@DavidZ Ok and that's good enough if the bin counts are large, I guess. In reality the bin counts ought to be Poisson.

@0celo7 Really?

although I don't know if that's required

@DanielSank Definitely seems like the standard thing to compute

@JohnRennie He's intentional abrasive. But when DS calls him out on it, he apologizes.

@Danu Sure but I don't understand what it means.

It's somehow the deviation of the observed data compared to the prediction, normalized by how big I expect that deviation to be.

@0celo7 You know full well I asked you several times if you were actually upset/offended by what I said/how I talk to you. You denied it every single time.

@0celo7 You're welcome to ignore Danu if he annoys/offends you.

By that reasoning I would expect the number I computed to be around 1, not 0.0008.

Do you want me to keep on asking?

@Danu and @0celo7, cool it, you're getting in the way

^ That

@ArtOfCode I think I will, he never tries to be constructive.

@0celo7 I think you just don't get the European sense of humour. I suspect that I have sometimes annoyed you without meaning to, and the Dutch sense of humour is even odder than the English.

@JohnRennie What the Dutch consider "humor" most people consider "abuse".

I grew up in Germany.

It's not "European humor"

Drop it already.

^ That

For the record, and lest I be accused of racism, I spent some time working in Holland and I loved it and the Dutch. Even though they can sometimes be the rudest sentient life forms in the universe :-)

Even ruder than the Nghrscfitrians?

@ACuriousMind now I'm worried that the Nghrscfitrians is some cool reference that I'm too boring to understand :-)

@JohnRennie No, I just mashed random keys ;)

@ACuriousMind nah, that makes too much sense to be random.

@DanielSank Does that mean you (also) see it that way?

@Danu I'm not sure I want to get into that discussion, at least not in this room.

heh

@ArtOfCode nice

@ArtOfCode You may be on to something

@DanielSank Normalized not by the expected deviation, but by the expected value itself.

@Danu Same thing :-)

@DanielSank So that'd be a relative error

@JohnRennie : it isn't numerology, it isn't devoid of meaningful content, and it really is simple. But I can't tell you about it here. Sorry about that. You could ask a question on the main site.

@JohnDuffield I imagine it would be closed if I asked it as a question here. We could discuss it in a separate chat room, though I must admit I don't know how to create a separate chat room ...

@JohnRennie I have faith in you. Use the power of clicking around on stuff.

Aha I've discovered how to create a separate chat room. Shall I do so?

@DanielSank it's the extensive training I received doing an experimental PhD :-)

@JohnRennie :-)

Actually, though it sounds trite, if you don't know what to do try something at random has turned out to be an awfully useful lesson to take from my PhD years.

@JohnRennie Only now? You really are a chat rookie ;)

@Danu the truth is I only join the chat when I run out of all other excuses not to get on with some work

seems appropriate

The single most important lesson I've learned in my thirty years of gainful employment since leaving university is that procrastination is an extraordinarily useful management tool.

It's extraordinary how many problems will just go away if you resist the temptation to worry about them.

@JohnRennie My colleague calls that "Productive Procrastination" (PP).

We use PP to great effect.

It is the first line of defense against "My codez don't work and I want you to figure out why".

Though it isn't always applicable. If you discover one day that you appear to be developing a third testicle I'm not sure I'd advise that you procrastinate before taking any action.

@JohnRennie Indeed. It's most appropriate in situations where my other favorite help-desk method, "Have you tried basic debugging", also applies.

Developing extra organs is not such a situation.

Aw, not "have you tried turning it off and on again?"

Lol

Dat appeasement

@DavidZ you can turn your testicles on and off?

@DavidZ I prefer "Have you tried basic debugging" because the answer is almost always "no", and the soliciter of my time leaves without further conversation and fixes the problem themselves.

@JohnRennie Guys, get a room.

@JohnRennie suppose I have two events in Minkowski spacetime

call them $p^\mu,q^\mu$ and set $p^\mu=0$ for convenience

@DanielSank ah, yeah, I guess if the people involved actually know what debugging is, that makes sense

and I have a curve $c^\mu(s)$ connecting the two, and $c'^\mu(s)$ is always a spacelike vector

@0celo7 hang on, let me write this down ...

@JohnRennie This isn't a crazy problem, but there's a trick that I'm missing

@JohnRennie I'll invite you to another room so I don't annoy the royal Dutchman

OK, so far so good

The central limit theorem is amazing.

Truly one of the most amazing mathematical things.

@DanielSank Funny, I was talking about this just a few days ago with one of my friends.

Just the best result

Literally "if you don't know what the fuck is going on, it's always this"

@Danu It's particularly neat that it applies to discrete Fourier transforms because the transform is itself a sum.

@Danu sounds like the statisticians' version of Fourier transforms

Somehow the phase part exp(i 2 pi n k / N) just doesn't get in the way of the central limit theorem.

@DanielSank Fourier transforms are your life, aren't they? ;) What do you mean by "it applies to discrete Fourier transforms"?

@Danu First of all, yes.

Second, suppose I have a time sequence of random numbers $x_n$.

These $x_n$ are Gaussian distributed with standard deviation $\sigma$. I denote this function $G_\sigma$.

In oher words, $P_{x_n}(x) = G_\sigma(x)$.

Make sense?

No, not really. What's $P$? How do you want to define your function?

$P$ means "probability density".

Oh, okay.

All right.

So the graph of $P$ is the Gaussian

Yes.

Now suppose we compute the discrete Fourier transform:

$X_k = (1/N)\sum_{n=0}^{N-1} x_n \exp(-i 2 \pi n k / N)$

$X_k$ is random.

However, we can compute it's statistics.

It turns out that the real and imaginary parts of $X_k$ are independent and both distributed as $G_{\sigma/\sqrt{2N}}$.

ah, yeah okay

The $1/\sqrt 2 $ comes from Pythagoras in the plane, right?

@Danu Well... what do you mean?

Never mind, continue

Ok now suppose that $x_n$ are randomly $\pm \sigma$

i.e. each $x_n$ takes the value $\sigma$ or $-\sigma$.

This is a very different distribution than the Gaussian we used before.

The weird thing is that the distribution of $X_k$ is the same, at least according to my calculation.

wut

What about $\pm n\sigma$ for $n\in\Bbb N$?

Well then you'd have $X_k$'s real and imaginary parts distributed according to $G_{n\sigma/\sqrt{2N}}$ I think.

Okay.

I think it's neat that the central limit theorem works even if you put a phase factor into the sum.

Yeah, I'm dumb, that'd just give a factor somewhere because $x_{\pm \sigma}=1/n x_{\pm n\sigma}$

Right? The central limit theorem says sums give Gaussians, and the width of the Gaussian depends only on the $\sigma$ of the summands.

It's interesting that if you throw in an $\exp(-i 2 \pi n k / N)$ into the sum nothing really changes.

I must say that it's not immediately interesting to me :P

Oh, I think I see what you're saying now

@Slereah So, turns out Lemma 7 is wrong for $n>1$. (thanks to @JohnRennie for a helpful discussion)

So I'm going to delete my post.

Now, you should try to do this for $x_n$ having some other distributions to check if it really holds

According to Wikipedia: "A curious footnote to the history of the Central Limit Theorem is that a proof of a result similar to the 1922 Lindeberg CLT was the subject of Alan Turing's 1934 Fellowship Dissertation for King's College at the University of Cambridge. Only after submitting the work did Turing learn it had already been proved. Consequently, Turing's dissertation was never published." For all you hsm buffs :)

@Danu I guess it works for the distributions for which the CLT holds.

@Slereah New conjecture! No spacetime with dimension $3$ or higher is spatially simple.

Now this is a more interesting problem!

It is also a warning against the allures of the canonical 1+1 spacetime diagram.

@Danu It's interesting because it means there may be a stronger statement than the usual CLT.

@DanielSank You guess, or you have good reason to believe (based on more than what you just told me)?

@Danu I guess :-)

@DanielSank I'm not entirely sure that the statement is actually stronger---it may well be equivalent.

@Danu Yeah, the CLT applies to cases where each summand is drawn from a different distribution.

This is probably just that, applied to the real and imaginary parts of the sum.

@all If I asked a question based on a false premise that none of the answers caught and didn't provide any interesting info, should I delete the question?

Yes

@0celo7 maybe not. You could post an answer yourself.

@DavidZ I think there is an answer, but I don't have it. (I am working on it.)

If the system lets you delete the question yourself, it's really up to you. I tend to lean in favor of keeping it around because other people may find it useful.

Actually, the answer would be a proof that the premise is false.

Just link it so people can see what you're talking about.

But what use would it be, other than don't make this mistake yourself?

^ That can be valuable.

Sorry, have to pay attention in class.

BBL.

^that can help

@0celo7 Isn't Minkowski?

@DanielSank true

Especially costly mistakes.

@Slereah simply connected yes, but what 0celo7 means is that it's possible to combine a series of spacelike vectors and end up with a timelike vector.

What's an example?

What's a smooth spacelike curve that links two causally related points

Oh wait

Is it gonna be like

A spirally curve

@Slereah got it :-)

And that's why we didn't see it in 2D

@Slereah Yup. Do you agree with my new conjecture?

Sounds like a reasonable enough conjecture

Hard to prove, though

Indeed.

Have to hunt for counterexamples first...

Right, I'm off. I'm currently halfway through reading a book about genitalia.

Nice!

Cheerio old chap

I mean

Since all spacetimes are locally Minkowski

It sounds reasonable enough that there will be at least a local violation

But what does that actually mean

Between two neighbouring points

Not a proof, but reasonable enough?

It warrants further study

I'm not sure if it implies anything interesting.

then again what does

Probably not, but spacetime topology is fun in itself.

It won't increase the GDP

(No, not enjoyable @ACuriousMind @skillpatrol)

@Slereah Sadly correct.

the proof is probably some local neighbourhood thing I guess

That could be a real issue in whether or not you succeed.

It comes into play when the going gets tough.

What are you talking about, old lady ;)

I mean

For the proof, you just need one counterexample

And the spiny spiral of spacelike wonder can be arbitrarily small

Below any meaningful curvature distance

NOT A PROOF

No

It's just a heuristic

To say that it's probably true

Dude, Visser may never come.

It's been like a week since it ordered it

@Slereah does Visser have any good stuff on casual structure

Hm

Or does he give you directions to the GR proof graveyards

There's the topological censorship theorem

But I don't think he proves it?

He discusses acausal spacetimes and all but nothing too spacetime topology

It's more about energy conditions, tidal forces, quantum fields

All that jazz

@Slereah Ah! Package for me in the mail room.

Finally the anthrax

NSA!!!!

@Slereah Ok, now to prove the thing in two dimensions.

Hm

How many spacetimes exist in 2D

There's the torus

Klein Bottle

Ok, I need to prove that in Minkowski space $\mathbb{R}^{1,1}$, the set $\mathbb{R}^{1,1}-(J^+(p)\cup J^-(p))$ has two connected components

Plane

Plane + toruses attached + holes removes + klein bottles attached

and that a spacelike curve starting at $p$, and $C^1$ spacelike everywhere will remain in a component

Is there anything else

This sounds almost like the Jordan curve theorem

mb I need algebraic topology

@Slereah No I think that's it

I don't know the classification of 2D non compact surfaces

I know it's some simplex business, but that's it

Oh, I don't think it's known

Is it?

It is known, I think

In 3D also, maybe?

4D isn't known tho

3D compact is known, at least, not sure about noncompact

Oh wait

Oh no, nvm

I thought about cylinders

But cylinder is just plane - a point

well in 2D compact you have to throw out all of the ones with nonzero $\chi$

Yeah

Do you know why

That's only the torus and klein bottle

Proof?

The proof is

IN THE BOOK

You know the one

What book?

Steenrod talks about compact $2$-manifolds?

It talks about compact n-manifolds

no the Lorentz proof

the other one

that the zero-$\chi$ compact $2$ surfaces are either tori or kleins

Oh

Hm

Probably a proof of that on math SE

or in Hatcher.

Is the torus globally hyperbolic

It's that time of month again...

@Slereah Not Visser!

Springer books get here damn quickly, wonder if there's a printing place in TN.

@BernardMeurer How is Khaled so fat?

He seems to work out a lot.

@Slereah Did you check that guy's metric?

@0celo7 Is that innuendo

Oh you accepted it

@Slereah btw it's on page 25 of SW, which came in the mail today.

Oh lordie tangent bundle on page 3

@Slereah random GR survey my advisor wants me to read

Of Star Wars?

Oh, Weinberg

Hm, do I have Weinberg

Weinberg?

No, Weinberg is Weinberg

What is SW

the only GR book in GTM

wot

Graduate Texts in Mathematicals

@Slereah Sachs and Wu

If you didn't already figure that out

I don't know who that is

It's a GR book...

I guessed yes

It has some neat proofs in it

@Slereah Does Steenrod talk about spinors on spacetimes

or just Lorentz metrics

i.e. does he prove the Stiefel-Whitney class thing

Steenrod is a math guy

He doesn't care about spinors

bullshit

I recall something about grassman bundles, though

And he might have done Stiefel?

I dunno

grassmann bundle is just the exterior bundle...

@Slereah Cartan was a math guy, too, and he was one of the first to work with spinors.

Seiberg-Witten theory is done by math people and it's with spin bundles

I don't think Steenrod even does connection on bundles

The math people just talk about them in a completely different fashion so you don't recognize all the mangled group theory physicists usually do when they say "spinor" ;)

@Slereah ok

What is the math word for "spinor", anyway

Section of a spin bundle, obviously.

Do they say "spin bundle", though

Yes.

Also is it even the spin bundle

The spin bundle is the rotation group one

Not the spinor field itself

It's a $G$-bundle where $G=\mathrm{Spin}(n)$.

Or maybe they get an associated bundle to that after complexifying it, I don't know.

It's in Jost.

There's a pretty common lack of clarity in terminology as to the difference between the spin group and spinor fields

I got so caught up in nor being able to do the algebra of Clifford stuffs

@Numrok Welcome

> A square is cut out of a copper sheet. Two straight scratches on the surface of the square intersect forming an angle θ. The square is heated uniformly. As a result, the angle between the scratches stays the same. Since all dimensions change by the same ratio, every "detail" of the square retains its shape. The scratches will extend in length, but they will still form the same angle.

That's not exactly a proof, physics prof.

I think it's because the heating is just a Weyl transformation.

@0celo7 I have no clue how he got that fat

But you know how passionate he is about McDonalds

@BernardMeurer his snap is full of him eating healthy shit and working out

@0celo7 That's just propaganda bullshit

Otherwise he'd be skinny

I think he has to be fat

for the Khaled brand

A skinny Khaled wouldn't be the same...

Agreed

also it's easy to look healthy on Snap

I could do that and still keep on having ice cream for dinner

@BernardMeurer I need an article from a Brazilian math journal

Not sure if it's online

Can you steal it from somewhere if I can't find it

I can try and get it for you

Sure, have a title?

Not right now, getting ready to leave

Alrighty

also I'll need a translation if it's in Brazilian

That's easy

And we speak portuguese

I know.

But it's probably butchered

Like the Brits butchered English

Swiss butchered German

Etc.

I can go on!

...sorry master :)

...of course you had to take the bait

Hi @yuggib

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ Hi

@ACuriousMind what bait

The bait you use for fishing