@Qmechanic This closure is a bit much, I think.

@Slereah dunno about obscure papers, but it sure felt good to cite Lanczos' variational principles of mechanics in a paper

hi @Kaumudi.H

congratulations on exams being over @Kaumudi.H

@Slereah I don't think so, the Euler characteristic for a $\mathbb{Z}$-sheeted covering space may be ill-defined

@ACuriousMind What's a good example

@Kaumudi.H food has that effect on me too :-)

@Slereah I was thinking of the double torus being covered by a plane with infinitely many handles attached, but there's a much simpler example where the formula will just fail although you can define the Euler characteristic: The circle covered by the line.

The issue is that an infinite cover will be non-compact, and the Euler characteristic is ill-behaved for non-compact manifolds

you're welcome @Kaumudi.H

@Kaumudi.H cool you should add JB to the mix

Hi guys, I don’t think I understand Griffith’s bit on the cancelation of electric fields in and outside a conductor. They say no external fields penetrate te conductor, for they are canceled at the outer surface by the induced charge there. However, we also have an induced charge on the inner surface, so wouldn’t this charge then create a field within the cavity?

As far as I can see, they seem to be saying that the outside charge is responsible for the cancellation; so how about the effects of the inner charge?

no the inner charge is responsible for the cancellation

I quote the paragraph:

"Similarly, the field due to charges within the cavity is cancelled, for all exterior points, by the induced charge on the inner surface"

there are two scenarios tho; when you have an outer charge and when you have an inner charge.

but we could take that scenario

it's the same question, it just goes the other way

they say that the conductor "communicates" the charge inside; that means that it creates an electric field, right?

yea

okay, so if we then go the other way

if we have an outside charge, the inner charge of the conductor will create an electric field in the cavity, right?

it's the same situation if you ask me, just reversed

no

inner charge= charge induced on the inner surface of the conductor

I say again

"Similarly, the field due to charges within the cavity is cancelled, for all exterior points, by the induced charge on the inner surface"

so it makes sense that charges outside the cavity are cancelled by outer surface

yes, but we get an induced charge on the inner surface

otherwise our conductor wouldn't be electrically neutral

shouldn't this inner charge then create a field in the cavity?

pick a point in the sphere, and calculate the field there

to do this, you would need to integrate over the induced components over the full sphere

this integration over the whole sphere yields 0 for any chosen point within the sphere

what do you mean by point in sphere? in the conductor or in the cavity?

cavity

pick any point in the cavity

and integrate the field over the conductor area

yea I know that argument

and the result is 0 at each point within the cavity

so

what it means is

any electric field at point A on the inner conductor, is cancelled out by other points on the conductor

@EmilioPisanty : Noted.

@Qmechanic though the question doesn't make sense anyway!

@Kenshin so when they say that the outer charge "communicates" the presence of the charge in the cavity, they don't mean that the outer charge creates an electric field, but just that the charge on its own is the communication?

For some reason I cannot simply pretend that heat is a figment of my imagination.

no

@ShaVuklia it is saying that the charge within the cavity doesn't communicate

instead, it tells the conductor

and the conductor then tells the world

the conductor creates a charge of +Q on the outside surface

ok but say you're sitting in a car, and the car is hit by lightning, then the outer charge of the conductor will cancel the lightning, however, we will also have inner charge that communicates the lightning to us. But I'm guessing this is not a problem; as long as you don't touch the metal of the car, you're safe?

where Q is hte charge within the cavity

@ShaVuklia I dunno about lighteninng but I think u have the right idea

fields outside the cavity will induce a field on the inside of the conductor

and charges inside the cavity will induce a field on the outside of the conductor

these are then used to "communicate"

rightt

okay then I get it!

:)

thanks

and of course, when a field is induced on the inside, there is an equal and opposite one on the outside

you mean when a charge is induced, no?

so if you have a charge of Q in the cavity, you get a charge of -Q on the inside of the cavity cancelling it out, and charge of +Q on the outside which communicates this

yea right it makes sense now

yeah easy stuff

just poorly worded by griffith here

haha:P

@Kaumudi.H I think I have the DVDs somewhere ...

They sent me the stuff

Sometimes, SMBC is just depressing :/

@0celouvskyopoulo7 What does "based" mean, anyway? :P

> Have adults just considered being honest and open?

Well I can pull this off to some extent so far without trouble

One of the most crucial criteria to be part of my social group is to be honest (although that does not necessary forbid necessary secrecy). Conversations tend to suffer fallouts very quickly whenever the group detect some dishonest behaviour

@ACuriousMind Based means based, nah mean

@0celouvskyopoulo7 noice

Guys, I don’t see how it makes physically sense that $\begin{align}f=\frac{1}{2\epsilon_0}\sigma^2\hat n\end{align}$. Didn’t they just say that “a patch cannot exert a force on itself”? It seems to be that in the case of a conductor, the conductor is in fact exerting a force on itself.

@ShaVuklia It's the rest of the conductor that exerts that force on the patch you're looking at, not the patch itself.

@0celouvskyopoulo7 Ever read Sternberg's book?

Shlomo?

No

It seems alright

Is it the curvature and physics one

Especially if you want really mathy GR

@ACurious oh, because we can neglect the effect of an infinitesimal patch, and just look at the effects of the surrounding? I'm assuming this surrounding should still be very close to the patch then

I have enough really mathy GR

No such thing

I started writing the book

What does this one do differently

I got about 90 pages out of the CTC book

Probably gonna put a bunch of stuff not in other books

Focus on non-globally hyperbolic spacetimes a bit

QFT defined

Maybe a bit of quantum gravity

I mean the Sternberg book

o

what does it do differently

Well it has a lot of bundle stuff

also the Petrov classification

well nothing new about bundles

and Petrov is probably in Stephani

it is

Stephani is big on Petrov

Are you talking about the Curvature one

The first chapter is "The principal curvature"

If that is what you mean

What's a good place to see the measure defined for integrals on manifolds

Amazon preview doesn't even have the index

can u send me a legal copy

@Slereah Jost

Lee

Either Lee

Straumann

Now if you want Lebesgue measure, well.

Maybe Federer, but it's probably best to do it yourself

Jost has manifolds?

Jost is a Riemannian geometry book...

Which book of Jost

Riemannian geometry...

Oh

I only know post modern analysis

@Slereah can u send me a legal copy of sternberg

Lemme see

or is it on russian servers

Gotta install skype on this pc

where are you looking at it?

It is probably on genlib

my computer

ok ill get it

Sternberg doesn't fuck around

First chapter doesn't even define manifolds

Wait, what's the title

Straight to the point

Why should he define what a manifold is

Semi-Riemann Geometry and General Relativity

P. Li doesn't define anything, you have to figure out wtf he's talking about

monographs are bad like that sometimes

Chapter 2 is "rules of calculus"

lel

He defines it by graded algebras

@Slereah Pretty sure I know everything there, modulo three or four sections.

By "rules of calculus" he means "exterior calculus" I'm sure.

yeah

Fuck me it's hot today

it's going to be 89 today

good thing I work in an office

@yuggib hola

What would you use to write the induced metric on a submanifold

$g |_\Sigma$?

$h$

@Slereah eeeeeehhhhhh

it's not really that

Well the symbol will be $\gamma$

that suggests you restrict $g$ to values on $\Sigma$

$g|_{T\Sigma}$ is better

because you restrict to tangent vectors along $\Sigma$ as well

Let's go with that

Hm, wait

Is it a good idea to use $\gamma$

@0celouvskyopoulo7 Is anyone who knows what a metric really gonna be confused by the lack of the $T$?

Am I gonna also use $\gamma$ for curves there

$\gamma$, $^{(3)}g$ or $h$

@ACuriousMind I would be...

I would contend that $g\rvert_\Sigma$ carries the same information to the average reader as $g\rvert_{T\Sigma}$.

@0celouvskyopoulo7 Would you be like "I don't know what they mean!" or like "If they wanted to restrict the metric, they should've written $g\rvert_{T\Sigma}$"?

Knowing you, I can see the latter, but I don't see that anyone would be really puzzled by how one restricts a metric

@ACuriousMind There are indeed two meanings, and I would be confused by which they meant. It would probably be clear from context, but why not just write $g|_{T\Sigma}$ for definiteness.

What's the meaning you could confuse it with?

If there are two different similar things here, then you might be right - I just don't know what the second one is

@ACuriousMind The metric is a map $g:M\to S^2(\Gamma(TM))$.

It makes sense to restrict its domain to $\Sigma\subset M$.

I don't know why one would do that, but hey.

I...would've never thought of that, and if anyone does that without explicitly telling the reader, they suck :P

@ACuriousMind Yeah, well, that was exactly what I thought when I read that.

@Slereah Just use $\iota^*g$

Because that's what you're "really" doing.

@ACuriousMind Ok sleepy man, does the parallel thing make sense now? It's crucial to this argument

@0celouvskyopoulo7 Yes, it's fine

What does "2eme et dernier envoi." mean

@Slereah

"2ème envoi."

aww yis

@0celouvskyopoulo7 hola

@0celouvskyopoulo7 second and last sending

@0celouvskyopoulo7 Not as evocative

Carlip says that in 2+1 dimensions the riemann tensor depends only on the Ricci tensor but he does not prove it

That's well known

See Petersen

it follows from the symmetries

Hm

Do I need the metric to define distributions?

On one hand, it appears on the volume form of the integral

@Kaumudi.H Well it was prolly more than my melting point.

On the other hand, you don't need to define distributions by integrals

I feel sorry for you. Try not to die.

page 86 of Petersen (2016)

(I also recommend reading the LOTR book instead of watching the movies. That's 100x better)

@Slereah What?

Distributions have nothing to do with integrals.

They're continuous linear functionals on $\mathscr D$

@0celouvskyopoulo7 You can define them as weakly convergent sequences

such that $\int D_n(x) f(x) dx$ converges to something

That's a bad definition.

Is it

And that's just saying $C^\infty$ is wk* dense in $\mathscr D'$ or something

@Slereah Yeah, because you need integrals

Look, distributions on manifolds are in many books, but I've lost hope you'll read any book I suggest, so whatever.

I guess that for manifolds, that might be an issue

I've got lots of books to read, my man

I haven't even finished reading all the Mister Men

Hm, I guess the question is

Is that definition also valid for manifolds

The canonical reference would be Hormander 1.

Let's see if I have Hormander in there

It's very technical because you have to worry about charts in all of your definitions.

Hoermander L. - Notions of Convexity

On compact manifolds it probably doesn't matter how you do things.

I'm guessing not that one

@Slereah "The Analysis of Linear Partial Differential Operators"

Let's go get it legally

How does Pauli Fierz work, anyway

How do you guarantee that the metric is of the correct signature

or even that it converges

Is $\text{diag}(1,0,0,0)$ not a valid solution to linearized GR

I guess it would not be a solution of the non-linear Pauli-Fierz equation

Is the only linear map with zero rank the zero map? (I know that is true for finite cases by rank nullity theorem, but I am not so sure about the infinite case)

@Secret what's the rank in infinite dimensions...

Dim im?

dimimimimim

sorry :P

What?

he said dimimimimim

and then sorry :P

I see the reading comprehension of the room is strong today :)

@ACuriousMind what is the jet bundle useful for, anyway

are there proofs easier with it than just using dumb old PDEs

@0celouvskyopoulo7 I am guessing that's the most logical way to interpret it, the dimension of the image of the linear map

I don't think it's about easier, but it's a better setting if you want a "space" where your DE live, I think. I dunno, never needed it much

I'm guessing maybe it's better for gauge fields?

I fairly rarely see the jet bundle used

Again, a minority of people actually working on PDE use it.

yeah, hence why I wonder what it's for

Hmm, I think I need to revise whether it is possible to start with a domain all of $\Bbb{R}$ and end up with an image that is some countable and discrete subset of $\Bbb{R}$ if the map $T$ is linear...

Sure

Oh, if it's linear

Probably not

outside of the obvious constant map

Ah yes, constant maps will map every point in $\Bbb{R}$ to a singleton, hence its image will be zero dimensional

which is unique to the infinite case because for the finite case, rank nullity theorem will guareentee that singleton is {0}

Just take a function that maps to two points $f(a) = A$, $f(b) = B$

Then if it's linear $f(a+b) = A + B$

This only works if $A = B = 0$

You can probably show it for arbitrarily many points

by recursion

The function must always include $0$ if it's linear by the existence of an inverse

Then you can show that by linearity it can't have any value other than $0$ by multiplying by a scalar

Right, so that concludes the zero map is the only map with rank zero even for the infinite dimensional case

$f(\lambda A) = \lambda f(A)$

Which can take arbitrarily many values if $f(A) \neq 0$

@Secret linear maps always have vector space images...

I wonder, can the image be $\Bbb{Z}$, wait no... cause the scalars are in $\Bbb{R}$ thus I must span a subspace of $\Bbb{R}$ "continuously"

No, the image must be an $R$-vector space.

Or just the zero vector.

I need to start bringing a waterbottle to class

to make it flip and amaze the crowd?

To hydrate

It's starting to heat up

@ACuriousMind $\phi^4$ we know is trivial, but do we know (or suspect) if QED is trivial yet

something something Landau pole