The h Bar

NeuroFuzzy

00:14

@0celo7 Wow, rude.

0celo7

00:25

@NeuroFuzzy Lol, nerd detected.

Only a nerd would defend a nerd.

@Slereah looked through the table of contents of Shouten, I have no clue what the book is about

I don't know any of these words

@Slereah you can send it to me if you don't want it

0celo7

01:56

@ACuriousMind He's hunting spiders again.

Kyle Kanos

02:26

@Danu Not unless you do it every day :D

0celo7

02:55

> Thus, if $u$ is a timelike vector with $\mathrm{Ric}(u,u)>0$, then in some sense the "average" sectional curvature for planes in the pencil of $u$ is negative.

@ACuriousMind Have you ever heard of "pencil" being used like this before?

All the planes contain $u$.

@Slereah If you bound the timelike sectional curvature above and below, the spacetime has constant curvature.

0celo7

03:11

@Slereah Wtf is up with the generic condition, it seems so random

DanielSank

04:12

Who wants to try an integral?

@ChrisWhite I've been there.

It's called The Internet.

Secret

04:52

https://www.sciencedaily.com/releases/2016/03/160316151619.htm

chemist make stuff to make quantum stuff

Joshua Lin

Just wondering, does anyone here know how good UCLA is for their undergrad programs (in like physics and stuff) -an australian in need

DanielSank

05:44

0

Q: Statistics of the product of two white noise Fourier amplitudes

Consider two sequences of random numbers \begin{align} A &= \{a_0, a_1, \ldots a_N\} \\ B &= \{b_0, b_1, \ldots b_N\} \, . \end{align} where each $a$ and $b$ value is independently drawn from a Gaussian distribution $$G_\sigma(x) = \frac{1}{\sqrt{2 \pi \sigma^2}} \exp \left[ -\frac{x^2}{2 \sig...

@JoshuaLin My fiance, who is now at OSU on a fellowship for astrophysics, went to UCLA.

@0celo7 I evicted a spider from our conference room a few weeks ago.

It was a black widow.

Joshua Lin

@DanielSank would you say that going over to UCLA for an undergrad degree would be a good idea? I'm currently at the Australian National Uni, but I just got an offer from UCLA so I'm no longer sure anymore... UCLA is ranked like higher in every respect, but I'm not really sure how the rankings affect the undergrad experience, so right now I'm super unsure about life

Danu

07:04

@DanielSank Yikes!!!

@JoshuaLin First of all, do you want to be a physicist? Are you sure about that?

Joshua Lin

@Danu I have absolutely no idea, I'm pretty interested in everything(science + mathsy) to be honest. At the time of application I applied to be a Physics major, but that was just because I know a little bit more about physics than anything else

Danu

@JoshuaLin So then, statistically speaking, you're probably not going to become a physicist (which is fine!).

In that case, I wouldn't worry so much about the ratings of your universities, but rather consider how happy you'll be to live where you study.

user54412

07:21

^ that

user54412

it's not like ANU is unheard of

user54412

07:34

@BernardMeurer My attendance in this room is completely unpredictable.

Joshua Lin

I think I'm a pretty happy chap, I don't really mind where I live too much. I moved down from Sydney to Canberra for a couple of months now to study for a bit at Aus National Uni, and everyone continuously says like "You'll get homesick soon" and all that stuff, but I'm pretty happy wherever I am I think

Its just that here at ANU, it feels like theres a part of me missing. Ive been thinking a lot lately, and I realised I derive satisfaction from three main things:
1. Having friends to talk to, party with, relax with, so on

user54412

well, I can't speak to what ANU is like socially

user54412

but I can tell you generically what big universities in the US like UCLA are like: most of the people sound like the ones you describe

user54412

most college students in the US, even the more prestigious schools, are there just to get a piece of paper that makes them more employable, or else they're there because they don't know what else to do with their time

Danu

I guess that holds everywhere

user54412

07:40

that said, when you have thousands or tens of thousands of classmates, there will always be some group you can fit into well

user54412

it's a matter of finding them

Danu

08:19

And now the queues are completely empty---I guess we're not so understaffed after all? :P

Slereah

10:27

So anyway

I have a physics question

What's a good video game to play this week end

yuggib

broforce

Slereah

Something more story-oriented, maybe

hubot

Hi!

I argue with my mom because I wanted to borrow from the library Resnick & Halliday's book.

I've version on pdf but I wanted to have paper version.

yuggib

@Slereah there's a lot of story in broforce

Slereah

Aren't all the missions SAVE THE PRESIDENT

hubot

10:42

Do someone know how to convince mum that she bought me Landau & Lifshitz's book and Resnick & Halliday's book?

yuggib

@Slereah no; you have also to do something something something unilateral

or americanize irakistan

ACuriousMind

11:01

@Danu The day is young - wait until the frequent reviewers run out of close votes or hit the 20 reviews.

@Slereah Did you play The Witcher 3?

Slereah

I never even finished the Witcher 1

Got bored

ACuriousMind

The 3rd is much better than the 2nd, and the 2nd was lightyears from the 1st

I never finished the first one, either. I liked the second, and now the third is my favourite game in a looong time

DeNiSkA

11:45

@Slereah you are a nerrrrrd!!

Slereah

12:04

So anyway

Where do you even get that idea for a manifold

David Z

12:24

@0celo7 don't do that

DeNiSkA

honestly saying!! your life is similar to that mine!(*my mom doesn't want me to build a kingdom of books*), so i don't have Resnick-halliday but i did mange somehow to do on iphone(*laptop was quite hard for studying*)
And for some topics which are very boring like sound waves , waves on string and for exercises of rotational motion which demanded high concentration i used to go to library!

@hubot to convince your mum you must go to parenting SE

0celo7

12:45

@Slereah Fallout 4

Slereah

Eh

Fallout 4 did not keep my interest

I think I'm gonna play some Don't Starve

Hey @0celo7

I'm gonna make an answer for Ellis

Any more things to ask?

0celo7

dude

I needed to ask you something last night

wtf was it

oh, do you understand page 69-70 in HE

Slereah

Dunno

Didn't read that bit

0celo7

wth

just send ellis Freire's proof

and ask him wtf "without boundary" means

then point out that technically the open disk is without boundary as a manifold

if he says some BS like "topological boundary" note that only clopen sets have empty topological boundary.

@Slereah Did you know that if $W$ is a nonnull vector that satisfies the generic condition, the sectional curvature vanishes for all planes containing $W$

@Slereah Do you know why $\sqrt{|x|^2|y|^2-(x\cdot y)^2}$ is the area of the parallelogram with sides $x,y$?

ACuriousMind

13:08

@0celo7 Draw it.

0celo7

@ACuriousMind huh?

@ACuriousMind Is there not some slick way to do this with forms?

like take $x\wedge y$, then...something?

ACuriousMind

@0celo7 The slick way is to write $(x\cdot y) = \lvert x\rvert\lvert y\rvert\cos(\theta)$ and use $\sin^2+\cos^2 = 1$.

0celo7

And $|x\times y|=|x||y|\sin\theta$?

@ACuriousMind What if I take the length of $\sharp\star(x^\flat\wedge y^\flat)$?

Slereah

email is sent

ACuriousMind

@0celo7 And...what do you think $x\times y$ is?

0celo7

13:21

@ACuriousMind that mess right there

ACuriousMind

So what are you asking?

0celo7

but I need this for the Riemannian case, not Euclidean

$\times$ is not really defined in the Riemannian case

I need the area of a parallelogram in $T_pM$

ACuriousMind

And again you ask a question without saying what you're actually trying to do :P

0celo7

sory

ACuriousMind

Of course you have to use the uglier expression already in the non 3D case, and hence also in the Riemannian case.

0celo7

13:23

but in the non 3D case, that above thingie does not give me a vector

do I have to take a determinant of something?

ACuriousMind

The inner product on vectors/forms extends to an inner product on k-vectors. Recall that $v\wedge{\star}w =\langle v,w\rangle\omega$ by definition (of either the r.h.s. or the l.h.s., depending on your preference).

(if expressed in terms of the component 1-vectors of the k-vectors, this is indeed a sort-of-determinant)

0celo7

uhh

yeah I know that

but now I'm confused as to what the actual area is, when we're not in 3D

@ACuriousMind Should I be taking the "form inner product" of $x\wedge y$ with itself?

ACuriousMind

Yes.

0celo7

It's not clear to me why that is the "area" though

or is that the definition

(then I'm not sure why we define area like that)

John Rennie

Words fail me, I just don't know what the young of today are coming to.

what doyou mean, i didn't get it? — shiamaa 48 mins ago

0celo7

13:35

@JohnRennie Hey!

You're one to talk

John Rennie

@0celo7 oops, I wasn't referring to your posts. Look at the link I posted.

0celo7

@JohnRennie I know you weren't

but you're attacking my kin

::snickers::

ACuriousMind

@0celo7 How do you define "area"?

0celo7

@ACuriousMind I take a ruler and measure

then multiply some stuff

ACuriousMind

Stop trolling. If you're asking "how do I compute the area in the general case", you must have a definition of "area in the general case", else that question doesn't mean anything.

0celo7

13:41

>stop trolling

Seriously?

Do you just call everything I do trolling now?

ACuriousMind

@0celo7 You mean to say that "I take a ruler and measure and then multiply some stuff" was meant as a serious reply as to what your definition is?

0celo7

@ACuriousMind Yes.

I have an intuitive notation of what area is.

ACuriousMind

As you're someone who always asks for "Proof?", surely you cannot believe that's a definition?

0celo7

@ACuriousMind well

area is like porn

I can't define it, but I know it when I see it

ACuriousMind

Math doesn't work like that.

0celo7

13:45

and I don't see it in $\sqrt{\langle x\wedge y,x\wedge y\rangle}$

@ACuriousMind I know

Is it so strange that I don't see why that square root corresponds to my intuitive notion of area?

ACuriousMind

No. But it is strange that you insist to be able to see that. You don't have an intuitve notion of area in 20D space. All you can do with geometric definitions that are outside our imagination is check that they properly reduce to the intuitive notions in the case we can imagine (like in this case, check that this gives the correct area for any parallelogram in 3D, and also if you project onto the plane spanned by the pg. and compute the area there by elementary means).

0celo7

Hmm, thanks

Not really convinced, but I have to go

Bernard Meurer

I bet you can see 20D on some heavy drug

0celo7

mb

John Duffield

@Danu : it's isn't that moment when you (explicitly) abandon coordinate invariance in GR. It's that moment when you realise that Eddington-Finkelstein coordinates were invented by Penrose. And in the next moment you realise you can't make a stopped clock tick by putting a stopped observer in front of it.

I've said already that Roger "parallel antiverse" Penrose makes a habit of appealing to Einstein's authority whilst flatly contradicting the guy.

DanielSank

14:05

@JoshuaLin There are lots of factors, obviously.

Going to a new country could be fun.

Bernard Meurer

@ChrisWhite Are you here now? :p

vzn

@JoshuaLin hi saw your blog, interesting. do you do other "simulations" not listed on the blog?

7 hours ago, by Chris White

most college students in the US, even the more prestigious schools, are there just to get a piece of paper that makes them more employable, or else they're there because they don't know what else to do with their time

@ChrisWhite reminds me of this article just ran across, quite an indictment of higher edu, alarming, devastating. from a humanities pov however, maybe case can be made the "hard" majors havent experienced same "malaise/ deterioration/ degeneration" etc... have seen some other signs of this...

Pass, Fail, An inside look at the retail scam known as the modern university / Srigley

0celo7

14:30

@JohnDuffield Lol!

DanielSank

15:00

@0celo7 Solve this problem.

ACuriousMind

15:32

@DanielSank Silly question: Have you given the integral to the CAS of your choice and tested if it can solve it?

DanielSank

@ACuriousMind Yes.

ACuriousMind

Aaaaaaaaand?

DanielSank

@ACuriousMind I tried Wolfram Alpha and it says "too long to compute" or some such thing.

If I make certain variables equal to 1, then the answer is one of the Bessel functions, but I need those variables to be variable.

DeNiSkA

@BernardMeurer how did you put strike on the text in your description?

DanielSank

This can't be so hard. I'm essentially asking for the noise level in a cross-correlation.

@ACuriousMind o_O

How?

ACuriousMind

15:45

Damn, mistyped :(

DanielSank

Ah.

ACuriousMind

With my mistype, I could get it to give me at least the indefinite integral, although it refused to compute the definite integral. Now I can't get it to do either, too

This indicates that the integral has no nice closed-form solution, though

I mean, what do you expect from an integral whose special case is already a Bessel function?

DanielSank

@ACuriousMind Uh, a hypergeometric function, probably.

ACuriousMind

Setting which variable to 1 gives the Bessel function?

DanielSank

The $\sigma^2 N$ bit.

...if I recall correctly.

Right now I'm trying to compute the mean square of the distribution, rather than the distribution itself.

This is going better...

Slereah

15:51

You'd be surprised how many weird integrals have a solution

ACuriousMind

Well...they have solutions in terms of functions which in turn are defined by other integrals

I never found finding such solutions particularly interesting

But I understand you didn't pick calculating this integral for fun ;)

Bernard Meurer

@DeNiSkA `---tsit

DanielSank

@ACuriousMind Indeed not.

Bernard Meurer

---strike---

Slereah

The special special functions are the worst

Bernard Meurer

15:53

~~strike~~

Slereah

Like the functions that are used for ONE thing

DanielSank

@ACuriousMind I don't really care whether or not the answer can be written in terms of the special functions we like. I'm not special-function-racist like so many people.

sine and cosine are special functions.

Essentially, I just want something I can plot. I'd even plot the integral computed numerically if it didn't depend on two parameters.

One parameter is acceptable (i.e. plot-able).

Slereah

what integrals are we talking about btw

DanielSank

1

Q: Statistics of the product of two white noise Fourier amplitudes

Consider two sequences of random numbers \begin{align} A &= \{a_0, a_1, \ldots a_N\} \\ B &= \{b_0, b_1, \ldots b_N\} \, . \end{align} where each $a$ and $b$ value is independently drawn from a Gaussian distribution $$G_\sigma(x) = \frac{1}{\sqrt{2 \pi \sigma^2}} \exp \left[ -\frac{x^2}{2 \sig...

statistics fourier-analysis noise

Slereah

Let me get

The Book

DanielSank

15:57

Now I'm finding that the mean square doesn't converge.

That can't be right.

Found mistake.

Woah.

ACuriousMind

@DanielSank Your general equation for $Z=XY$ doesn't mesh with what you write specifically for $C$ thereafter.

DanielSank

@ACuriousMind In what way?

Slereah

There's an integral for $e^{-ax^2 - \frac{b}{x^2}}$

Close but not quite

Let's push on

ACuriousMind

$$P_Z(z) = \int_{-\infty}^\infty P_X(x) P_Y(z/y) \frac{1}{|z|}dz \, .$$ and \begin{align}
P_{|\tilde{C}_k|}(z)
&= \int_{-\infty}^\infty P_r(r) P_r(z/r) \frac{dr}{|r|} \\
&= \int_0^\infty P_r(r) P_r(z/r) \frac{dr}{r} \\
&= \left( \frac{2}{\sigma^2 N} \right)^2 z \int_0^\infty \exp \left[ - \frac{ r^2 + (z/r)^2 }{\sigma^2 N} \right] \frac{dr}{r} \, .
\end{align}

DanielSank

What's the problem?

ACuriousMind

16:01

Wait

nvm

0celo7

What is going on here

ACuriousMind

I hate integrals :D

DanielSank

@0celo7 RTFL

0celo7

Sigh...BEE uses relativistic mass.

DanielSank

@ACuriousMind Awww come on.

0celo7

16:02

Damn mathematicians. They're just as bad about physics as physicists are about math.

DanielSank

They're fun.

ACuriousMind

@DanielSank I have already made two silly mistakes trying to help you here, I'm really not good at this

DanielSank

@ACuriousMind Heh, ok.

Well, I think I just found that $\langle |\tilde{C}_k|^2\rangle = (\sigma^2 N)^2$.

The dimensions are not what I expect, though.

I don't like a square radius being proportional to $\sigma^4$.

0celo7

@Slereah There's so much spacetime crap we don't know

DanielSank

@ACuriousMind does the stuff I've written on Math.SE look correct to you?

0celo7

16:05

There are whole papers analyzing curvature bounds and curvature conditions

This is some crazy shit

Slereah

Hey

@DanielSank

DanielSank

@Slereah Yo.

0celo7

@DanielSank Why should I?

You've never been too keen to help me.

Slereah

I found the integral for $e^{-\frac{a}{x^2} - \mu x^2} \frac{1}{x^2}$

So far not $\frac{1}{x}$ though

DanielSank

@Slereah Dang.

Oh, ha I'm dumb.

Slereah

16:08

I did find $x^{\nu - 1} e^{-\frac{\beta}{x} - \gamma x}$ though

DanielSank

$|\tilde{C}_k|$ is already a squared quantity.

Slereah

You might be able to switch variables to make it that?

DanielSank

Of course it's square involves $\sigma^4$.

@Slereah I tried that but it just moves the problem around.

0celo7

>it's

Slereah

Well from what I've seen of all the results

It's probably something involbing exponentials, Bessel functions or Gamma functions

DanielSank

16:10

@Slereah That's like saying "it involves functions", heh.

ACuriousMind

@DanielSank Yes

Slereah

Well gee sorry for trying to help :p

DanielSank

@Slereah Awww, I was just teasing.

I appreciate your help!

I think you're right about the Bessel functions.

Slereah

Did you try to differentiate under the integral sign, maybe?

DanielSank

@Slereah Have not tried that yet.

0celo7

16:13

No one cares about me, sniff, back to reading about geometry

Hah, the book discards Λ because it was written in 1996.

I wonder how much damage the Λ actually does to spacetime topology.

Slereah

I posted your bloody proof to Ellis :p

0celo7

Good. Not my proof though.

Slereah

Yeah

We'll see what he says about it

0celo7

Is Ellis a math person or physics?

Slereah

Physics I think

All the things I've seenby him are GR stuff

0celo7

16:18

There are plenty of math people who do GR

Oh great now we're Lorentz transforming the curvature tensor

Hyperbolic trig yay

Leuchte

Hello everyone

Why would you use a mechanical alarm clock rather than the one on your phone?

DanielSank

@Leuchte Battery doesn't run out?

Phone alarm sounds are annoying?

Leuchte

Couldn't you just change the alarm sound on your phone?

DanielSank

Rather break a 5 dollar mechanical alarm clock instead of your 300 dollar phone in the rage ensuing from being awoken from sleep?

See note about battery, and rage-proof-ness.

Leuchte

Is there any aesthetic side to it? e.g. it is easier to press a real button than look on your screen and touch it with your finger

DanielSank

16:28

@Leuchte Probably for some people. I dunno.

@ACuriousMind You can integrate $z$ first and find $\langle |\tilde{C}_k|\rangle = \pi \sigma^2 N$.

@Slereah ^

Slereah

o

DanielSank

@Slereah Not sure if that's enough info for the physics problem I'm trying to solve though.

0celo7

@DanielSank why would you break anything

You're such a violent person..,

0celo7

16:50

@Slereah if Ric(v,v)>=0 for all timelike vectors, is it also true for v null?

Slereah

I'd say yes?

Unless the metric isn't continuous

Well, the ricci tensor

Hm wait

No

Otherwise the NEC and the WEC would be the same

0celo7

BEE claims its true, but I'm sceptical.

They don't give a proof

Slereah

If it's true the proof is probably a convergent sequence to a null curve I guess?

0celo7

Oh. Yeah.

Slereah

Wait

The WEC does imply the NEC

I'm just being dumb

So I'd say true, yes

0celo7

16:55

Can you always find a convergent sequence of timelike vectors that go to a null vector?

That's probably a nontrivial theorem in HE.

But BEE assumes you've read HE, they keep taking theorems from it

This shit's intense

Slereah

A timelike vector with an angle to the light cone that is divided by 2 every step?

0celo7

Thanks.

Slereah

I dunno

Oh wait that's a vector not a curve

I think Sanchez has arguments of that type in his paper

about curves converging to null curves

0celo7

Ok but the timelike tangents will converge to a null tangent.

Slereah

Yes

0celo7

17:00

So it works. A little strange they didn't prove that, though.

Maybe they'll do it in chap 3.

Slereah

Wait

I don't think that you can always make a timelike curve converge to a null curve

Unless the lightcone is the boundary of the chronological future

I dunno

0celo7

I'll investigate when I have access to HE and Wald

So why does WEC imply NEC

Slereah

Well, I think if you have a null curve, you can always describe it as the limit of a converging sequence of timelike curves

So as long as everything is smooth it should work

Danu

@0celo7 and @Slereah, what would you guys think about a room of your own for this kind of stuff?

0celo7

I wouldn't like it.

Slereah

17:10

Why

0celo7

@Slereah Why wouldn't I like it or why does Danu want us to have a separate room

Slereah

Why a separate room

ACuriousMind

I'm guessing because he doesn't want to read pages of this when reading the chat log :P

Slereah

Doesn't he want to read physics in the physics room

I can post more memes, if he desires

0celo7

we can talk about his love life

Slereah

17:17

The Room Anyway, how's your sex life HD

►

0celo7

@ACuriousMind are we as bad as JD?

At least we say different things every time

ACuriousMind

@0celo7 What?

I was just offering a guess that Danu means that pages of stuff that only you two care about might be more appropriate in another room. But perhaps he just hates you :P

Slereah

>only you two

Gee I'm sorry we only care about general relativity :p

Sir Cumference

Question...

anyone know about neutron stars and degenerate gases?

ACuriousMind

...what do you want to know about them?

Sir Cumference

17:24

Well, when a neutron star compresses and the Fermi energy rises, will all the neutrons rise to a higher energy level?

Physics Meta

0

Q: Asking moderators favors when looking at questions

Some of my questions were on-hold and closed. As everyone knows, it comes a yellow rectangle where it explains what it means and why is your question not so good. For example: Sometimes, this rectangle doesn't explain enough why was the question closed or put on-hold. Is it okay to write in t...

discussion comments moderators

ACuriousMind

@SirCumference Are you asking whether the energy of the lowest lying state rises or stays the same? (I think it stays the same, but I don't know about neutron stars)

Sir Cumference

Yep

Really?

Then how can all the energy levels be filled to the Fermi energy when a star compresses?

If the Fermi energy rises

ACuriousMind

The Fermi energy is just the difference between the lowest and the highest occupied state at zero temperature. Isn't saying "the Fermi energy rises" the same as saying "the Fermi gas compresses", since more fermions distributed on the same states means higher states must get occupied?

Sir Cumference

Sorry if I'm misunderstanding it, but when the Fermi energy rises, doesn't that mean that most of the fermions will gain energy?

If the Fermi energy rises, then the highest occupied state would probably be higher, right?

Or the lowest state would be lower...any way, the difference in highest and lowest must be greater, right?

ACuriousMind

17:34

@SirCumference No. It just means the difference between the lowest and the highest occupied state changes. But a priori you can't say whether that is because the individual levels shifted or because you just have more fermions occuping states whose spacing is still the same.

Sir Cumference

But there aren't necessarily more fermions in the star, right? So why would more fermions be distributed on the same states?

Sorry if I'm not understanding you

Slereah

Doesn't pressure just generally change the energy levels of a system

Sir Cumference

It does

ACuriousMind

@SirCumference The density of states of a Fermi gas is constant. If you increase the density of the fermions, then you get a rising Fermi energy simply because you need to distribute more fermions on the same states.

@Slereah What? The states know nothing of pressure, pressure is a thermodynamic quantity.

Slereah

Well states know of the volume in which they are constrained!

Sir Cumference

17:39

Alright, idiot question here, but why would you need to distribute more fermions on the same states if you increase the density of the fermions?

I need to read more into this

ACuriousMind

@SirCumference Just look at the computation of the Fermi energy for a 3D fermi gas in a box. The number density of the fermions appears very straightforwardly.

0celo7

17:54

@Slereah everyone else does quantum BS

Sir Cumference

All right, and with a rising Fermi energy, higher energy levels will be occupied, and pressure increases?

Since the neutrons now move more?

Slereah

Let's talk about quantum states in GR

0celo7

We're the only people who care about the only physics that actually matters

No.

Sir Cumference

@0celo7 Cough neutron stars cough

Pretty important

ACuriousMind

@SirCumference Ehhhhh, I'd say that the pressure comes from elsewhere (gravitational attraction), I don't see what it has to do with neutrons "moving" more (don't say that, quantum objects don't really move about). And "particles moving" would be temperature, not pressure, anyway

Sir Cumference

17:57

Fuck it, my source sucks then

It's feeding me lies

Doesn't pressure cause gravitational attraction?

ACuriousMind

...what?

Sir Cumference

sigh...

I heard in GR, pressure can be a source of gravitational curvature in spacetime

ACuriousMind

Oh, that

Sir Cumference

I don't know what to think now...

ACuriousMind

Yes, pressure is a part of the stress-energy

Sir Cumference

17:59

Oh thank god

Alright, so gravity is causing the pressure, which causes gravity, which...

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