The h Bar

@EmilioPisanty In case you really do like measure theory: Let $(\Omega,\Sigma,\mu)$ be a measure space, and $P:\Sigma\to\Sigma$ an "equimeasure transformation," that is, $\mu(PB)=\mu(B)$ for each $B\in\Sigma$. Consider the linear operator $T:L^2(\Omega)\to L^2(\Omega)$ given by $(Tf)(x)=f(Px)$, $f\in L^2(\Omega)$. Then, using some measure theory + functional analysis, we may apply the "mean ergodic theorem" which says, among other things, that $L^2(\Omega)=\overline{R(I-T)}\oplus N(I-T)$.

It also says that for any $f\in L^2(\Omega)$, $$\lim_{n\to\infty}\frac{1}{n}\sum_{m=1}^n T^mf=f_0\in L^2(\Omega)$$ exists and $Tf_0=f_0$.

That is, we can average out $T$ and obtain a fixed point.

Jaime Gallego

12:28 AM

::imagines @BernardoMeurer programming in octal just with the front switches of a PDP-8::

Bernardo Meurer

@JaimeGallego Have more respect, I would use a PDP-11 :P

0celouvsky

where did @BenNiehoff go, anyway

Jaime Gallego

Why @BernardoMeurer

We could stick with Holleritz tabulators and not these newfangled programmable gizmos anyway

Back in the good ol' days, computers were still people!

0celouvsky

12:53 AM

@BernardoMeurer halp

Bernardo Meurer

@JaimeGallego Because I like them better :P

@0celouvsky What's up?

0celouvsky

@BernardoMeurer I have a capacitor and inductor hooked up in series to an AC source

I needs to find out the frequency I need so that it's like an open circuit

for a short circuit I set the combined impedance to zero

what do I do for an open circuit?

$Z=\infty$?

it won't get to $Z=\infty$

Bernardo Meurer

I have no clue :/

I only take circuits next semester

0celouvsky

ayy

Anonymous

@YashasSamaga LOL :-P So many people made the same mistake :-D...Anyhow, the chemistry paper was bogus, physics was good and maths was lengthy...I was so bored while solving the chem paper that I almost slept

Anonymous

1:01 AM

@YashasSamaga What? You gave the B Arch paper too? Going to be an architect now? :-P

Anonymous

I am getting around 250, how much are you getting?

Jaime Gallego

@0celouvsky If it's series, then the frequency can go either to zero or to infinity. f=0 means the capacitor won't let current pass. Conversely, the inductor does not like f->infinity and "chokes" it.

0celouvsky

@JaimeGallego but what impedance corresponds to an "open circuit" here

Clearly zero impedance is the short circuit, it acts just like a wire

So I solve Z(omega)=0

But what is Z supposed to be for the open circuit?

infinity? my only option is omega = infinity in that case

and that doesn't make much sense

or zero, but thats not very good either

Jaime Gallego

Why doesn't it make much sense?

0celouvsky

what the hell is omega = infinity supposed to be

that's definitely not physical

and the whole AC analysis fails in the omega -> 0 limit

I mean, I should be able to run DC current through the setup

So shouldn't act like an open circuit

Jaime Gallego

1:10 AM

No, capacitors block DC

0celouvsky

@JaimeGallego ah. because i = Cv'

thanks

0celouvsky

1:50 AM

@Secret So I've been thinking about that proof (which I took from a functional analysis book) and I'm not happy with it. I read another one and I don't like it either. I'm considering writing my own, from scratch.

The difficulty is due to there being two topologies floating about.

It's very difficult to keep track what is compact/continuous in which topology.

Slereah

2:11 AM

So anyway

Are manifolds really locally euclidian

0celouvsky

@Secret My approach is as follows: 1. If two metric topologies on a set have the same convergent sequences, they are equal. 2. Any norm topology on $\Bbb R^n$ has the same convergent sequences as the usual topology. 3. If two norms generate the same topology, they are equivalent.

1. Is more or less clear, and we already talked about it I believe.

2. Is just the statement that weak convergence and strong convergence are equivalent in finite dimensions, and I can prove that pretty easily.

3. Is well known in arbitrary normed vector spaces, so it should not be a problem to prove.

@Slereah what does that mean?

are you having trouble sleeping again?

Slereah

for a small enough neighbourhood, do we have $$\int_\gamma d^nx \sqrt{\langle \gamma'(\lambda), \gamma'(\lambda)\rangle} = |\gamma(\lambda_1) - \gamma(\lambda_0)|$$

For a geodesic $\gamma$

$|\gamma(\lambda_1) - \gamma(\lambda_0)| + \mathcal{O}(\mathfrak{something})$

I'm guessing depending on the size of the neighbourghood

it sounds reasonable enough but who knows

0celouvsky

jesus christ I got 90A

that's probably wrong, crap

Slereah

very good grade

0celouvsky

90 amperes

@Slereah oh, uh

there's an MSE post on this

Slereah

2:19 AM

That was why I was wondering about analytic manifolds 'cause that would make it easier to prove

possibly

0celouvsky

let's see if i favorited it

0

A: Asymptotic behaviour of the length of a curve .

Let $\delta >0$ be small so that $\gamma$ lies completely in the normal coordinate centered at $y$. In this normal coordinate, we write $$\gamma(0) = 0, \ \ \ g_{ij}(x) = \delta_{ij} + O(|x|^2).$$ Write $v = \dot\gamma(0)$. Then we have $$ \gamma (t) = tv + O(t^2), \ \ \ \dot\gamma = v+ O(t)$...

Slereah

The $g_{ij} = \delta_{ij} + \mathcal O(|x|^2)$ is quite mysterious

is that a generic feature of normal neighs?

(I am a horse now)

0celouvsky

Yeah it is

Because $\partial g(0)=0$

Slereah

ic

0celouvsky

Note the common confusion here

$g=\delta+\mathcal O(x^2)$ is not saying $g$ is analytic

Taylor's formula holds because $g$ is $C^\infty$

But $g$ doesn't have to equal the full power series

Slereah

2:26 AM

Yeah that I remember

0celouvsky

@Slereah how can you beat this classic cover?

it's clean

no fuss

straight to the point

has a cute horse

Slereah

0celouvsky

that seems like a book for insane people

Slereah

great design for the cover, though

just making wormholes at the beach

0celouvsky

that's a woman, right?

or is it a giraffe

Slereah

2:38 AM

the book does not specify, but I assume, yes

0celouvsky

2:53 AM

yo what's $\int_0^L \sin^2(n\pi x/L)\,dx$

DanielSank

What's going on in here?

@0celouvsky Oh come on.

Use your noggin.

0celouvsky

@DanielSank it's $L/2$

I must have made a mistake elsewhere

Slereah

3:13 AM

@0celouvsky Convert $\sin$ to $e$

should be easy enough

0celouvsky

@Secret Theorem. Let $||\cdot||_1,||\cdot||_2$ be two norms on $\Bbb R$. There exist constants $c,C>0$ such that $$c||v||_1\le ||v||_2\le C||v||_1$$ for all $v\in\Bbb R^n$.

*Proof.* Clearly it is enough to prove this with $||\cdot||_1=||\cdot||_e$, the Euclidean norm, since this relation is transitive. Let $e_1,\dotsc, e_n$ be a basis of $\Bbb R^n$, and write $\Bbb R^n\ni v=\sum_{i=1}^nx^ie_i$ so that $||v||_e=(\sum_{i=1}^n |x^i|^2)^{1/2}$. Our first estimate is
$$\sum_{i=1}^n|x^i|\le \sqrt{n}||v||_e. $$
To see this, note that the LHS is an $\ell^1$ norm, and $||\cdot||_e$ is an $\ell^2$ norm, so by H\"older's inequality,
$$\sum_{i=1}^n|x^i|=\sum_{i=1}^n 1\cdot |x^i|\le \left(\sum_{i=1}^n1\right)^{1/2}\left(\sum_{i=1}^n|x^i|^2\right)^{1/2}=\sqrt n||v||_e.$$

Next,
$$||v||_2=||x^1e_1+\cdots +x^ne_n||_2\le\left(\max_{1\le i\le n}||e_i||_2\right)\left(\sum_{i=1}^n|x^i|\right).$$
Letting $C=\sqrt n\cdot \max_{1\le i\le n}||e_i||_2$, we find that
$$||v||_2\le C||v||_e.$$

This inequality shows that $f:v\mapsto ||v||_2$ is Lipschitz continuous in $\Bbb R^n$ with the standard topology. Let $S$ be the usual unit sphere in $\Bbb R^n$. It is compact in the Euclidean topology, hence $f$ attains its infimum $c$ in $S$: there exists a $v\in S$ with $f(v)=m$. Since $v\ne 0$, $c>0$. Let $w\in\Bbb R^n$ be nonzero, and set $v=w/||w||_e\in S$. Hence $||v||_2=f(v)\ge c$, or
$$||\frac{w}{||w||_e}||_2\ge c\implies ||w||_2\ge c||w||_e.$$

And the proof is complete.

That's the best version I know of.

Slereah

So anyway

what kind of fiber are continuous spin fields

I still don't know

0celouvsky

what?

Slereah

Continuous spin representation fields have this weird type of field with a continuous range of index

I am wondering if they can be described by a fiber bundle

0celouvsky

oh christ no

what is a continuous index anyway

it's just a function dude

Slereah

3:27 AM

The field transforms as

$$U(\Lambda)\hat \varphi_{\theta_n}(x) U^{-1}(\Lambda)\rightarrow \hat \varphi_{\Gamma_{mn}(\Lambda)\theta_n}(\Lambda x)$$

With $\Gamma$ a spinor Lorentz transformation

The index itself is a spinor

0celouvsky

@Slereah lol

Slereah

yeah it's fairly shit

Also does the complete atlas of a manifold have the same open sets as the projections of the local trivialisations?

fairly sure it does but I couldn't say why

0celouvsky

well that's what we mean by a smooth vector bundle

TanMath

3:43 AM

hey @DanielSank, are you available?

DanielSank

4:02 AM

@TanMath I'm cleaning my room. Availability is relative.

Slereah

reading about vertical lifts on manifolds makes me think of this

Secret

(Uh... this question turned out to be more lengthy than expected, I guess I will post on the main instead)

I want something more illustrative:

Let's take classical correlations vs entanglement as examples. For classical correlations, the observables are already determined even before a measurement, and thus knowing information of one subsystem will immediately let you know the corresponding information in another subsystem.

For correlations in entanglement, suppose I take the singlet bell state. Then upon preparing the bell state, the correlations of the 2 subsystems are already establis…

0celouvsky

The "Frechet-Kolmogorov" theorem

who even knows this much functional analysis

Ah, this isn't even a crazy theorem

Slereah

What's a crazy theorem

0celouvsky

@Slereah idk, pick some horrible estimate from PDE theory

something like that

or that

Slereah

4:14 AM

I'm not reading that

0celouvsky

exactly.

ah, the nuclear space of A. Grothendieck

finally something engineering related

Slereah

I should read this paper

It seems relevant to my interests

0celouvsky

haha wtf

what are those ones without arrows?

wormholes?

what on Earth is a virtuous spacetime

Slereah

A spacetime without sin

0celouvsky

@Slereah what paper?

I might enjoy it too

Slereah

4:19 AM

"Causal structure in spacetim"

By Brandon Carter

He defines spacetimes to be $C^1$

The madman

0celouvsky

do you have the paper?

where did you find it

Slereah

I don't remember

It's just in my files

0celouvsky

ah, got it

it's a long paper, wow

Slereah

yeah

0celouvsky

@Slereah Oh, here's a crazy theorem

A Banach space is reflexive if and only if it is locally weakly sequentially compact.

Slereah

4:24 AM

that paper has a lot of random causal theorems

0celouvsky

page 380 and 381 seems like what you're looking for

@Slereah How's progress on the book?

@NeuroFuzzy long time no see

Slereah

It's alright

60 pages so far

NeuroFuzzy

@0celouvsky Yee! What's up!

Slereah

though the tough part is actually knowing what I'm talking about

Many things to learn

NeuroFuzzy

I haven't been on the website so much. A bit burnt out answering questions about special relativity haha

Slereah

4:27 AM

Virtuous is just that there's no CTC

Well, no CCCs

No CTCs is almost virtuous

though then I'm not sure why in the diagram acausal implies virtuous

NeuroFuzzy

Accepted an offer for a university far off from my alma mater of UCSD, which goes by the strange name of UCSD. (My only acceptance letter and a really good offer at that!)

0celouvsky

@NeuroFuzzy trying to figure out if there's an easy way to see that $L^2$ is locally weakly sequentially compact

I know Jost does it for Hilbert spaces in general

wondering if Yosida has the proof

@NeuroFuzzy congrats

@Slereah link?

Slereah

No link right now

Just a smattering of things

NeuroFuzzy

@0celouvsky Ahhh more analysis!

0celouvsky

who doesn't like a good bit of functional analysis at 0:36 am

Slereah

4:39 AM

What's the difference between $$$$ and \begin{equation}\end{equation} in latex, anyway

0celouvsky

can't see shit

Slereah

The double dollar sign-double dollar sign versus begin equation end equation

0celouvsky

usually the latter puts an equation number

NeuroFuzzy

@Slereah or rather {equation*}?

0celouvsky

who even knows how to prove the Fourier expansion for $\Bbb R^n$ :o

I guess you'd prove it for the smooth functions as usual, then use density of bump functions in $L^2$

terrible

amazon.com/Classical-Fourier-Analysis-Graduate-Mathematics/dp/…

@Slereah oooo boi

Slereah

4:53 AM

I've been using greek indexes and $g

$g$ for the group structure of the bundle

Bad idea

Everything looks like the metric tensor now

0celouvsky

The group should be G, not g...

Slereah

Well the group is $G$

but $g_{\alpha\beta}(p) \in G$

For $p \in V_\alpha \cap V_\beta$

0celouvsky

You should be using abstract index notation for tensors anyway

Slereah

Also Steenrod is a bit confusing because I think his definitions are a bit dated

Like he calls fiber bundles for just $\{E, X, \pi\}$

And with local trivializations, he calls them coordinate bundles

Is that a terminology still used?

DanielSank

Anyone want to try to get the PSE logo in this?

0celouvsky

4:59 AM

@Slereah no

We always assume there's a local trivialization compatible with the manifold structure

Slereah

Well his fiber bundles aren't even necessarily over manifolds

Just topological spaces

Although he does mention the other definition of fiber bundles that includes the trivialization

He calls it the EHRESMANN-FELDBAU definition

0celouvsky

Hmm ok

Slereah

then again when he wrote that bundles were like

ten years old or so

Probbly not too much standard terminology

0celouvsky

Jesus. I'm off, cheers.

Slereah

I wonder what's the original bundle paper

Let's see

later

mathnet.ru/links/132e475702d36500a94e38e2b125f4c5/sm5499.pdf

that's the one

Secret

5:20 AM

Some problem solving musings:
Imagine you know the answers to a problem, and your task is to work out what is the problem that is being asked

This happens to me sometimes when I knew what is the solution I am seeking, but less clear on what is the problem I am trying to solve. As a result, the ultimate "work backwards" occurs to figure out the problem that is being asked

uhoh

1

Q: How to find more information about Chang'e-2's actual 3D orbit in space, since it's not in JPL Horizons?

edit: A simple web search in Chinese may turn up something, but I don't know anyone personally who is both both fluent and interested in space exploration. This is a pretty cool orbit - there must be something somewhere! I tried to find Chang'e-2 in the JPL Horizons database but coulnd't find it...

+100 Bounty expiring in one day. Anything??

Secret

5:38 AM

Cool, the proof looks fine (and you have used compactness) except this part is not very clear to me:

$$||v||_2=||x^1e_1+\cdots +x^ne_n||_2\le\left(\max_{1\le i\le n}||e_i||_2\right)\left(\sum_{i=1}^n|x^i|\right).$$
Remind me which inequality we use here to split up the eucledian norm to the product of the longest basis vector and the 1-norm of the components?

user228700

5:53 AM

@JohnR: Morning :-)

Sir Cumference

@Kaumudi.H How was the test?

John Rennie

@Kaumudi.H Morning. Sorry for the slow response, I was on the supermarket web site ordering groceries - an absolutely vital task!!

user228700

@SirCumference Hi. Wasn't as great as I'd hoped. I'm preparing for the next one now :-)

Sir Cumference

@Kaumudi.H In a year?

user228700

Huh? Noo. In a few weeks.

user228700

6:05 AM

@JohnRennie Ah, OK :-)

Sir Cumference

Oh, I heard that the test is only given every year o_O

user228700

@SirCumference There are other tests as well, you know :-P

Sir Cumference

@Kaumudi.H Oye facepalm

user228700

:-)

Sir Cumference

Welp, good luck :)

I gotta head to sleep

'Night

user228700

6:07 AM

Thanks. Night! :-)

Anonymous

6:34 AM

@Kaumudi.H How much are you scoring? My maths went bad....phy and chem were good

Anonymous

I'm getting around 250...

Anonymous

One guy from my center is getting 325

Anonymous

:-P

John Rennie

My kettle has just gone bang and appears to be dead. Oh no. This could be a critical failure!

Fortunately, like the good experimental scientist I am, I have established that it's possible to boil water in a pan :-)

Slereah

Hm

How to note the Lorentz group

$O$ or $\mathrm{O}$

user228700

6:47 AM

@blue I haven't checked properly yet. I didn't do great so I'm trying not to think about it.

user228700

@JohnRennie :-) Hehe.

John Rennie

@Kaumudi.H it's no laughing matter. Caffeine withdrawal symptoms can be fatal!

user228700

:-) Yes, of course!

John Rennie

Actually I feel quite sad because I've had that kettle for over 20 years. It's older than you are :-) And now it has passed on.

I wonder where the best place to buy a new kettle is. Amazon probably ...

user228700

@JohnRennie Wow, that's crazy...

John Rennie

6:50 AM

That's the thing about getting ~~old~~ middle aged. All the stuff you own is ~~old~~ middle aged as well :-)

user228700

Hehe, yeah :-)

John Rennie

Wow, £5.49 for a new kettle. That's cheap.

Secret

@Slereah $O(1,3)$, you need to notate the signatures to distinguish it from other types of orthogonal groups $O(m,n)$

user228700

@JohnRennie eBay or Amazon?

Slereah

I am aware, yes

I'm just talking typography

John Rennie

6:59 AM

@Kaumudi.H Argos

It's kind of a general retailer in the UK and sells all sorts of stuff - including kettles.

www.argos.co.uk

user228700

Ohh, OK...

John Rennie

Whether the kettle is any good I'll find out when I collect it, but ... well ... it's a kettle. All it has to do is boil water :-)

Anyway I need to go. I have to take my car to the garage for it's annual check-up. Back in a bit.

user228700

OK bye :-)

Secret

Older papers use $O$, more modern papers tend to use the upright $\mathrm{O}$

personally, upright O looks better

Anonymous

7:22 AM

@Kaumudi.H Oh :-P That feeling :-)

Anonymous

Even I was scared to check the answer key

Anonymous

But it is good to analyze your mistakes so that you can avoid them in future exams. :-)

Anonymous

Anyhow, all the best for advanced

skill patrol

9:04 AM

@JohnRennie "Annual?" You should get the oil changed at-least twice a year.

Emilio Pisanty

9:19 AM

Yo @Qmechanic what's with migrating this one?

0

Q: Two waves interference animation

I would like to plot (or to find) an animation where two waves at different frequencies, propagating with different velocities, sum together $$w_1 (x,t) = \cos(\omega_1 t - k_1 x)\\w_2 (x,t) = \cos(\omega_2 t - k_2 x)$$ and generate the resulting signal $$w(x,t) = \cos(\omega_1 t - k_1 x) + \c...

plotting physics

OP specifically asks for

> 2. Is there a free tool / software / site to generate such an animation?

Mathematica is nowhere near free

John Rennie

9:38 AM

I think therefore jam? :-)

Bernardo Meurer

@JohnRennie Good morning :)

John Rennie

Morning. Sorry I wasn't around last night.

Sunday evening I'm comatose after eating a massive lunch :-)

Bernardo Meurer

Nah, come on, it's weekend :)

I branched the code

made some changes already but miniscule

just created a Game struct pretty much

and then I wanted to discuss with you a bit the linked list deal

John Rennie

OK ...

Bernardo Meurer

Should the linked list be a struct inside the Board struct?

John Rennie

9:51 AM

I would put it in the game struct.

Bernardo Meurer

Hmm

John Rennie

After all, the undo history is part of the way the game is designed not the board.

Bernardo Meurer

Yeah

Then gameInit would receive a board and initialize the linked list with that

after this every move we add a node to the list and perform the move there

no more buffers

John Rennie

Yes, that's the way I would do it.

Are you going to create a separate struct/class for the list?

Anonjohn

@E

Bernardo Meurer

9:55 AM

I wanted the linked list struct to be inside the game struct

There's no reason to expose something no one else uses

Anonjohn

@EmilioPisanty I am beginning to think that there exists a critical internuclear distance where the there are no bound states. This looks to happen around $a_0$, continuing along my variational path, i have it at $<0.8 a_0$

Bernardo Meurer

@JohnRennie Sounds reasonable?

John Rennie

Yes.

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