But people with weird ideas especially, it seems from this group. It's like they kind of know that they're saying wacky stuff...

@Obliv Particles turn out to be extraordinarily elusive objects. It's remarkably difficult to point to anything in QM or QFT and say here's a particle.

@acuriousmind Are there any interactions that are not reversible? Such as $A + B \to C$ but $C \to D + E \to ...$

so that $A + B$ cannot be attained again

No.

@Obliv Not in terms of particles, no (although one might need external energy for one direction buit not the other)

There are cases where the reverse reaction is improbable, but not impossible.

@Danu to some extent it makes sense that those who experience rejection would feel more need for validation

Okay so then theoretically could there exist a state of the universe where all of the matter exists as one type of particle?

I think you'd have to clarify what you mean by that.

For example the net baryon number of the universe is non-zero (we think) so the universe couldn't have only electrons.

That would violate baryon number conservation.

And lepton number as well, though perhaps you'd allow a mixture of electrons and positrons. In any case it would be so wildly improbably that only the most fanatical of pedants would regard it as a possibility.

@JohnR It seems there are more observed rules about particle interactions than I imagined. So even if there are no irreversible interactions, that doesn't mean that every particle can be changed into particle $A$

@Obliv Correct. For example a quark can't turn into an electron.

But there aren't that many rules and mainly they're obvious. For example a quark turning into an electron would violate conservation of electrical charge.

@EmilioPisanty that reminds me, saw this at an airport book store recently, reminded me of DS, looked fun, awardwinning/ bestselling,kinda wanna read it sometime. (also know another young octopus fan) amazon.com/Soul-Octopus-Surprising-Exploration-Consciousness/dp/…

There are a handful of other conserved quantities.

@vzn I read that and it was very enjoyable if a bit lightweight.

@JohnRennie wow what a coincidence cool ok thx for minireview. will have to dig up something else if in the mood for something heavyweight :P

I might even have an epub of it lying around ...

@Rahat you were interested in the difference between photons and light beams?

lol

:-) We weren't really talking about octopuses. vzn mentioned a book about octopuses that I have read and which I think is very good.

It seems @DanielSank's love for octopodes is infectious

Argument about the plural of octopus ensues :-)

I think most agree that octopi is wrong.

octopi is definitely false because it's not a Latin ending

The trouble is that the word octopus is half Latin and half Greek.

So there aren't any hard and fast rules for deciding what its plural should be.

@JohnRennie What half is Latin, exactly?

podes (c) ACM, is the Greek plural of foot

The octo half.

@JohnRennie octo-/okto is both Latin and Greek for eight.

That's what I do, so I just use octopuses on the grounds that at least everyone will know what I mean :-)

@ACuriousMind OK, fair enough, though at least we can agree that octopi is wrong :-)

@Rahat: did you want to talk about photons? I have to go off and fold my washing soon (and no, that isn't a euphemism).

@Rahat Which video are you referring to?

@Rahat both photons and waves are useful approximations. As a rough rule light acts like a wave when it is propagating and it behaves like a photon when it is exchanging energy with something.

@JohnRennie when does it act like both ?

@JohnRennie it is?

There's a part of physics called quantum optics that delas with this sort of issue, though it can get very complicated very quickly. The Wikipedia article is worth a read though.

@0celo7 Yes, it is definitely wrong.

Forming a plural with i is a Latin plural and the pus bit of the word is Greek not Latin.

So

You're not a linguist

What is the official English plural?

@JohnRennie Can you say something about XBMC aka Kodi?

@HariPrasad I can't think of any cases where light acts as a photon and a wave simultaneously. That doesn't mean there aren't any ...

English is not Latin, FYI

@0celo7 octopuses, as a quick Google would have told you.

According to?

@Rahat if you work out the light intensity needed to have only one photon's worth of energy present at a time then you'll find it is absurdly low.

I'm on mobile, can't google

You'd need specialist kit to detect such low light levels. Realistically it isn't a home experiment.

@0celo7 oxforddictionaries.com/definition/english/octopus

@Rahat when you have a light beam in your expt the energy is delocalised just as if it was a wave. However if you detect the light, e.g. by a photomultiplier tube, then that detection works because the light exchanges one photon's worth of energy at a point. So the detection process localises the energy.

In fact a photomultiplier tube works using the photoelectric effect, which is one of the standard experiments people trot out to show that photons exist.

Photons do exist, but I think they are best understood as the mechanism by which light exchanges energy with something else e.g. the photoelectron.

@JohnRennie I don't trust that

@0celo7 that's the Oxford University Press you're distrusting. It's hard to think of any more authoritative source on the English language, though what you Yanks do with the language is anyone's guess.

@JohnRennie It's British

The Isles are British, the language is English.

Hate to break it to you, but you lost the authority over your language in the war

LOL

Though it's fun to joke about, I am not at all a language zealot. I really like the way English has evolved in lots of different countries and I like the way those innovations are working their way back in British English

It makes the language much more fun.

One of the strengths of English is that it's such a mongrel language with inputs form loads of different sources.

where does "horse" come from, anyway

sounds nothing like the German or Latin words

@Rahat: See? Now we're talking about linguistics. And you thought we were serious physicists :-)

I'm not a physicist

@0celo7 I'm pretty sure that's of Anglo-Saxon origin

I think the equivalent modern German word is Ross, which you'd have to concede is similar.

We use Pferd as the everyday word for horse, though

But yes, Ross and "horse" sound as if they're cognate

I think the English word cognate with Pferd is palfrey, which in modern English refers to a type of horse.

That's what Googling tells me, too ;)

@Rahat Goodnight. And I really must get back to the folding.

@JohnRennie lol

Too bad that's a fake

Fake? A fake? Never!

@ACuriousMind Given a symplectic situation $(M^{2n},\omega,H)$, and an energy hypersurface $\Sigma$, can I use $\omega$ to construct a measure on $\Sigma$?

wtf would "folding the laundry" be an euphemism for, anyway?

I know one can put a measure on $M$ via $\mu(U)=\int_U \omega^{\wedge n}$.

But this requires some trickery, I think

Can one do it on $\Sigma$?

$\Sigma$ is regular, i.e. $\mathrm dH\ne0$.

@0celo7 By "energy hypersurface" you mean a level set of $H$?

@ACuriousMind Yes, a regular one.

I think $\omega^{\wedge n}$ should somehow reduce to a $(2n-1)$-form on $\Sigma$.

I've forgotten how to show a volume form induces a measure in the analytical sense

But one should at least be able to obtain a volume form on $\Sigma$, right?

@EmilioPisanty Um, so... I really love octopuses. I would ask that you kindly not direct me to statements about them being tortured for the sake of curiosity.

@bolbteppa Answer to your first question: I know a little bit, but not very much. An elliptic curve is an algebraivc curve over a field, which is described by polynomial equations (a one-dimensional case of an algebraic variety).

@DanielSank hahaha

@Danu Downloaded.

@vzn That is an excellent book.

@ACuriousMind As well it should be.

@DanielSank What's the name of that octopus ?

@bolbteppa The modularity theorem is a link from these elliptic curves to modular forms. It is the Taniyama-shimura conjecture which says that all elliptic curves must satisfy the definition of modularity. This is proved.

@0celo7 I don't think it does.

@bolpeppa $a^n+b^n=c^n$ is related to $y^2=x(x-a^n)(x+b^n)$ because that curve can be constructed from the solutions of the first equations.

@bolpteppa

@ACuriousMind Are you sure? Why can't we solve $\omega^{\wedge n}=\mathrm dH\wedge \sigma$?

At least in some neighborhood of $\Sigma$...

Then set the volume form on $\Sigma$ equal to $\iota^*\sigma$

@PhysicsGuy do you have any idea how to get from $a^2 + b^2 = c^2$ to $y^2 = ...$? Looks a bit crazy :p

@0celo7 Well...you may be able solve that (I don't know!), but why should I accept that $\sigma$ as the "volume form"?

Or what modularity really means

@ACuriousMind Let me work on it.

@0celo7 My problem is not whether or not that $\sigma$ exists, my problem is why that would yield a meaningful notion of volume on the level set (as opposed to picking any other random top-dimensional form)

@ACuriousMind I think you get a unique pulled back form

unique with what property?

Let's take a step back: Why do you need such a volume form to begin with?

I'm doing physics, this should make you happy

I need to put a measure on $\Sigma$, clearly.

...why do you "need" to put a measure on $\Sigma$?

@ACuriousMind To prove something

Involving measures

You're not being very cooperative here :P

I'm trying to understand Poincare recurrence

I'm trying to evolve $\Sigma$ and I'm looking for recurrence points

"almost all" points are recurrent

So I have to do some measure stuffs

Halt

Oh god

The "almost all" refers to a neighbourhood of a point in $M$, right?

I also have to prove it's invariant under the flow in the hypersurface

@ACuriousMind I'm...not sure.

And why are you "evolving $\Sigma$"? Since it's a level set of $H$, it will be invariant under the time evolution, won't it?

I.e. points in $\Sigma$ shouldn't leave it, ever (energy conservation)

They will travel around in $\Sigma$

Yes. But $\Sigma$, itself, is not a Hamiltonian system, so Poincaré recurrence doesn't directly apply to it afaics

I think that it is if you do some Trickery

What the hell is Poincare recurrence then

Precisely.

I love how "AFAI[stuff]" just gets expanded to accommodate everything nowadays :D

@0celo7 It can't - it's odd-dimensional

@Danu afaik, afaics...what else?

@ACuriousMind I should rephrase. I don't care if it's a Hamiltonian system.

@ACuriousMind afaict

I'm fairly sure the theorem applies to it, with some trickery.

Well, not knowing what the theorem says, I can't tell that for sure, but the lack of an obvious measure on $\Sigma$ doesn't bode well I'd say :P

@ACuriousMind There is one

I described it above

You should be able to construct it in each chart

It's not obvious at all to me why you should pick that as the measure on $\Sigma$.

I'm fairly sure when you pull it back on chart overlaps, you get agreement

It's a measure, but what makes that a good choice?

@ACuriousMind Because it's invariant under the flow restricted to $\Sigma$

I need to think about why that's true

I'll accept that if that makes it unique up to scalar multiples.

@ACuriousMind Scalar (constant) or function?

@0celo7 Scalar. If a function is allowed you'll have to justify which function to pick.

Oh come on :(

@ACuriousMind It's very natural though

What can be more natural than $\omega^{\wedge n}=\mathrm dH\wedge\sigma$

What's natural about it?

Solving an equation of differential forms for a factor doesn't feel very natural to me

It uses the Hamiltonian

@ACuriousMind Note that $\omega^{\wedge n}=\mathrm dH\wedge\sigma'$ defines the same volume form on $\Sigma$

Well, $\omega^{\wedge (n-1)}\wedge\mathrm{d}H$ would also use the Hamiltonian and give a form of the right dimension :P

@0celo7 what? What's $\sigma'$?

You madman

@ACuriousMind Any other form that solves that equation.

@ACuriousMind No. It does not give a volume form on $\Sigma$.

Yes, it gives zero, but it still looks just as natural to me! :P

...what

Mine does not give zero...I think...

Well, $\mathrm{d}H$ pulled back to $\Sigma$ should be zero, or am I wrong?

@ACuriousMind Exactly, that's why yours is wrong.

My volume form is $\iota^*\sigma$.

@0celo7 Yeah, but it's as natural an object to try as yours in my eyes. I'm saying I'm still not convinced at all why $\sigma$ should be a "natural" notion of volume form in this context, but then I'm not convinced you should be looking for a measure on $\Sigma$ in the first place...

:/

I have a partial proof in these Hungarian notes

Let me work on fixing their proof

@ACuriousMind I can show that the points which do not recur form a set with empty interior

But we've seen that's not the same as measure zero

@ACuriousMind btw Poincare recurrence should work on any manifold, not just symplectic ones

symplectic just gives a natural flow and measure...

@ACuriousMind Is $\Bbb N_0$ standard?

I see I got another four downvotes yesterday. Including another one for this, taking my answer to -7.

@HariPrasad Which?

@ACuriousMind I don't think I have to look at $\Sigma$.

@ACuriousMind I think I have a proof...

I think this is fairly straightforward, actually

Let $\varphi_t$ be a complete flow on a compact manifold $M$ equipped with a measure $\mu$.

Compactness implies $\mu(U)<\infty$ for any $U\subseteq M$.

Let $B_\alpha$ be a countable basis of $M$, and let $B_\alpha'$ be the subset of $B_\alpha$ which does not return to $B_\alpha$.

lol

"fairly straightforward"

[messages removed]

:D

I know that feeling.

@Danu My prof pulled out an ergodic theory book to loop up the proof

It's more straightforward than that :P

@DanielSank OK, point taken. If it helps, it wasn't (completely) pointless torture - they were researching the capacities of the internal neural system in the arms, which is enough for a recently-amputated arm to grasp for and grab food, without any direction from the central neural system. It's pretty interesting stuff and it tells us a lot about octopuses and the possibilities for neural systems in general, but unfortunately you obviously can't do it without amputations.

@yuggib I hear you know some measure theory

Once the arms died they used them to extract as much research as they could and that's what they came up with.

@Danu The issue is making the physics terminology precise.

@0celo7 yes, but not from the "integration of differential forms" point of view

:/

because I don't care about geometry

I'm trying to figure out exactly what kind of measure you get that way

Presumably a Borel measure

@yuggib what does Borel measure do for me

borel stands for the sigma algebra of sets the measure is defined upon

i.e. the sigma algebra generated by open (closed) sets

what is that

@yuggib I think I can define the "volume" of any open set

is that good enough?

I'm not sure if there's a good way of finding the volume of intersections/unions

a measure is usually an assignment $\mu: \Sigma(X)\to \mathbb{R}^+_{\infty}$, where $X$ is a topological space, $\Sigma(X)$ a sigma algebra on $X$, and $\mathbb{R}^+_{\infty}$ the positive extended reals

finite ones, sure

but probably not infinite

at least, I dunno how to integrate over an infinite union of things

that satisfies $\mu(\emptyset)=0$, $\mu(\bigcup_{n\in\mathbb{N}}\sigma_n)=\sum_{n\in\mathbb{N}}\mu(\sigma_n)$

$\sigma_n$?

Shouldn't that be $\coprod$

for mutually disjoint sets $(\sigma_n)_{n\in\mathbb{N}}\subset \Sigma(X)$

@0celo7 finite/infinite what?

Right, I can show $\mu(\bigcup_{n=1}^k\sigma_n)=\sum \mu(\sigma_n)$ by induction

I don't know how to get infinite unions...

@0celo7 it is a definition

@yuggib finite intersections/unions

a measure is usually defined to be countably additive

@yuggib I mean given my definition of measure.

Aaaand another downvote for my answer. That's nine in all. Whilst Lub got six upvotes for his. Whistle, whistle.

I have a function which I think is a measure

But I'm having trouble with countable additivity.

if it is only finitely additive, it is a less manageable measure

also you need to prove its features on a sigma algebra, not just on open sets

download.springer.com/static/pdf/402/…*~hmac=f09fea500853942f53cc134d198ee7‌​41043df00289e96a56408698cb0025d03e

@yuggib halp

is your function defined only on open sets? if yes, it's not enough

@yuggib Yeah

It's $O\mapsto\int_O\omega$.

where $\omega$ is the volume form

is it defined also on closed sets?

Just sets that are closures of open sets...but I'm not sure about that.

you should extend your definition to a sigma algebra in a way that preserves countable or finite additivity in order to define a measure

Maybe you have to pull back the Lebesgue measure in each chart

and somehow show that doesn't depend on the chart

i.e. to closed sets, and countable intersections and unions of open and closed sets

That sounds horrible, honestly.

@yuggib If $C$ is closed, and $C^\circ\ne0$, then is $\overline{C^\circ}=C$?

wikipedia says that a volume form defines a borel measure

I know what Wiki says

I'm trying to show that

@0celo7 No

@ACuriousMind I know

THENWHYDOYOUASK!!!!

I figured it out after I asked, clearly.

"Do we feel spacetime only because we are not moving at the speed of light?"

Guys do any of you feel spacetime

I don't feel spacetime

Yeah

Should I ask a doctor

In my butt, it's not very pleasant

is it a black hole

@yuggib what the heck is a sigma algebra anyway

@0celo7 then you have to understand the definition of volume form and corresponding integration because evidently it can be done on the whole sigma algebra

@Slereah I think your local drug dealer is the better person to ask if you want to feel spacetime.

@yuggib I understand the definition

I gotta go, it is a collection of sets that is closed under complementation and countable union and intersection

wth

I need this book "geometric measure theory"

@Slereah pls send Steenrod

Aaaand now my answer has ten downvotes. And not one explanatory comment.

Two more days until ICHEP ^^

Time to go downvote it then

@Slereah Can one downvote twice?

In your heart, yes

Not on the board though

Not unless you've got lots of friends.

@bolpteppa I have always told you. It is not a simple term forming for a linear equation. An elliptic curve can be constructed from $a^n+b^n=c^n§.

@John Of course it has

@PhysicsGuy Is that something you can justify or just read can be done?

Oh boy

It's like all my Christmases have come at once

arxiv.org/abs/1603.01154

@ACuriousMind How do "local homology properties" distinguish the boundary of $\Bbb R^m_+$?

Aaaand now my answer has eleven downvotes. And still not one explanatory comment.

I guess I'm learning something about real physicists here.

John, you're a meme

I'm not the meme round here Ryan. The etymology of the word is "that which is imitated". That's not me.

Let's move this discussion to the trash?..

@Slereah What are you doing today

Coding a vidyah gehm

Tonight's research for me is: write a formal discussion for dense flows on the torus, understand the flux tube thing for diff forms, and prove Frobenius' theorem for PDEs

Wonder if the library has a copy of MTW

who has it checked out

communist

it's a month overdue...

@EmilioPisanty I find it rather difficult to accept "we learned stuff" as a justification for dismembering a live, intelligent creature.

And yes, I know we do horrible things to mice in the name of medical science.

I admit that the only real difference here is that I like octopuses.

@DanielSank My reaction at the time was "man, biologists are messed up"

and I stand by it

@DanielSank Humans > mice.

Well yeah... but that argument could extend to the medical research industry.

Are we all messed up for sacrificing mice to enhance our medical science?

Not rhetorical question. I think it's difficult to come to a conclusion.

I usually take a purely biological position on ethics: we defend our existence as a species at all costs, and in some cases yeah, we kill other animals to do it (eating, medical research, etc).

However, I have a lot of emotional attachment to octopuses, for whatever reason.

Anyway, the point is, please do not draw my attention to things like that :)

(smbc-comics.com/comic/2014-08-06 to give back to the googleyness of that site)

It makes me rather sad.

@DanielSank Is weird Japanese porn to blame?

@EmilioPisanty Might want to edit the starred post.

@DanielSank Sure, I only brought it up because I knew you had an octopus 'thing' but did not quite understand what it was. I know better now.

@EmilioPisanty No worries.

My thing is that they're awesome.

@DanielSank how do you mean? I'd rather keep it oneboxed if you mean including the link.

@EmilioPisanty oh ok

w/e

Octopuses are a neat case of context dependent intelligence.

They seem very intelligent when tested in certain ways, but very dumb in others.

@DanielSank Aren't humans like that too

@DanielSank There is utilitarian justification for the usage of animals for medical purposes (animal suffering now helps prevent potentially limitless human suffering in the future), although utilitarians might be caught in a place where their argument would also permit the usage of humans. There's no justification for such research on animals that's done purely for curiosity's (or cosmetic's) sake.

A lot of that makes sense when their lifestyle is taken into account.

@ACuriousMind Oh? So you are absolutely sure that better cosmetics don't prevent suffering?

:)

I'm not kidding.

The line is blurry no matter where you draw it.

@ACuriousMind A true utilitarian would harvest organs from one human to save two others from certain death.

I'm 100% sure you can't classify use cases into justified and unjustified in a consistent way that will make you comfortable with the results.

@0celo7 There's no "true" utilitarian, there are several major variants of that breed of philosophy. The extreme single-case based utilitarianism you're referring to is but one of them.

@ACuriousMind Of course, a utilitarian is poorly defined without a utility function.

It's constructing the utility function where all the difficulties come in.

My utility function right now is making this damn measure theory work

Funny aside: a quotation from a classmate in my college days:

Paraphrased: "Dan, I'm taking a course in mathematical microeconomics. It's interesting, but the math is getting a bit hard. I liked that we proved that communism doesn't work by invoking the implicit function theorem."

@DanielSank No, indeed you can't. That's been the problem of moral philosophy for millenia now, and I certainly don't think we're going to solve it in my lifetime

What the hell does recurrent even mean?

@0celo7 Many of your chat messages evoking deja-vu in me are perfect examples of recurrence :P

He didn't define that properly, so it's worthless

@DanielSank That's a great quotation because it makes me immediately curious what the hell they did in that course

As is all of physics, as I'm coming to realize

@ACuriousMind Unfortunately I didn't take it.

I think we need a finite time such that the flow brings the point back into itself?

Or is it a limiting thing where at $t=\infty$ we bring the point back into itself

Ach! PRL rejected paper.

~Sigh~

One of the referees said the result was "too obvious". If that were true, why did it take our field almost ten years to figure it out?

Plea to all theorists here: If you read a paper where people figure out a problem experimentally, and it seems obvious after reading the paper, that probably means they wrote the paper well.

If it's not obvious after reading the paper, there's a problem.

Hi everyone:)

Let's say I want to find out the top researches about a very specific topic in a given country or continent. How do I approach it to get fast results. It don't have to be an "exact" ranking but I want to get an impression of the research field and know the players in the league or the most influencial research groups...

Skimming a lot of papers on arxiv.org is not practical due to the lack of time and also often the knowledge to judge the importance of the paper. Background - I want to look for places where I could/should apply for a phd. Since it is still some years to go and I haven't decided on my research field, I was looking for a quick way to orient myself in different fields.

@bolpeppa It is an obvious thing. A proof through a counterexample. If a^n+b^n=c^n has a solution for n>2, then a created elliptic curve with these numbers (in that special relation) would be non-modular. It is just the proof, that solutions can not exist in that way.

@Ocelo7 Thats cool. I had been busy with Frobenius theorem, too (some time ago). Then I moved on to Ed Wittens interpretation of supersymmetry with Morse-theory.

@DanielSank sounds significant/ substantial & would be interesting to hear more. re octopuses saw a live one few wks ago at birch aquarium, not easy (they seem shy) but recommend it :) aquarium.ucsd.edu

@GPhys are you going?

@DavidZ Can I ask about that big suspension for 'voting irregularities'? What exactly does that mean?

If it means 'too much serial up/downvoting' then I'll have to start being more careful myself.

@bolpeppa Also, how can it be, that you dont know about the most famous and important problem in mathematics ?

@vzn I've been there.

Seen the octo.

:D

@vzn No, but some friends are

@DanielSank so what do you think is the best octopus viewing spot? (uh, for non scuba divers!)

@GPhys saw this panel re "popsci", looks interesting. what are you interested in/ following at the mtg? ichep2016.org/media_lunch

@vzn CMS 2016 run results :P (I only have access to ATLAS right now)