The h Bar

@0celo7 sour grapes?

:P

The other guy got twice as long!

1:44 AM

nature.com/articles/s41586-018-0038-x

Yes, we have drumhead entanglement now

nature.com/articles/s41586-018-0025-2

It's incredibly crowded in there

Bernardo Meurer

@JohnRennie Can you give me an openBSD instance on your server as a VM?

2:08 AM

sciencedaily.com/releases/2018/04/180423135054.htm

We are fortunate to have time sort of "flowing" in a sense such that for anything transient, we just need to make the right probe, and then wait until the detection took place

We don't quite have that luxury for spatial things as space (except near a black hole) is not directiomal

sciencedaily.com/releases/2018/04/180423155051.htm

Wait, if stirring fluid is already a computation, then what physical process is not computation??

7:13 AM

Morning

@enumaris i actually have worked with autoencoders

Tho I don't know what a variational one is

7:52 AM

@JohnRennie^

@Slereah morning

@s.patroller I have always pronounced scone to rhyme with gone, but then both my parents are from north of the great scone divide.

scone

:-)

@JohnRennie oop north?

8:25 AM

@Slereah My dad was from Northern Ireland and my mum from Middlesborough

Though I spent my formative years in Khartoum in Sudan, which is a wee bit south of the UK :-)

(how you pronounce the Arabic for scone I have no idea)

By a strange coincidence Michael Atiyah also spent his early childhood in Khartoum, though he had returned to the UK by the time I was born.

Kyle Oman

Gaia satellite data release 2 is out, with accurate motions for more than a billion stars. Fun time to be an astronomer, or even around astronomers for that matter! esa.int/esatv/Videos/2018/04/Gaia_second_data_release_-_replay (from 29:00 or so is the more relevant part)

3

Mithrandir24601

@JohnRennie And you're exactly right :P :)

8:52 AM

How can a black hole even form in a finite time for people at earth if everything in stellar core would go slower and slower as it collapses due to gravitational time dilation?

@Allahjane Things do not seem to go slower for observers inside the black hole

please read the question again

"for people at earth"

@Slereah

Well what does "black hole forming" mean for the people at earth

what do you expect to see corresponding to the formation of a black hole

We just see a collapsing star very slowly collapsing

a star collapsing under supernova, the way its believed how most blackholes form in space

yes , if we were never inside a black hole how did the existing blackholes formed in the first place? won't the stars's mass collapse slower and slower due to increase in gravitational time dilation at the centere of the collapsing star

so for an outside observer it would take infinite time for that mass to collapse to a point where singularity forms

@Allahjane you are quite correct that event horizons take an infinite time to form for outside observers.

9:03 AM

so how do we explain existing black holes? if any of them are real

When astronomers talk about black holes they mean objects so dense that they woul form an event horizon given infinite time.

so technically saggitarius A* is not really a black hole but a lot of matter packed so closely that it appears to be a black hole

Well if you fell into it you'd get a pretty good idea that it really is a black hole!

but of course that is hard to check

that's my question

and there's the whole problem that Schwarzschild black holes aren't really "real" black holes, which are much more complicated

9:05 AM

if it really is a black hole, then time dilation forbids its formation or growth in finite time

by stellar means ofc

@Allahjane it depends on how you define the term black hole. That sounds like playing with words, but the archetypal black hole - the Schwarzschild metric - is a mathematical ideal that cannot exist in reality. So the term black hole is actually somewhat vaugely defined.

damn so all those documentaries I saw on youtube lied to me

That's pop science for you. Real physics is rather more complicated than they make out.

Documentaries are made by people who don't know physics for people who don't know physics

Some elements will get lost in the shuffle

I'd guess when most physicists say black hole they mean something like the Oppenheimer-Snyder geometry, and that does exist in real life. So in that sense black holes really do exist.

9:08 AM

I mean Schwarzschild because who cares for stellar black holes

Also OS-black holes also aren't realistic

The real black holes also need some Vaidya sort of section

@Slereah that's why I said something like the Oppenheimer-Snyder geometry

hah so we're not even sure if black holes even exist but we've derived a bunch of theories on it like hawking radiation , no hair theorm dual reality and what not

Hi, I'm trying to figure out if a question is on topic for physics stack exchange or not. Would this be a good place to find out?

sure

You don't get points for posts in community wiki threads?

Whack.

Alright, so the question is loosely speaking "What would the universe be like towards heat death?" Specifically, I'd be wanting to know what a civilization trying to survive in the universe might experience.

9:17 AM

sunburn! duh

@Allahjane nope, Hawking's original treatment of the radiation specifically took into account the formation stage of the black hole i.e. it was nearer to the OS metric than the Schwarzschild metric.

plus there's tons of papers on various variations on Hawking radiation

not very fun papers but they do exist

@Rithaniel jokes aside, I don't think any civilisation would be able to survive near the time of heat death as matter would likely start to disintegrate into sub atomic particles due to intense thermal energy.

So unless they find a way to generate a force field of sort to make a liveable bubble they'll all get sunburns

So, I think this might be a little bit too speculative, but, at the same time, many questions on the exchange tend to be seeking speculation on topics. So, I'm up in the air on if I should ask it here or not.

heat death isn't really the "intense thermal energy" era

9:22 AM

@Allahjane heat death is when there is "no heat left," though.

oh

well there is plenty of heat, just spread over the universe

To put it in colloquial terms.

that's kinda opposite of what I explained lol

then they'll all get frostbites

and liveable bubble would need to keep enclosed space from expanding too fst

maybe around some exotic black hole

So, would it be a good question to post? Of course, I'd give more specifics as to what I'm looking for in the body of the question. I'm most concerned with posting it in the wrong place.

9:24 AM

you might need to make it a bit more specific , otherwise it'd be more of a sci fi and speculation rather than known physics

@Semiclassical duh, got it

say you've got a 1D $k$-dependent hamiltonian $$H(k) = \begin{pmatrix} 1& k \\ k & -1 \end{pmatrix}$$

then the eigenvalues are $E(k) = \pm \sqrt{1+k^2}$

but more simply, those eigenvalues are the implicit-function projections (a.k.a. "solutions") of the characteristic polynomial

$$ E^2 = 1+ k^2$$

a.k.a. your wonky single-Riemann-sphere polynomial thing

so the natural 2D extension is to $$H(k_x,k_y) = \begin{pmatrix} 1& k_x-ik_y \\ k_x+ik_y & -1 \end{pmatrix}$$

and $$ E^2 = 1+ k_x^2+k_y^2$$

though I guess if you want nonzero genus then you need at least one more band

speak english you barbarian

so maybe for a bigger genus, in 1D

$$H(k) = \begin{pmatrix} k & 1 & 2 \\ 1& 0 & 1 \\ 2 & 1 & -k \end{pmatrix}$$

with characteristic polynomial $$ E^3-(6+k^2)E+4=0$$

which I imagine you can happily extend to $$ E^3-(6+k_x^2+k_y^2)E+4=0$$ if you're in the mood for 2D stuff

so I guess the question is, what's the topology of that complex manifold?

@Allahjane I am speaking English ;-)

9:41 AM

would barbarians speak Vulgar Latin?

@s.patroller yes

cf the Gauls

9:51 AM

wait, simpler version. $$H(k) = \begin{pmatrix} 2 & -k & 0 \\ -k & -2 & k \\ 0 & k & 0 \end{pmatrix}$$ with characteristic polynomial $$ E^3 -2(2+k^2) E +2k^2=0$$

10:56 AM

@Allahjane I'm afraid this is what happens when you poke your nose into a den of physicists :-)

11:10 AM

We didn't even bring any nlab talk yet

Let's see how they define black holes

nothing fun

boo

where's the category of black holes

12:26 PM

Hey Arxiv updated its search engine

2

cool beans

I tried to find a picture to reply but they're all great, I can't choose

:-D

Hi, I'm your new meteorologist and a former software developer. Hey, when we say 12pm, does that mean the hour from 12pm to 1pm, or the hour centered on 12pm? Or is it a snapshot at 12:00 exactly? Because our 24-hour forecast has midnight at both ends, and I'm worried we have an off-by-one error.

Semiclassical

1:33 PM

@EmilioPisanty yeah. you're still doing complex manifolds, instead, it's just one which lives in CC^3 not CC^2. But my knowledge of such is...not great. (Something something Kahler manifolds, I think?)

I think there's a few people in the math chat who may know more (e.g. Danu or Ted Shifrin)

1:58 PM

@Semiclassical hmmmm, interesting

> In mathematics and especially differential geometry, a Kähler manifold is a manifold with three mutually compatible structures: a complex structure, a Riemannian structure, and a symplectic structure.

how do you get a symplectic structure?

@EmilioPisanty with a symplectic form

or is it just the complex structure with a fancy name?

It's a $2$-form that's skew symmetric

@Slereah so how do you induce a natural symplectic form on $$ E^3 -2(2+k_x^2+k_y^2) E +2k_x^2+k_y^2=0$$

?

I don't know what you're doing so I dunno

2:00 PM

i.e. when seen as a 2$\mathbb C$D manifold embedded in $\mathbb C^3 \ni (E,k_x,k_y)$

@Slereah well, I want to understand what type of manifold that restriction gives

Sympletic manifolds need to be even dimensional

@Slereah real-dimensional or complex-dimensional?

I'm not sure

I don't deal a lot with complex manifolds

The standard symplectic manifold is $T^* M$

@Slereah I know

the question was how you induce a natural symplectic structure on a manifold $M= \{(x,y,z)\in \mathbb C^3 : F(x,y,z) =0 \}$

where $F(x,y,z)$ is complex-valued and analytic

Sounds like an @0celo7 question

or @BalarkaSen

2:05 PM

if we're pinging people let's throw @Danu into the mix

And @JohnDuffield

@Slereah oh fcs

Q: what prevents a manifold to be symplectic?

I guess @ACuriousMind would know, too

I hope he doesn't take umbrage to the fact that I thought of JD before him

what are you doing that requires a symplectic Kahler manifold

From descriptions, I think it would be even complex dimension

I don't know if any even-dimensional manifold admits a symplectic form, though

"Typically if you figure out a new way to build symplectic manifolds via surgery, it's worth writing a paper about."

ouch!

7

Are there any obstructions known which prevent an even dimensional orientable manifold from being symplectic? I am a novice in this area so I unfortunately I cannot make the question more precise. What I have in mind is raw and is as follows:from the point of handle body theory you can put a symp...

sg.symplectic-geometry

@EmilioPisanty Ask Mike Miller on the math chat, he will know for sure.

@Slereah You need to learn some complex geometry.

Probably

Especially if I hope to do some more string theory

They are fond of complex structures

Pretty much all string theory is done on complex surfaces

My bank paperwork has gone through

Finally, free money

2:23 PM

/

Just kidding, it's money for the same amount of money plus 20k

hullo @EmilioPisanty

@EmilioPisanty Even real

not even complex

The point is that the existence of a non-degenerate two-form on a real vector space implies even-dimensionality

You want to have a natural symplectic structure on $\{F(x,y,z)=0 \}\subset \Bbb C^3$ for any analytic complex function $F$?

@Slereah No, not every even-dimensional manifold admits a symplectic structure

Sad!

that's very easy to see, since non-degeneracy of $\omega$ implies that $[\omega^k]\in H^{2k}(M;\Bbb R)$ is non-zero.

Hence, no even-dimensional sphere except for $S^2$ admits a symplectic structure, for instance.

What about $S^0$

2:31 PM

of course it does

it's 2 points; its tangent spaces are points; there are no non-zero tangent vectors

it's a trivial case

Is orientability gonna be an issue

You should read Huybrechts' book @Slereah, if you're serious about learning some basic complex geometry.

Of course, but who cares about non-orientable manifolds?

Usually there's nothing fun on a non-orientable manifold

@Danu America.

Iraq was a non-orientable manifold

@EmilioPisanty Not every space of the form you mention is even a manifold, let alone symplectic.

($F=xy$)

2:37 PM

He presumably means a regular level set.

What is something that is not even a conifold or a causal (forgot exact name) thingy?

guess that will be just a set with a topology?

@Secret a varifold

By the way, @Slereah, what I said earlier only applies to compact manifolds, of course

why do math people only ever deal with the comfortable objects

They always do compact manifolds and bounded operators for every theorem

I'm working on non-compact manifolds at the moment

Their topology is usually very boring.

2:42 PM

What about the Loch ness surface

I am interested in unbounded operators, but so many things can go wrong with them

So for topological questions compact and oriented is much more interesting usually

You gotta have Poincare duality

I don't live in a compact spacetime my man

I gotta have my non-compact manifolds

Riemannian geometry is interesting for non-compact things too

what's a bounded example of Riemann manifold that is not compact?

attempt to think about it just give me hyperbolic planes after hyperbolic planes, which is non riemannian

2:46 PM

The disk

I work on that

the disk?

(=non-compact symmetric space)

@Slereah except for reed and simon is 85% unbounded and there's a shitload of work on noncompact geometric analysis

that is good

for some reason QM is really unwilling to deal with unbounded operators even though that's most of the important ones

2:52 PM

what world are you living in

America

that's a Rammstein song

I dunno, whenever I find some QFT axiomatic thing it's always starting from a $C^*$ algebra

Who will speak for the unbounded QFT operators

Like the FIELD OPERATOR

I think that's because they're considering everything to be exponentiated

yeah that's what they usually do

I guess it's easier to deal with

2:56 PM

if you want the unbounded goodness, see Reed and Simon or Davies

True

although Davis cheats a lot by exponentiating as well