@0celo7 Card(A) = Card(B)

I think [A] = [B] is also used

So after thinking on it a bit, the spacetime I have in mind has exactly 0 Killing vectors

It sounds challenging to solve PDEs on it

proof?

Well the Krasnikov tunnel is created at a time $t$, so that's time out of there

Ahhhh, I can take a rational sequence on the radial line

Since it's a finite length Krasnikov tunnel $z$ also goes out

@Slereah just because your coordinates are fucked doesn't mean there's no killing vectors

Obviously it is not symmetric in $\rho$

And since I have two of them that's rotational out

you wouldn't tell from $dr^2+r^2 d\theta^2$ that the plane has 3 of them

yeah but I use a thing called physical intuition tho

@Slereah but there might be some fucked up space-time one

I doubt it

It's basically Minkowski everywhere

I'm going to prove Rigorously that $\Bbb R^n$ is second countable.

So either I solve the wave equation on a spacetime with 0 Killing vector, or...

I have no idea really

What is the alternative

I always walk in on here straight out of context

I guess if I pretend that the transition is discontinuous, that could work out to a solution

But then again that means some deltas in the wave equation

Not sure it's a bright idea

Glorious.

Maybe I could try to do a time evolution from the partial Cauchy surface

I dunno

@ChrisWhite It does seem that the Ubuntu pdf viewer is the problem.

But it's hard to tell what's going on because as you say some readers may pick up slack.

That said:

If I build my document, the links do not work in the Ubuntu pdf viewer, but they do work if I view the pdf in Chrome.

At the same time, if I open the arxiv version in Chrome, the links do not work.

So I think this is some horrible arxiv issue.

Doesn't work for me in chrome

fucking LaTeX won't render matrices anymore for me

@0celo7 are you using amsmath?

\usepackage{amsmath}

it should work.

I am under an impression that orbital angular momentum has quite nothing to do with " orbital", since the only way we measure the eigenvalues of angular momentum is via magnetic field. The only thing seems related is if we find the position of a eletron of a H atom, it seems to be localed in some sort of (say) shells. Am I correct?

I mean in quantum mechanics

I think I get it now, the eigenfunction of orbital angular momentum can be written in terms of spherical coordinates, that's why it is called orbital angular momentum? (while spin can't)

@Shing it is called orbital since it's the equivalent of the angular momentum due to the electron orbitting, clasdically

@knzhou depends on the extent of the edit. The fact that it happens to change the answer from being incorrect to being correct doesn't really matter, for deciding whether to accept an edit. If it's wholesale replacement of the answer with a different one, definitely reject, but if it's just fixing a coefficient in a formula or something, that's usually fine.

@yuggib you around?

Hello everyone. I think I'm back

Welcome back, @Jim!

Where have you been?

@0celo7 yep

@Danu haven't had a job. Without one, I lack the focus necessary to contribute to this site

@Jim So you found one, now?

@Danu yup, I'm the new physics lab coordinator at university of toronto: Mississauga

:)

gotta say. I really missed all you guys

@yuggib ok say I have a function from $\Bbb R^0$ to $\Bbb R^n$, how do I define its derivative?

hmmm.... just noticed Kyle Kanos retired. That's too bad

@Jim hey

you've missed some stuff

@Jim D'aww haha

@KyleKanos, we remember your years of service well. And while your virtual body may no longer be with us, your spirit lives on.... In your actual body because, let's face it, you're not actually dead (as far as we know)

@0celo7 yeah, I'm catching up on all that now

that we know of

@0celo7 morbid but true

@yuggib I have reason to believe it should be possible to define

If we use a sequential definition it would seem it's identically zero

i.e. a sequence of difference quotients

but all of the difference quotients are zero

@0celo7 if you don't have a topology on $\mathbb{R}^0$, it's not possible

so, what's your definition of $\mathbb{R}^0$ and its topology?

@yuggib the singelton with the trivial topology, I suppose.

@0celo7 trivial is the only possible one so yeah

Also what is the norm on R0

@0celo7 then you have a problem for another reason, there is no group/monoid operation

to define the derivative with the limit

@yuggib what

you can't define $x+h$ in $\mathbb{R}^0$

because there is no closed addition operation on the set with one point

apart from the trivial one if you want

but then the derivative as usually defined is always zero

trivially

I doubt that there is a non-trivial derivative for such functions

also because the functions are rather trivial themselves

they are just identifications of points on $\mathbb{R}^n$ (i.e. you could label them with the points of $\mathbb{R}^n$)

@Jim Welcome back! We missed you too <3

@0celo7 The straightfoward way is to say that the tangent bundle of $\mathbb{R}^0$ is just $\mathbb{R}^0$ again, so there is only one possible derivative - the zero map.

@ACuriousMind what

@yuggib All I need is for the Jacobian to have rank less than $1$

But if I "define" it like that I won't be convinced I can do induction on it

@0celo7 the point is that it is essentially "silly" to make a big fuss to define the zero map

I need to do induction on $k$ in $f:\Bbb R^k\to\Bbb R^n$, starting at $k=0$

calling it derivative is not worth it

why?

@yuggib Sard's theorem.

can't you start at $k=1$?

I understand the proof completely except for this detail.

@yuggib No one has done that...

I wouldn't know how to do it, the proof is very complex.

@0celo7 still it's perfectly possible

@0celo7 Well, $\mathbb{R}^0$ is just a point, and the point is zero-dimensional - so it's tangent space is also the zero-dimensional vector space $\mathbb{R}^0$, so the tangent bundle is $\mathrm{pt}\times\mathbb{R}^0$. The derivative of $f: \mathrm{pt}\to M$ is a linear map $\mathrm{D}f: \mathbb{R}^0\to T_{f(\mathrm{pt})} M$, and there is only one such linear map - the zero map.

I mean, do you know it works for $k=1$?

@yuggib No.

You're supposed to "see" that it works for $k=0$ and then there's a nasty proof that the $k-1$ implies $k$

Took me a few hours to sort through the details

I don't see a direct proof of $k=1$ being feasible.

@ACuriousMind Ok, I guess this words.

(of course, $\mathbb{R}^0$ is the point because the empty product is the terminal object ;) )

^huh

still, it seems not the good way to do it

probably there is a reasonable direct proof it works for $k=1$, and then the induction works

@0celo7 Well, how would you have shown that $\mathbb{R}^0$ is the point?

Ok I've checked Milnor, GP, Lee, Sternberg

...let's see what Bredon says

@ACuriousMind that "huh" was "literally no clue what you just said"

> The proof will be by induction on n, and the case n= 0 is true.

>:(

T_T

@Qmechanic: Why was my flag on this post marked as declined if you deleted the answer?

then it seems it is done like that

@yuggib 5 topology books say it's trivially true for $n=0$

FIVE

Ok I know of a sixth

@ACuriousMind how is it you know category theory but not measure theory

I'm not sure how one even defines $M^0$ if not categorially

@0celo7 I mean, obviously it's true because it's trivial

I don't like a proof that works constructively from a trivial statement

@0celo7 Because I find one fascinating and the other not, simple as that

it means that the statement is essentially trivial by itself

Sard's theorem is hardly trivial...

@yuggib Well, the art of proving something is showing why it is trivial, is it not? ;)

@ACuriousMind : Dunno. I think @David Z handled your flag.

I really need to learn the Baire category theorem

@Jim : Ah, Jimself. Welcome back!

@ACuriousMind more or less ;-)

@yuggib Aha, Hirsch seems to prove it without induction.

@yuggib Hmm, it seems to be possible to argue Sard's theorem directly for $k<n$ by using measure theory and then using this special case to prove the general case.

@0celo7 I see

And I know it's true that the image of $f:\Bbb R^0\to\Bbb R^n$ has measure zero

always

...that might be all one needs, actually

by the way, @0celo7, @ACuriousMind , etc... nobody asked questions for the next AMA ;-P

@yuggib because none of us have a clue what you do

@ACuriousMind like to pretend, but let's get real

@0celo7 since it must be only a point, indeed

@0celo7 yes you do

@yuggib I'll ask my advisor when I meet him today

@yuggib well what do you do?

nothing that I would be able to ask questions about

@Qmechanic thanks, great to be back

@0celo7 it depends; you ask a lot of questions about mathematics...only it is the wrong one ;-P

@yuggib I'm stupid.

I proved the following before I proved sard:

> Let $U\subset \Bbb R^n$ be open and let $f:U\to\Bbb R^k$ be smooth with $n<k$. Then $f(U)$ has measure zero in $\Bbb R^m$.

> Proof. Define the map $F:U\times\Bbb R^{k-n}\to\Bbb R^k,(x,y)\mapsto f(x)$. This map is smooth and $F(U)=f(U)$. By Proposition 7.3, $U$ has measure zero in $\Bbb R^k$, so $F(U)=f(U)$ has measure zero in $\Bbb R^k$.

@ACuriousMind <3

@yuggib So I think one can use that to improve the proof...

Because unless I'm mistaken, in the proof all you need is that it's true for $n-1$ and that implies $n$

But I proved that directly there.

...wait how is this proof not trivial

I give up.

@yuggib I'm very interested in analysis but I'm bad at it

@0celo7 ask something about it then

;-P

@ACuriousMind I dismissed the flag as invalid because the post is an answer. So I thought, anyway. Apparently @Qmechanic disagrees.

@yuggib how do I get better

@0celo7 patience

how can I be patient when others are so much better

Is here any researcher?

@DavidZ Hm, ok.

@0celo7 others who?

@DeNiSkA researcher of what?

@yuggib ACM, etc.

@yuggib physics

@0celo7 that hardly matters to what you will be able to learn and do

@yuggib why do people always say that

@DeNiSkA physics is pretty vague, however count me out ;-P

it affects motivation

@0celo7 why? I am sure nobody was born knowing analysis, and therefore you have to be patient and learn it

it does not mean you'll never be good

but ACM was already good

by this time in his life

@DeNiSkA well, there are so many factors that may affect such choice

@0celo7 Why do you think you are "bad", and why would it matter if I was better?

@yuggib like?

@deniska read about both majors, what kind of life that will lead and come to a choice. You will have at least 1-2 years to decide (most credits will transfer if you decide to change majors)

You're struggling with the proofs of rather hard theorems. That's how one learns in math.

@DeNiSkA patience, willingness to do research/teaching, interest in a more applied job,...

The most important thing is to not have skewed expectations for either side.

@ACuriousMind but you struggled with them earlier

@0celo7 it's neither a race nor a contest

@yuggib easy for you to say

@0celo7 I learned math rather late

Me, too :D

@0celo7 That's both false and irrelevant.

I don't believe you

About 1.5 years after I joined SE, I think.

@0celo7 Hint: He didn't struggle with them ;)

@yuggib well i like research and even teaching, but money drags me to engineering

@Danu ok so he learned them earlier and didn't struggle

this place is really helpful but destroys any confidence I had

@0celo7 hey is noether's theorem taught in most lagrangian mech. texts?

why would you ask me

I'm not a physicist

@Obliv Yes

u read zee and he formulated lagrangian mech. from scratch, no?

@DeNiSkA surely the academic path is rather difficult to pursue nowadays

@danu thanks

@DeNiSkA so you need a lot of passion and commitment

@0celo7 For a troll you're surprisingly easy to troll

@Danu I'm not a troll...

@0celo7 and that's not what your advisor said 8)

@yuggib i am landing on so many '????'

@Obliv He's German, something probably got lost in translation.

@ACuriousMind Physiker can translate to engineer, right

@0celo7 Wouldn't say so, no. Ingenieur = engineer, AFAIK (that's how it is in Dutch, at least)

I know the German word for engineer

I could never spell it though

Damn French

du bist ein physiker! deutch sounds so cool

du bist ein blödkopf

ikr

whats a blodkopf

@0celo7 of course germans cannot use the accents properly...

@Obliv dunno

@0celo7 No, Danu's right, that would be Ingenieur exclusively.

@DeNiSkA I did a PhD intending to go into academia then I got a job as an industrial scientist instead, which was fun and paid well. So you could go do your degree then PhD then decide what you want to do.

"Physiker" and "physicist" are really one-to-one matches to me

@0celo7 "Sie sind" - be polite now :-)

@JohnRennie Sie sind alt

Better?

@Obliv Literally "a silly head".

It's a mild insult.

sounds much more insulting than silly head, but okay :p

maybe it's because i yell it in a really deep rugged voice in my head

I don't know of any good German insults

My go-to was pretty nuts though

@0celo7 dummkopf means dumb head i think

@0celo7 mittleren Alters :-)

I know

@JohnRennie You're beyond that if your smilies have unironical noses.

What's an ironical nose?

@ACuriousMind god you're old

@0celo7 my (14 year old) niece criticises my text messaging style on similar grounds

ironical nose -> <|:^)

with a party hat

^

or a nose after a good zinger

you old people don't get it

i don't think ACM is that old, is he?

23

that's like 5 years older than us @0celo7 he's still a 90's kid

@JohnRennie "what you want to do" this is the toughest thing to do! And teacher is the least reputed job in my nation :''''(

@Obliv he's said many times he's out of touch

@JohnRennie Good to know @0celo7 shares the aesthetics of a 14 year old girl ;)

you forgot the ironical nose @acuriousmind

@0celo7 Yeah, I don't understand what you young whippersnappers are up to today

@Obliv My smileys never had and never will have noses. :P

lol @0celo7

8^(

@ACuriousMind what

do they even have yoga pants in Germany

why did you remove that?

It would look slightly obscene if I wore yoga pants @Obliv

alright we're done with that.

No it would not

@DeNiSkA Well what did you want to do when you were eleven years old? If you wanted to be a scientist fulfill that dream and do a science degree. If you wanted to be an enginerr then fulfill that dream and do an engineering degree.

I could try on a pair of my gf's and send you a pic @Obliv

But she's like half my size, it would be doubly obscene

@0celo7 as tempting as that offer may be, I would prefer to not have to bleach my eyes. Thanks for the offer though.

np

@JohnRennie I don't think dreams of a 11-year-old are a good thing to go by :P

@0celo7 How are you going to explain what you need that for to her?

@ACuriousMind Well I guess my options are to disclose the existence of 0celo7 or say I have a yoga pants fetish

That's a sticky wicket

@Danu wong, wrong wrong. At age eleven I was desparate to go to Cambridge and become a scientist. And that's what I did and it was the best six (degree + PhD) years of my life.

I had no idea at all what I wanted to become when I was 11 years old I think. I know I wanted to become a garbage man earlier :D (Don't ask me why, I have no real clue)

I think I...liked their trucks?

@johnRennie I didn't have any such aspirations at age 11. I developed them in my senior year of high school :p it varies from person to person I guess.

omg me too @acuriousmind

it was the trucks.

@JohnRennie hmm, my society right from the age of 10 wants a boy to become an engineer, so i think my life upto 15 was full of engineering, but now at 16 i have confusion

@JohnRennie Lucky you :P

I wasn't interested in physics or mathematics much, at that age.

@DeNiSkA if you're only 16 you don't have to decide for at least a year, and both physics and engineering require the same school exams. So have fun for a year then worry about it.

@Danu I was a nerd before nerds existed :-)

@Obliv I wanted to be a fireman when I was very young. But I guess that's fairly standard for boys.

@JohnRennie hmm, that sounds correct that both requires same exams, so i think i should ask this question again after 6 months

I think I'm doing this wrong. If I'm to prove that if $|x| = n < \infty$ then $|x^a| = \frac{n}{(n,a)}$ so I said let $|x^a| = j < \infty$ then $(x^a)^j = x^n = 1$ so $aj = n$ but this shows $a$ is a multiple of $n$.

@MAFIA36790 BTW, what did you pursue?

@DeNiSkA that's what I would do. And in the mean time do lots of physics and maths and see how much you enjoy it. You might decide you prefer engineering after all.

good advice :-)

oh cool !!!

@MAFIA36790 Welcome to south India?

@MAFIA36790 guilty as charged. At that age I was designing spaceships!

@MAFIA36790 actually I only learned about GR in my 40s, after I quit my job as a scientist to become a computer nerd. A shame really as I think the basic principles are simple enough to be learned when I was a lot younger.

@JohnRennie "computer nerd" are you computer science engineer?

But then I think the books available in the 70s weren't as good as the ones available now. These days authors seem to go to a lot of trouble to make their books accessible. Back in the 70s it was more elitist. Authors seemed to take the view that if your were too stupid to understand their books you shouldn't even be trying.

@DeNiSkA no, I went into server and network management. That was really good fun as well. I still do it part time now.

@Danu I wish you explained that troll comment

Sure I kid around sometimes but I don't try to infuriate people

@MAFIA36790 I'm doing well. no meds. still have a tube in my lung.

I barely feel it though

@JohnRennie "I quit my job as a scientist" seems like getting jobs in UK is quite easy?

@Danu What's the definition of a cohomology with two "arguments" like $H^n(T^n,T^n\setminus\text{pt})$? I mean I know the basics of (co-)homology like de Rham cohomology, but there it's only one space in the brackets, like $H^n(T^m)$.

@Bass It's relative homology

I'd collapse my lung 5 more times than deal with a broken heart. ;'(

Thanks ACM.

@3075 careful what you wish for

@DeNiSkA in the UK all companies are desparate for good staff. In the IT world, which is what I know best, getting really good staff (without having to pay a fortune) is exceedingly difficult. If you're good there are plenty of jobs available.

@Obliv 5 collapsed lungs for her, I'd do anyday.

even 7.

@3075 I don't think sacrificing lungs for a girl is very romantic. In any case, you'll feel better with time. Try not to think about her too much, it'll be more painful this way.

@3075 why don't you make a move

@johnR have you any idea what would happen physics.stackexchange.com/questions/214/… if you tried 'rolling' at the end of the fall after jumping? Kind of like how people that do parkour roll after high jumps to reduce damage done to the body?

@Obliv I suspect the parkour roll is more for show than anything else. The idea is to spread the acceleration over a longer distance to reduce the magnitude of the acceleration. But at best the distance available to you is only your height, and in a falling lift I doubt that's enough to make a difference.

Indeed, relative, @Bass. Note that the particular example you gave is the "localized" cohomology group.

By excision, it senses only the behavior at the point

All manifolds of fixed dimension have the same localized homology

(And it doesn't depend on the point either)

To see this, use the homeo. to Euclidean space.

@Danu That's sufficiently vague it covers 90% of manifold theory proofs ;P

No, this follows directly, after using the homology of spheres

But I see your point ;)

hi guys. What is the definition of dijet transverse momentum and dijet invariant mass?

@Obliv lol I know I was only comparing it to something that I could relate to.

@0celo7 idk how. :(

@3075 ask her to see a movie???

@0celo7 he said she has a boyfriend now, + they're going to different uni's in the fall so I don't think there's many moves to make.

@Obliv oh right

basically 1 more day left.

@3075 you could always ask her if she is interested in becoming a Mormon :D sorry

@Obliv why would I ask her that?

they're okay with multiple spouses I think

@Danu how come we don't hear about your love life anymore

@Obliv I think that's multiple wives per man, not multiple husbands per wife.

@johnR tomato tomato

that doesn't really work in text

I could ask to go to the library?