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7:00 PM
holy crap the inf could be attained, what is life???
Samsung Galaxy Note 7 is officially declared dead :(
@ACuriousMind Argh! Wtf!?
If I have a compact set $K\subset U\subset\Bbb R^n$, $U$ open
@0celo7 what if the homogeneity of a linear map is greater than degree 1? does it still count as a linear map
I have an open cover $\mathcal B$ of $U$, then I get a finite subcover $\mathcal B_K$ on $K$
@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?
7:11 PM
I find a Lebesgue number of $\mathcal B_K$
stop asking me questions I'm mad
@ACuriousMind Is this also a Lebesgue number for the union of elements in $\mathcal B_K$?
It's probably not, but who knows?!
@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?
Noether energy momentum tensor is not symmetric, get out of here
I know it's not.
7:13 PM
I didn't say it was.
@bl00 Once again, you seem to be asking me whether a certain strategy will work instead of simply trying it.
I mean I tried it.
does it make sense though?
to do that?
@0celo7 I don't understand the question.
@bl00 Well, what did you find?
7:16 PM
@ACuriousMind clearly $\mathcal B_K$ is an open cover of $\bigcup_{B\in\mathcal B_K}B$, right?
@0celo7 Yes
is $\delta$ also a Lebesgue number for this cover?
Q: Why does my daily reputation not grow beyond 200?

THE LONE WOLF.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?

@0celo7 If the $\mathcal{B}$ consists of balls, then yes, otherwise no.
@ACuriousMind this is for the electromagnetic stress energy tensor.
7:20 PM
I found that the metric term becomes negative
@ACuriousMind I need to know why o.o
but idk about the field strength tensor term
it's balls, yes
@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.
@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?
7:26 PM
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.
Trivial has a different meaning in ACM's world...
@Obliv Are you talking about the sequence $a_n = \frac{1}{n}$ or about the series $a_n = \sum_{i=1}^n \frac{1}{i}$?
@ACuriousMind I have to show it.
I know the electromagnetic field tensor is antisymmetric.
7:29 PM
@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?
@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.
@Obliv the LoL guy is a weeb, too
Not really surprising
@0celo7 your roommate?
the one that brought in his overpriced razer gear? @0celo7
yeah not surprising.
@bl00 huh?
My roommate plays rugby, not LoL
7:35 PM
oh I know who you're referring to now.
@ACuriousMind I know that.
Hm, didn't crop right, sec
Fug, I see
@bl00 then what is there left to show?
@ACuriousMind ok but can I find a lebesgue number for the balls?
Maybe a different one
7:39 PM
@ACuriousMind I think I'm missing something, by what you said, it should be symmetric...
@0celo7 Well, the closure of their union is compact if they are finitely many, right?
the whole expression I mean
@bl00 Yes. The EM stress tensor is symmetric.
so I'm trying to prove it's antisymmetric?
@ACuriousMind sure
7:40 PM
I don't get it.
@bl00 wat
I constructed mine from the noether current for spacetime translations.
is that one also symmetric?
The EM field strength tensor is antisymmetric. The EM stress-energy tensor is symmetric. What is your problem?
yes I know that, the problem says to show it's antisymmetric.
7:42 PM
I re-read the problem 5 times, I'm 100% sure it says to show it's antisymmetric.
@ACuriousMind were you going somewhere with this?
@0celo7 Yes - can't you just apply the Lebesgue lemma to the closure of $\mathcal{B}_K$ to get that Lebesgue number?
what I got was $-F^{\mu\alpha}\partial^\nu A_\alpha + 1/4 g^{\mu\nu} F_{\epsilon\delta}F^{\epsilon\delta}$ @ACuriousMind
@ACuriousMind with what covering?
did I derive it right?
7:44 PM
I lied btw, I don't have a covering of $U$
@0celo7 Ehhh, damn :D
you're saying that is 100% symmetric? @ACuriousMind
@bl00 Oh, yes, the pure Noether tensor might not be symmetric.
okay how do I show it?
It would be interesting if the Noether tensor turned out to be antisym.
7:46 PM
I flipped the indices
Did you compute the Noether SET @bl00
@ACuriousMind This is the problem btw
the damn $T$ tensro
7:47 PM
what is it
what I said
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
I used the vector field $A_\alpha$ and the massless EM lagrangian. and plugged it into the noether current
7:48 PM
@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).
@ACuriousMind I guess that's what i'll do xD
So the whole expression isn't symmetric.
doesn't look antisym to me
@ACuriousMind So, I have a bunch of balls on $K$ for which the property holds
@ACuriousMind I'm writing it out, I'll lyk what I find.
it's not obvious to me without writing it out and simplifying
7:51 PM
i.e. for each $a\in K$ I can find $r,c>0$ s.t. $||x-y||<r\implies ||f(x)-f(y)||\ge c||x-y||$
nvm they don't have to be equivalent
@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
$\partial^\alpha \partial^\nu A_\alpha A^\mu$ is antisymmetric under exchange of $\mu, \nu$ right?
@ACuriousMind sound good?
@bl00 proof?
because if you change the indices it becomes a different thing.
7:56 PM
that's not what antisym means.
I know
but that's all I could think of
how do I prove that it's antisymmetric?
because it has to be
@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).
8:00 PM
then what makes that stress-energy tensor antisymmetric?
the second term is symmetric
the first term has to be antisymmetric then
Nothing. The Noetherian result simply is not symmetric.
so who is antisymmetric here?
but I need to show it
8:01 PM
If the problem states it's "antisymmetric" it's simply wrong, or you have read it wrong.
@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 $K$"?
given $U\supset K$, can I find $N$ inside $U$ that covers $K$ with "uniform thickness"?
@0celo7 I don't understand the question
@ACuriousMind ?
dkm please
8:09 PM
@ACuriousMind Given $\epsilon>0$, can I find an open set $N\supset K$, $N\subset U$ such that $x\in K, y\in U-K, d(x,y)<\delta\implies y\in N$.
I was reading not symmetric as anti-symmetric.
@bl00 You misread "not symmetric" as "anti-symmetric" 5 times???
@0celo7 IF your $\delta$ is supposed to be an $\epsilon$, then sure, just take $\left(\bigcup_{x\in K} B_\epsilon(x)\right) \cap U$.
8:14 PM
@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 $U$?
That's not my problem
but not equal to $U$
@0celo7 If $U$ is open and $K$ closed, then yes.
@ACuriousMind yeah $K$ is compact and $U$ is open
why can we do that?
8:19 PM
@ACuriousMind Is that an idk ugh or a idc ugh?
That's an "thinking about my idea I'm not sure anymore" ugh
@ACuriousMind I'm thinking it should be true if $K$ is compact
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
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
@0celo7 Well, my problem isn't whether the inf is attained or not - it's whether it can be zero or not
8:23 PM
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
Ah, sure.
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
Yep, that's the issue
Isn't that already a counterexample to what you want to show?
Ah, no, the line isn't compact
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
this problem set is asking to show like a billion things are or are not symmetric traceless and gauge invariant.
8:27 PM
i have like 3 pages of equations
there has to be an easier way to do this without writing it all out every time
@ACuriousMind ok playing 31 hours of New Vegas in 4 days was NOT GOOD
weak. I could pull that off in 1 day. @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
who is will lol
8:43 PM
Is your name not Will?
no lol
oh, it was worth a shot
@0celo7 I can solve it.
go for it
8:43 PM
i came to ask what are good websites to read up on current research/developments in physics
@WDUK the reference frame
i want to read papers but they too complex for me at moment to understand
@0celo7 a compact metric space is a space that is closed & bounded with a given metric right
the hell is an open cover
@Obliv yes play 31 hours in a day.
8:46 PM
a metric space is always closed, trivially
but not bounded
bounded actually has no content for the most part
is physorg a good website for it ? or is that bit lacklustre?
in $\Bbb R^n$ it does
what about the euclidean metric space @0celo7 that's not closed or bounded
of course it's closed
@WDUK it's pretty good.
8:49 PM
oh damn mixed that up. closed means limits exist on the space right? thought it meant finite @0celo7
closed means the complement is open
@Obliv why don't you learn this stuff before blindly attempting to do the problem without taking the class.
this problem is pretty hard
@ACuriousMind aka Dr. AI doesn't seem to know it
ACM is a Dr.?
a Dr. of Love
8:51 PM
Q: Exactly how does an oscillating electric field produce an oscillating magnetic field?

Shahe AnsarLet'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...

Flemming's left hand rule?
@bl00 I was joking, and no thanks. Have to learn PhD level calculus rn
@Obliv btw you should read those calc 3 notes I linked sircumference.
they're pretty good
8:54 PM
quantum calculus? didn't know that was a thing
pretty much anything you can think of is a thing
that wasn't at you lol
women like wussi
that's the plural btw
@0celo7 what do you think a tangent space looks like
can I see one in the night sky?
8:58 PM
it's a space of operators, no clue
@Obliv no
you cannot see such a thing
therefore the tangent space doesn't exist because you can't see it. q.e.d.
nice try
open cover is a partition of a topological space with each member being an open set
@0celo7 true or false?
@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
ahhhhh I think that works
9:09 PM
@Obliv the diameter is the sup of $d(x,y)$ for $x,y\in X$
Ok, writing my proof now.
good luck
closed - open = closed, right?
idk i would think only if the former is larger than the latter
9:16 PM
this proof is wrong
wtf am I doing
@0celo7 having trouble proving your balls?
They're too damn big
I need small balls for this
and noone's in /36 wow that sucks.
math chat
9:21 PM
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.
@0celo7 that is so dense holy shit
@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
:32977562 the problem statement is way above
what is epsilon in this problem? @0celo7
9:31 PM
I wrote it as era
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
@0celo7 so what is $\partial \bar{B_K}$? guessing $B_K$ is a ball of metric $K$?
I'll write it up later and explain
9:56 PM
@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"?
How the heck is that physics?
10:13 PM
@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
@ACuriousMind I'm proud of this.
@Obliv ^
@ACuriousMind check it pretty please :)
@0celo7 is this for a class ur taking?
intro analysis 1
graduate intro analysis?
or undergrad?
10:22 PM
@bl00 2 days!!!
@ACuriousMind I don't believe you.
@0celo7 oh forgot about it.
skype me
in class
do you want me to write it on Skype?
@0celo7 sure
10:36 PM
btw how can I show that this really big tensor is gauge invariant without multiplying out everything?
what really big tensor
@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.
@0celo7 EM stress energy minus $F^{\sigma\mu}\partial_\sigma A^\nu$
probably not then
probably not what?
10:40 PM
probably not gauge invariant
it is
I swear I read it correctly this time.
says to prove it's gauge invariant
let me check
10:42 PM
$-F^{\mu\alpha}\partial^\nu A_\alpha + 1/4 g^{\mu\nu} F_{\epsilon\delta}F^{\epsilon\delta} - F^{\sigma\mu}\partial_\sigma A^\nu$
that is gauge invariant
I am in the process of proving it
but I have like 30 terms and it looks horrible
you didn't tell me there were two terms
it's a 2-line calculation
10:42 PM
I know the middle term is gauge invariant
oh I'm retarded.
I did the same dumb thing as last time
why would I write it all out when most of it is obviously gauge invariant?
10:47 PM
so dumb
11:07 PM
@Obliv now it is correct
11:21 PM
@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?
@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
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.
11:44 PM
@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.
Just watched the Cygnus launch live!

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