The h Bar

Skyler

6:00 AM

electroscopy definition "Measurement with an electroscope."

Sir Cumference

Huh

Skyler

doesnt sound like much of a field for a discipline to be based around

Sir Cumference

Still can't find any definition for "iohydrogen"

@Slereah that a real thing?

Slereah

not that I've heard of it

Skyler

@SirCumference probably doesnt exist, might be a waste of your time

Sir Cumference

6:02 AM

Or the "five extended fundamental forces"

@Skyler Yeah, probably gibberish

@JohnRennie @Jim Well when discussing the Einstein universe, my professor basically said the following (which is still unclear to me):

Skyler

so @Slereah, do you see the stupid little mistake I have to be making there

Sir Cumference

> If gravity perfectly counters the expansion, then the Universe is unstable, because if something 'kicks' the Universe to get it expanding, it'll keep expanding nonstop. If something 'kicks' the Universe to get it contracting, it'll keep contracting nonstop.

Still have no idea what would "kick" the universe to get it expanding or contracting nonstop

Skyler

i just dont remember how you go from your initial value to a solution

John Rennie

@SirCumference I assume he's talking about Einstein's model where the cosmological constant exactly balances out the gravitational attraction of all the matter. Yes?

Sir Cumference

@JohnRennie Yep

G. Bergeron

6:06 AM

@Skyler Take the Fourier transform of the initial conditions

John Rennie

@SirCumference The key point is that the cosmological constant is, well, constant. It doesn't change as the universe expands or contracts.

DanielSank

@Skyler heh, well... Sometimes it takes me years to solve a problem.

Sir Cumference

@JohnRennie So what he said didn't make much sense?

DanielSank

Often, step one is break it down into pieces, and then figure out what math I need for each piece.

John Rennie

@SirCumference but the gravitational attraction of the matter does change. If the universe expands a bit the density of the matter decreases and that means it's gravtiotational decreases.

This means that the (unchanged) cosmological constant now has a bigger effect than the matter, so the universe starts to expand.

Skyler

6:08 AM

@G.Bergeron so just start with $F\{0\}=F\{C(x)\}$

John Rennie

And the more the universe expands the more the density of the matter decreases and the lower becomes its gravitational attraction.

Sir Cumference

@JohnRennie So we get a feedback loop

G. Bergeron

@Skyler Yes

John Rennie

@SirCumference Yes. The more the universe expands the faster it expands.

Skyler

since the LHS is $\frac{dc}{dt}$

G. Bergeron

6:09 AM

Then you apply the diffusion equation

Skyler

ok, that seemed too easy

god i cant believe i wasted so much time

G. Bergeron

And each mode.... Ok you got it

Sir Cumference

@JohnRennie But what about regarding contraction?

G. Bergeron

Laplace transform is the same

Sir Cumference

What you described never works the other way around?

John Rennie

6:10 AM

@SirCumference and the same applies the other way round for contraction. Contraction increases the matter density so it gravitates more, but it still leaves the CC unchanged.

Skyler

@G.Bergeron would you say its the better method here since were given an initial condition?

G. Bergeron

but it's only relevant when there is dissipation

Skyler

rather than BC

Sir Cumference

@Skyler What are you talking about?

@JohnRennie Ah

Skyler

oops

G. Bergeron

6:11 AM

I think he was talking to me

Sir Cumference

@JohnRennie But what would cause that the Universe to expand or contract suddenly in the first place?

John Rennie

@SirCumference so now the matter is compressing the universe more than the CC is expanding it, and the contraction gets faster, which makes the matter density even higher, and we get another feedback loop.

Skyler

@G.Bergeron like in diffusion problems right =P

G. Bergeron

@Skyler You always need the boundary conditions to solve a PDE

John Rennie

@SirCumference The point is that the Einstein universe is fundamentally unstable in that even the tiniest, tiniest change causes it to either expand or contract. For example thermal fluctuuations could cause this.

G. Bergeron

6:12 AM

The question is (in the linear case), can the PDE, as an operator, be diagonalized by a complete set of solutions forming a basis of ''function'' space.

Skyler

@G.Bergeron so am I just using the words "initial value" incorrectly

Sir Cumference

@JohnRennie Wouldn't it only be affected by the shifting in the distribution of matter?

Skyler

i remember back in the days of ODEs talking about IVP vs BC

Sir Cumference

After all, gravity here is basically decided by the mass density, right?

John Rennie

@SirCumference the equation for the Einstein universe assumes that matter is uniformly distributed. Technically it assumes isotropy and homogeneity.

G. Bergeron

6:14 AM

If yes, then you can decompose any initial conditions on that basis, and since you know the action of the operator (really, the PDE) on each ''basis element'', it's easy to evolve the initial condition

@Skyler Sorry I meant to say initial conditions

Sir Cumference

@JohnRennie All right

Btw, one more question

John Rennie

@SirCumference in the real universe matter isn't uniformly distributed, and as you say even the smallest fluctuation would destabilise an Einstein universe.

Sir Cumference

Can the Big Crunch only happen in a closed Universe where $\rho > \rho_{\text{crit}}$?

G. Bergeron

@Skyler The boundary conditions has to do with what will be the basis functions while the initial conditions has to do with, well, the actual solution your system has out of the many possible ones.

@Skyler In diffusion there is not necessarily dissipation, in general not.

I meant dissipation as in: some energy leaves the system

John Rennie

@SirCumference That is certainly true for a matter dominated universe. However it depends on the equation of state of the stuff in the universe. For example Andre Linde has proposed that there could be a crunch far in the future due to supergravity effects.

Skyler

6:17 AM

@G.Bergeron so just $0 = k^2 \hat{C} (k)$

G. Bergeron

As an example, think about an RC system... Some energy will be dissipated by the R which is not modelled by the equation

So they will lead to "imaginary Fourier variable" or the Laplace transform

@Skyler ?

No

Sir Cumference

@JohnRennie Well, realistically speaking, in a Universe where gravity is strong enough to contract space to a Big Crunch, won't most of that gravity be caused by matter?

G. Bergeron

Yes

but there is the time evolution

so not zero

Skyler

@G.Bergeron but isnt the fourier transform of the 2nd derivative of C just $(-ik)^2 \hat{C}(k)$

John Rennie

@SirCumference yes, realistically speaking, the condition you give is sufficient.

With a small correction due to dark energy ...

Sir Cumference

6:20 AM

@JohnRennie But technically speaking, you wouldn't be able to determine the curvature of a Universe where a Big Crunch would take place, right?

Solely by knowing the deceleration parameter, etc

John Rennie

@SirCumference I'm not sure why you say that. I'm fairly sure that isn't true ...

Sir Cumference

@JohnRennie Well, you explained earlier that the curvature of the Universe doesn't tell us the sign of the deceleration parameter, right?

Or perhaps it was @ACM who explained that...

G. Bergeron

@Skyler $k^2 \hat{C}(k) = - s \alpha $

Skyler

@G.Bergeron OH WAIT, I need to find $\hat{C}(k)$ first by just fourier transforming

G. Bergeron

yes

Sir Cumference

6:24 AM

@JohnRennie I'm using the formula $q=\frac{1}{2}(1+3w)\left(1+k/(aH)^2\right)$

Where $q$ is the deceleration parameter, $k$ is the curvature of the Universe and $w$ is the equation of state

G. Bergeron

Everything relies on separating the variables then seeing that the space part is solved by functions of the form $e^{-i k x}$ and time part by functions of the form $e^{-s \alpha t}$

Skyler

@G.Bergeron you lost me at the $-s\alpha$

John Rennie

@SirCumference I prefer $q = \frac{\ddot{a}a}{\dot{a}^2}$ because I'm more familiar with working with that form, but the two forms are equivalent.

Skyler

@G.Bergeron hmm

John Rennie

@SirCumference and the point is that it doesn't depend on the sign of $\dot{a}$ because that term is squared.

Sir Cumference

6:26 AM

@JohnRennie So as you can see, knowing the curvature $k$ doesn't tell us the sign of the deceleration parameter if we don't know $w$

G. Bergeron

Suppose the solution is of the form $u(t,x)=X(x)T(t)$

Sir Cumference

@JohnRennie That works too

G. Bergeron

Shove that in the heat equation and realize that $X''(x)/X(x)=T'(t)/T(t)\alpha$

John Rennie

@SirCumference So while $q$ tells us the sign of $\ddot{a}$ it doesn't depend on whether the universe is expanding or contracting.

Sir Cumference

@JohnRennie Wait, I'm sorry?

G. Bergeron

6:28 AM

so $X''(x)=-\lambda X(x)$ and $T'(t)=-\lambda \alpha T(t)$

Sir Cumference

You mean the sign of $q$ doesn't tell us the sign of $\ddot{a}$?

Skyler

@G.Bergeron btw, this problem is generally seperable as well isn't it

John Rennie

@SirCumference I didn't say that did I? It doesn't tell us the sign of $\dot{a}$.

Skyler

so basically Laplace Fourier show the obvious seperability relation, but sometimes its not so obvious

and thats the power in this analysis

John Rennie

It does tell us the sign of $\ddot{a}$

Sir Cumference

6:29 AM

@JohnRennie Oh, all right

G. Bergeron

@Skyler Well, yes it's a property of the heat equation

Actually, it comes from the symmetries of the equation

Sir Cumference

@JohnRennie So we can't actually know whether the Universe is expanding or contracting via the deceleration parameter; we can only know if it's speeding up or slowing down?

John Rennie

@SirCumference Yes

Sir Cumference

Oye

John Rennie

But why the focus on the deceleration parameter? That is just one out of lots of things we could experimentall measure.

G. Bergeron

6:31 AM

@Skyler Laplace/Fourier thingy is there because of the existence of many many second order separable differential equations

Sir Cumference

So wait, is $\dot{a}$ related to $k$ (or $\Omega$, etc)?

@JohnRennie I feel dumb, the darn thing is called the "deceleration parameter"...

G. Bergeron

Fourier basis functions are the general solutions to $\nabla^2 f = \lambda f$

In cartesian coordinates

They are not always practical to work with if the domain has weird boundary shape

Skyler

@G.Bergeron for this basically my Heaviside can just be simplified into just limiting the integration bounds to $[-\frac{a}{2}, \frac{a}{2}]$

when transforming the time invariant

John Rennie

@SirCumference $\dot{a}$ is given by the first Friedmann equation, and it depends on lots of things including the curvature and the density.

G. Bergeron

I'm not sure from your domain, but it may even be only the fourier series, not transform

Sir Cumference

6:34 AM

@JohnRennie Wait, hold on. If $q$ tells us the sign of $\ddot{a}$, and $q$ were negative, wouldn't we know that $\dot{a}$ would eventually decrease below zero and the Universe would collapse?

G. Bergeron

If it is finite

Skyler

@G.Bergeron infinite domain

G. Bergeron

And I miss the last subway and had to grab a cab >:(

@Skyler So transform it is

Skyler

@G.Bergeron so I should just transform a square wave right

G. Bergeron

It's all done here en.wikipedia.org/wiki/Heat_equation

Yeah

For the shock solution, though, this will be an entirely other beast

Because non-linear

John Rennie

6:37 AM

@SirCumference no. A negative $\ddot{a}$ just means that $\dot{a}$ is decreasing, but it doesn't necessarily mean $\dot{a}$ will ever reach zero or become negative. Remember that the magnitude of $\ddot{a}$ decreases as the universe expands so it isn't a constant deceleration.

G. Bergeron

But I'm not sure they can arises here...

@JohnRennie Ideas for food? :)

Sir Cumference

@JohnRennie So $\dddot{a}$ can be negative?

John Rennie

@SirCumference In a universe without dark energy, or where the matter density dominates over dark energy, $\ddot[a} < 0$.

Until we discovered dark energy we thought our universe was like this.

DanielSank

Ok, @Skyler, what's up?

Skyler

@G.Bergeron have a sweet tooth

G. Bergeron

6:39 AM

I really want to eat pierogies! It's been in my head since that picture... Not sure where to get them, though.

Skyler

@DanielSank I think G helped cover most of it

John Rennie

@SirCumference oops sorry, I only just noticed that was three d's. Did you mean $d^3a/dt^3$?

Sir Cumference

@JohnRennie Yep

G. Bergeron

@Skyler I don't. Fatty, salty and savory for me

@Skyler I don't really like sugar

Skyler

hey guys

what are some spanish style chicken spicings

DanielSank

6:41 AM

@Skyler k

Skyler

im going to have it with refried beans, rice, cheese

G. Bergeron

@Skyler There remains the possibility of shocks, but I'd think they're not possible in this case

John Rennie

@SirCumference Erm, well, in a universe without dark energy $\ddot{a}$ is negative but it's magnitude decreases with time, so $\dddot{a}$ is positive.

Skyler

@DanielSank i could use help deciding on dinner tho

Sir Cumference

@JohnRennie Wait what

Skyler

6:42 AM

got any more cool recipes to throw my way

G. Bergeron

@Skyler ...hmm

DanielSank

@Skyler Depends, what ingredients do you have, how much time do you have?

John Rennie

@SirCumference that didn't seem controversial ...

G. Bergeron

@DanielSank Listening 8D

Sir Cumference

@JohnRennie I thought right here you explained that if $\ddot{a}$ decreases but never reaches zero, $\dot{a}$ will never be less than 0?

Skyler

6:42 AM

@DanielSank basically the main thing im missing is a spice blend for chicken to go along with mexican styledd side

Sir Cumference

Sigh...I might be confusing myself

John Rennie

@SirCumference I don't think I said that ...

DanielSank

@Skyler Sounds like you don't need my help...

G. Bergeron

avocado, spinach, brocoli, rice, chicken meat balls, all the spices and stuff in furnished kitchen

Skyler

@DanielSank i dont know spicing for mexican food AT ALL

Sir Cumference

6:43 AM

@JohnRennie Wait, so what did ya mean by this?

6 mins ago, by John Rennie

@SirCumference no. A negative $\ddot{a}$ just means that $\dot{a}$ is decreasing, but it doesn't necessarily mean $\dot{a}$ will ever reach zero or become negative. Remember that the magnitude of $\ddot{a}$ decreases as the universe expands so it isn't a constant deceleration.

Skyler

especially when it comes to meat

DanielSank

@G.Bergeron Ok here's what I'd do:

John Rennie

@SirCumference NB: it doesn't necessarily mean $\dot{a}$ will ever reach zero or become negative

DanielSank

Steam broccoli and spinach. Cook rice with cumin, then after the rice is done, stir in some avocado. Put the veggies on top of the rice and if you have it, fry an egg and put that on top and eat it all.

Skyler

@G.Bergeron do you have butter or parmessan?

Sir Cumference

6:44 AM

@JohnRennie I know

DanielSank

Chicken meatballs optional.

G. Bergeron

@Skyler Yes

Skyler

are the meatballs preseasoned

Sir Cumference

So I asked if $\dddot{a}$ could be negative in that case

G. Bergeron

@Skyler not really

John Rennie

6:45 AM

I didn't say $\dot{a}$ will never be less than 0

G. Bergeron

@DanielSank Miam!

DanielSank

@G.Bergeron Wat?

Sir Cumference

Woah, hold on

I gotta get everything straight

G. Bergeron

@DanielSank French expression I guess

Skyler

something i often do with chicken is I take my rice or pasta , and stir in grated parmessan and butter while its still steaming hot

steam broccolis, cook chicken

save avocados for another meal

Sir Cumference

6:46 AM

@JohnRennie All right, now I got what you're saying

G. Bergeron

But I like the idea of the fried egg over it all

DanielSank

@Skyler No no, use the avocado as fat for the rice.

Skyler

@G.Bergeron its basically an alfredo but with rice

Sir Cumference

Basically, again, $q$ doesn't tell us $\dot{a}$, right?

G. Bergeron

@DanielSank Yes, much avocado = many goodness

Skyler

6:47 AM

@DanielSank hmm...ive never thought about trying to sub out the fat for avocado in an alfredo

that could actually work

John Rennie

@SirCumference it doesn't tell us the sign of $\dot{a}$

Sir Cumference

@JohnRennie Yeah, sorry

Skyler

@DanielSank we should try this when we meet up in january

G. Bergeron

@DanielSank Do you think the avocado would go well with extra butter?

Or coconut oil

John Rennie

@G.Bergeron everything goes well with extra butter :-)

DanielSank

6:48 AM

@G.Bergeron Coconut is a gamble.

G. Bergeron

@JohnRennie True

DanielSank

I wouldn't pair it with cumin, but that's your choice.

G. Bergeron

@DanielSank Yes, it's quite fragrant

DanielSank

@Skyler Dude, next time we meet up we're making my special magic chili.

Skyler

@DanielSank true true

DanielSank

6:48 AM

Multiple people have said it was one of the best dishes they've had.

G. Bergeron

Usually I just put a super dollop in coffee if I'm to skip breakfast

DanielSank

Learned how to play Piano Man on guitar this evening.

Skyler

@DanielSank do you think Pão de Queijo rolls would go along with it

G. Bergeron

@DanielSank I have two vegetarian brothers, might propose it to them

DanielSank

@Skyler dafuq are those?

John Rennie

6:49 AM

@SirCumference can I suggest a simple analogy? I won't bother if you think you have things under control ...

G. Bergeron

It's the recipe on you profile, right?

DanielSank

@G.Bergeron Yeah, it's designed for vegetarians.

Sir Cumference

@JohnRennie I do, but it might help

DanielSank

@G.Bergeron o_O It's on my profile?

Skyler

@DanielSank amazing heavenly brazillian cheese rolls, we talked about it once

or twice

@DanielSank hey, actually, spanish rice

with a fresh tomato

(found a big one I left at home)

any idea what I should do there

G. Bergeron

6:51 AM

@DanielSank No... but then where did I see it...

Ah! On your description for the AMA?

DanielSank

@G.Bergeron Possibly...

John Rennie

@SirCumference suppose you site on the surface of the Earth and throw a stone outwards i.e. to increasing $r$. Then $\dot{r}$ is positive and $\ddot{r}$ is negative. And $\ddot{r}$ chages with time because the Earth's gravity decreases with $r$. Make sense so far?

G. Bergeron

yes : meta.physics.stackexchange.com/questions/8981/…

Sir Cumference

@JohnRennie Yeah, so $\dddot{r} \ne 0$, right?

Skyler

@DanielSank i feel like you mentioned spanish rice very recently

John Rennie

6:52 AM

@SirCumference correct.

The calculations for an expanding universe are actually conceptuallt very similar.

G. Bergeron

Gosh! Seeing the picture now makes me super hungry! Gotta go fetch something... Coming back later

DanielSank

@Skyler Me? Nah. Don't think so.

@G.Bergeron Ciao.

John Rennie

@SirCumference The stone will fall back if you don't throw it hard enough, or it will speed off to infinity if you throw it really hard. Likewsie a universe can recollapse or expand forever.

Skyler

@G.Bergeron later

Sir Cumference

@JohnRennie Yup

DanielSank

6:54 AM

Bye, everybody.

Sir Cumference

But in the latter case, $\ddot{r}$ wouldn't be positive, right?

Sigh, wait, I'm thinking of throwing a stone until it reaches space and just keeps going

John Rennie

@SirCumference yes. You would have to introduce some repulsive force to simulate dark energy for $\ddot{r}$ to be positive.

Sir Cumference

Oh, all right

John Rennie

$\ddot{r}$ is of course just $-GM/r^2$ in this analogy.

i.e. the gravitational acceleration.

Sir Cumference

Makes sense

Thanks a ton by the way

John Rennie

6:59 AM

You're welcome :-) This stuff can be massively confusing until you get your head around it.

2

G. Bergeron

There is a God! Somebody cooked a meal out of the above ingredients already! 8D

Sir Cumference

@Qmechanic Uh, buddy

2

How on earth did you get 95 silver badges

G. Bergeron

lol

Sir Cumference

@MAFIA36790 @BernardMeurer @heather @Anyone who might care

Found this on Obliv's profile

> Sorry to those that have got to know me at the h-bar. I will one day return but right now things are changing so I can't with good conscious remain active there. Thank you for everything :)

Jesus, everyone's leaving ;(

I'm guessing it's because @0celo7 got banned...

Literally everyone I give 100 rep to leaves

ffs