@AndrasDeak ...in an attempt to not waste more time on the SE network you have come to chat? Not sure I follow your logic there, but welcome

Well it's not going well so far:D

@heather Ok, so I claimed that $\sup A=u$ is a number such that $u-\epsilon$ is not an upper bound for any $\epsilon>0$.

So in this case, we have $u=1$.

So what's up with $1-\epsilon$?

> Passing by some construction, the grad student sees a hole in the road smaller than the one in his dissertation's theory chapter.

lol

Lego Grad Student is spot on

I guess $\epsilon$ must either be infinitely small or 0, right?

$\epsilon>0$

just a liiiiiittle bit too close to the mark to be comfortable, though

@0celo7, be back in a moment

What does "infinitely small" mean

@0celo7 nothing

@EmilioPisanty it's a response to a message, look above.

@0celo7 yeah

sorry, I'm just still ticked off about this one

So I want @heather to answer

@EmilioPisanty you have your PhD, what's the issue?

@EmilioPisanty Oh.

oh, and it got tagged mathematical-physics too?

no way in hell that's staying

:(

It's so cold

at least "object infinitely far from a lens" can be made sense of:P

@AndrasDeak not really

more than the others:D

as long as the lens has zero diameter

if you mean "a collimated beam" then you just say that

@EmilioPisanty interesting fact: the derivative of a function is continuous on a dense set

@EmilioPisanty not enough infinities:P

@EmilioPisanty I talked about that question with OP a bit further upthread

@0celo7 even for a nowhere-differentiable function?

@ACuriousMind yeah, I saw that

@EmilioPisanty how are you gonna take the derivative of a nowhere diff'ble function??

@0celo7 precisely the point

@EmilioPisanty If $f:\Bbb R\to\Bbb R$ is differentiable on $\Bbb R$, then $f'$ is continuous on a dense set $\subset\Bbb R$.

I DON'T mean $C^1$, obviously.

@0celo7 yyeaahhh, that's not too much of a stretch

that doesn't rule out $f'$ being discontinuous on a dense set though

or wait

That is true. There's a slightly stronger statement.

maybe it does?

probably not

@EmilioPisanty $\Bbb Q$ and $\Bbb R-\Bbb Q$ are both dense.

@0celo7 yeah, but continuity is not a local statement

$g$ continuous on a dense set could conceivably rule out $g$ being discontinuous on a dense set

It depends on small balls, what else would "local" mean

@0celo7 ok, not a pointwise statement

...yeah it is. Continuous on $A$ means continuous at all $x\in A$.

pffff

@EmilioPisanty Not as I stated it above

You can have a function discontinuous on just $\Bbb Q$.

@EmilioPisanty The strong statement is that $f'$ is continuous on a $G_\delta$-dense set.

@EmilioPisanty If you want some fun, think about this problem: let $f_n:[0,1]\to\Bbb R$ be a family of continuous functions. Is it possible for such a family to converge pointwise to $f(x)=0$ for $x\in\Bbb Q$ and $f(x)=1$ otherwise?

the limit is the Dirichlet function, right?

That's the name for it

But can you find such a sequence?

Oh, I understand the question, but it's way over my head:) I just keep forgetting whether Dirichlet is the indicator of rationals or irrationals

@EmilioPisanty Thank you Emilio

@0celo7 what's a $G_\delta$-dense set?

@0celo7 surely yes

not quite a trivial construction

but it sounds doable

Countable intersection of dense sets I think. The terminology is horrible

G delta dense is not necessarily dense

I can tell you more when I get home.

@0celo7 that's bad notation right there

@EmilioPisanty it's not :(

@0celo7 is it actually?

oh well

any deep reasons?

Baire category theorem

man, it's been too long since I took analysis classes

I knew that one

once

WHAT IF THE MARS CIVILIZATION FALLS INTO THE WRONG HANDS?

Asking the real questions

Nice link from worldbuilding....oh.

Wtf ACM

@ACuriousMind love the tag

and here I thought the real threat was polluting Europa with germs

maybe we can give it some love and make it sink out of front-page visibility though

guess I should gather 24 rep

It is a good question.

Look at America

@0celo7 let's not

Hey @ACuriousMind, while we're shooting the breeze

Okay, who of you jokers upvoted that? :P

I'm on mobile

"the spin angular momentum of the EM field generates the rotations of the polarization vector"

↑ can that be made sense of for classical EM?

I mean, it can

but how exactly?

classical EM and spin?

@AndrasDeak yeah, it's a thing

@EmilioPisanty I would guess that, in the Hamiltonian description, the Poisson bracket of it with the polarization vector gives an infinitesimal rotation of that polarization vector. But that's really a complete guess

@EmilioPisanty guess I'll stick with *shrug* then:P

@ACuriousMind Poisson bracket over what variables, though?

or, more importantly, the Poisson bracket of which quantity

ah right, I used to be aware of this, but then QM overwrote everything spin-related with Hilbert spaces

@EmilioPisanty Mhhh. On the infinite-dimensional phase space spanned by the four-potential $A$ and its conjugate, which becomes $E$ after some gauge choice.

And I meant the Poisson bracket of the spin angular momentum with the polarization vector.

@ACuriousMind goodness, yeah, that's gonna be hell

The orbital angular momentum is so simple, though

@EmilioPisanty Yes, the description of EM as a constrained Hamiltonian system is not particularly nice

Wth is spin angular momentum

just exponentiate $\partial/\partial\theta$ and you're done

@0celo7, okay, I'm back. Infinitely small: so close to zero it is zero in practicality but infinity messes with it and makes it not zero. (Sorry about the delay, thought I only had to do one chore but turned into around 10.)

@EmilioPisanty I suppose you could start with What is spin?, Am. J. Phys. 54 (1986) 500, by Hans C. Ohanian. Recent question on the paper at: physics.stackexchange.com/questions/285222/….

@0celo7 There's a part of even classical angular momentum of the EM field that's not dependent on the choice of origin, so it's not orbital angular momentum.

@dmckee ooooh, nice

@ACuriousMind ????

Proof?

@0celo7 it's like that for everything

including mechanical systems

take e.g. the Earth

and calculate its total angular momentum

part of it comes from its COM momentum having a nonzero moment

e.g. $\mathbf L = \mathbf r\times \mathbf p$

part of it comes from the Earth spinning, and that doesn't change if you change your origin

Is it not because of its magnetic moment? Or is that excluded?

If it includes r, it's not spin

The earth is a spinning ball though

@0celo7 well, that's the thing, it sort of depends on how you split up your system

@0celo7 so?

Electrons don't spin

@EmilioPisanty What do you mean by this? What's $\theta$?

@ACuriousMind angle about the rotation axis

So, uh, how does that show which rotation orbital angular momentum generates?

@0celo7 Point is, every system has a total angular momentum $\mathbf J$ which can be split into two components, one which depends on the coordinate origin and one which doesn't. If the system has momentum $\mathbf p$ and you move your origin by $\Delta \mathbf r$, then $\mathbf L$ will change by $\Delta \mathbf r\times \mathbf p$, but that doesn't mean that $\mathbf J = \mathbf r\times \mathbf p$.

technical question: is it expected that the chatjax++ userscript doesn't convert new messages, only existing ones once I reload the page?

@ACuriousMind i.e., with an axis understood, if $L=-i\frac{\partial}{\partial \theta}$, rotations by an angle $\Delta \theta$ are implemented by $\exp(i\Delta \theta L)$

@heather Are you still here?

We need to talk about open intervals a bit

@EmilioPisanty Angular momentum in the Noether sense?

@0celo7 what other sense is there?

$r\times p$.

i.e. what other constructions are provably different

@EmilioPisanty That is true for all rotations/translations: The transformation that shifts a variable $k$ by $\delta k'$ is $\exp(-\delta k' \partial/\partial k)$. I'm not following why you mentioned this in the context of the orbital angular momentum. The orbital angular momentum is something like $\int \vec E\cdot (\vec r \times \nabla) \vec A$, what has it to do with $\partial/\partial \theta$?

@ACuriousMind I need to think about it a bit more

@EmilioPisanty $G_\delta$ dense -- countable intersection of open dense sets.

but it has to come down to something similar

$\Bbb R$ is not $G_\delta$ dense, this is Baire's theorem.

So you can't have a sequence of continuous functions which converges to a function discontinuous on all of $\Bbb R$

@0celo7 what's that got do do with Baire's theorem?

@0celo7 is it not continuous on irrational points?

@AndrasDeak no

ah

every irrational point has rationals arbitrarily close to it

right

@EmilioPisanty A pointwise limit is continuous on a $G_\delta$ dense set

Clearly the empty set is not $G_\delta$ dense

So if you want a sequence that converges to the Dirichlet function, you're out of luck

it's continuous nowhere

But any such limit is continuous on a $G_\delta$ dense set

yeah, you lost me