good

thats better

mmmm some actual physics

go go gadget ACM

answer

@enumaris What I was talking about wasn't physics?

elliptical PDEs? -.-

lots of things are elliptic PDEs

@user400188 What page in Schwartz?

Page 5, (25 if you have the same pdf as me)

I once and only once encountered an elliptical integral

it was for trying to solve for the EM field of some distribution of charges

that has nothing to do with elliptic PDE

forgot all the details

so I dunno what elliptic PDEs are then lol

Laplace's equation is the simplest and most obvious one.

o

what makes them elliptical?

wikipedia discusses the classification here: en.wikipedia.org/wiki/…

interesting way to classify it

@user400188 Okay, so, as you correctly observe, in a true $L\to\infty$ limit this doesn't make a whole lot of sense because the $E_n$ all go to zero. But that's fine - in the passage to the continuum limit, we have to think in terms of densities instead of actual values: The meaningful quantity is not actually $E(\omega)$ - note that it's infinite since it has $L^3$ in it, but the energy density $E(\omega)/L^3$.

In that computation, Schwartz is not actually taking a true limit $L\to\infty$, he's just saying that we treat $L$ as so large that the $E_n$ become continuous, i.e. we get an integral, but it's still finite otherwise the $L^3$ wouldn't be there

ayyyy, finally some hand wavy maths, now this feels like home

This is always what scares me the most

don't you dare make it a rigorous definition

How can $L$ be so large that $E_n$ becomes continuous, without $L$ being infinite?

It can't, it's a trick :P

yesssss

accept it, you know in your heart that you want to do hand wavy math, that's why you became a physicist

krank

@ACuriousMind Yo

@user400188 Here's a way to arrive that the formula without taking this weird limit: The energy density is the integral of the energy of a state times the density of states. If you compute the density of states, you find it is independent of $L$, so you can evaluate this formulation of the integral without having $L$ show up at all.

Suppose I have family of operators $L_t\in \mathscr B(X,Y)$ between Banach spaces such that $t\mapsto L_t$ is continuous

Then you can formally (and trivially, since it doesn't occur) take the limit $L\to\infty$.

And suppose each is boundedly invertible

Ah that makes more sense

and suppose I solve $L_t\psi_t=\gamma_t$, for some family $\gamma_t\in Y$ with $\gamma_t\to 0$

must then $\psi_t\to 0$ in the topology of $X$?

I don't have the slightest idea

I might if I thought a bit about it, but I'm about to go to bed

ok, good night

Good night @ACuriousMind.

I guess one needs to know if $t\mapsto \psi_t$ is continuous in a sense

I haver to leave too. Bye @everyone

Thanks, good time-of-day to you too

@ACuriousMind I bet we have $t\mapsto (L_t)^{-1}\gamma_t$ being continuous, and it $\to 0$.

@ACuriousMind Ah, I do think bounded inversion is continuous so this makes sense.

@ACuriousMind Toilet math: $$\| \psi_t\|= \| (L_t)^{-1}\gamma_t\|\le \| (L_t)^{-1}\|\cdot \|\gamma_t\|.$$ Now since $t\mapsto L_t$ is continuous in $\mathscr B(X,Y)$, the set of norms is bounded. But apply this and use the same logic on the inverses to conclude that the norms of the inverses are bounded. Thus $$\|\psi_t\|\le C\|\gamma_t\|\to 0.$$

However, one must use the fact that $L_t$ never hits the zero operator, which is true for $t$ close to $\infty$.

waddup physics fellows

@goodnight No one here knows any physics

Ah

You're just the guy who changes name every moon

who?

IceLord, Blue, 3507, so on

wait

those are the same person!?

?

Come on

who is that

wait ur saying this isn't the krusty krab

A young physics major at UoT ;)

idek what physics is

i'm doing aboriginal studies

:p

That just makes you even more of a physics major tbh

alright

waddup are you at waterloo

you're not doing math or physics?

lol

I am :p

ok

non-stem people are not allowed here

@goodnight Nope, I'm living in Santa Barbara

tfw ur a leaf

Embracing my ultimate failure

@BernardoMeurer work?

Nope

CC

oh right

did you apply to waterloo

The lowest of the low :)

@goodnight Dude what is with these names

apps close in jan

I'll apply for transfer

@BernardoMeurer did you watch community

But I'll finish the year here

Nope

Did not watch it

well you don't need to since you're basically living it

that's not a bad thing

@BernardoMeurer but applications

are due in JAN

even for transfers

that's two months out

what is your issue?

I am aware. That means I'm not applying to waterloo this year

??

I honestly don't even know what I'm doing with my life anymore

living it?

I suppose

@0celo7 well it takes a long time

my life has no meaning since ACM became a coder

@goodnight ??

kek

@BernardoMeurer Ok I will call you on Sunday

let's talk

can I talk too

Who are you?

you tell me

@0celo7 Thanks

dude are you fried

he's been over this a billion times

@SirCumference I say this in the nicest possible manner, but you don't seem to be the sharpest tool in the shed.

I'm so confused...