_YX$

that's...backwards

Slereah

Which theorem is that btw

that Cauchy surfaces are defined by null curves

0celouvskyopoulo7

@Slereah Is the Eiffel tower closed?

@Slereah It's in Wald

Slereah

I've never been to the Eiffel tower

0celouvskyopoulo7

16:44

page 205, theorem 8.3.7

Slereah

Many things are in Wald

Let's see

Ah thanks

Well under that definition, I guess AdS is indeed not globally hyperbolic

It's... it's actually just like the Space Invader scenario

The curves fuck off to infinity

and never cross the spacelike hypersurface of $t$

0celouvskyopoulo7

Yeah.

So the surface is Cauchy iff every null geodesic that ever existed crosses it exactly once (roughly)

Slereah

and yet it has no singularities, I think

So the old "no CTC or naked singularities" chestnut isn't quite true

0celouvskyopoulo7

Yeah

I think what I just said should be the rule of thumb

Slereah

I mean I guess that just looking at it, you can tell that the causal diamond probably won't be compact

since $J^+$ is going off to infinity

If you patch it together with a $J^-$ going off to infinity on the same surface, then it obviously won't be compact

u learn a new thing every day

0celouvskyopoulo7

16:50

Christ

those names lol

Slereah

Is that Stephani?

0celouvskyopoulo7

Yvonne.

Slereah

Stephani does have the Tolman spacetime, btw

0celouvskyopoulo7

yeah but not in the form of Yvonne

$S^{11}$ has 992 smooth structures.

@BalarkaSen Why?

Slereah

more importantly who counted them

0celouvskyopoulo7

16:56

oh my god

992 -> 9(92) -> 9(9+2) -> 9/11

Balarka Sen

@0celouvskyopoulo7 Group of exotic structures on S^n is isomorphic to a quotient of the stable homotopy group IIRC

Slereah

I should have guessed it wasn't Stephani because Stephani never uses special characters

Balarka Sen

assorteddetails.wordpress.com/2015/05/06/the-dimension-barrier

Slereah

I don't think they ever use mathcal

0celouvskyopoulo7

And the Polish invented topology, so Poles did 9/11

Invade Polen now

Slereah

16:57

oh wait, they do for the Lie derivative

Balarka Sen

The sequence is apparently $0 \to \theta_n \to \pi_n^S/J_n \to \theta_{n-1}^b \to 0$

Slereah

why do we know every sphere exotic structure except for $S^4$

We live in four dimensions

We could count them by hand

Balarka Sen

because h cobordism theorem

only works for dim > 4

gotta have dinner now, see ya

0celouvskyopoulo7

Ahhh too much to read

You can tell this book was written by a very old person

No fucks given about formatting

Slereah

Very slightly off center

Quite infuriating

0celouvskyopoulo7

17:02

"compact R. manifolds with negative Ricci do not admit killing fields"

That theorem gets me every time

Slereah

the only ones I can think of are the multiple toruses

So it sounds plausible I guess?

0celouvskyopoulo7

any compact surface admits a metric of curvature $-1$

oh that's what you just said

Slereah

What about the torus?

Isn't it $\int R = 0$

0celouvskyopoulo7

@Slereah The proof goes like this: Take the derivative of the Killing eq. to get $\nabla^i\nabla_i X_j+R^i{}_jX_i=0$. Contract with $X^j$, integrate, and use the divergence theorem: $$\int (-|\nabla X|^2+Ric(X,X))=0$$

So if $Ric<0$, this is a contradiction.

Slereah

That is quite a nice short proof

0celouvskyopoulo7

17:06

@Slereah Oh, you need genus at least two

Slereah

Well yes, that's why i said multiple toruses :p

0celouvskyopoulo7

I wonder if $S^3$ has a metric of negative curvature. Probably violates Priessman if true...

Maybe not

Who knows

Slereah

I guess proving Minkowski space globally hyperbolic isn't too hard with null geodesics

It's gonna be just straight lines

Just take slices $t = 0$

0celouvskyopoulo7

they're straight lines, yeah

use Wald's theorem.

Slereah

How bad is causality on AdS, anyway

Unlike with closed timelike curves, I don't think you have information popping out of nowhere

it's more that it is lost, really

You're just gonna have solutions in the future that don't depend on $J^-$

not all of it, anyway

I dunno, I should look it up

but not today

I need some bed time

0celouvskyopoulo7

17:14

it's only like 7PM where you are

Slereah

yeah but I slept like 4 hours

0celouvskyopoulo7

Ok, on to chapter 6

The Local Cauchy Problem

I don't think I'm ready for this...

Slereah

I hope you buffed up on your Sobolev

0celouvskyopoulo7

I am a Sobolev

she's just doing moving frames

this is trivial

0celouvskyopoulo7

17:58

@Slereah aaaaand she lost me with symmetrizable hyperbolic systems

Secret

Last night dream:

$$\Large{\mathbf{K}^i\vcenter{\small{j}}_{_k}=S^i T\vcenter{\small{j}}U_k=(((S)S)\cdots S)TT\cdots T(U \cdots(U(U)))}$$

(Obviously cannot be a tensor, as tensors are associative (or are they?))

Sha Vuklia

guys, in the book they call $\phi$ the "fase of the motion"

i don't really understand what that means

oh shit

i understand

it's kind of like the fase of the driven motion i think

i guess, i donno

actually no, i don't see how this makes sense

well, I guess it's just a phase that belongs to the particular solution

but if anyone has anything useful to says that might help my intuition, be my guest

Sir Cumference

One of my biggest petpeeves with LaTeX is when people use $<<$ instead of $\ll$

Also when people use $d$ instead of $\mathrm{d}$

0celouvskyopoulo7

$d$ is fine

Sha Vuklia

^ agree

Sir Cumference

18:10

It makes it look like a variable...

Sha Vuklia

yes, but latex should have dealt with that

0celouvskyopoulo7

then you clearly don't understand the context.

Sir Cumference

@ShaVuklia Ya mean a specific "\d"?

Sha Vuklia

yes

Sir Cumference

Agreed

Also a "\degree" command for angle degrees

Sha Vuklia

18:11

lol a-gree

Sir Cumference

You have to crudely put "\circ" in an exponent

Sha Vuklia

that's ridiculous :P

(no irony)

John Rennie

@ShaVuklia when you apply a driving force to the system it oscillates at the same frequency as the driving force.

Sir Cumference

So wait, is \mathrm preferred for units, or \text?

0celouvskyopoulo7

text so you can put in spaces.

Sir Cumference

18:14

Then what's \mathrm for?

John Rennie

@ShaVuklia: but that oscillation can lead or lag the driving oscillation i.e. the peaks of the oscillation do not coincide with the peaks of the driving force.

0celouvskyopoulo7

$\mathrm e^{\mathrm i\theta}$

Sha Vuklia

@John well sure, because eventually your homogeneous solution will go to zero, so your motion just follows the driven frequency

Sir Cumference

@0celouvskyopoulo7 What's the point?

Sha Vuklia

right, makes sense @John

but for all we know our original motion also has the same angle?

John Rennie

18:15

@ShaVuklia The phase angle is the angle by which the motion of the oscillator leads or lags the driving force.

0celouvskyopoulo7

@SirCumference If you don't understand what I'm saying, all hope is lost I'm afraid :(

Shame

Sir Cumference

Wait what

John Rennie

So if the phase angle is zero then the motion and the driving force are exactly synchronised.

That will happen at the resonant frequency.

Sir Cumference

Is there some hidden reason why you used mathrm?

DanielSank

Oooooooh, are we talking about LaTeX?

I did a thing.

Sir Cumference

18:17

Also, I think we can all agree that $cosx$ is the worst

Sha Vuklia

ok but that's just stupid @Sir

DanielSank

@SirCumference Yeah, it's bad.

Sha Vuklia

it takes no effort to write "\cos x"

John Rennie

:37330466 If you plot the (sinusoidal) driving force and the (sinusoidal) motion of the oscillator on the same graph the two sine waves will in general be out of phase. The phase difference is what $\phi$ is in your book.

Sir Cumference

Too many people do that, @ShaVuklia

They don't know about \cos

DanielSank

18:17

@ShaVuklia ...but it takes knowledge.

@SirCumference That.

Sir Cumference

Or \log, etc.

Sha Vuklia

KNAWLEDGE @DanielSank

DanielSank

The one everyone messes up is subscripts.

Sir Cumference

Really, \d needs to be a thing

Sha Vuklia

@John okay, thanks, I didn't realise that!

Sir Cumference

18:18

@DanielSank ^^

DanielSank

$$F_{external}$$ NO!

$$F_\text{external}$$ YES

0celouvskyopoulo7

what?

Sha Vuklia

LOL

ofc

0celouvskyopoulo7

$$F_\mathfrak{external}$$

Sha Vuklia

hahahahaha

DanielSank

18:19

:-(

Sir Cumference

The worst is when people think "5^-25" will give them $5^{-25}$

Instead they write $5^-25$

Sha Vuklia

hahahah really

0celouvskyopoulo7

literally no one thinks that

Sha Vuklia

i've never seen that before

0celouvskyopoulo7

^

Sha Vuklia

18:19

who do you hang out with, @Sir ?:P

Sir Cumference

Guy does it here:

57

A: Why are protons heavier than electrons?

There are multiple reasons why protons are heavier than electrons. As you suggested, there are empirical and theoretical evidence behind this. I'll begin with the empirical, since they have important historical context associated with them. As a preface, this will be a fairly long post as I'll be...

I had to edit it for him

A post with 57 upvotes...

Sha Vuklia

lol dude

you edited like 80% of that post it seems XD

Sir Cumference

Yeah :P

Sha Vuklia

ofc you are frustrated

or annoyed :P

Sir Cumference

It's just cringy

Sha Vuklia

18:21

lol someone should make a latex cringe compilation

0celouvskyopoulo7

how about the book I'm reading

Sir Cumference

Another petpeeve is when people write $ .... $ on its own line, instead of $$ .... $$

0celouvskyopoulo7

Sir Cumference

@0celouvskyopoulo7 faints

0celouvskyopoulo7

Sha Vuklia

18:23

I usually write something like $\cos(kt)\, dt$

like, I add a little space

to make it look less like a variable

Sir Cumference

Ok good

@ShaVuklia Just use $\d$

It's superior in every way

Sha Vuklia

hahaha

0celouvskyopoulo7

Sha Vuklia

well it is

OOOOMMGGGG

faints then dies

Sir Cumference

@0celouvskyopoulo7 Ew

Sha Vuklia

18:24

what is wrong with people? why can't they just $\LaTeX$

0celouvskyopoulo7

Sir Cumference

@0celouvskyopoulo7 Jesus who are u reading?

Sha Vuklia

stop it please @0

i can't handle this anymore :P

0celouvskyopoulo7

@ShaVuklia The person who wrote this book is extremely old and very famous. She doesn't need to have good formatting.

@SirCumference Y. Choquet-Bruhat.

Sha Vuklia

oh it's a she?

0celouvskyopoulo7

18:25

I'm sure you know the name

Sha Vuklia

i expected better from a woman :P

like, we're famous for paying attention to details, just saying

DanielSank

The other that annoys me a lot is using \frac inline.

Sir Cumference

Reminds me, I always wonder why women seem to have better handwriting than guys

DanielSank

This is stupid: $H = - \frac{1}{2} \hbar \omega \sigma_z$.

Don't do that.

0celouvskyopoulo7

@DanielSank Disagree.

DanielSank

18:27

Do this: $H = -(1/2)\hbar \omega \sigma_z$.

0celouvskyopoulo7

No.

Avantgarde

their hands are softer

DanielSank

@0celouvskyopoulo7 Yeah, but you're wrong all the time anyway.

Sir Cumference

@DanielSank Or misusing \over (e.g. $H = 5 \over 6$ gives $H = 5 \over 6$)

Sha Vuklia

hahahah yea men are too rough for neat handwriting

0celouvskyopoulo7

18:27

@DanielSank I am never wrong.

Avantgarde

my handwriting looks pathetic

DanielSank

My handwriting is complemented often...

0celouvskyopoulo7

@SirCumference Literally no one does that.

Sir Cumference

@0celouvskyopoulo7 I've seen it done

0celouvskyopoulo7

Stop projecting your own Latex failures onto the rest of us :P

Sha Vuklia

18:28

hahahaha ^

Avantgarde

lol

Sir Cumference

Oh come on D:

I've seen too much on Astronomy

0celouvskyopoulo7

It's a 750 page book, I get the feeling the editor was not thorough at all

Sir Cumference

The worst is when people post an image of LaTeX instead of writing it themselves

DanielSank

Sir Cumference

18:31

Still don't know how to write script/cursive...

DanielSank

Can you read it?

Sir Cumference

Not like it's useful tho

@DanielSank Sorta

0celouvskyopoulo7

of course

Sir Cumference

._.

0celouvskyopoulo7

who can't read it?

DanielSank

18:32

@SirCumference It is. It's faster.

Sir Cumference

.-.

DanielSank

It was useful in college.

Avantgarde

i thought cursive is taught

Sir Cumference

@DanielSank Dang, I kind of wish I learned it...

DanielSank

@Avantgarde Not all schools are identical.

@SirCumference Meh.

Sir Cumference

18:33

I might still have time, it could save me from the pain of taking notes in print

Avantgarde

yeah. i guess they are here, at least in that respect

Sir Cumference

Especially when the profs talk at 90 words per minute

0celouvskyopoulo7

@SirCumference You need to listen slower.

Sir Cumference

@0celouvskyopoulo7 ?

0celouvskyopoulo7

What is unclear?

Sir Cumference

18:36

How does one "listen slower"?

0celouvskyopoulo7

Ah, run in circles to dilate your time.

Sha Vuklia

guys, sorry to interrupt

but

I don’t understand why $a$ and $v$ are small if $\omega_d\approx 0$; apparently because they are proportional to $\omega_d^2$ and $\omega_d$, respectively - but I don’t see how come?

this is the given situation

Balarka Sen

@0celo Gonna pull an all-nighter Riemannian geometry man

come at me Riemann

0celouvskyopoulo7

@BalarkaSen dC or GHL?

Balarka Sen

combined

0celouvskyopoulo7

18:40

Do they have roughly the same structure?

Balarka Sen

More or less so far, yeah. GHL has more stuff, doCarmo is lucid - that's all the difference I can percieve.

0celouvskyopoulo7

3.G.1 in GHL is one of my favorites

Balarka Sen

huh I see

Avantgarde

@ShaVuklia what is the expression for $x(t)$?

0celouvskyopoulo7

@BalarkaSen According to Petersen, every orientable 3-fold is parallelizable. I thought you said it was only true for compact ones.

ACuriousMind

18:50

@ShaVuklia It comes from the solution of the equation - if you look at the solution $x(t)$ then you'll see that $\dot{x}$ and $\ddot{x}$ are propertional as they claim. You can't read it off from the equation itself.

Sha Vuklia

ohh

i have to differentiate?

Avantgarde

@ShaVuklia yeah so x(t) has omega_d dependence coming from the cosine

yes

Sha Vuklia

oh right

so a priori, we don't know what $\ddot x$ and $\dot x$ depend on? or what they're proportional to

Balarka Sen

@0celouvskyopoulo7 Did I? I don't find it unbelievable.

I don't know a proof either way, I don't think.

0celouvskyopoulo7

Well, I've searched long and hard for the proof. Because this is the second time I've seen the claim.

Qiuaochu seems to think it's only true for compact ones.

Avantgarde

18:52

not until one has the equation of motion, that is, x as a func of t

0celouvskyopoulo7

And Milnor-Stasheff only has a sketch for compact ones.

Sha Vuklia

ok thanks @Avant

0celouvskyopoulo7

Petersen references MS, so I don't know what's going on there.

Avantgarde

welcome

Balarka Sen

oops

well i dunno

Sha Vuklia

18:54

why is the phase zero tho if $x$ is proportional to $\cos\omega_d t$?

oh wait

it's an approximation

so we kind of assume $\cos \omega_d t$ is constant (i.e., equal to 1)?

no actually, that doesn't make sense to me

oh wait, with phase they mean this phase of motion?

ACuriousMind

@ShaVuklia Because if the phase wasn't zero, it would be $\propto\cos(\omega_d t + \phi)$.

Sha Vuklia

yea I'm still a bit confused about this phase, but I think it will sink in over the days @ACurious

Avantgarde

F ~ cos omega_d t, if \phi = 0, then x is also cos omega_d t

0celouvskyopoulo7

@ACuriousMind What's a "Wedderburn theorem"?

Avantgarde

if phi is nonzero, position and force will be out of phase

Sha Vuklia

18:59

right

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