The h Bar

@0celo7 I'm uncertain. If I was writing a question of that kind I would either want the whole normal force acting on the thrower or would ask explicitly about the change of the normal force acting on the thrower.

Unfortunately I don't know that the language there is universally understood to have a unique meaning.

0celo7

@dmckee someone asked one of the profs on Blackboard and he clarified as normal force

@ACuriousMind are all linear transformations invertible?

mapping from n dimensions to n

ACuriousMind

12:55 AM

What have you tried in finding the answer to that question? :P

Because there is a completely equivalent question I'm quite sure you know the answer to

0celo7

uh

I don't really know how to approach it

I told you, my linear algebra class is a joke

ACuriousMind

Do you know that there is a bijection between linear maps $\mathbb{R}^n\to\mathbb{R}^m$ and $n\times m$ matrices?

I.e. every matrix defines a linear map and every kinear map has a matrix?

0celo7

yes, yes, of course

that we did cover

ACuriousMind

Are all $n\times n$ matrices invertible?

0celo7

::smashes head against keyboard::

uh, just forget I asked that

ACuriousMind

12:58 AM

Told you you knew the answer :P

0celo7

ok, so in the "all inner product spaces are isomorphic" proof

we set $\phi(e_i)$ equal to the vector in R^n with 1 in the ith slot, right?

how does one show that $\phi$ has an inverse

ACuriousMind

You, uh, just define it in the same manner?

0celo7

what guarantees that the matrix of $\phi$ is invertible

ACuriousMind

@0celo7 Why are you worrying about that? Just define the inverse $\psi$ by $\psi(x_i) = e_i$, where I denoted by $x_i$ the vector with a one in the ith slot.

Now it's evident that $\psi\circ\phi$ and $\phi\circ\psi$ are both the identites, and thus $\phi$ is an isomorphism.

0celo7

I just am!

ACuriousMind

1:09 AM

@0celo7 Well then, strictly speaking, you don't have the notion of "invertible matrix" since the map is not from a vector space to itself.

And the "matrix" of $\phi$ is rather silly: In the natural way to write it down, it looks like the identity matrix (but it isn't, because it's between different spaces).

0celo7

alright, I get it

all inner product spaces are isomorphic, proof recorded in the holy notebook

what should I write down next?

ah, the unitary matrix thing

@ACuriousMind So I tried checking what matrix commutes with the complex structure

and I found it's just GL(n,R)

I think

I commuted a matrix with four blocks ABCD with the complex stucture and got A=B=D=-C

so that's one block dof, so unless the blocks are complex I got GL(n,R)

Oct 21 at 16:00, by ACuriousMind

As for why those are those are those who preserve the complex structure, just write the complex structure as $\left(\begin{matrix}0 & -E_n \\ E_n & 0\end{matrix}\right)$ and just examine which block matrices $\left(\begin{matrix} A & B \\ C & D\end{matrix}\right)$ with $A,B,C,D\in\mathrm{GL}(n,\mathbb{R})$ commute with it.

ACuriousMind

1:30 AM

@0celo7 I get $B=-C$ and $A=D$, so two $\mathrm{GL}(n,\mathbb{R})$ matrices.

0celo7

...

how did I get D=-C?

that's not even what I have written in my matrices?

smh

see what I mean D:

@ACuriousMind other problem, why do these have to be GL(n,R)?

isn't GL(n,R) invertible? it has to be by defintion of a group

Slereah

it is

0celo7

but I see no reason why A,B have to be invertible

ACuriousMind

@0celo7 Yes. Non-invertible matrices have no chance of preserving the structure

0celo7

@ACuriousMind ok, why?

@ACuriousMind the hermitian inner product defined under that exercise is not antilinear in one of its arguments, is that an issue?

ACuriousMind

1:53 AM

@0celo7 Uh. Thinking about it, we're actually only interested in the group of matrices preserving the structure. Sorry, of course non-invertible matrices can commute with the structure.

0celo7

@ACuriousMind ok, I figured something like that

ACuriousMind

@0celo7 Why would it be an issue?

0celo7

the unitary group preserves a Hermitian inner product

the one he defines isn't a hermitian inner product

it's not antilinear in an argument

ACuriousMind

Is it linear in that argument?

0celo7

it's linear in both

@ACuriousMind I can't get this to work with that antilinearity in the "standard" inner product!

@ACuriousMind yeah, I'm having trouble with this proof :(

ACuriousMind

2:12 AM

@0celo7 Did you remember that the complex scalar multiplication is defined as $(a+b\mathrm{i})x = ax + bJx$ for $J$ the complex structure?

0celo7

complex structure outta nowhere

uh, I did not

first I'll prove "all unitary groups are isomorphic"

then do that one

notation should be obvious: $$\langle x,y\rangle_V=\langle U_V,x,U_Vy\rangle_V=\langle\phi U_Vx,\phi U_Vy\rangle_W=\langle U_W\phi x,U_W\phi y\rangle_W$$

$\phi:V\to W$ was proven to exist earlier this evening

@ACuriousMind make sense?

so $U_V=\phi^{-1}U_W\phi$

user685252

in Mathematics, 8 mins ago, by Martin Sleziak

We are currently mitigating an attack. Standby for updates.

Tweeted by StackStatus on October 30, 2015 at 1:48 AM

0celo7

@ACuriousMind sorry, I don't see how that helps

my proof of "all unitary groups are isomorphic" never uses the properties of unitarity, antilinearity or hermiticity

ACuriousMind

@0celo7 Write the Hermitian product as $y^T x + \mathrm{i} y^T J x$. Now using $(a+b\mathrm{i})x = ax + bJx$ and $J^T = -J$, you can show it's antilinear in the first and linear in the second argument.

Also: We are under ATTACK?!

2

0celo7

@ACuriousMind ok

DanielSank

2:23 AM

@ACuriousMind Yeah.

Happens all the time.

0celo7

that's not clear at all from the defintion though

wait, if $Jx=\mathrm{i}x$, then $y^Tx-y^Tx=0$ o.o

ACuriousMind

@DanielSank First time I hear of it

DanielSank

@ACuriousMind I mean, websites are attacked all the time.

ACuriousMind

@0celo7 No, you are not allowed to do that. In $\mathrm{i}y^T J x$, the $y^T,J,x$ are treated as a real matrix and as real vectors. Inside that formula, complex scalar multiplication is not defined.

0celo7

o...k, so not sure what to do now

also my unitary isomorphic proof is flawed somewhere

I think...

I never use any of the properties of the complex inner product in the proof

all I use is the general invariance group defintion

ACuriousMind

2:29 AM

@DanielSank Hm, I guess so. Didn't notice anything odd on the site, though

@0celo7 That's because the same proof works to show that all groups that preserve the same kind of inner product are isomorphic, e.g. it also shows all orthogonal groups are the same.

0celo7

> it also shows all orthogonal groups are the same

yes, that's true

ACuriousMind

(of the same dimension)

0celo7

@ACuriousMind that's implied

so, about this antilinear thing

user685252

in English Language & Usage, 4 mins ago, by cornbread ninja 麵包忍者

We are back online and monitoring all systems.

Tweeted by StackStatus on October 30, 2015 at 2:23 AM

0celo7

@ACuriousMind ok, I need a bigger hint for this problem

I'm not sure what I can and can't do now

ACuriousMind

2:43 AM

Just compute $\langle (ax+bJx,y) \rangle$ and $\langle x,ay+bJy\rangle$ where $\langle -,-\rangle$ is the Hermitian product.

0celo7

NO

I WAS DOING THAT

DON'T HELP ME

;_;

I'm a confused kid

ACuriousMind

You, uh, told me you needed a bigger hint?

0celo7

@ACuriousMind I know, but I had figured it out

now the victory is less good

actually I got antilinear in the second

ok, after a bit of derping I figured it out

ACuriousMind

As usual, then ;)

0celo7

$\langle x,y\rangle =(x,y)+\mathrm{i}[x,y]$. $\langle \eta x,\beta y\rangle=\eta\bar\beta\langle x,y\rangle$

You agree?

ACuriousMind

2:55 AM

yes

0celo7

but...

you said antilinear in the first

that was my derping

ACuriousMind

It, uh, doesn't matter in which it is antilinear

0celo7

;_; I spent so long trying to reproduce your result

@ACuriousMind I'm flipping through problems in Arnold

"find 1-forms on the plane that are not differentials of any function"

I thought the plane had trivial cohomology

ACuriousMind

So?

0celo7

I can't read :)

thought it said closed

0celo7

3:18 AM

@ACuriousMind how exacly is the limit on the top of page 81 chart independent?

ACuriousMind

@0celo7 The "prove this!" there is meant to say you should figure that out yourself :P

0celo7

@ACuriousMind honestly, I've tried

I don't see what property of charts is needed

ACuriousMind

That they're diffeomorphisms.

0celo7

ok, so the charts are diffeomorphisms between the domains

is $\phi$ the image of a curve in $M$ under a chart?

ACuriousMind

Yes.

0celo7

3:27 AM

well, no

it says on the previous page near the bottom $\phi(t)\in M$

ACuriousMind

@0celo7 On the previous page, $M$ is embedded. Here it is not, hence the need for a chart.

0celo7

ah

ok, notation

$\phi$ is the curve

chart is $h$, other chart is $h'$

ugh no clue what to do :/

@ACuriousMind I don't see what diffeomorphism has to do with anything

0celo7

3:51 AM

@ACuriousMind are the charts somehow linear or something?

I'm completely lost

ACuriousMind

4:06 AM

@0celo7 Observe that $\lim_{t\to 0}\frac{(h\circ \phi)(t)-(h\circ \psi)(t)}{t} = (h\circ \phi)'(t)-(h\circ \psi)'(t)$ and the use the chain rule for $(h'\circ \phi)' = (h'\circ h^{-1}\circ h\circ \phi)'$

(where primes except for the one on $h'$ denote usual differentiation)

0celo7

ok, I figured out I need to get the equation after "observe that"

but I don't see how I get it

ACuriousMind

$0 =t-t$

0celo7

o.o

what

ACuriousMind

Add $+t-t$ to the numerator.

0celo7

I know that's what you meant

I still don't see it

ACuriousMind

4:10 AM

You do realize that $(h\circ \phi)'(0)$ is defined as $\lim_{t\to 0}\frac{(h\circ \phi)(t) - t}{t}$?

0celo7

@ACuriousMind no

ACuriousMind

But that is the definition of a derivative!

0celo7

@ACuriousMind huh?

I don't see how that is of the form $[f(x)-f(0)]/x$

ACuriousMind

No, wait :D

Godamnit, why it Arnold define this so weirdly

0celo7

Russian

ACuriousMind

4:12 AM

If he had just taken the equality of the derivatives as the definition instead of that limit, it'd be so much easier.

0celo7

:)

I have 5 books here that do it that way lol

Arnold is the ugly duckling

ACuriousMind

Well, we can always throw the towel and just Taylor expand

0celo7

D:

HERESY

this isn't physics, taylor expansions are forbidden

@ACuriousMind ok, then what?

ACuriousMind

No, it's rigorously allowed.

0celo7

are chart transtitions just linear maps on vectors?

@ACuriousMind is that it? just taylor expand?

ACuriousMind

4:21 AM

We Taylor expand $f := h'\circ h = f(h(\phi(0))) + Df(h(\phi(0))(h(\phi(t))-h(\phi(0)) + r(h(\phi(t))(h(\phi(t))-h(\phi(0))$ where $\lim_{t\to 0} r(h(\phi(t))) = 0$ by Taylor's theorem.

0celo7

lol screw that

ACuriousMind

Hm, few brackets missing. Anyway, doing that carefully yields the chart independence for Arnold's weird definition, since a bunch of terms drop out due to $\phi(0) = \psi(0)$ and the Jacobian $Df$ is linear

0celo7

aight

thanks pal, will look at it in physics lecture tomorrow

@ACuriousMind shouldn't you be asleep?

is this a record?

ACuriousMind

What? No, this is quite far from a record :P

In autumn/winter the sun comes up much later to remind me to go to sleep :D

0celo7

you're a daemon

dude, your sleep schedule is identical to mine and we live 6 hours apart :D

diff eq test on Tuesday is gonna suck

Variation of crap and Laplace transforms and reduction of order

and pursuit curves isogonal trajectories

gulp this might be pretty tough

@ACuriousMind I'm off, I still have a reflection ;P

ACuriousMind

4:31 AM

g'night

DanielSank

4:52 AM

@0celo7 What have you learned in college so far?

skull petrol

5:43 AM

He's learned that Mathematics is more of an obstacle than a tool at this point.

Slereah

8:38 AM

Soon he will learn the evils of mathematics and become a real physicist

Physics Meta

8:57 AM

0

Q: Why is the main page of PSE looking so weird in my Opera browser?

Pictures would say better: Then this; notice a large bar in the right part of the page; the conspicuous part is the appearance of hot questions at the bottom: What is going on? I'm using Opera 30.0.1835.88; I then checked the Chem.SE page It can be easily seen the ads are at their prop...

Danu

10:52 AM

!@#$/^ Google it — Joel 14 hours ago

lol hemad

0celo7

12:30 PM

@DanielSank What?

user685252

He's asking about the big picture.

The so called view of the forest, not the trees :-)

0celo7

12:57 PM

@user685252 Uh, I'm way too lazy to do anything with my life ;_;

user685252

1:13 PM

Well then, you will be pushed aside by those who are not.

aka survival of the most competitive @0celo7

dmckee

1:51 PM

Seriously, while being smart will let you skate though (roughly) undergrad, after that brains alone won't carry you: diligence trumps them regularly in grad-school and the work place. (Honest geniuses get a little bit of a break, but the merely very smart get no such consideration.)

3

0celo7

@dmckee I'm diligent when something challenges me

Y'all have seen, I've done a lot of work to teach myself stuff, but that stuff was interesting.

0celo7

2:14 PM

@ACuriousMind lol

@ACuriousMind Ok, we might have to revisit the Galilean thing. I've managed to confuse myself again by reading what I wrote in the past D:

My past self confused my present self, lol

ACuriousMind

Is there a way to search one's favourited questions, and only those?

user685252

The starred ones?

ACuriousMind

Yep

I mean, I'm favouriting mainly questions I want to refer to later, and it is kinda annoying to scroll through the list searching for them when I want to do that

user685252

2:31 PM

There is a "favourites" option.

Loong

@ACuriousMind infavorites:mine

user685252

You have 57

0celo7

He doesn't want to scroll through all 57

Talk about lazy :P

ACuriousMind

@Loong Thanks

@0celo7 That's already three pages - that is inefficient to scroll through when I know the exact keyword I'm looking for

Also, I figured there must be a way otherwise I would have no idea why Qmechanic favourites so many things

0celo7

And solving quadratic equations is inefficient as well :)

user685252

2:37 PM

66

Q: The Role of Rigor

The purpose of this question is to ask about the role of mathematical rigor in physics. In order to formulate a question that can be answered, and not just discussed, I divided this large issue into five specific questions. 1) What are the most important and the oldest insights (notions, results...

mathematical-physics mathematics research-level

Now that^ is worth following

:-)

0celo7

@ACuriousMind What is $r$?

ACuriousMind

@0celo7 Some function that goes to zero like that. There a explicit formulae for the Taylor error term, but they're not needed here.

I also realized we can do this an easier way:

0celo7

@ACuriousMind Oh the error! Silly me.

ACuriousMind

Just add $h(\phi(0))-h(\psi(0))$ to the original statement to get that it's just the difference between to derivatives. This works since $\phi(0)=\psi(0)$

Then what I originally wanted to do with $+t-t$ works.

0celo7

Uh how is $h(\phi(0))$ defined? I thought $\phi$ was in $R^n$.

Slereah

2:40 PM

Hm

I want to ask a question

0celo7

I thought $\phi$ is the image of some $\gamma:I\to M$ under $h$.

Slereah

But that would mean typing it on my phone currently

0celo7

here or main site?

ACuriousMind

@0celo7 Oh, yeah, I chose to interpret $\phi$ as the curve and denote what Arnold writes $\phi$ as $h\circ \phi$ Call it whatever you want

0celo7

@ACuriousMind great

make the shitty notation more complicated :)

Slereah

2:42 PM

Main site

0celo7

@Slereah What phone?

Slereah

Wouldn't want to ask it here, it's a PHYSICS question!

Not math :p

0celo7

lel

I'm going through the exercises in Arnold

don't worry, physics will be in there

@ACuriousMind difference between two derivatives?

in the limit or something?

ACuriousMind

@0celo7 Remember that the chain rule gives the chart independence once we know that Arnold's $\lim_{t\to 0}\frac{\phi(t)-\psi(t)}{t}$ is the same as $\phi'(t)-\psi'(t) = 0$?

user685252

6 hours ago, by Slereah

Soon he will learn the evils of mathematics and become a real physicist

0celo7

2:45 PM

Right, oh oh you want to add $-\phi(0)+\psi(0)=0$?

ACuriousMind

@0celo7 Yes

@user685252 Nooooooooooooooooooooooooo

0celo7

ahhhhhh

@user685252 I'm not even a physics student ffs