The h Bar

0celo7

19:00

holy crap the inf could be attained, what is life???

user116211

Samsung Galaxy Note 7 is officially declared dead :(

0celo7

Good.

@ACuriousMind Argh! Wtf!?

If I have a compact set $K\subset U\subset\Bbb R^n$, $U$ open

Obliv

@0celo7 what if the homogeneity of a linear map is greater than degree 1? does it still count as a linear map

0celo7

I have an open cover $\mathcal B$ of $U$, then I get a finite subcover $\mathcal B_K$ on $K$

user218912

@0celo7 how can I show that a tensor is not symmetric? do I just flip the indices and show it isn't the same anymore?

0celo7

19:11

I find a Lebesgue number of $\mathcal B_K$

stop asking me questions I'm mad

user218912

:(

0celo7

@ACuriousMind Is this also a Lebesgue number for the union of elements in $\mathcal B_K$?

It's probably not, but who knows?!

user218912

@ACuriousMind if I want to prove that the regular energy-momentum tensor is not symmetric do I just flip the indices in the noether current definition of it for spacetime translations?

0celo7

Noether energy momentum tensor is not symmetric, get out of here

user218912

I know it's not.

user218912

19:13

I didn't say it was.

ACuriousMind

@bl00 Once again, you seem to be asking me whether a certain strategy will work instead of simply trying it.

user218912

I mean I tried it.

user218912

does it make sense though?

user218912

to do that?

ACuriousMind

@0celo7 I don't understand the question.

@bl00 Well, what did you find?

0celo7

19:16

@ACuriousMind clearly $\mathcal B_K$ is an open cover of $\bigcup_{B\in\mathcal B_K}B$, right?

ACuriousMind

@0celo7 Yes

0celo7

is $\delta$ also a Lebesgue number for this cover?

Physics Meta

0

Q: Why does my daily reputation not grow beyond 200?

Today, I asked a question on physics SE which is still being massively upvoted by people, but there is a problem,till 40 upvotes i got 5 reputations for each and got a badge of 200 reputations in a day ,but after that i'm not receiving any reputation. Can anyone tell me why is it happening?

ACuriousMind

@0celo7 If the $\mathcal{B}$ consists of balls, then yes, otherwise no.

user218912

@ACuriousMind this is for the electromagnetic stress energy tensor.

user218912

19:20

I found that the metric term becomes negative

0celo7

@ACuriousMind I need to know why o.o

user218912

but idk about the field strength tensor term

0celo7

it's balls, yes

ACuriousMind

@0celo7 On second thought, I think you can't even show that for balls.

@bl00 I don't know what "the metric term becomes negative" means.

Obliv

@acuriousmind a cauchy sequence is one where all elements of the sequence get close to each other right? So the series $\sum \frac{1}{n}$ does not fit this description because $|a_m - a_n| < \epsilon$ does not hold true, where $\epsilon$ is really small? $|a_{m+1} - a_m| < \epsilon$ but this is not sufficient. Am i right?

ACuriousMind

19:26

also, I don't really know what you are doing. The EM stress-energy is (proportional to) $F^{\mu\sigma} {F^{\nu}}_\sigma - \frac{1}{4}\eta^{\mu\nu} F^2$. It's rather trivial to see whether or not this expression is symmetric.

0celo7

Trivial has a different meaning in ACM's world...

ACuriousMind

@Obliv Are you talking about the sequence $a_n = \frac{1}{n}$ or about the series $a_n = \sum_{i=1}^n \frac{1}{i}$?

Obliv

series

wait

sequence.

user218912

@ACuriousMind I have to show it.

user218912

I know the electromagnetic field tensor is antisymmetric.

0celo7

19:29

@ACuriousMind I'm pretty sure my whole approach is wrong then

No I take that back

Wtf why is this problem so hard

@ACuriousMind Why do you think that?

ACuriousMind

@bl00 But...what is there to show? $\eta^{\mu\nu} = \eta^{\nu\mu}$ by definition and $F^{\mu\sigma}{F^{\nu}}_\sigma = {F^{\mu}}_\sigma F^{\nu\sigma}$ because I can raise/lower dummy indices without changing anything.

@0celo7 Lemme draw a picture.

0celo7

FREEHAND CIRCLE TIME

@Obliv the LoL guy is a weeb, too

Not really surprising

user218912

@0celo7 your roommate?

Obliv

the one that brought in his overpriced razer gear? @0celo7

yeah not surprising.

0celo7

@bl00 huh?

My roommate plays rugby, not LoL

user218912

19:35

oh I know who you're referring to now.

user218912

@ACuriousMind I know that.

ACuriousMind

Hm, didn't crop right, sec

0celo7

Fug, I see

Maybe

ACuriousMind

@bl00 then what is there left to show?

0celo7

@ACuriousMind ok but can I find a lebesgue number for the balls?

Maybe a different one

user218912

19:39

@ACuriousMind I think I'm missing something, by what you said, it should be symmetric...

ACuriousMind

@0celo7 Well, the closure of their union is compact if they are finitely many, right?

user218912

the whole expression I mean

ACuriousMind

@bl00 Yes. The EM stress tensor is symmetric.

user218912

so I'm trying to prove it's antisymmetric?

0celo7

@ACuriousMind sure

user218912

19:40

I don't get it.

ACuriousMind

@bl00 wat

user218912

I constructed mine from the noether current for spacetime translations.

user218912

is that one also symmetric?

ACuriousMind

The EM field strength tensor is antisymmetric. The EM stress-energy tensor is symmetric. What is your problem?

user218912

yes I know that, the problem says to show it's antisymmetric.

user218912

19:42

I re-read the problem 5 times, I'm 100% sure it says to show it's antisymmetric.

0celo7

@ACuriousMind were you going somewhere with this?

ACuriousMind

@0celo7 Yes - can't you just apply the Lebesgue lemma to the closure of $\mathcal{B}_K$ to get that Lebesgue number?

user218912

what I got was $-F^{\mu\alpha}\partial^\nu A_\alpha + 1/4 g^{\mu\nu} F_{\epsilon\delta}F^{\epsilon\delta}$ @ACuriousMind

0celo7

@ACuriousMind with what covering?

user218912

did I derive it right?

0celo7

19:44

I lied btw, I don't have a covering of $U$

ACuriousMind

@0celo7 Ehhh, damn :D

user218912

you're saying that is 100% symmetric? @ACuriousMind

ACuriousMind

@bl00 Oh, yes, the pure Noether tensor might not be symmetric.

user218912

okay how do I show it?

0celo7

It would be interesting if the Noether tensor turned out to be antisym.

user218912

19:46

I flipped the indices

0celo7

Did you compute the Noether SET @bl00

user218912

SEM?

0celo7

@ACuriousMind This is the problem btw

EMT

SET

whatever

user218912

what?

0celo7

the damn $T$ tensro

user218912

19:47

yes

0celo7

what is it

user218912

what I said

0celo7

where?

user218912

4 mins ago, by bl00

what I got was $-F^{\mu\alpha}\partial^\nu A_\alpha + 1/4 g^{\mu\nu} F_{\epsilon\delta}F^{\epsilon\delta}$ @ACuriousMind

user218912

I used the vector field $A_\alpha$ and the massless EM lagrangian. and plugged it into the noether current

ACuriousMind

19:48

@bl00 I'm not sure what you think there is to show, either - the second summand is symmetric, the first isn't (if you really want to "show" something, just write out the definition of $F$ and observe the term isn't the same after you switch indices).

user218912

@ACuriousMind I guess that's what i'll do xD

ACuriousMind

So the whole expression isn't symmetric.

0celo7

doesn't look antisym to me

@ACuriousMind So, I have a bunch of balls on $K$ for which the property holds

user218912

@ACuriousMind I'm writing it out, I'll lyk what I find.

user218912

it's not obvious to me without writing it out and simplifying

0celo7

19:51

i.e. for each $a\in K$ I can find $r,c>0$ s.t. $||x-y||<r\implies ||f(x)-f(y)||\ge c||x-y||$

$c(a)=c$

Obliv

nvm they don't have to be equivalent

0celo7

@ACuriousMind The fact that $y\in U$ at the end is blowing my mind!

@ACuriousMind !

I take $\mathcal B_K$, then find a precompact open set in it

s.t. the closure is contained in it

then I have to find a tubular neighborhood inside of that

But I can apply Lebesgue to the tubular neighborhood

user218912

$\partial^\alpha \partial^\nu A_\alpha A^\mu$ is antisymmetric under exchange of $\mu, \nu$ right?

0celo7

@ACuriousMind sound good?

@bl00 proof?

user218912

because if you change the indices it becomes a different thing.

0celo7

19:56

that's not what antisym means.

user218912

I know

user218912

but that's all I could think of

user218912

how do I prove that it's antisymmetric?

user218912

because it has to be

ACuriousMind

@bl00 It's neither antisymmetric nor symmetric (as choosing the example $A^0 = (x^1)^2$ and the rest of the $A^i = 0$ shows).

user218912

20:00

then what makes that stress-energy tensor antisymmetric?

user218912

the second term is symmetric

user218912

the first term has to be antisymmetric then

ACuriousMind

Nothing. The Noetherian result simply is not symmetric.

user218912

so who is antisymmetric here?

user218912

but I need to show it

ACuriousMind

20:01

If the problem states it's "antisymmetric" it's simply wrong, or you have read it wrong.

Wait

0celo7

@ACuriousMind What's the precise definition for "I want $\mathcal N$ around $K$ such that no element of $K$ is more than $\epsilon$ away from $K$"?

or

given $U\supset K$, can I find $N$ inside $U$ that covers $K$ with "uniform thickness"?

ACuriousMind

@0celo7 I don't understand the question

user218912

@ACuriousMind ?

user218912

oh

user218912

dkm please

0celo7

20:09

@ACuriousMind Given $\epsilon>0$, can I find an open set $N\supset K$, $N\subset U$ such that $x\in K, y\in U-K, d(x,y)<\delta\implies y\in N$.

user218912

I was reading not symmetric as anti-symmetric.

ACuriousMind

@bl00 You misread "not symmetric" as "anti-symmetric" 5 times???

user218912

yes.

user218912

sorry

ACuriousMind

@0celo7 IF your $\delta$ is supposed to be an $\epsilon$, then sure, just take $\left(\bigcup_{x\in K} B_\epsilon(x)\right) \cap U$.

0celo7

20:14

@ACuriousMind Second problem

It's supposed to be just balls in that union

no intersection

so can I chose $\epsilon$ small enough so that union is entirely in $U$?

ACuriousMind

That's not my problem

0celo7

but not equal to $U$

ACuriousMind

@0celo7 If $U$ is open and $K$ closed, then yes.

0celo7

@ACuriousMind yeah $K$ is compact and $U$ is open

why can we do that?

ACuriousMind

Ugh

0celo7

20:19

@ACuriousMind Is that an idk ugh or a idc ugh?

ACuriousMind

That's an "thinking about my idea I'm not sure anymore" ugh

0celo7

@ACuriousMind I'm thinking it should be true if $K$ is compact

ACuriousMind

My plan was to take the neighbourhoods $B_\delta(x)$ around each $x\in\partial K$ that by definition must lie inside $U$ and then take the minimum over all those $\delta$

I'm struggling to see why that minimum is forbidden to be zero, though

0celo7

for general closed sets it's certainly false

@ACuriousMind maybe construct a continuous function in that spirit, then use that a continuous function on a compact set attains its inf

ACuriousMind

@0celo7 Well, my problem isn't whether the inf is attained or not - it's whether it can be zero or not

0celo7

20:23

ok, suppose the inf is attained and is zero

that's a contradiction as it would mean that point is not in an open set

ACuriousMind

Ah, sure.

0celo7

but if the inf is not attained, all hell breaks loose

take a line, then some thin tube along it that shrinks down onto it

the inf is never attained but you clearly never have a tubular neighborhood

ACuriousMind

Yep, that's the issue

Isn't that already a counterexample to what you want to show?

Ah, no, the line isn't compact

0celo7

So we need: $\Delta(x)=\sup_{\delta\in\Bbb R}B_\delta(x)\subset U$

Then we need $\inf_{x\in K}\Delta(x)$

lol this is horrible

user218912

this problem set is asking to show like a billion things are or are not symmetric traceless and gauge invariant.

user218912

20:27

i have like 3 pages of equations

user218912

there has to be an easier way to do this without writing it all out every time

0celo7

@ACuriousMind ok playing 31 hours of New Vegas in 4 days was NOT GOOD

Obliv

weak. I could pull that off in 1 day. @0celo7

0celo7

maybe it was 3

but still

I have no clue how to do this problem

I don't know what to do

@WDUK Hello Will

WDUK

will?

who is will lol

0celo7

20:43

Is your name not Will?

WDUK

no lol

0celo7

oh, it was worth a shot

WDUK

haha

Obliv

@0celo7 I can solve it.

0celo7

go for it

WDUK

20:43

i came to ask what are good websites to read up on current research/developments in physics

0celo7

@WDUK the reference frame

WDUK

i want to read papers but they too complex for me at moment to understand

Obliv

@0celo7 a compact metric space is a space that is closed & bounded with a given metric right

the hell is an open cover

0celo7

no

user218912

@Obliv yes play 31 hours in a day.

0celo7

20:46

a metric space is always closed, trivially

but not bounded

bounded actually has no content for the most part

well

WDUK

is physorg a good website for it ? or is that bit lacklustre?

0celo7

in $\Bbb R^n$ it does

Obliv

what about the euclidean metric space @0celo7 that's not closed or bounded

0celo7

of course it's closed

user218912

@WDUK it's pretty good.

Obliv

20:49

oh damn mixed that up. closed means limits exist on the space right? thought it meant finite @0celo7

0celo7

no

closed means the complement is open

user218912

@Obliv why don't you learn this stuff before blindly attempting to do the problem without taking the class.

0celo7

this problem is pretty hard

@ACuriousMind aka Dr. AI doesn't seem to know it

user218912

ACM is a Dr.?

0celo7

yes

a Dr. of Love

user218912

20:51

oh

DHMO

1

Q: Exactly how does an oscillating electric field produce an oscillating magnetic field?

Let's say we have a capacitor which is connected to a sinusoidal voltage source, that means that the electric field within the capacitor is a sinusoidal function(assuming that the capacitor is a parallel plate capacitor ), but how does this model generate any magnetic field? If it does, then d...

electromagnetism visible-light electric-fields

Flemming's left hand rule?

Obliv

@bl00 I was joking, and no thanks. Have to learn PhD level calculus rn

0celo7

link.springer.com/book/10.1007%2F978-1-4613-0071-7

I should read this one day

user218912

@Obliv btw you should read those calc 3 notes I linked sircumference.

user218912

they're pretty good

Obliv

20:54

quantum calculus? didn't know that was a thing

0celo7

pretty much anything you can think of is a thing

user218912

proof?

user218912

:(

0celo7

that wasn't at you lol

women like wussi

that's the plural btw

Obliv

@0celo7 what do you think a tangent space looks like

can I see one in the night sky?

0celo7

20:58

it's a space of operators, no clue

@Obliv no

$T_pM=\mathrm{Der}\,C^\infty_p(M)$

you cannot see such a thing

user218912

therefore the tangent space doesn't exist because you can't see it. q.e.d.

0celo7

nice try

JD!!!!

flagging

Obliv

open cover is a partition of a topological space with each member being an open set

@0celo7 true or false?

0celo7

wait

@Obliv no

a partition implies no overlap

@ACuriousMind Ok pelase come back, I have an idea

Clearly $\overline{\mathcal B_K}$ is Kompact

So is $\partial\overline{B_K}$

So define $D:\partial\overline{\mathcal B_K}\to\Bbb R$ by $y\mapsto d(y,K)=\inf\{||y-v||\mid v\in K\}$.

This is continuous

And is never zero

So its inf, which it attains, is never zero

take the inf/2

USE THAT FOR THE BALL RADII

ahhhhh I think that works

Obliv

nvm

0celo7

21:09

@Obliv the diameter is the sup of $d(x,y)$ for $x,y\in X$

Obliv

yeah

0celo7

Ok, writing my proof now.

Obliv

good luck

0celo7

closed - open = closed, right?

Obliv

idk i would think only if the former is larger than the latter

0celo7

21:16

wait

this proof is wrong

wtf am I doing

ahhhhh

Obliv

@0celo7 having trouble proving your balls?

0celo7

They're too damn big

I need small balls for this

Obliv

and noone's in /36 wow that sucks.

0celo7

what?

Obliv

math chat

0celo7

21:21

Since $f$ is continuously differentiable on $U$, it is strongly differentiable on $U$. For each $a\in U$ we may find $r_a,c_a>0$ such that $x,y\in B(a,r_a)$ implies $||f(x)-f(y)||\ge c_a||x-y||$. We may form a covering $\{B(a,r_a)\mid a\in K\}$ of $K$. Since $K$ is compact, we may extract a finite subcover $\mathcal B=\{B(a_1,r_1),\dotsc, B(a_N,r_N))\}$. Let $\delta>0$ be a Lebesgue number of $\mathcal B$. Let $c=\min\{c_1,\dotsc, c_N\}$. If we take $x,y\in K$ with $||x-y||<\delta$, it is contained in some member of $\mathcal B$, $B(a_i,r_i)$. Then

ok, this MIGHT work

too damn hard lol

yeah it works

fuck yes

I just have to write it so it doesn't sound autistic

@Obliv Now that is an amazing proof honestly.

Obliv

@0celo7 that is so dense holy shit

0celo7

@Obliv I guarantee I'm overcomplicating this

But hey, I just proved one can oncstruct epsilon neighborhoods around compact sets inside of open sets

At least in Rn

Obliv

oh

0celo7

:32977562 the problem statement is way above

Obliv

what is epsilon in this problem? @0celo7

0celo7

21:31

I wrote it as era

Eta

Like I said, I need to fix the notation

I just wrote down what I had in mind and on my paper

The notation is not good or consistent

But the idea is there

Obliv

@0celo7 so what is $\partial \bar{B_K}$? guessing $B_K$ is a ball of metric $K$?

0celo7

No

I'll write it up later and explain

DanielSank

21:56

@DavidZ I do not think this is a physics question and I wonder why you think otherwise.

The question is literally "how do you define convergence of series of operators"?

0celo7

How the heck is that physics?

ACuriousMind

22:13

@DanielSank The question is much more subtle that you give it credit for (but I'll concede that the question doesn't do much to show that). Making the Fourier transforms/fields/mode expansions of QFT rigorous is much more difficult than defining the "convergence of a series of operators" and is a question that is specific to QFT

0celo7

@ACuriousMind I'm proud of this.

@Obliv ^

@ACuriousMind check it pretty please :)

user218912

@0celo7 is this for a class ur taking?

0celo7

yes

intro analysis 1

user218912

graduate intro analysis?

user218912

or undergrad?

0celo7

22:22

under

@bl00 2 days!!!

DanielSank

@ACuriousMind I don't believe you.

user218912

@0celo7 oh forgot about it.

user218912

skype me

0celo7

in class

do you want me to write it on Skype?

user218912

@0celo7 sure

user218912

22:36

btw how can I show that this really big tensor is gauge invariant without multiplying out everything?

0celo7

what really big tensor

ACuriousMind

@DanielSank There are many pitfalls in trying to make QFT rigorous. In this case the issue is that one might be tempted to just extend the idea of the Fourier transform to vector/Hilbert space valued fields, but there are issues with the boundedness of the field operators. The correct way (that is, the way currently favoured my most axiomatic treatments of QFT) out is the realize that quantum field need to be rigorously defined as operator-valued distributions rather than functions.

And then one simply shifts the Fourier transform onto the test function, and doesn't need to worry about convergence at all.

user218912

@0celo7 EM stress energy minus $F^{\sigma\mu}\partial_\sigma A^\nu$

0celo7

probably not then

user218912

probably not what?

0celo7

22:40

probably not gauge invariant

user218912

it is

0celo7

proof?

user218912

I swear I read it correctly this time.

user218912

says to prove it's gauge invariant

0celo7

hmm

let me check

user218912

22:42

$-F^{\mu\alpha}\partial^\nu A_\alpha + 1/4 g^{\mu\nu} F_{\epsilon\delta}F^{\epsilon\delta} - F^{\sigma\mu}\partial_\sigma A^\nu$

user218912

that is gauge invariant

user218912

I am in the process of proving it

user218912

but I have like 30 terms and it looks horrible

0celo7

you didn't tell me there were two terms

it's a 2-line calculation

user218912

what?

user218912

22:42

how

user218912

I know the middle term is gauge invariant

user218912

oh I'm retarded.

user218912

:(

user218912

I did the same dumb thing as last time

user218912

why would I write it all out when most of it is obviously gauge invariant?

user218912

22:47

so dumb

0celo7

23:07

ok

@Obliv now it is correct

DanielSank

23:21

@ACuriousMind How much does that really change the basic idea though.

My guess is basically "it doesn't".

And besides, you can define the distance between operators in terms of the test functions, right?

ACuriousMind

@DanielSank There's no good notion of distance on the space of tempered distributions (=distributions with a Fourier transform, really). There are several topologies you can choose on the space of distributions (be they scalar-valued or operator-valued) with differing notions of convergence. The one of use in QFT is often the so-called weak*-topology, which does not come from a notion of distance.

I think the topology induced by the notion of distance on the space of test functions would be the strong topology, in which many things crucially don't converge which we would like to be convergent.

You could also choose the norm topology, which is even worse

But the norm topology would be the only one which really defines a notion of distance between operators

DanielSank

Do all these non-convergences go away if we regularize the sum?

If so then I claim this entire issue to be a total red herring.

ACuriousMind

23:44

@DanielSank I'm not sure.

However, I don't see how the regularization would be permissible here - we're talking about Fourier transforming an arbitrary field, so you would have to worry about a) finding a regulator that respects all symmetries the field is supposed to transform under and b) showing that the results are independent of the regulator

Also, note that if you regularize the Fourier transform, it loses its unitarity property - Fourier transforming twice doesn't give you back the original function! Which mostly destroys the nice properties we want to use it for.

heather

Just watched the Cygnus launch live!

nasa.gov/launchschedule

Transcript for