Then you're probably trying the wrong thing

Who told you?

prof

@ACuriousMind He said: integrate against a bump function

@0celo7 Yeah, but you're probably trying the wrong thing with the "closeness", then. Or you mean something by "closeness" that I do not understand.

basically, I need to smooth out the coordinates of a timelike curve $x^i(t)$ to $\tilde x^i(t)$

so that $\dot{\tilde x^i}(t)$ is a timelike vector everywhere

and it is known that $x^i(t)$ is Lipschitz

well, what I have is that $$|\tilde x^i-x^i|(t)\le K\int_\mathbb{R}\phi(t-u)|u-t|\,\mathrm{d}u=K\int_\mathbb{R} |u|\phi(u)\,\mathrm{d}u$$

then if I pick the thingie with compact support on $[-1,1]$ I can get the further inequality

$$\le 2K\int_\mathbb{R} \phi(u)\,\mathrm{d}u$$

So by adjusting the overall scale of the bump function I can make the curves close.

@ACuriousMind Does that make sense?

Now what I want to do is calculate $\dot{\tilde x}$ so I can calculate $$g_{ij}\dot{\tilde x^i}\dot{\tilde x^j}$$

and see if I can make it negative

@0celo7 The inequality is rather obviously wrong, isn't it? Choose $\phi = 0$ and you get $\tilde{x} = x$, but $\tilde{x} = 0$.

@ACuriousMind Hrm, $|\int f|\le \int |f|$, right?

I assumed $\int_\mathbb{R}\phi=1$ in my derivation of that inequality, whoops

So I'm completely lost now

Good time to take a break :)

@0celo7 Yes

Because Cauchy Schwarz

Well this is pooey

Dunno what do do now

@ArtOfCode Huh?

Ok what if I mess around with the support of phi...

Now that might work!

We had some flags just now. I'm not going to sugar-coat this, we all know who I'm talking about. No matter if someone annoys you, no matter how much you disagree with them, you still gotta be nice to them.

we did?

If they're a problem, raise a custom moderator flag explaining the problem and someone will deal with it.

Huh

Wonder who got flagged and for what

You're also within your rights to ask a Physics mod to deal with it, if you think that would help.

Aaand I'm back, did you miss me? ;)

@ACuriousMind Do you think messing with the support of the bump function could let me sharpen that inequality?

I don't have the slightest idea

How are you? @ArtOfCode

@ACuriousMind According to my prof this is a trivial problem

He was shocked I couldn't do it myself :(

I think he has not yet realized that your knowledge is...unconventional :P

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ not bad, not bad

Remember the first month or so where I also was regularly astounded that you didn't know things one usually learns before one does the "advanced" math you do?

With unconventional knowledge comes unconventional skill :P

I can thouch the thing in the back of my throat with my tongue

Does that count

no

@ACuriousMind Turns out he's teaching my analysis class next semester

He's going to cover bump functions

@Slereah wow, Penrose is literally a pamphlet

@0celo7 I'm afraid you've been outrun

Somebody answered the question already!

oh, I forgot about it

I don't know if that's correct

@FenderLesPaul I hate GR

You should really see someone about these mood swings :P

@ACuriousMind what

I've never liked GR

what does "like" mean, anyway

none of this has any meaning

bah

you don't even have proof that was me

@ACuriousMind I really don't like GR, but I dislike it the least

Why...don't you do something you actually like then instead?

No one forces you to do physics at all

Or math, for that matter

I'm not doing physics

Mathy GR isn't really physics

bump functions are not physics

@ACuriousMind The only things I enjoy would be GDP raising

Also there's a difference between enjoying something and it being fun.

I don't follow

Why are you doing GR/math/whatever if you don't enjoy it?

And what's the difference between something being enjoyable and it being fun?

@ACuriousMind Enjoying something means enjoying the idea of it

GR is worthless, there's nothing to enjoy

Do you enjoy having fun?

define "enjoy"

fun is worthless

define "fun"

define "worth"

define "define"

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ Something is fun if I'd rather be doing it than nothing.

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ Fun is a primitive, you can't define emotions.

@ACuriousMind Of course I can.

@ACuriousMind positive operator on the GDP

Okay, are we being serious here or are we joking around?

@ACuriousMind Half.

I can go with both, I just need to know which it is :P

I think I like my definition of fun.

define "both" :P

@ACuriousMind @Slereah Penrose's smoothing proof: just smooth it out and check it with a reference metric

WTF

define "Sᴋᴜʟʟᴘᴇᴛʀᴏʟ"

Undefinable

0/0

$0/0 = \Bbb R$

Obviously

Since $\forall x . 0\times x = 0$

Yup.

Just solve $0/0=x$!

@ACuriousMind My algebra prof said you're wrong

:O

$ua=1\implies au=1$ even in a noncommutative ring.

He didn't know/want to do the proof, he'll give it to me on Friday

My master is not wrong

What master do you serve,dark minion

rigour

ACM is not that rigorous

-_-

what

@0celo7 Consider the ring of linear operators on real polynomials in one variable. Take $A$ as indefinite integration (with some fixed integration constant $c$) and $B$ as differentiation. Then $BA = 1$ but $AB \neq 1$ since $A$ does not recover the constant term correctly.

@ACuriousMind I'll give you his email

mb you can trade algebraic geometry tips or something

Hmm...I actually agree with you.

Well, we'll have to see what his proof is!

I think he was thinking it's true in all rings because it's true in the matrices

because he told me to look at the proof for matrices

Yeah, it's true for matrices because they're operators on finite-dimensional spaces. My counterexample relies on injectivity for an operator on the polynomials not implying surjectivity on them.

ok, ACM knows more algebra than a tenured algebra wizard

I give up

ACM is just too strong

I wonder if I can pick the coordinates in such a way that $g_{ij}(x(t))x^i(t)x^j(t)$ is negative on some small interval

OH

@ACuriousMind Is the derivative of a bump function necessarily negative at some point

Yes

Hmm, then what if I just try to calculate the Lorentz norm of $\tilde v^i=\int \phi'(t-u)x^i(u)\,\mathrm{d}u$...

That gives me...$$g_{ij}\tilde v^i\tilde v^j=\iint \phi'(t-u)\phi'(t-w)g_{ij}x^i(u)x^i(w)\,\mathrm{d}u\,\mathrm{d}w$$

well that's useless!

I don't think he thought this through properly

It's nontrivial

@ACuriousMind Does it seem nontrivial to you?

What does that mean

I'm asking you, on my hands and knees, for help

Do you not want to help me or actually don't know

Get up of your hands and knees

I really don't know anything about smoothing timelike $C^1$ curves or whatever you're trying to do

In my world, everything is always as smooth as it needs to be

@0celo7 you hate GR?

I don't know what I used to see in you

I think we're growing apart

@FenderLesPaul we're both cheating on each other

That's true

we're not the Underwoods

also true

@ACuriousMind I'm trying to prove that $p\ll q$ and $q\ll r$ implies $p\ll r$.

i.e. transitivity

I also don't know about or care for causal relations

But do to that I need to smooth the kink in the curve connecting $p$ and $q$ union the curve connecting $q$ and $r$

This has nothing to do with GR!!!!

I need to smooth a curve

I'd personally prefer to keep the kink in the curve

(I'm sorry.)

You're a German BDSM club manager.

That's natural.