The h Bar

Okay, here's my question — and it might sound idiotic/nonsensical, but here goes:

@EmilioPisanty hmm, I think the ugliest were probably in my math classes ... but I'd be uncertain. Most of the time we quickly took some limit to obtain integrals :D

@Sanya yeah, but you normally have some finite radius of convergence which stops at some singularity in the complex plane, but you can do an analytical continuation elsewhere, right?

@EmilioPisanty this looks like a Laurent series to me and not like a normal series, so no, I'm not completely sure of that either

12:16 AM

@Sanya yeah, but you can split it up in two so it's fine

focus on $f(z)=\sum_{n=0}^\infty z^{2^n}$ to start with, it's easier

obviously $f(z)$ converges if $|z|<1$

but what happens at $f(1)$?

@Slereah Does the Big Rip only happen in a Universe in which $\dot{H} > 0$?

but isn't the trouble usually that the two sums that you split it into have different "areas" of convergence in the complex plane and you thus have only an intersection where you can obtain any values?

Where $H$ is the Hubble constant

but yeah, your $f(1)$ should be divergent I think - 1+1+1+1+1+...

@heather Onto the next problem then

12:18 AM

@Sanya for this specific series the relevant variable is actually $q=e^{i\pi^2/\kappa\tau}$, and it's easy to show that it converges if $|q|<1$. The question is what happens at the boundary.

@Sanya yeah. now what happens at $q=-1$?

also at $q=\pm i$

@BernardMeurer, yep

@Slereah Actually, ignore that. Better question is: does the Big Rip only happen in a universe where $\ddot{R} > 0$, or can it happen in a universe where $\ddot{R} < 0$ but $\rho < \rho_{\text{crit}}$?

My god this is bad beer

Sheesh

@EmilioPisanty -1 should also diverge as all the exponents are even; the $i$ should for $n>1$ do the same as the exponent divided by two is always something even and $-1$ to the power of something even should be +1 if I'm correct

@BernardMeurer Never tried one :/

12:22 AM

@Sanya precisely

but you can step the argument up to $\sqrt{i}$

@SirCumference You're american, right?

@BernardMeurer Yep :(

Yeah just wait a bit I guess

i.e. numbers of the form $e^{in\pi/4}$

Not everyone is blessed with being born in a country where no one gives one damn

12:23 AM

and to their square roots, at $e^{in\pi/8}$

@EmilioPisanty but even +1-1+1-1+1-1+... wouldn't converge, would they now?

Never touch Bud Light

@Sanya yeah, but this isn't that

Sigh...I'm really freaking out

I got a test coming up

Still got lots of cosmology confusion

for any number of the form $e^{in\pi/2^k}$, after the $(k+1)$th term you just get $1+1+1+1+1+\cdots$

12:24 AM

yeah, that's clear to see

@SirCumference Have a beer

still - that's not the complete unit circle

@dmckee Wait, you know GR, right?

@Sanya no, but it's dense on the circle

yep

so on a dense set on the circle, it diverges

not a good sign

12:26 AM

this means that for any $z$ in the circle, any $M>0$ and any $\epsilon>0$, I can find you a $z'$ such that $|z-z'|<\epsilon$ and for all sufficiently big $N$, $\mathrm{Re}\left(\sum_{n=0}^Nz'^{2^k}\right)>M$.

you tell me what the chances are of getting an analytical continuation out of that

nil, even if we would find some weird spot on the circle where it actually converges

Plumbus: How They Do It | Rick and Morty | Adult Swim

Very interesting for the physics fanatics:

►

Makes you see the beauty in the things we use every day, I had no clue how a Plumbus was made!

@Sanya you can't

@Jim OH MY GOD

You're a cosmologist!

$f(z)$ is continuous on the interior of the disk

12:31 AM

Can you answer a question?

I've been looking for a cosmologist everywhere

if it converged for a point on the circle then it would at the very least be bounded on an arc around it

PLEASE COME BACK CHRIS WHITE

@EmilioPisanty gotta love complex functions, I actually seem to remember faintly that there was a theorem like that - but I like how their local properties always influence the global ones

@Jim Plz be real

And here

anyway - getting back to the sum from $-\infty$ to $\infty$ - what we said implies it converges nowhere, does it?

12:36 AM

I'm so confused

there are middle schoolers on Math SE who know advanced math

what is even real anymore

@SirCumference, meow-mix, right?

@heather yeah, among others

jesus

@SirCumference, yeah, I've talked with him, he knows a bunch of cool stuff.

@0celo7 could you pease be unbanned?

I need help with GR

GRAAHHHHHH

I'm fducked

I'll get to bed ... see you all and thanks @EmilioPisanty - I'll read your full answer tomorrow :)

12:41 AM

@Sanya, have a good night

@BernardMeurer I see little chances, I'm afraid, and I don't think he'll see your message no matter how many stars it gets

I did write him in the end, but I didn't hear back

@EmilioPisanty change.org

We'll make a petition

@EmilioPisanty I wrote to him too, but had no response

IIRC Daniel wrote him and also did not hear back

It's a shame, I liked him a lot

@BernardMeurer that's interesting to note. Still the man wants out, and we gotta respect that.

I hear all these mentions of Chris White...I assume he was awesome/amazing?

12:47 AM

@heather Knew more GR than anyone else I know here

also cosmology

@EmilioPisanty It's like when that really good beer ends though, you'll try pouring the bottle at least twice more before throwing it away

why isn't he around anymore?

user228700

Does anybody know any organic chemistry? I've a quick question about the relation b/w the stability of a molecule and the dipole moment. Namely: is there any relation?

No one knows

Something probably happened, but all we know is that he deleted his SE account

@heather He was a great guy, he taught me what I know about computing very large numbers

12:49 AM

@heather He came up with one of my favorite quotes

Feb 2 at 5:34, by user54412

One thing to note is that comet-hunting used to be at least 50% of astronomy (back a couple hundred years ago). Now there might not be 50 people in the world who research comets.

I quoted him on my profile

Good times...

He seems like a nice guy; wish I could've met him. When did he leave?

@heather Very soon after he got his PhD

how many months/years ago?

Right after he defended his thesis more or less

Months

i probably only missed him by a little then ='( I joined physics.SE 5 months ago according to my profile

I wonder if, now that he's a mod, ACM knows what happened to cause him to leave

12:53 AM

AFAIK the mods don't know

It doesn't matter too

The point is he's not here & we miss the man

@heather I think he left in September

Why did that link to ACM?

This headache is driving me insane

I need glasses

but I also need a Game Boy Advance

Life's tough

@BernardMeurer The hell?

It's almost 2017

So?

I've been nostalgic lately

I miss being a kid

Being a phony-adult sucks balls

@BernardMeurer Welcome to the club :(

I don't feel 18

12:59 AM

I'm almost 19, and it's a v. large turd

This year was horrible

even YouTube Rewind for this year was horrible

^

2012 was the only good one

@BernardMeurer Perfect end to this awful year: Trump becomes President

Just what we needed as the icing on the cake

Meh screw Donald Trump

A: The position-representation matrix elements of the propagator for a particle in a ring

1:24 AM

@Sanya there we go

0

You wouldn't think it, from how easy it is to pose this question, but it is ridiculously nontrivial. As it happens, it is entirely impossible to find the position-basis matrix elements of this propagator. So far you've done good, and the identification $$ U\left( t_{2},t_{1}\right) =e^{\frac {-...

c'monnnnnn, stack exchange, show me what you can do. that's a killer answer that you won't find anywhere else. shower it with praise and fake internet points! praise be!

user228700

Does anybody know the relation b/w solvation and inductive & mesomeric effects?

user228700

@BernardMeurer When did it release? :-o

@JohnRennie Could you help with some cosmology confusion?

user228700

1:45 AM

Does anybody know anything about the hyperconjugation effect?

user228700

(No, no luck at the CSE chat. That place is dead)

@Kaumudi Today.

user228700

Nothing can compare to the one they made last year.

user228700

I had that in mind so yeah, it sort of did suck this year :-/

That was pretty great yeah

1:58 AM

@Kaumudi This is how I feel with cosmology ;-;

user228700

But this year, India was represented. FINALLY!! so I'm super pumped about that!

user228700

@SirCumference :-(

Meh, Brazil get's represented every year, I usually think it shouldn't be there though

@BernardMeurer Didn't you say you were originally from India?

Huh? I'm Brazilian

I have never even touched an Indian I think

2:00 AM

Ok I dunno what I'm talking about then...

Wait

No, that's right actually

@BernardMeurer Wait what

user228700

@BernardMeurer That's a weird thing to say to establish that you have no ties with India :-P

I don't think I have physically met an indian

@Kaumudi Well, it's weird though isn't it :P

@Kaumudi Oh btw, what's your avatar?

2:02 AM

The closest human I have had physical contact with to an Indian as a Pakistani, and that's arguably not that close (or very close depending on your political views)

Come on folks, I'm about to dominate the star board

user228700

@SirCumference The Pizza John.

@Kaumudi Huh?

user228700

Google it.

user228700

> "When John Green (Author and half of Vlogbrothers) was shaving his Movember beard (into a Very Creeper Mustache) while discussing the political, social, and economic situations in Pakistan someone screenshotted and then added the word pizza under. There are pizza john shirts available for sale at dftba records."

Ok, so what about @BernardMeurer? What's your avatar supposed to be?

2:04 AM

@SirCumference It's a picture of me

Oh...I feel dumb...

user228700

@SirCumference His face? O.o

user228700

@SirCumference :-P

Taken under the californian sun earlier this year :)

user228700

2:39 AM

@TheStackExchange: How the heck are we s'posed to memorize the first terms of the Maclaurin series of $(1+x)^{1/x}$ ?!

user228700

It's very random, much like that of $\tan x$ :-(

...why would you memorize terms of Maclaurin series? :P

Alas, while memorizing $\tan(x)$ is hard, memorizing $\sin$ and $\cos$ is not, and then you can use $\sin(x) = \cos(x)\tan(x)$ to deduce the $\tan$ coefficients from that if you ever really need it and cannot just ask WolframAlpha or similar sources.

@ACuriousMind You know GR?

@SirCumference Some of it. Just ask your question - you lose nothing by having asked it even if no one present answers it.

@ACuriousMind Well I lose the effort I put into typing the equations...

You know the Einstein field equations?

2:52 AM

Yes.

You familiar with the Friedmann–Lemaître–Robertson–Walker metric?

Yes.

Ok, so here goes

@ACuriousMind I'm trying to understand what would lead to a Big Rip. Does it only occur in a universe where $\ddot{R} > 0$, or in a universe where $\rho < \rho_{\text{crit}}$ regardless of $\ddot{R}$?

@SirCumference ...what are $R$ and $\rho$?

@ACuriousMind $R$ is the cosmic scale factor and $\rho$ is the mass density of the Universe

$\rho_{\text{crit}}$ is the critical density

2:58 AM

Well, $\ddot{R} > 0$ is not sufficient for a big rip, assuming you mean that $\lim_{t\to t_\text{rip}} R(t) = \infty$ for some finite $t_\text{rip}$ by that.

@ACuriousMind Well, let me rephrase: is $\dot{H} > 0$ sufficient for a Big Rip?

I'm 99% it's not

So what causes it? Is it simply determined by the mass density of the universe?

Well, $\dot{H} > 0$ is necessary, it's just not sufficient

The rip needs not only a universe that expands faster and faster, it needs a universe that reaches infinite scale factor at finite time.

@ACuriousMind Yes, so what would allow that?

Wait, is there a formula that relates $\dot{H}$ with $\ddot{R}$, or am I crazy?

3:04 AM

@SirCumference Uh, any matter/energy content of the universe for which the Friedmann equations yield a solution such that $R(t)$ becomes infinite at finite time?

@SirCumference ...is your definition of $H$ not simply $\frac{\dot{R}}{R}$?

@ACuriousMind Sigh, I forgot that that's a definition

I've been thinking only of $H = \frac{V}{D}$

@ACuriousMind What would $\rho$ have to be?

@SirCumference According to Wiki, a perfect fluid with $w < -1$ seems to do the job, I'm too lazy to check the math

@ACuriousMind So $\rho$ doesn't matter?

...$w = p/\rho$ is the ratio between pressure and density.

All right, nvm

So then what's the fate of a universe in which $\rho < \rho_{\text{crit}}$?

3:09 AM

Sigh...what's your definition of $\rho_\text{crit}$?

@ACuriousMind As in the mathematical value?

$\rho_{\text{crit}} = \frac{3 H^2}{8 \pi G}$

No, I meant as in "Why do you call it $\text{crit}$?"

Because I'm referring to the critical density?

I'm not sure I understand what you mean

What's "critical" about it?

@ACuriousMind The ratio between the mass density of the universe and the critical density ($\Omega$) tells us the curvature of the universe

The way I learned it in class (though I'm not sure how accurate this is) is that the critical density is the mass density that a flat Universe will have

3:18 AM

Yes, okay. Why do you think that $\rho < \rho_\text{crit}$ has something to do with the fate of the universe then, when you clearly already know that what it determines is the shape of the universe?

@ACuriousMind Because a universe in which $\rho < \rho_\text{crit}$ will continue expanding faster than escape velocity

In the same way that $\rho > \rho_\text{crit}$ would lead to a Big Crunch

Okay, so you also know that a universe with $\rho > \rho_\text{crit}$ will stop expanding, while a universe with $\rho < \rho_\text{crit}$ will never stop expanding. What's the question?

@ACuriousMind The big factor is the deceleration parameter $q_0=\frac{\Omega_{m,0}}{2}$

Which is, like the curvature of the Universe, determined by the mass density

@ACuriousMind I'm asking whether or not a universe with $\rho < \rho_\text{crit}$ will end up in the Big Rip scenario

Even if $\ddot{R} < 0$

@SirCumference Not necessarily. It will expand forever, but whether it rips or not depends on the exact functional form of $R(t)$.

@ACuriousMind So does anything necessarily differentiate the fate of a universe with $\rho < \rho_\text{crit}$ and one with $\rho = \rho_\text{crit}$?

@ACuriousMind All right, recap. What factors besides $\ddot{R} > 0$ are necessary for a Big Rip?

3:26 AM

@SirCumference That $R(t)$ becomes infinite at finite $t$.

@ACuriousMind Does it matter whether $\rho < \rho_\text{crit}$ or $\rho = \rho_\text{crit}$?

@SirCumference Yes, because the universe with $\rho = \rho_\text{crit}$ has a decreasing expansion rate. I think it cannot rip. The other can rip, but need not necessarily.

@ACuriousMind Can you show me what the function $R(t)$ looks like and how $R$ can become infinite within a finite time?

@ACuriousMind So in reality, the Universe has $\rho < \rho_\text{crit}$, since $\ddot{R} > 0$?

I remember hearing that our Universe is indeed flat, though?

3:51 AM

@SirCumference Look for example here for a comprehensive list of solutions. For $w < -1$, you get $R(t) = (1 / (1 + \frac{3w+3}{2}t ))^{2/(3w-3)}$, which, since $w$ is negative, blows up at some finite $t$ since the denominator goes to 0.

user228700

@ACuriousMind For my exam :'-( We're not allowed anything. Not even a pen- they give us a pen and a few sheets of paper.

@Kaumudi We gotta use a four-function calculator that can't even do exponents

And yet the formulae we work with have exponents

user228700

@ACuriousMind Yes, but I've somehow managed to remember the coefficients for now: 1, 2, 16, 272, 7936... Geez.

user228700

@SirCumference U're bragging. We get nothing.

@ACuriousMind I still don't see how $\rho = \rho_\text{crit}$ implies a decreasing expansion rate. I've heard our universe is flat, yet $\ddot{R} > 0$.

@Kaumudi You have to exponentiate on your own?

user228700

3:56 AM

Yeah.

What kind of formulae you gotta use?

In one case we had to use the Friedmann equations

@SirCumference My bad, it doesn't always. The implications only hold if you assuming a matter/radiation filled universe. But we have dark matter and possibly other stuff floating around, so actually, $\rho$ only determines the shape. If we allow for more matter types we have no relation at all between the shape of the universe and the rate of expansion.

@ACuriousMind But for a simple universe, the deceleration parameter is $q_0=\frac{\Omega_{m,0}}{2}$, right?

no idea

I don't have cosmo math memorized

Shouldn't that imply that curvature is directly related to rate of expansion?

@ACuriousMind :/

4:01 AM

what's a "simple" universe?

I suspect you're assuming "radiation-dominated" or "ordinary-matter-dominated" there

@ACuriousMind Yes, the latter

I should've been clearer

Yeah, so you're missing the essential ingredient of the Big Rip and accelerating flat universe models: dark matter and dark energy.

"Dark energy" is precisely that: The energy density needed to have our universe's expansion accelerate although it is nearly flat.

@ACuriousMind Ok, hold on

$\frac{H^2}{H_0^2} = \Omega_R a^{-4} + \Omega_M a^{-3} + \Omega_k a^{-2} + \Omega_{\Lambda}$

and $\frac{\dot{H}}{H^2}=-(1+q)$

There seems like there should be a way to relate $\Omega$ with the deceleration parameter

Through algebra

Though I'm too lazy to do it myself

@ACuriousMind Actually, forget that messiness

$q=\frac{1}{2}(1+3w)\left(1+K/(aH)^2\right)$

And K is a representation of the curvature of the Universe

This, as well as what I said above (about those two formulae), should indicate the deceleration is related to the curvature, right?

Sure, $q$ is obviously a function of $w$ and $K$.

But you can also see that $K$ itself doesn't control $q$ - a flat universe still can accelerate or decelerate however it likes if the stuff in it has the right $w$.

@ACuriousMind All right, but what about in the first two equations?

4:13 AM

What about them?

Shouldn't we be able to relate all those Omegas with the deceleration parameter, and thus relate curvature with expansion rate?

You already used all the Friedman equations to arrive at that expression for $q$, you have no independent equations left.

Oh crap

@ACuriousMind Wait, are you looking at the first two equations?

What "first two"?

$\frac{H^2}{H_0^2} = \Omega_R a^{-4} + \Omega_M a^{-3} + \Omega_k a^{-2} + \Omega_{\Lambda}$ and $\frac{\dot{H}}{H^2}=-(1+q)$

Can't we relate $q$ with the Omegas?

4:16 AM

...at the point where you wrote down the equation in terms of $w$, you already chose all of the $\Omega$s to be zero except for one

I.e. you declared one of the energy types "dominating".

So $q$ is already related to the $\Omega$s in the formula you wrote.

Sigh...I'm an idiot

All right, so we've established that in reality, the expansion of the universe is not related to the curvature of space

So the Big Rip scenario has literally nothing to do with whether or not $\rho < \rho_\text{crit}$?

Yes, I believe so.

Now does the Big Crunch have to do with whether or not $\rho > \rho_\text{crit}$?

It does in an ordinary-matter dominated universe, of course, but how about one with all those other factors?

Have we not established that the acceleration and the curvature are independent?

Yes, but how could the Big Crunch happen in an open universe?

Wait, I realize that that is a stupid question

4:21 AM

By $R(t)$ becoming zero at finite time

Q: How can gravity lead to the Big Crunch scenario?

Yeah...

@ACuriousMind Guess I'll need a new answer with completely different mathematics in my older question

5

According to modern cosmology, space is expanding, causing proper distances (but not comoving distances) to increase between galaxies. In the Big Crunch hypothesis, gravity halts and reverses the expansion of the Universe, causing all matter to collide and eventually form a single black hole. Thi...

4:49 AM

why dont we add some kind of button near send/upload that toggles clientside LaTeX on/off

currently I have a bookmark that if im ever on another computer I need to go and hunt down

ooy

how are you doing @DanielSank ?

5:04 AM

@Skyler Just use chatjax++

Dunno why you are using the bookmark

bam got it

@SirCumference that being said, still doesnt help when people are on public computers though

or phones

DanielSank

@Skyler Good!

edit: No idea what I meant to type there.

@Skyler Well, you could read the official reason

18

A: LaTeX on Stack Overflow?

This is implemented on http://math.stackexchange.com -- you can check it out there. It will never be on Stack Overflow, though, as it is an extremely heavy dependency. (See also Nick's investigations about impact in November 2013.) Info here: TeX math markup is sorely needed

Though if you read the comments on that answer, you'll see how little sense it makes to disable LaTeX

@DanielSank was trying to figure that out too

i figured "I wanna play you"?

but decided to wait a sec since i knew an edit was coming

@Skyler But still, I don't get why you're using the slow bookmarklet instead of the plugin

5:09 AM

speaking about playing, i can do fri sat or sunday

@SirCumference well im not anymore, at least most of the time

occasionally i use it when im on slack to get clientside latex

but i just added the greasemonkey script

@SirCumference that being said, couldn't one easily implement a client side button so that the slow latex stuff doesnt happen on the server end

similar to how the bookmarks work

but on the actual webpage

DanielSank

@Skyler That's a safe assumption, because I do.

@Skyler Go write a post about it on Meta SE

Though I wouldn't expect the staff to ever look at those...