Mathematics - 2020-11-27 (page 2 of 2)

6:01 PM

@love_sodam oh, that's an even nicer observation than what I was trying to do, just compare degrees of both sides and you have a contradiction

Semiclassical

denoting the surface current density as $\mathbf{K}$, i guess the integral should be $$\mathbf{A}(\mathbf{x}) = \frac{\mu_0}{4\pi}\int_S \frac{\mathbf{K}(\mathbf{x}')}{\| \mathbf{x}-\mathbf{x}'\|} dS'$$

Balarka Sen

yeah; you can compute $K$ explicitly because $I$ amount of current is entering and exiting through those filaments

Semiclassical

yeah

you want to write $\mathbf{K}$ in Cartesian components tho

Balarka Sen

yeah haha

oh man

love_sodam

@Thorgott How can I get the degree of $F(\beta)$ over $F$? well $x^{p^2}-a$ has $\beta$ as a root but I don't know if it's irreducible or not

Semiclassical

6:04 PM

like, it's okay to leave $\mathbf{x}'$ in spherical coordinates

Thorgott

it is irreducible

Semiclassical

(or cylindrical, not sure which is easier here tbh)

love_sodam

@Thorgott why is that?

Semiclassical

but definitely need to find $K_x,K_y,K_z$

Balarka Sen

right agree

Semiclassical

6:05 PM

still less gross than Biot-Savart tho

for the derivation of $\mathbf{A}$ for a loop current: physics.usu.edu/Wheeler/EMarchive/Jch5Notes.pdf

they do it in terms of a distributional $\mathbf{J}$ but probably easier to write using $\vec{I}$ tbh

Balarka Sen

yeah

Semiclassical

one really annoying thing on this point is that Mathematica seems to suuuuck at elliptic integrals

Balarka Sen

thanks for those notes

Semiclassical

using Assumptions helps but

Balarka Sen

i'll try to emulate the computation

Semiclassical

6:08 PM

still terrible

yeah

i would be shocked if the integrals are computable in closed form to be clear

Balarka Sen

maybe the guy didnt look and gave a harder problem than he intended

i can ask

Semiclassical

Do you have the problem statement, out of curiousity

Balarka Sen

yes, one second

Semiclassical

oh

that's better, actually

it's got azimuthal symmetry

still painful but not as bad

Astyx

symmetry tells you the field is rotating along the axis

Balarka Sen

6:15 PM

huh? it just goes up

Astyx

colinear to $u_\theta$

There's a rotational symmetry around the z axis right?

Balarka Sen

like, $K$ is upto scale just the unit tangent field along the longitudes right?

Semiclassical

right. but that means that the magnetic field should end up being circumferential, i think

Balarka Sen

@Astyx ok yeah thats what you mean

Astyx

I mean the magnetic field

Balarka Sen

6:18 PM

Ahh

got it

Semiclassical

yeah, this isn't the worst

Balarka Sen

OK, so maybe I can compute it. I'll try after dinner

love_sodam

@Thorgott I think somehow irreducibility of $x^p-a$ implies the irreducibility of $x^{p^2}-a$ right?

Balarka Sen

Thanks!

Semiclassical

later

Astyx

6:19 PM

bon appetit

Semiclassical

oh, yeah, K's magnitude will change as it moves along the sphere, eh

gross

Balarka Sen

yeah

Astyx

Isn't there an easy formula that gives the magnetic norm as a function of the current through a loop ?

Semiclassical

on axis, yes

off-axis, there is a closed-form but it's in terms of complete elliptic integrals

Astyx

Since there's rotational symmetry, the integral of the field around a circle is just 2 pi R |B|

And the current in the circle is either 0 or I

Semiclassical

6:22 PM

hmmmm

Astyx

So the mag field is 0 inside the sphere, and whatever the formula gives everywhere else

like $\mu_0 I$ IIRC

Semiclassical

that could be right. problem is justifying it

Thorgott

@love_sodam yeah, this follows from the general theory of when polynomials of the form $x^n-a$ are irreducible

Astyx

It's Gauss' theorem

Thorgott

I believe you can cook up some ad hoc argument with norms

Semiclassical

6:24 PM

well, it's more than that

you have to justify why certain components vanish

Astyx

planar symmetry

Means the vectors have to be along $u_\theta$

Edward Evans

"Wenn du dich unmittelbar der kategorientheoretische Paradigmatik hingeben willst, dann wäre wohl ein Wechsel in die Homotopietheorie, Topostheorie oder (höhere) Kategorientheorie empfehlenswert." The algebraic number theory 2 course is getting too nerdy for my taste

Tobias Kildetoft

higher category theory is great

love_sodam

@Thorgott I've never learned such general theory of $x^n-a$. You mean primitive $n$th root of unity?

Edward Evans

By nerds, for nerds

Tobias Kildetoft

6:26 PM

(but just up to 2, not any higher than that or it gets weird)

And not too general. There should be some additional structure so you can say nice stuff

Astyx

Oh I might be saying nonsense

Edward Evans

there's a dude that wrote a huge post on the course forum complaining about how algebraic number theory ignores category theoretic language and that loads of the statements in the lecture could be simplified by category theory

Tobias Kildetoft

Actually, let's focus on fiat, or at least finitary :)

Edward Evans

and the course tutor was like

"dude stfu"

Balarka Sen

lol

Edward Evans

6:28 PM

literally said "if you wanna go down that rabbit hole post it on MathOverflow or MathStack, leave it out here pls"

Thorgott

no, I mean the irreducibility of $x^n-a$

this is classic and I'm afraid there's no completely easy way of getting to it

Edward Evans

the guy is knowledgable af though so that's cool

love_sodam

@Thorgott I found this: math.stackexchange.com/a/133598/668308

You mean this?

Thorgott

yeah, that's the general result

but this specific case should be easier

love_sodam

Well $p^2$ is not prime so what do you mean specific?

Edward Evans

6:32 PM

you were mistaken

Thorgott

yes lol

urgh, gotta think for a sec

no wait, I believe I wasn't mistaken

Edward Evans

idek what you wrote, I just saw you write if I'm not mistaken and then delete it

Thorgott

$N(\beta)^p=N(\alpha)=a$, where the last step is explicit computation, and this contradicts irreducibility of $x^p-a$

love_sodam

what is N?

Thorgott

norm

love_sodam

6:36 PM

what norm?

Edward Evans

det of mult

Thorgott

Field norm

In mathematics, the (field) norm is a particular mapping defined in field theory, which maps elements of a larger field into a subfield. == Formal definition == Let K be a field and L a finite extension (and hence an algebraic extension) of K. The field L is then a finite dimensional vector space over K. Multiplication by α, an element of L, m α : L → L {\displaystyle m_{\alpha }\colon L\to L} m α...

they even compute what I just said as example down in the article

love_sodam

I've never seen that before in the lecture.

Any other way without that field norm?

Thorgott

dunno

Astyx

I meant ampere's theorem, not Gauss'

Balarka Sen

6:45 PM

thats like dual Gauss hah

Ted Shifrin

Revolting!

Balarka Sen

Isn't it?

Astyx

Yeah it's all pretty much the same

Balarka Sen

$\nabla \times B = \mu_0 I$ is dual $\nabla \cdot E = \rho/\varepsilon_0$

EM duality

Astyx

And then the fact that all planes containing the z axis are symmetry planes for the current is sufficient to deduce that the magnetic field on those planes is perpendicular to those plane

Balarka Sen

6:48 PM

yeah so that says eg $B = 0$ along the axis?

Astyx

yup

Balarka Sen

actually maybe $B = 0$ inside the sphere as well

Astyx

Oh, you meant that, but still yes

Balarka Sen

no, why can't the field like rotate

inside the sphere, around the $z$-axis

why is $B = 0$ off of the axis inside the sphere?

love_sodam

I saw Lang algebra and other textbook, it seems the proof is very long. I'm gonna use this without proof as the lecture already provided :)

Astyx

6:50 PM

First prove that the field in cylindrical coordinates is only along $u_{\theta}$

so $\vec B = B_0 u\theta$. Then integrate along circles centered on the z axis, and you get $B_0$

gotta go eat

cya

love_sodam

Oh no, If I don't prove that it's just one line

Balarka Sen

@Astyx ok but dont you want to prove $B_0 = 0$

inside the sphere

Semiclassical

presumably once you prove that $\vec{B}$ is circumferential, you appeal to Ampere's law

i'm a bit dubious that helps tho

maybe it does tho

Balarka Sen

what's the loop?

Semiclassical

around the z-axis

arbitrary height and radius

if B is circumferential, that's just B*2pi*R

Thorgott

6:57 PM

there is no long proof necessary for this particular result, the argument I gave above works

Semiclassical

and then one computes the current enclosed, which is easy

Thorgott

only requires one explicit norm computation and multiplicativity of the norm

Balarka Sen

hmm fine

Semiclassical

also, you may be able to duplicate the argument for the vector potential

volume current density : magnetic field :: magnetic field : vector potential (up to some multiplicative factors maybe)

so you'd now have a source which is zero inside the sphere and known outside

Balarka Sen

yeah fine

weird trick

love_sodam

7:06 PM

@Thorgott Yes it would if I know that field norm..

Rather change the question to $x^p-\alpha\in F(\alpha)[x]$ is irreducible

Balarka Sen

that is direct actually

$x^p - c$ is irreducible over $K$ iff it has no roots over $K$

love_sodam

Oh I can't. I assumed the existence of $\beta$

$\beta^p=\alpha$ I assumed

Balarka Sen

i dont know what these symbols are

love_sodam

do you know the question?

Balarka Sen

no; i just saw your last statement about changing the question

love_sodam

7:14 PM

Ok, the question is if $f=x^p-a\in F[x]$ is irreducible where $char(F)\neq p$ for odd prime $p$ and if we denote $\alpha$ as a root of $f$, then $\alpha\in F(\alpha)-F(\alpha)^p$. So, assume there is $\beta\in F(\alpha)$ such that $\beta^p=\alpha$. Then we get $F(\alpha)=F(\beta)$. Now we derive contradiction by showing $x^{p^2}-a$ is irreducible over $F$ ($\beta^{p^2}-a$ so degree argument derive the contradiction)

Astyx

@Balarka if you apply Ampere along a circle in the sphere, the current through the circle is zero so $2\pi \mu_0 B_0$ (or whatever the formula is) is 0

Balarka Sen

yeah got it @Astyx

love_sodam

And I found some general result about the polynomial of the form $x^n-a$ but the proof is very long and I didn't learn that. @Thorgott used field norm but I also didn't learn that.

Astyx

It's $B_0(r)$ actually

Well $B_0(r,z)$

Anyway

love_sodam

7:41 PM

Seems hard to prove without those..

Oh, I just found the statement and the proof in some textbook lol

schn

8:05 PM

What is the inverse of the function $f(x,y)=\frac{x}{\sqrt{y}/c}$, where $x$ is any real number, $y$ any positive real number and $c$ a constant?

Astyx

@schn What is $f(\lambda x, \lambda^2y)$ ?

porridgemathematics

8:15 PM

hi, im a bit confused about something to do with presentation matrices, could anyone take a look? mathb.in/47577

we are assuming M is finitely presented

schn

@Astyx The function, or?

Astyx

What does it evaluate to ?

schn

@Astyx Yes, sorry, it evaluates to $f(x,y)$.

Astyx

So the function is not injective

schn

@Astyx Correct.

Astyx

8:26 PM

Does it have an inverse?

schn

Well, no.

Is there such a thing as an inverse with respect to $u$ or $v$?

Sorry $x$ and $y$.

Astyx

If you fix one of them, yes, most of the time

not for x=0 for instance

schn

Alright, thanks!

porridgemathematics

love_sodam what does $\alpha \in F(\alpha) - F(\alpha)^p$ even mean?

Astyx

That $\alpha$ is not a p-th power

porridgemathematics

8:31 PM

oh setminus

if $\alpha = \beta^p$, then $\beta$ is a root of $x^{p^2} - a$, and $\beta \in F(\alpha)$, but $x^{p^2} - a$ is the minimal polynomial of $\beta$ in $F[X]$, and so $F(\beta) \subset F(\alpha)$ has dimension $p^2$ as a vector space over $F$, but $F(\alpha)$ has dimension $p$ as a vector space over $F$, so isn't that a contradiction?

oh wait nvm

Thorgott

yes, it is, that's the argument we discussed above

showing irreducibility of $x^{p^2}-a$ was the issue

porridgemathematics

8:48 PM

right, I sort of just assumed that, also im not sure why I wrote $F(\beta) \subset F(\alpha)$

Astyx

Because $\beta \in F(\alpha)$ if we assume $\alpha$ to be a power

porridgemathematics

oh of course. lol

weirdly that is why i wrote it in the moment

but i sort of blanked out while starting at what iwrote afterwards

*staring

Astyx

Yeah that stuff happens lol

porridgemathematics

anyone care to help me out with something related to modules?

its probably pretty basic

mathb.in/47577

Astyx

I looked at it but I'm not sure i understand what the text is saying

porridgemathematics

8:54 PM

ah, apologies its using verbage from my course

i can add some defintions

Astyx

My understanding is you're taking the free Z module generated by three elements and quotienting by the relations of the matrix

porridgemathematics

uh yes exactly

Astyx

So in this case, you get $\Bbb Z^2\oplus\Bbb Z/2\Bbb Z$

Because you're essentially quotienting by only one relation: 2a = 0

porridgemathematics

and thats iso to $$\mathbb{Z}^3 / <(2,0,0)>?$$

Astyx

Yes

porridgemathematics

8:57 PM

oh, thats what I wrote down below :/ so maybe I do get it?

is my justification in terms of the generators at the bottom correct?

Astyx

Maybe you're confusing collumns and rows

If you assume there are two generating elements and three relations instead, you get $\Bbb Z/2\Bbb Z\oplus \Bbb Z$

porridgemathematics

oh, so what im saying is for some reason i was led to think (assuming the columns are relations, and the rows the generators), that ^ is what you should get, but at the bottom i reasoned that the first thing you wrote is what you should get

so just to be clear, if we change the matrix to $$\begin{matrix} 1 & 0 \\ 0 & 2 \\ 0 & 0 \end{matrix}$$

then this should be iso to $$\mathbb{Z} / 2\mathbb{Z} \oplus \mathbb{Z}$?

one too many $

oh dang

But yes

9:00 PM

sorry

$$\mathbb{Z} / 2\mathbb{Z} \oplus \mathbb{Z}$$

Astyx

Because one of the relations is a=0

porridgemathematics

right

@Astyx the reason I was confused was because of this imgur.com/a/DQUCFL6

there matrix has rows as the basic relations, and number of columns equal to the number of generators

so shouldn'

shouldn't their module be iso to $$\mathbb{Z}^3 / <(8,4,8), (4,8,4)>

and then by the same argument as what we just did, shouldn't that be iso to $$\mathbb{Z} / 4\mathbb{Z} \oplus \mathbb{Z}/12\mathbb{Z} \oplus \mathbb{Z}$$

$$\mathbb{Z}^3 / <(8,4,8), (4,8,4)>$$ * sorry

Astyx

I agree with you

porridgemathematics

hmm, so weird that this error would come up, anyway thanks!

schn

9:19 PM

Is it possible to find the points $(x,y)$ in the set $\{(x,y) : -\infty \leq x/ \sqrt{y} \leq t \}$, where $t$ is a fixed constant?

Or give bounds to $x$ and $y$?

Solved it.

Vasting

10:23 PM

Suppose I have a concave function f where f(0)=f(1)=0 and the maximum f achieves is 1 on the interval [0,1]

Must it be the case that there is a 1/2 * 1/2 square that fits between f and the x-axis?

Astyx

no

unless you place a condition |f'|<= 1/2

Vasting

why is that?

Ted Shifrin

@Astyx: That doesn't seem right.

Are you assuming the function is strictly concave?

Vasting

Yes

Astyx

I can't read

Ted Shifrin

10:26 PM

So it has no piecewise linear portions.

Astyx

I didn't see that the function was concave

Vasting

yeah i think that's safe to assume

im not sure if that would change anything

Ted Shifrin

Yeah, I could go up and down very quickly and then just stay horizontal.

Vasting

ah yeah

strictly concave then

geocalc33

does anyone have an example of $$\bigg(\int_0^1 f(x) dx\bigg)^2=\int_0^1 \int_0^1 f(x,y) dxdy $$?

Ted Shifrin

10:28 PM

You can certainly give an easy example yourself.

Vasting

wait - in terms of going up and down very quickly and staying horizontal - that's not concave no?

Ted Shifrin

Oh, you're right, @Vasting.

And if I stay horizontal at the top, there's plenty of room for your square.

Vasting

yeah i think so

from random examples ive drawn it seems true

Ted Shifrin

There certainly has to be a point $x$ so that $f(x)=f(x+1/2)$. And I think concavity tells us that the height there cannot be $<1/2$. But that's what has to be proved.

Vasting

Yes, interesting. I've tried the kind of mirrored approach where we know $f(x)=f(y)=1/2$ for some $x<y$, with the goal of showing that $|x-y|\geq 1/2$.

Balarka Sen

10:34 PM

Certainly a triangle with base [0, 1] and height 1 fits. It seems you can fit a square of area 1/4 inside such a triangle.

Ted Shifrin

Yes, @Balarka is right for the first one by concavity. Then the other should just be a standard differential calculus question. Nice thought.

Vasting

Oh yep, that doe sit

Thanks @BalarkaSen!

Ted Shifrin

I was trying to make the square touch the graph at the top, but that's not required by @Vasting's question.

Vasting

should not have missed this

Balarka Sen

In fact there will be a unique square which is "periscribed" on the triangle, so you can directly compute: 1/2 = x^2 + x(1 - x)/2 + x(1 - x)/2

Which says x = 1/2 :)

Neat question

Astyx

10:41 PM

the triangles suffice

Ted Shifrin

Yes, Balarka definitely out-geometried me on that one!

Balarka Sen

I didn't think I'd get it! Just luck

Ted Shifrin

No, I was being an idiot. Vasting already called me out on it.

It can do it at the top, but then there's no problem whatsoever.

Astyx

oh ok sorry

Balarka Sen

I have a question but the premise is so obviously pretentious that I feel it shouldn't be publicly asked.

Ted Shifrin

10:52 PM

That doesn't sound like you, a @Balarka :)

Balarka Sen

Hah

Here's a better question maybe

There's a hyperbolic $n$-gon in $\Bbb H^2$ with all angles $\pi/2$, for any $n \geq 5$. This can be seen by intermediate value theorem; an ideal $n$-gon has all angles $0$, start shrinking it; if you have shrunk it very close to a point it's a regular Euclidean $n$-gon, all angles $>\pi/2$ since $n \geq 5$.

Somewhere in the middle our equality must be attained.

I was thinking of an alternative idea which feels much harder to implement, but somehow I should also know this? Namely, pick infinitely many copies of a Euclidean $n$-gon, and paste them by isometries along edges so that at each vertex exactly $4$ of them meet. This gives a simply connected simplicial complex homeomorphic to $\Bbb R^2$ such that at each vertex you have angle defect $2\pi - 4(1 - 2/n)\pi = (2/n - 2)\pi < 0$, a uniform negative constant for $n \geq 5$.

This is a combinatorial hyperbolic plane. Can you "uniformize" this object, whatever it is, to $\Bbb H^2$, so that the edges map to geodesics?

This is what the uniformized tessellation will look like.

Ted Shifrin

I'm lost.

Balarka Sen

I can explain if you tell me where I went pretention-overboard maybe :)

Ted Shifrin

Oh, no longer regular. Where are you getting angle deficit?

And how do you control angles when the polygons are irregular?

Balarka Sen

So I have made a simplicial complex by pasting 4 copies of the regular Euclidean $n$-gons at each vertex. So total angle at the vertex is $4 \cdot (n-2)\pi/n$, right?

Ah the pasting cannot be done inside $\Bbb R^3$; I'm doing an abstract pasting of the edges by isometries.

It's a metric simplicial complex maybe.

All my $n$-gons in the simplicial complex are isometric to the regular Euclidean $n$-gon.

Ted Shifrin

11:05 PM

That can't be right. Conformally equivalent, maybe?

Balarka Sen

That seems better. But why not isometric? It's a simplicial complex with a metric -- how is the metric determined? The faces are isometric to the regular Euclidean $n$-gon. Seems fair, right?

Ted Shifrin

But the picture you end up drawing there's a conformal map, not an isometry, obviously.

Balarka Sen

Yes, agreed, so what I am saying is if I define the object, let's call it $X$, like I said, then the angle defect at vertices is well-defined, $2\pi - 4\cdot (n-2)\pi/n < 0$, a uniform negative constant. So there should be a "conformal map" $X \to \Bbb H^2$

Sending the canonical triangulation of $X$ to the $(4, n)$-tessellation of $\Bbb H^2$, if that's what they call it

Ted Shifrin

I'm confused. On the sphere, I detect positive curvature by summing the angles of the triangle. At each vertex, I have a usual triangulation with angles summing to $2\pi$.

Balarka Sen

But the trick is to treat the triangles as Euclidean triangles instead of spherical triangles, right?

Ted Shifrin

11:11 PM

Ah, OK. So flattening them gives positive angle defect. I see.

Balarka Sen

Right.

Ted Shifrin

OK, now I'm less confused.

Where were you?

Balarka Sen

Right, so now what I am wondering is these kind of "metric triangulations", where each triangle is isometric to a Euclidean triangle, carries curvature but concentrated at the vertices. How do you uniformize these fellas?

To get actual geodesic triangulations of $S^2, \Bbb H^2$, etc

Ted Shifrin

Ah, one of my former colleagues thought about these things.

Balarka Sen

Oh, interesting. Do you have a reference?

Ted Shifrin

11:15 PM

He was doing discrete Dirichlet problems to do uniformization. Look up Sa’ar Hersonsky.

I spent months being a sounding board, but that was 6 years ago and I don't berember.

Balarka Sen

Nice! Thanks, let me see if I can find some works of his.

Ted Shifrin