Mathematics - 2021-06-10 (page 2 of 2)

Xander Henderson

20:01

You are on your own for that.

copper.hat

comes with a clean klein bottle

More Anonymous

@copper.hat Break the bottle and use the shards to free the mobius band

Xander Henderson

@copper.hat That's good. I don't know where my Klein bottle cleaner is.

robjohn

I keep all my money in a Klein bottle in my basement

copper.hat

mine has a non occidentiable surface

PM 2Ring

20:06

Sadly, Google can't find an image of a Klein-Gordon gin bottle.

Ted Shifrin

@MoreAnonymous i used to pass for one.

More Anonymous

@TedShifrin Used to?

Ted Shifrin

Retired bum now.

More Anonymous

@TedShifrin Ah ... okay I wish I did more mathematics

I did physics

Ted Shifrin

Nothing wrong with that! What was your geometry question?

More Anonymous

20:17

@TedShifrin I think I can go from the line element to another coordinate invariant quantity. I'm not sure what this quantity is

or why this works

Was hoping a light discussion might clear things up

Are you still interested?

Ted Shifrin

You mean a Riemannian or semi-Riemannian metric? What precisely do you mean by line element?

More Anonymous

@TedShifrin either will do. Physicists often write line element as $ds^2 = -c^2 dt^2 + dx^2 + dy^2 + dz^2$ for example is line element of a photon in flat sapcetime

Ted Shifrin

Yes, it is standard lingo. So we're talking about the metric tensor.

More Anonymous

$ds^2$ is the line element

Ted Shifrin

No, $ds$ is, but it's a density …

More Anonymous

20:25

@TedShifrin Why $ds$ thought of as a density?

Ted Shifrin

Because it is orientation-independent, so not a differential $1$-form on a curve.

More Anonymous

@TedShifrin Interesting

I'll have to think about that

Ted Shifrin

OK ;)

More Anonymous

But proceeding with my question. lets I manage to find the transformations that enable me to take $ds^2 \to \lambda ds^2$. For example $ds^2 = dx^2 + dy^2$ becomes $\lambda^2 ds^2 $ when $x \to \lambda x $ and $y \to \lambda y $

Ted Shifrin

This is called a conformal change of metric.

More Anonymous

20:30

Ah I see ...

Didn't know

Xander Henderson

@MoreAnonymous Gosh, I kind of wish I had done my physics. I did mathematics.

More Anonymous

Now all those coordinates which participate in the transformation are replaced with their integrated variable $dx \to x$ and $dy \to y$ and I get $K = x^2 + y^2$

Ted Shifrin

Conformal transformations can be much more interesting than just scaling.

Xander Henderson

Honestly, I am fairly close to applying for a physics masters. That would let me teach physics here (there is no one within 90 miles of me who is qualified to teach even introductory college physics).

More Anonymous

Those which do not participate in these transformations are replaced with $0$

Ted Shifrin

20:32

What you just wrote I do not understand.

More Anonymous

@TedShifrin can you reply to that message so I know from where

Ted Shifrin

Crazy, Xander.

More Anonymous

@XanderHenderson Where is this place?

Xander Henderson

@MoreAnonymous Northeastern Arizona.

The nearest physicist is 90+ miles to the west, at Northern Arizona University. The next nearest physicist is likely in Tempe, which is 175 miles south of here.

Ted Shifrin

Your integrated thing doesn't make a well-defined function even though the metric tensor is.

In Euclidean space, sure, but …

Xander Henderson

20:35

@TedShifrin Yeah, you can rotate, too! :P

More Anonymous

@TedShifrin Yea its an algorithm I don't know why this produces a coordinate invariant quantity but it does

(I dont know of any counter example)

Ted Shifrin

What does coordinate invariant mean? You are not allowing general coordinate changes, clearly, so …

More Anonymous

@TedShifrin I mean if I started with the metric $ds^2 = dx^2 + dy^2$ or $ds^2 = dr^2 + r^2 d \theta^2 $

The algorithm produces the same result in both cases

(related by a coordinate transformation of r and thetha to x and y)

Ted Shifrin

You haven’t answered my question.

You need to understand why tensors really do define coordinate-invariant notions, but most things don’t.

More Anonymous

So this is what I meant by coordinate invariant

@TedShifrin So the output of $ds^2 = dx^2 + dy^2$ becomes $K = x^2 + y^2$ while that of $ds^2 = d r^2 + r^2 d \theta^2 $ becomes $K = r^2$. Since $K$ is coordinate invariant (as conjectured) $r^2 = x^2 + y^2$

the quantity $K$ which is the output of the algorithm is coordinate invariant and does not depend on the coordinates of the metric used

Ted Shifrin

20:43

You're wrong. Try different changes of coordinates.

More Anonymous

I'll let you pick?

I tried $x = tu$ and $y = t/u$

it works here too

Ted Shifrin

Try $u=x$, $v=x+y$.

Thorgott

I don't think your algorithm makes much sense, but why does the second case not yields $r^2+r^2\theta^2$ if you just "omit the ds"

Ted Shifrin

$r\,d\theta$ is not exact.

More Anonymous

@Thorgott The algorithm says if a coordinate does not participate in taking $ds^2 \to \lambda^2 ds^2$ then we put $0$. For polar coordinates $r \to \lambda r$ but $\theta \to \theta $ it undergoes the identity so we put $0$

*put $ d \theta $ = 0

Thorgott

20:49

I read that sentence 3 times and still don't have a clue what any of it means

Ted Shifrin

Pull back the metric under $f(x]=\lambda x$, @Thor. What part of the pullback is homogeneous of degree $2$ in $\lambda$?

More Anonymous

So the algorithm is this

1. find all coordinates which scale $ds^2 \to \lambda^2 ds^2$.

In our example that would be $r\ to \lambda r$ then $\lambda^2 ds^2 = \lambda^2 (dr^2 + r^2 d \theta^2 ) $

2. Now since $\theta$ was untouched we put $d \theta = 0$
3. Since $r$ underwent $r \to \lambda r$ we replace $d r \to r$
4. Hence we get $K = r^2 $

Ted Shifrin

Let me know how my example turns out.

More Anonymous

@TedShifrin Sure penning it

first

Thorgott

@TedShifrin I don't follow? Both are in the polar case. They're diagonal coordinates after all.

More Anonymous

20:58

@TedShifrin Seems to work

I'll latex the solution?

One moment

Ted Shifrin

Certainly that linear transformation does not preserve $\|x\|^2$.

I need to go out for my walk, but I'll be back later.

More Anonymous

$(dv - du)^2 = dy^2$ and $du^2 = dx^2$

Then

$ = dx^2 + dy^2 = ds^2 = (dv - du)^2 + du^2 = dv^2 + 2 du^2 - 2 dv du $

Now we notice $v \to \lambda v$ and $u \to \lambda u$ gives us $ds^2 \to \lambda ds^2 $ So we replace both $du \to u$ and $dv \to v$. Thus:

$ v^2 + 2 u^2 - 2 uv = (x+y)^2 + 2x^2 - 2(x+y) x = x^2 + y^2$

@TedShifrin lemme know if you notice some mistake?

@TedShifrin also if you could mention when you would be back from your walk? Its kind of late here

Ted Shifrin

22:07

@MoreAnonymous Mechanically replacing the $2$-tensor $du\otimes dv$ with $uv$ makes no mathematical sense. All right, try the metric $cos^2 y\,dx^2 + e^x\, dy^2$.

I can understand integrating the exact 1-form $dx$, but other things you're doing are mathematically not meaningful.

Govind75

23:30

Anyone know what $\mathbb{H}$ identified with $\mathbb{R}^4$ means? Does it mean to just replace quaternions with a vector representation in $\mathbb{R}^4$

Thorgott

It means that you think of $\mathbb{H}$ as $\mathbb{R}^4$

usually (read: almost exclusively) by identifying $1,i,j,k$ with the standard basis in $\mathbb{R}^4$

as far as I'm concerned, they're literally equal as sets, but I guess your preferred convention may vary

Govind75

Is this something that I have to do, or can I just proceed with $\mathbb{H}$

Essentially, I am following a proof and it states the above before proving a bunch of stuff.

Thorgott

that is a way too open-ended question

I can't tell you what you have to do

Govind75

true

Thorgott

but let me ask you, what is $\mathbb{H}$ to you?

Govind75

23:37

$\{a + bi + cj + dij| a,b,c,d \in \mathbb{R}\}$ with rules on multiplication

I understand the analogy with $\mathbb{R}^4$ but is there any actual need to do that ever?

Ted Shifrin

It is $\Bbb R^4$ as a vector space. What analogy?

Govind75

sorry poor phrasing from myself

Thorgott

Ok, that's just notation, but what kind of things are $i,j$? What space do they live in? My point is that you may as well define $i=(0,1,0,0)$ and $j=(0,0,1,0)$ (and then define the multiplication, which is the important additional structure).
Anyway, whether there truly is a need to do something or not is a philosophical question I can't entertain. But $\mathbb{H}$ and $\mathbb{R}^4$ are so naturally identified that it won't make a genuine difference. It's just a matter of notation, I'm almost 100% certain even without even knowing what proof you're looking at.

Govind75

oh yh, tysm

leslie townes

23:58

when i was a child i behaved as R^4

Govind75

You know the group homomorphism: $$\phi ; S^3 \rightarrow SO(3) ; q \mapsto \phi_q$$
Where $$\phi_q; \mathbb{H}^* \rightarrow \mathbb{H}^*; x \mapsto qxq^{-1}$$ for a non-zero pure quaternion $q$.
How does it actually work, I read somewhere it's to do with these loops which collapse?

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$Mathematics$ Mathematics