The h Bar

0celouvskyopoulo7

3:02 PM

How does Kevin Gates even work out his two passions

Trap music and Islam

they're pretty much incompatible

Slereah

is it

0celouvskyopoulo7

no drinking in Islam

John Rennie

Event Horizon Telescope

The Event Horizon Telescope (EHT) is a project to create a large telescope array consisting of a global network of radio telescopes and combining data from several very-long-baseline interferometry (VLBI) stations around the Earth. The aim is to observe the immediate environment of the Milky Way's supermassive black hole Sagittarius A*, as well as the even larger black hole in Messier 87, with angular resolution comparable to the black hole's event horizon. == Overview == The EHT is composed of many radio observatories or radio telescope facilities around the world to produce a high-sensitivity...

0celouvskyopoulo7

trap = drug house

Slereah

Not all drugs are alcohol

John Rennie

3:03 PM

@Slereah Results due in a few months.

0celouvskyopoulo7

pretty sure no drugs period

Slereah

@JohnRennie woo

0celouvskyopoulo7

what are they looking for?

aliens?

Slereah

Alcohol is globally banned (and even then there are interpretations), but drugs vary

0celouvskyopoulo7

where are the aliens at, anyway

@Slereah are skrippers allowed?

Slereah

3:04 PM

I do not know

John Rennie

@0celouvskyopoulo7 That nice Mr. Wikipedia gives a good explanation.

Slereah

In some places, the interpretation is that "alcohol" refers to wine in the qu'ran

So beer is fine

you may recall that a lot of cannabis and heroin comes from muslim countries

Obviously not everyone has that attitude towards drugs

Balarka Sen

drugs are great

0celouvskyopoulo7

I would think American rap music is seen as degenerate by Muslims

@BalarkaSen implying you do drugs

Slereah

In France a lot of muslims went into rap

Balarka Sen

3:07 PM

no but they're great

0celouvskyopoulo7

it's seen as degenerate by basically everyone

@BalarkaSen you're just being edgy

@BalarkaSen Do you know the co-area formula?

Balarka Sen

yes. doing drug would be mass culture; denying drug as bad would also be mass culture

Slereah

What will the EHT test for, exactly?

Balarka Sen

so i am thinking of it as great but not doing it

that's avant garde

0celouvskyopoulo7

@BalarkaSen lol

mathvc_

3:08 PM

@Slereah i guess that's not true

Slereah

Well all religions are false

But them's the facts

Balarka Sen

@0celouvskyopoulo7 nope

Slereah

Apparently the GR test will be observing the black hole's photon sphere

0celouvskyopoulo7

Aha. This is stupid. He defines $\Omega(t)=[|f|>t]$, so you have to do the integral backwards

Lame

Balarka Sen

hey, is a small patch on the pseudosphere isometric to H^2?

0celouvskyopoulo7

3:10 PM

@BalarkaSen Pseudosphere?

Slereah

$H^2$ is the cover of the pseudosphere

John Rennie

@Slereah The rotation of the black hole should make the shadow it cases slightly asymmetric. This is expected to be just about observable by the EHT.

Balarka Sen

a surface in R^3 which is everywhere K = -1

John Rennie

@Slereah Since this is a direct test of the strong field region of GR, and I think it's the first test of the strong field regime, it will be an important result even if it seems a slightly unerwhelming one.

Slereah

that would be nice

John Rennie

3:12 PM

It would be the first direct evidence for a black hole.

0celouvskyopoulo7

@BalarkaSen Then Sam is right, the cover is $H^2$.

So your little patch is isometric to an open subset.

Balarka Sen

@Slereah @0celo Hmm. Why does that not break the Hilbert's theorem?

Slereah

Please, let's not be too informal

Balarka Sen

am I misremembering Hilbert's theorem?

Slereah

Call me Dr.

0celouvskyopoulo7

3:13 PM

Certainly not isometric to the whole thing, you'd have to blow it up which would scale $K$.

Slereah

Dr. Toboggan.

0celouvskyopoulo7

I don't know which of Hilbert's theorems you want

Slereah

Which one is the Hilbert theorem

Balarka Sen

@0celouvskyopoulo7 Oh. A small patch on H^2 is not isometric to H^2?

I guess that makes sense

0celouvskyopoulo7

Isometric? I really doubt it because you'd have to scale it

An isometry is length preserving

Slereah

3:15 PM

^me

0celouvskyopoulo7

How are you going to preserve lengths if you blow up a small set to the whole thing

Does that make sense?

Danu

1

Q: Gauge Redundancy-Fredholm Operators

Below is my attempt to understand gauge redundancy: Let's consider some starshaped region $S$ of a smooth 4-dimensional manifold $M$. The space of 1-forms I will denote $\Lambda^1(S)$. Then the equations of motion for (say) electrodynamics, are in fact nothing but a linear differential operator ...

Is this comment by Lawrence B. Crowell crazy or is it me?

Balarka Sen

@0celouvskyopoulo7 yeah

Danu

Hey 0ce did you ever encounter "homotheties" in Riemannian geometry?

Do they mean isometries + scaling or something else?

0celouvskyopoulo7

Conformal maps with constant conformal factor or something else?

Danu

3:19 PM

You mean scalings :P

But sure

so you've seen them defined before?

0celouvskyopoulo7

@BalarkaSen Well now I feel silly because I don't understand that theorem of Hilbert. $H^2$ is in $\Bbb R^2$ as the half-plane. I guess it can't be put into $\Bbb R^3$?

@Danu Yeah, but just in exercises.

Danu

If it can be put in $\Bbb R^2$ then also in $\Bbb R^3$

Balarka Sen

@0celouvskyopoulo7 not isometrically

0celouvskyopoulo7

@Danu I mean an isometric immersion.

@BalarkaSen Huh

Danu

@0celouvskyopoulo7 My comment is still true :P

compose it with the obvious isometric embedding

0celouvskyopoulo7

3:21 PM

I guess $H^2$ isn't in $\Bbb R^2$ isometrically.

Danu

nope :P

Slereah

Hm, to show that the distance function on a Riemannian manifold is $d(p,q) = 0 \to p = q$, do I just show that if the curve ever has a non-zero tangent, its integral will be $>0$?

0celouvskyopoulo7

@Danu So what's up with homotheties?

I'm sure it's in do Carmo but the index is crap so I can't tell you where.

Slereah

Ah, the proof is in O'neill

The proof uses... the Hausdorff property!

Danu

@0celouvskyopoulo7 Metrics always come in families

0celouvskyopoulo7

3:26 PM

@Slereah What page?

Danu

so you often get uniqueness up to homothety

Slereah

136

0celouvskyopoulo7

@Danu What problems are you doing where you have to solve for a metric?

Danu

I never solve for any metrics :P

0celouvskyopoulo7

Then what uniqueness are you talking about?

Danu

3:27 PM

and I'm not doing any problems. Just understanding classifications of Einstein metrics on some spaces

Slereah

Back when I did that year of math, I did a presentation on Riemannian manifolds, and showed that for geodesics the distance was the shortest

0celouvskyopoulo7

@Danu ok

Slereah

I remember the professor saying that the rigorous proof involved jets

I wonder what that's about

0celouvskyopoulo7

What?

It...doesn't.

Slereah

Well I only vaguely remember

Maybe I remember wrong

Danu

3:28 PM

@Slereah Of course it must, since you need unique limits.

It indeed does not @Slereah.

0celouvskyopoulo7

Geodesics don't always minimize distance

But they are critical points of the length functional

And any length minimizing curve is a geodesic

Danu

they locally minimize distance also

which is important

0celouvskyopoulo7

Yes

Which is why geodesic balls are so important

Danu

Do you know what Jacobi fields are good for?

0celouvskyopoulo7

@Danu No, you need to find disjoint geodesic balls containing each of $p$ and $q$

Danu

3:30 PM

They just seemed ugly and boring in Milnor's book on Morse theory, but I didn't read the applications part of the book

@0celouvskyopoulo7 Doesn't matter. You will always need uniqueness of limits.

0celouvskyopoulo7

I am telling you what the proof he is referring to uses explicitly. No use arguing with me on that.

@Danu Yes

Danu

and I'm telling you why it must rely on the Hausdorff property

0celouvskyopoulo7

Limits has nothing to do with it though. One uses the definition of Hausdorff directly...

Danu

lol

you can keep thinking that

but without uniqueness of limits it's obviously false

and the regularity condition on manifolds that ensures it is the Hausdorff property

hence it must rely on that property

0celouvskyopoulo7

Uniqueness of limits is equivalent to Hausdorff (assuming some other axiom) so this is a pointless discussion.

Danu

3:33 PM

So why start it?

0celouvskyopoulo7

You started it

Danu

LOL

3 mins ago, by 0celouvskyopoulo7

@Danu No, you need to find disjoint geodesic balls containing each of $p$ and $q$

0celouvskyopoulo7

That wasn't me.

Danu

Bye

0celouvskyopoulo7

So what do you want to know about Jacobi fields?

Slereah

3:42 PM

@0celouvskyopoulo7 it's not

There are non-Hausdorff spaces with unique limits

"The co-countable topology on an uncountable set X (like the reals) (the only closed sets are X and all sets that are at most countable) is such a space.

The only convergent sequences are eventually constant and so the limits of convergent sequences are unique. But all non-empty open sets intersect, so X is in fact anti-Hausdorff."

ANTI HAUSDORFF

it's the least Hausdorff

0celouvskyopoulo7

4:00 PM

T1 + first countable + unique limits doesn't imply Hausdorff?

Slereah

I do not know

12

Q: Unique limits of sequences plus what implies Hausdorff?

It is known that there are non-Hausdorff spaces which admit unique limits for all convergent sequence (see here) and it is also known that unique limits for nets implies Hausdorff. What I am wondering is, if there is a (somehow weak) condition which one should add to "unique limits of sequences"...

sequences-and-series gn.general-topology limits-and-convergence

Apparently first countable will do

0celouvskyopoulo7

Manifolds are first countable

I hope that's not only for hausdorff ones

Non hausdorff is terrible! Who would make such a thing

Slereah

Well they're still manifolds

So the neighbourhood is still a subset of $\Bbb R^n$

Well, one like that exists, anyway

0celouvskyopoulo7

4:26 PM

@Slereah Does that make it first countable though...

Ben Niehoff

4:42 PM

@Danu Homotheties are constant scalings, yes

@Danu And Jacobi fields tell you about geodesic deviation!

0celouvskyopoulo7

He probably doesn't see why that's interesting

DanielSank

Hi everybody.

0celouvskyopoulo7

hello

Ben Niehoff

I don't see any earlier context...how did these things come up?

DanielSank

@0celouvskyopoulo7 Your name is some kind of joke on Russian and Greek?

Ben Niehoff

4:45 PM

and ocelots

DanielSank

Μου αρεσει

Мне нравится

0celouvskyopoulo7

@DanielSank I'm multicultural.

DanielSank

@0celouvskyopoulo7 Then tell me what I just said.

0celouvskyopoulo7

Beats me.

DanielSank

You happen to have picked to two languages I learned in college ;-)

Obviously, I approve.

0celouvskyopoulo7

4:46 PM

It's also French.

DanielSank

You should change your avatar to a picture of spanakopita.

Ben Niehoff

so what did you say, @DanielSank ?

DanielSank

Spanakopita

Spanakopita (/ˌspænəˈkɒpɪtə/;: σπανακόπιτα, from σπανάκι, spanáki, spinach, and πίτα, píta, pie) or spinach pie is a Greek savory pastry. The traditional filling comprises chopped spinach, feta cheese, onions or scallions, egg, and seasoning. The filling is wrapped or layered in phyllo (filo) pastry with butter or olive oil, either in a large pan from which individual servings are cut, or rolled into individual triangular servings. While the filo-dough recipe is most common, many recipes from the Greek islands call for a crust made of flour and water to form a crunchier, calzone-like exterior in...

0celouvskyopoulo7

Uh, no?

DanielSank

@BenNiehoff It says "I like it" in Greek and then Russian.

Ben Niehoff

4:47 PM

do Russians eat Spanakopita?

DanielSank

Ok, then how about a matryoshka:

Sir Cumference

Is there any proof that our universe is infinite in size?

DanielSank

Matryoshka doll

A matryoshka doll (Russian: матрёшка; IPA: [mɐˈtrʲɵʂkə], matrëška), also known as a Russian nesting doll, or Russian doll, is a set of wooden dolls of decreasing size placed one inside another. The name "matryoshka" (матрёшка), literally "little matron", is a diminutive form of Russian female first name "Matryona" (Матрёна) or "Matriosha". A set of matryoshkas consists of a wooden figure which separates, top from bottom, to reveal a smaller figure of the same sort inside, which has, in turn, another figure inside of it, and so on. The first Russian nested doll set was made in 1890 by Vasily...

Sir Cumference

I forgot to realize that being flat ≠ being infinite

DanielSank

@BenNiehoff Probably not so much. I was just picking something.

0celouvskyopoulo7

4:48 PM

no proof

Ben Niehoff

how could you possibly prove the universe is infinite?

DanielSank

^

Sir Cumference

@BenNiehoff I dunno, cosmology maths

Slereah

@BenNiehoff take an infinite ruler

measure it

Sir Cumference

Well I don't think the Lambda-CDM model assumes an infinite universe, does it?

Slereah

4:50 PM

it does not

0celouvskyopoulo7

A large enough flat torus would appear to be infinite for all practical purposes.

Sir Cumference

@0celouvskyopoulo7 Yeah, only now it hit me

Slereah

although of course

That might be different for big bang cosmology

Sir Cumference

Wait, misread that

Ben Niehoff

also since there is a cosmological horizon (in an expanding universe), then there is definitely a finite size to the observable universe

Slereah

4:51 PM

Since the size of the torus will be arbitrarily small

Sir Cumference

Dammit. So every time I've answered a "how big is the universe" question with "infinitely big", I've been making a premature assumption

0celouvskyopoulo7

You've been making a fool of yourself, yep.

Ben Niehoff

the correct answer is "we don't know"

Slereah

So scenarios of cosmology probably play out differently if the universe is different or not

0celouvskyopoulo7

@Slereah Yeah probably.

We'd need a quantum cosmologist for that

Ben Niehoff

4:52 PM

but the observable universe is something like 46 billion light-years across

Sir Cumference

@BenNiehoff Yeah, that's easy enough

Slereah

92 I think

or something

92

Radius

who knows

4:52 PM

How do people come up with 92

It's only like 14 old

Slereah

Redshift

The expansion, baby

Sir Cumference

@0celouvskyopoulo7 Hubble flow

0celouvskyopoulo7

So it's calculated, not measured directly

Slereah

All the famous GR people answered me, but know who didn't?

Sir Cumference

@0celouvskyopoulo7 Yep

Ben Niehoff

4:53 PM

@0celouvskyopoulo7 we have a pretty good idea of the age (like 13.6-ish?), so you just calculate using GR

Slereah

The particle data group

0celouvskyopoulo7

@SirCumference Implying you know the calculations.

Slereah

Where's my book dang it

0celouvskyopoulo7

@BenNiehoff Yes, I recall that now

Sir Cumference

@0celouvskyopoulo7 Hey, I'm taking a cosmology class

0celouvskyopoulo7

4:53 PM

But that's assuming GR is right.

I doubt the validity of any physics outside of a small ball around the Earth

Ben Niehoff

so far we haven't found any evidence that GR is wrong

Sir Cumference

@BenNiehoff Uh, it breaks down at the earliest periods of the universe

0celouvskyopoulo7

@BenNiehoff lmao

They keep telling us the Earth is round and you believe them on GR?

Ben Niehoff

sure, yes, it encounters problems at singularities!

Phase

@0celouvskyopoulo7 Honestly these sheeple need to break the conditioning

GR fails to explain the dome

Ben Niehoff

4:55 PM

and, in my opinion, horizons (even low-curvature ones)

0celouvskyopoulo7

GR can't explain Norton's dome

Phase

GR fails to explain why the sun is only a few thousand miles away

0celouvskyopoulo7

If the Earth is round why are there no pictures of it?

Sir Cumference

Anyway, I'm waiting for the day when we actually know what the Big Bang was

Phase

@SirCumference ::begins rant on semantics of 'knowing'::

god I hate those people

Ben Niehoff

4:56 PM

well, you might be waiting long enough to learn what the Heat Death and/or Big Crunch are!

Sir Cumference

@BenNiehoff I'm actually hoping for a Big Rip

That stuff is so cool

All distances in the universe becoming infinite in a finite time

Ben Niehoff

how does that happen? dark energy just gives you exponential expansion

Sir Cumference

@BenNiehoff Phantom energy, my friend

If the universe's equation of state is less than -1, then phantom energy (which increases in density as the universe expands) will cause a Rip

0celouvskyopoulo7

insanity

why can't people do something useful?

Sir Cumference

The actual formula for the scale factor shows that it will asymptote at a certain time

0celouvskyopoulo7

4:59 PM

who gives a shit what happens with the universe in a billion years

gigantic waste of time

Sir Cumference

@0celouvskyopoulo7 Nihilist

0celouvskyopoulo7

I'm going to leave because my next comment would get me banned.

Have a bad day.

Ben Niehoff

so the guy whose favorite topic is "mathematical rigor" is concerned about whether what someone else is doing is a waste of time? :P

2

Sir Cumference

@0celouvskyopoulo7 It's just cool that our universe could work this way

0celouvskyopoulo7

Then you should pay for it

Sir Cumference

5:01 PM

Is it not cool that the observable universe would shrink to zero?

0celouvskyopoulo7

No

Sir Cumference

OK, then the scale factor would reach infinity — also meaning that all distances everywhere become infinite?

0celouvskyopoulo7

It's all boring.

Sir Cumference

...

0celouvskyopoulo7

And a huge waste of money and time

Ben Niehoff

5:02 PM

I think a lot of cosmology is boring, too, but that doesn't make it a waste of time

peterh

@SirCumference I am curious, what will do the Big Rip with the black holes

Sir Cumference

Pure maths isn't?

@BenNiehoff Dang, I don't get you people

0celouvskyopoulo7

It's also a waste of time

Danu

@BenNiehoff So why do I care about that?

Ben Niehoff

@Danu about what?

peterh

5:03 PM

@0celouvskyopoulo7 No! It is important, to understand the world in which we live.

Sir Cumference

@peterh I don't know that much about it. It's a nonstandard cosmology, though it's still really cool

Danu

@BenNiehoff You can see what message I replied to by clicking the small arrow at its start

0celouvskyopoulo7

@peterh He's not talking about the world in which we live.

Sir Cumference

I do know that the scale factor reaches infinity in a finite time, just as the observable universe's size reaches 0

Danu

I care about invariants and stuff like that

Sir Cumference

5:04 PM

Thanks to an energy species called "phantom energy"

The math is also pretty cool

Danu

so what do I do with Jacobi fields

Sir Cumference

@0celouvskyopoulo7 I may be

0celouvskyopoulo7

@Danu They're an integral part of Morse theory

Sir Cumference

We haven't ruled out phantom energy's existence

Ben Niehoff

@SirCumference I dunno, seriously, cosmology talks bore me to tears. If I have to look at a graph of the CMB power spectrum one more time, I don't know what I'll do

Danu

5:04 PM

Cosmology = relativistic chemistry

Sir Cumference

@BenNiehoff Yeah, but it gives you insight on the birth, the evolution, and the future of the universe

Danu

God that was by far the most boring course in my master's program

0celouvskyopoulo7

Who cares!?

What happens on some star a billion lyears away has no relevance to us

Sir Cumference

@0celouvskyopoulo7 Every cosmologist ever?

Ben Niehoff

@Danu What was the context of your question? Geodesic deviation contains all of the curvature data, by the way

0celouvskyopoulo7

5:05 PM

I don't think you understand the meaning of the phrase.

Danu

@BenNiehoff Sure, but I don't need Jacobi fields for that.

Sir Cumference

@0celouvskyopoulo7 At the very least, the math behind the Big Rip is cool

You see how the scale factor asymptotes at a certain time

peterh

@0celouvskyopoulo7 Sorry I didn't follow the talk.

Danu

I mean, Jacobi fields don't even mean anything without a metric.

0celouvskyopoulo7

@Danu You're not going to get invariants from Jacobi fields. If that's all you care about then you don't care about Jacobi fields.

Sir Cumference

5:07 PM

@peterh I might be talking about our universe

Danu

Once I have a metric what do I care about other things giving me curvature data?

Sir Cumference

We haven't ruled out a Rip yet

0celouvskyopoulo7

And you presumably don't care about Morse theory because the Morse index theorem basically uses geodesics/Jacobi fields 100%.

Ben Niehoff

@Danu I did ask you for the context of your question. Maybe you could let me know rather than telling me my answers are useless

0celouvskyopoulo7

And he's ignoring me

Whatever

He likes being combative

Sir Cumference

5:09 PM

Wait, is an open universe necessarily infinite?

Ben Niehoff

depends what you mean

0celouvskyopoulo7

What does open mean here?

Ben Niehoff

it's infinite in the future

Sir Cumference

@0celouvskyopoulo7 Negatively curved

@BenNiehoff Huh?

0celouvskyopoulo7

@SirCumference yeah I'm puzzled too.

Danu

5:10 PM

@BenNiehoff I am asking in the context of having read about them in a book (Milnor's book on Morse Theory) but not seen any applications. I didn't mean to imply your answer is useless.

Ben Niehoff

well what did the book say?

0celouvskyopoulo7

@SirCumference so you're asking if there's a compact 3-manifold with constant negative curvature?

Sir Cumference

@0celouvskyopoulo7 Wait, actually the calculations seem pretty easy

0celouvskyopoulo7

I don't believe for a second you derived the FLRW metric.

Sir Cumference

@0celouvskyopoulo7 I believe we've done it in class

Ben Niehoff

5:13 PM

yeah, the metrics used in cosmology are pretty easy

Danu

@BenNiehoff I stopped reading before the applications. It was used in the parts I read to find critical points of the energy functional on path space, which I find not very interesting at all.

Ben Niehoff

because you assume a bunch of symmetries

Sir Cumference

@BenNiehoff I haven't done much with Lambda-CDM, but I imagine it's harder

0celouvskyopoulo7

@SirCumference You'd have holes in your space, at least

Ben Niehoff

@Danu So, Jacobi fields contain the same information as the curvature (on a Riemannian manifold)

0celouvskyopoulo7

5:14 PM

Cartan-Ambrose-Hicks implies $\pi_1\ne 0$

Sir Cumference

Oh, misread you

Ben Niehoff

instead of measuring infinitesimal holonomies, you can measure geodesic deviations

measure them in all directions, and you can reconstruct the Riemann tensor

Danu

Yeah, I understand.

0celouvskyopoulo7

I think it's possible to have negative curvature and be compact...

Sir Cumference

ugh, be back in a few hours

Danu

5:14 PM

But that's not intrinsically interesting to me

0celouvskyopoulo7

Maybe I'm being silly

Ben Niehoff

Seifert manifolds, @0celouvskyopoulo7

0celouvskyopoulo7

I don't know what that is.

Ben Niehoff

it's the 3d analogue of the negatively-curved Riemann surfaces

i.e., you take H^3 and quotient it by a discrete group

0celouvskyopoulo7

is that necessarily compact?

Ben Niehoff

5:17 PM

it can be...isn't that all you care about?

one example is to take a dodecahedron and identify opposite faces with a 1/5 twist

0celouvskyopoulo7

I was looking for a concrete example for @SirCumference

Ben Niehoff

you can put a constant-negative-curvature metric on that

0celouvskyopoulo7

@BenNiehoff ...why do you know that?

Ben Niehoff

(solid dodecahedron, that is)

Danu

This is in Thurston's book on 3-manifolds ^^

Ben Niehoff

5:19 PM

@Danu, do you have an intuitive understanding of what all of the curvature tensors mean? Because the easiest path to such an understanding is to use Jacobi fields

0celouvskyopoulo7

27

Q: Compact surfaces of negative curvature

John Hubbard recently told me that he has been asking people if there are compact surfaces of negative curvature in $\mathbb{R}^4$ without getting any definite answers. I had assumed it was possible, but couldn't come up with an easy example off the top of my head. In $\mathbb{R}^3$ it is easy ...

dg.differential-geometry

Danu

I'm fine with my understanding of curvature tensors, yes

Slereah

aren't all surfaces of genus $>2$ of negative curvature

Danu

They have a constant negative scalar curvature metric, yes

Ben Niehoff

@Danu Then I daresay you've probably already applied Jacobi fields implicitly

Danu

5:22 PM

Maybe I'd like to know more about sectional curvature except for the interpretation in terms of the lowest order correction to distances between exponential geodesics

(and yes, I want sectional directly, not through equivalence with $R$ :P )

Ben Niehoff

Isn't sectional curvature just the Gauss curvature of a geodesic surface?

Danu

I don't know what a geodesic surface is, but probably.

Ben Niehoff

a "totally geodesic submanifold" of dimension 2

Slereah

hey look, it's our friend @JohnDuffield

Edlothiad

Sorry for interrupting, I was redirected here from EE.SE with the possibility that you guys could answer my question, let me know if this isn't on topic.
I've been reading about Gravitational Models and they mention something about "geopotential models of the Earth consisting of spherical harmonic coefficients complete to degree and order 360". Would asking a question about such be on topic here?
"20th degree and order EGM2008 model is used to calculate gravity gradients" This is another example.

Phase

5:32 PM

First assume a flat stationary earth as depicted in the bible

Ben Niehoff

@Danu So yes, I think that is exactly right. To find the sectional curvature in the direction of two vectors u and v at a point P, extend the subspace $u \oplus v$ into a family of geodesics, which will form a surface (is this always true? I think so, without torsion anyway). The Gaussian curvature of this surface at P is the sectional curvature.

Danu

There is definintely something to prove there, but OK I'll take it ^^

0celouvskyopoulo7

@BenNiehoff yes.

Ben Niehoff

And then I guess torsion precisely measures the failure of geodesics emanating from a point to form surfaces ;)

or at least, part of the torsion

there are different pieces that measure different things

Edlothiad

@Phase Sorry was that to me?

Phase

5:46 PM

Yeah it's a joke, I'm not entirely sure about your question, maybe try another SE?

Or just use Wiki

Geoid

The geoid is the shape that the surface of the oceans would take under the influence of Earth's gravity and rotation alone, in the absence of other influences such as winds and tides. This surface is extended through the continents (such as with very narrow hypothetical canals). All points on a geoid surface have the same gravitational potential energy (the sum of gravitational potential energy and centrifugal potential energy). The geoid can be defined at any value of gravitational potential such as within the Earth's crust or far out in space, not just at sea level. The force of gravity acts...

There's a section about Harmonics with EGM models

Edlothiad

Okay thanks

I'll take a look, but guess this isn't the right place

have a nice day :)

Phase

You too

Slereah

fuck

I think I'm getting sick

That will teach me to go to a class with other people in it

One of them was sick

Why do sick people come to class and spread their germs

Phase

Because you've used $sin(x) = x$ for small $x$ in your life

Karma never misses a soul

Slereah

Well you used sin(x) instead of \sin(x)

I'm afraid you're next

Phase

5:57 PM

I live with that knowledge every day

Not only have I used $sin(x) = x$, I've made many many ironic proofs that say that $\pi$ approximates to 3, and so does e, so $\pi = e$

Slereah

I only use natural mathematical units

$e = \pi = 1$

Phase

I use natural natural units

for a given function $f(x)$ where $x$ is in the domain of real numbers, the values the function maps x to can be rewritten as $x\pi ^n$, and as $\pi$ can be approximated as $1$, $f(x) = x$ for all $f $

And this is why i'm going to die young

AccidentalFourierTransform

6:58 PM

I just had a brilliant idea

you know how in the large $N_c$ limit you can take $\mathrm{SU}(N)\approx \mathrm{U}(N)$ and neglect the fermions?

well, you can use that to solve QED exactly

you just take $N_c=1\gg 1$, and neglect matter

you get a free theory, and solve it

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