The h Bar

12:00 AM

@peterh Yeah, I've done my reading on them already. I really wish they sold domains on their TLD, meur.er would be an awesome site

peterh

@BernardMeurer For that you are probably not enough rich. What about info? There is a meurer.info, but it is used by xing.com/profile/Harald_Meurer

Bernard Meurer

nah, .info is lame

peterh

And the good old .org?

Bernard Meurer

.xyz, .com, .org. .io are nice

.org is being used

peterh

I know quite... interesting site on .info

uhm, sad

Bernard Meurer

12:05 AM

all it has is a dumb site with almost nothing in it

I was considering .io and .cc

peterh

.io ! .xyz looks like the trashbox of the net. But .io is wonderful.

user218912

12:27 AM

@0celo7 hi

0celo7

@IceLord Busy

user218912

not asking for help...

0celo7

@IceLord I need like seven books.

I have a problem...

user218912

alright

user218912

i'll buy you one of them

user218912

12:41 AM

if you help me

0celo7

not now

I have a really hard analysis problem...

user218912

can you just tell me what $\frac{\partial A_\mu}{\partial A_\nu}$ is?

user218912

ACM won't tell me.

user218912

:(

0celo7

why won't he?

user218912

12:42 AM

i couldn't figure it out all day

0celo7

b/c it's trivial?

user218912

because he says i should figure it out

user218912

yes

0celo7

...are you bribing me

$E\subset\Bbb R^n$ linear subspace, $\Bbb R^n-E$ connected $\Leftrightarrow\dim E\le n-2$.

Solve that, then sure.

user218912

:|

user218912

12:44 AM

nah i'm a physics guy

user218912

but i'm bad at physics...

user218912

so help me plz

0celo7

the proof is approaching a page!

that's never good

user218912

why? aren't a lot of proofs like 500 pages long?

0celo7

no proof is 500 pages long

a homework proof that is a page is pretty long

user218912

12:46 AM

@0celo7 en.wikipedia.org/wiki/…

user218912

:o

user218912

@0celo7 anyway are you going to tell me what it is please?

user218912

if you don't I'll have to ask my TA and look like a dumbass in front of him

0celo7

what are your ideas on it

I figured it out before coming to PSE

else I'd have some sympathy

user218912

no idea.

user218912

12:57 AM

since these are the same vector fields just with different indices

user218912

so i don't know how to deal with that

0celo7

I don't know if I should help.

I generally trust @ACuriousMind

user218912

ahhhh

user218912

i'll figure it out eventually but i'll be wasting so much time

user218912

so please tell me

0celo7

1:00 AM

what is $\partial x/\partial x$

user218912

$x$ is a vector?

0celo7

$x$ is a number

wait

what do you think $A^\mu$ is

user218912

a vector field

0celo7

no

fucking physicists

$A^\mu$ is a number

user218912

;-;

user218912

1:02 AM

okay

0celo7

it's $A^\mu e_\mu$ that's a vector field

user218912

so then it's just 1?

0celo7

@IceLord What is?

user218912

$\partial x/ \partial x$

user218912

and same for the "vector field"

user218912

1:03 AM

am I being stupid

0celo7

write it out.

@IceLord a little, write it out.

DanielSank

@0celo7 Yes well as I recall your institution has some issues with that.

0celo7

Proof:

($\Leftarrow$) We will show $\Bbb R^n-E$ is in fact path connected. It is sufficient to show this for $\dim E=n-2$, for any vector subspace of dimension $n-3$ or less can be realized as a subspace of an $n-2$ dimensional subspace. Let $p,q\in \Bbb R^n-E$. Let $L\subset\Bbb R^n$ be the straight line from $p$ to $q$. (We also use the notation $L=\overline{pq}$.) We set up coordinates $x^1,\dotsc, x^n$ with the origin on $E$ and with $E$ spanning the first $n-2$ coordinates, that is, $E$ is the level set $x^{n-1}=x^n=0$. Since $p,q\notin E$, at least one of the last two coordinates of $p$ and …

@DanielSank this was to people in my department!

DanielSank

@0celo7 Yes,

your institution has issues.

user218912

$\frac{\partial}{\partial A_\alpha}\big[ \frac{m^2}{2}\eta^{\omega\xi} [A_\omega A_\xi ]\big] = m^2$ ? @0celo7

0celo7

1:12 AM

Christ

No, what is $\partial A_\mu/\partial A_\nu$

@DanielSank one of the profs will think I'm crazy if he gets the email late

I discussed with him what I said in the email this morning

user218912

@0celo7 $1$?

0celo7

NO

he probably got it a few hours ago...

user218912

fml

0celo7

@IceLord what is $\partial x^i/\partial x^j$

you'd better know this

Zee explains this

user218912

in GR or QFT?

0celo7

1:15 AM

No. Figure it out yourself.

You're not learning anything if you look this up

user218912

i don't even know what to look up

user218912

if i did I wouldn't be asking for help xD

user218912

this looks really familiar

user218912

i probably learned it before

0celo7

@DanielSank do you know how to help him without giving it completely away?

user218912

1:21 AM

or... how about you just tell me and tell me the reason and i'll learn

user218912

so next time i won't have this issue

0celo7

@IceLord let's think concretely, $\Bbb R^2$ with coordinates $(x,y)$

@IceLord Because as much as I hate to admit it, pain is best way to learn.

@IceLord What is $\partial y/\partial x$ in these coordinates?

DanielSank

@0celo7 What's the question?

Hi, @Milou.

0celo7

@DanielSank What is $\partial A_\mu/\partial A_\nu$?

DanielSank

@0celo7 What's $A$?

0celo7

1:24 AM

@DanielSank Vector potential

user218912

@0celo7 just the definition of the partial derivative.

DanielSank

Haha, ok then I have no idea. Notation unclear (keep in mind I don't do this stuff every day).

user218912

what do you mean what is it?

0celo7

@IceLord Compute it!

user218912

for what?

user218912

1:25 AM

you didn't give me $y$

0celo7

@DanielSank How about: coordinates $x^i$ on $\Bbb R^n$, what is $\partial x^i/\partial x^j$?

DanielSank

@0celo7 Also, I'm somewhat reluctant to spend time helping someone with QFT when they haven't gone through quantum yet.

0celo7

@IceLord oh my god

DanielSank

@0celo7 Bad physicist notation.

That's what it is.

;)

@0celo7 Dude, be nice.

0celo7

@DanielSank ...as your mathematical overlord I declare that as perfectly good notation.

DanielSank

1:26 AM

Seriously. If you're getting frustrated just stop. Don't put people down, even implicitly.

0celo7

What's wrong with it?

(the notation, not my attitude)

DanielSank

@0celo7 Are $x^i$ functions?

0celo7

@DanielSank Yes

The coordinate functions

DanielSank

So then you're taking the derivative of a function with respect to another function?

Your argument is invalid.

0celo7

yes

nope

DanielSank

1:27 AM

I have no idea what that means.

Enlighten me.

(FYI I was planning to write a blog post about this issue, that physicists confuse variables and functions all the time)

0celo7

@DanielSank First, can I point out you do that every time you do the chain rule.

@DanielSank Also how're you gonna tell me you don't understand this. The Euler-Lagrange equation are literally derivatives wrt. functions.

DanielSank

@0celo7 False.

0celo7

$\partial L/\partial q$ is the derivative wrt. the function $q=q(t)$.

DanielSank

@0celo7 That's rather different.

Oh that.

0celo7

@DanielSank This is exactly what we're doing here

DanielSank

1:29 AM

I get very confused when I actually try to think about what that notation means.

0celo7

@DanielSank Do you want me to explain?

DanielSank

The derivative of a one variable function is obvious.

user218912

@DanielSank that doesn't make any sense. if I was good with quantum and index notation I wouldn't need help...

0celo7

Yes

user218912

I'm asking for help because I'm lacking...

DanielSank

1:30 AM

The derivative of a multivariable function is a linear transformation.

Milou

@DanielSank hi! I'm sorry late to reply, but got distracted

DanielSank

@0celo7 I think of a function $f: \mathbb{R}^n \rightarrow \mathbb{R}^m$.

The derivative, $Df$, is a linear transformation.

It's matrix representation is an $m \times n$ matrix.

0celo7

@DanielSank Sure

DanielSank

Ok, here's the thing: to me, linear transformations exist apart from any basis.

(You've explained this to me before, but I always forget)

(I have to rederive this argument every time)

Now, we can invent coordinate functions $x^i$.

user218912

@0celo7 what's wrong with this?

0celo7

1:33 AM

@IceLord I gave you $y$

DanielSank

$x^i: \mathbb{R}^n \rightarrow \mathbb{R}$, right?

0celo7

You have a bad habit of not reading...

@DanielSank yes.

DanielSank

Oh good, I'm retaining information. That's a good sign.

user218912

@0celo7 $(1, 0)$?

Bernard Meurer

How does one prove that $\mathbb{R}$ is infinite?

DanielSank

1:34 AM

Ok so for a point $p$ in $\mathbb{R}^n$, we get some values $x^i(p)$.

0celo7

@BernardMeurer $x\in \Bbb R\implies x+1\in\Bbb R$.

DanielSank

Those are the coordinates of the point $p$.

Bernard Meurer

@0celo7 Ah, because it's inductive

makes sense

0celo7

@BernardMeurer Don't know what that means, but if you do, ok.

@DanielSank Yes

DanielSank

I suppose we can say something like $f(p) = f(x^1(p), x^2(p),\ldots,x^n(p))$ or some such nonsense.

0celo7

1:36 AM

Yes

user218912

is this discussion related to my problem?

0celo7

@DanielSank I think you're making this too hard. You're viewing $\Bbb R^n$ as a manifold.

DanielSank

This is slightly annoying notation to me, but fine, whatever. Now $f$ means a function of a point or a function of a bunch of numbers.

@0celo7 You're right. Should I not do that?

Bernard Meurer

How does one define a function $c(\mathbb{L})$ that yields the number of elements in $\mathbb{L}$? Would a sum work?

0celo7

If you view $\Bbb R^n$ as a vector space (as one does in analysis), then $p=(x^1,\dotsc, x^n)$ literally. By definition of $\Bbb R^n$.

Bernard Meurer

1:37 AM

$\mathbb{L}$ being some set

DanielSank

@0celo7 Yeah ok fine.

0celo7

@IceLord Maybe.

@IceLord If I tell you what it is will you figure it out?

DanielSank

@IceLord It will be. @0celo7 is educating me.

0celo7

@BernardMeurer No.

DanielSank

Ok, @0celo7 I've got $f$ and it eats $n$ real numbers and poops out $m$ real numbers.

Very good.

Now what?

0celo7

1:38 AM

@BernardMeurer Do you have to construct this function?

Also what is $\mathbb L$?

Bernard Meurer

@0celo7 Yes, I'm curious to know now

0celo7

You're in the "let's use weird notation because it looks cool" phase.

A set is $S$. Or $X$. Not $\mathbb L$.

Bernard Meurer

$\mathbb{L}$ is some set, can be anything

Okay then

$X$

0celo7

@BernardMeurer There are some standard notations for things

Fields are $k$ or $\Bbb K$

(for instance)

if you say "let $\mathfrak Z$ be a field" people will look at you weird

@BernardMeurer Ok, is your set $X$ finite or infinite? Or are you asking how to tell?

If it's infinite, then $|X|$ (standard notation for your $c$) is not easy to define.

@DanielSank What are we trying to do again?

user218912

@DanielSank him saying things like that doesn't put me down, at least he is helping me. your ideology that I shouldn't be helped because I'm lacking in an area is what puts me down.

DanielSank

1:43 AM

@0celo7 Understand what $\partial A^\nu / \partial A^\mu$ means.

0celo7

@BernardMeurer A set $A\subset\Bbb N$ is said to be finite if $A$ has an element $a\in A$ such that $a\ge b$ for any $b\in A$.

Make sense?

Bernard Meurer

@0celo7 $X$ is some set, I can't tell whether it's infinite or not, I'm looking for a general solution

0celo7

@BernardMeurer There isn't a general solution.

DanielSank

@IceLord Dude, I think it's silly to spend time helping you with QFT when you haven't gone through quantum! That's not putting you down, it's trying to help.

Disagree all you want, of course.

Bernard Meurer

What's $B$?

0celo7

1:44 AM

A typo!

Bernard Meurer

Ah :)

user218912

@DanielSank well I told you I'm in the course and I can't drop anymore.

user218912

so you saying that is not helping.

Bernard Meurer

Thinking

0celo7

@DanielSank Ah, well. One has to be somewhat pragmatic with that one.

Not sure if that's the right word.

It's $\delta_{\mu\nu}$.

Bernard Meurer

1:45 AM

finite?

DanielSank

@0celo7 I just want to understand what the notation means.

@0celo7 Yes yes, but why?

user218912

@0celo7 I just found that.

user218912

why is it that.

DanielSank

A wild typo appears. @BernardMeurer tries logic! It's not very effective.

0celo7

@DanielSank Because $\partial A^\mu/\partial A^\mu=1$ (no sum) by definition of the derivative, and $A^\nu$ does not depend on $A^\mu$ for $\nu\ne \mu$.

This can be made rigorous.

user218912

1:46 AM

@0celo7 DUDE

user218912

this is what I said

Bernard Meurer

@DanielSank lol

user218912

but you said

0celo7

But it involves vector bundles.

user218912

the last term is not 0

user218912

1:46 AM

so I thought i was wrong

Bernard Meurer

@0celo7 So did you really mean finite?

0celo7

@IceLord It's not

It's $\delta_{\mu\nu}$ contracted with stuff

user218912

oh...

0celo7

@BernardMeurer A set is finite if it has a maximal element, yes I think I agree with that.

user218912

right

Bernard Meurer

1:47 AM

OOOH

Okay

that makes perfect sense!

0celo7

@BernardMeurer WAIT

Now let's look at $X$

Bernard Meurer

Wait

what about a minimal element?

0celo7

We say that $X$ is finite if there is a bijection $X\to A\subset\Bbb N$ for some finite $A$.

@BernardMeurer Axiom of the natural numbers is that any nonempty set has a minimal element.

Bernard Meurer

bijection?

Oh, we're dealing with $\mathbb{N}$, forgot that

0celo7

@BernardMeurer One-one, onto.

@DanielSank Ok, I will try to make it rigorous

Bernard Meurer

1:50 AM

$\mathbb{N}$ is the intersection of all inductive subsets of $\mathbb{R}$ right?

0celo7

When we're doing field theory on a manifold $M$, we're really taking fields to have values in a vector bundle $E\to M$.

We can do calculus on $E$.

user218912

@0celo7 so does it become $\frac{m^2}{2} \eta^{\omega\xi} \delta_{\omega\alpha} \delta_{\xi\alpha}$?

0celo7

@DanielSank That's why we need covariant derivatives for field theory on curved spacetime.

So $\partial A^\mu/\partial A^\nu$ means we're doing $\partial x^\mu/\partial x^\nu$ on the vector bundle

The coordinates $x^\mu$ are physically identified with the field values $A^\mu$

@IceLord perhaps with a factor of 2, not sure.

I haven't worked it out

@BernardMeurer No clue, I haven't taken PhD set theory.

user218912

@0celo7 why a 2?

Bernard Meurer

This isn't PhD stuff dude, it's just analysis

0celo7

1:52 AM

@BernardMeurer What does "inductive" mean?

@BernardMeurer It's not analysis.

@IceLord product rule

user218912

so then there would have to be an $A$ too right?

Bernard Meurer

A set is said to be inductive if $x\in X\Rightarrow x+1\in X$

0celo7

I think that $\frac{1}{2}$ needs to get canceled.

@BernardMeurer What? That last $\in$ makes no sense.

user218912

I knew it

user218912

10 hours ago, by IceLord

@ACuriousMind do I also use product rule?

Bernard Meurer

1:54 AM

I think that's the definition of an inductive set, my notes are far

user218912

but I thought ACM said no.

Bernard Meurer

X is said to be inductive if for all x in X, x+1 is in X

0celo7

@IceLord Proof?

@BernardMeurer x contains zero?

I think you've got at least one typo.

Bernard Meurer

X

NO

Wait

user218912

@0celo7 he said use whatever derivative rule or something like that.

user218912

1:55 AM

thought that implied it's not product rule.

0celo7

Uh, $\cup$ is not the notation for "and"

Bernard Meurer

the 0 thing doesn't need at all

0celo7

@IceLord ACM does not know the American word for product rule.

user218912

it's or.

0celo7

He knows it as Leibnitz rule.

user218912

1:56 AM

oh ok lol

0celo7

@IceLord No, that's $\vee$.

And is $\land$.

@BernardMeurer hmm?

user218912

that's in logic language.

0celo7

Then I can produce an induct set $\{2,3,4,\dotsc\}\ne\Bbb N$.

So your claim is wrong

user218912

@0celo7 why is it product rule because $A_\mu$ is just a number as you said

0celo7

@BernardMeurer Please tell me you're not including $0$ in $\Bbb N$.

Bernard Meurer

1:57 AM

It's just $$\text{A set is said to be inductive when }\forall x\in X\Rightarrow x+1\in X$$

0celo7

@IceLord Product rule works for "just numbers"

user218912

true

0celo7

@BernardMeurer Then $\Bbb N$ is certainly not the intersection of all inductive subsets of $\Bbb R$.

user218912

idk what i'm saying

Bernard Meurer

That doesn't prove my claim wrong?

0celo7

1:58 AM

@BernardMeurer Don't use logical quantifiers like $\forall$ when typing a simple definition.

They're perhaps acceptable in long, tedious computations, but it's always best to type out the words.

Bernard Meurer

1

Q: $\mathbb N$ as the intersection of all inductive subsets of $\mathbb R$

I read in an undergraduate real analysis textbook that the set of the natural numbers $\mathbb N$ is defined as the intersection of all inductive subsets of $\mathbb R$. However, I'm having trouble understanding what this means, specifically what an inductive subset of $\mathbb R$ is referring t...

elementary-number-theory

0celo7

Clearly you're using a different definition of inductive...

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