Mathematics - 2019-06-12 (page 3 of 4)

DogAteMy, the easiest place to start is the moment of inertia tensor. We typically teach moment of inertia about a single axis, but what data do you need to compute it about an arbitrary axis?

No, that's contravariant, unless by gradient you mean the exterior derivative or differential.

Subhasis Biswas

@TedShifrin does it all start from transformation of coordinates ?

Ted Shifrin

6:31 PM

Not necessarily, but that's how physicists think of it.

Akiva Weinberger

I heard the phrase "a tensor is something that transforms like a tensor" many times long before I had any idea what that meant

I still only somewhat know

Subhasis Biswas

@TedShifrin $ \frac {\partial f}{ \partial x_i}$

the nature of transformation determines whether or not it is contravariant or covariant.

Ted Shifrin

Those are the coefficients of $df$. In Euclidean space with the usual boring inner product, it happens that they're the coefficients of the gradient, but that's an accident. So you really have to specify ...

Subhasis Biswas

correct me if I am wrong .

Akiva Weinberger

I have this vague idea that, like, the gradient of the function $f(x,y)=x$ is the constant vector field $\hat\imath$, but

if I shear the plane,

Ted Shifrin

6:34 PM

We confuse vector fields and 1-forms constantly in vector calculus, but we're using the underlying boring metric on Euclidean space to do so.

Adeek

@AkivaWeinberger .... Shalom

Akiva Weinberger

the vector field stays the same, but the function becomes $f(x,y)=x-y$ whose gradient is different

@Adeek Salaam

Ted Shifrin

So what's the correct definition of the gradient, DogAteMy?

Adeek

I am really having philosophical troubles with the empty set. I have to give introduction to set theory for my class.

Akiva Weinberger

One of these is a covariant tensor and the other is a contravariant tensor

Adeek

6:34 PM

But I really don't like the axiom posulating the existence of empty set.

Subhasis Biswas

vacuous truth?

Ted Shifrin

What class, Adeek?

Akiva Weinberger

I think the vector field is a covariant tensor and the gradient is contravariant?

Adeek

Because you could say meaningless shit such as existence of fairy tails or gods etc

Ted Shifrin

I'm sure @Alessandro can give you explanations of what goes wrong if you disavow the empty set.

Adeek

6:35 PM

@TedShifrin Set theory

Akiva Weinberger

And that's why they react differently to shearing

Ted Shifrin

DogAteMy: No, vector fields are contravariant.

Akiva Weinberger

Oh

Adeek

@TedShifrin Do you know if there is other axioms of set theory that doesn't postulate existence of empty set?

I mean other system of axioms.

Ted Shifrin

The gradient is defined by $\nabla f(p)\cdot v = df(p)(v)$. So it depends on the dot product.

Akiva Weinberger

6:36 PM

@Adeek Do you know why $\{\}\ne\{\{\}\}$?

Ted Shifrin

I'm totally the wrong person to ask such things, Adeek.

Adeek

Yeah @AkivaWeinberger but I just have troubles philosophically with these issues.

Akiva Weinberger

@TedShifrin So if we have no inner product, we can't lower the indices

or raise them

One of the two

The empty set is just an empty box

Ted Shifrin

So, to turn @SubhasisBiswas's $\partial f/\partial x_i$ into the gradient we need $$\sum g^{ij}\partial f/\partial x_j.$$

Akiva Weinberger

$\{\{\}\}$ is just a box with an empty box in it

Mhm @TedShifrin

Rithaniel

6:37 PM

Well, philosophically, I think of the existence of the empty set to be equivalent to the existence of the number 0.

Subhasis Biswas

@TedShifrin that's just $\delta^i_j$

Akiva Weinberger

- which raises the indices?

Adeek

Yeah but you could also represent it in a different philosophical way @AkivaWeinberger

Ted Shifrin

But that's misleading, @SubhasisBiswas. If you change coordinates, it won't be.

Adeek

existence of nothingness

Subhasis Biswas

6:38 PM

@TedShifrin but it does behave like a tensor, right?

Rithaniel

Can you have a box with nothing in it? I don't see why not.

Adeek

yeah @Rithaniel

Subhasis Biswas

$\delta_{ij}$ or $\delta^{ij}$ doesn't, i guess

Adeek

but you could say things such as nothingness can just come out of existence

Akiva Weinberger

In ZFC, all mathematical objects are represented as "pure sets"

Ted Shifrin

6:39 PM

DogAteMy and @SubhasisBiswas: The actual tensors are $$df = \sum \frac{\partial f}{\partial x_i} dx_i \quad\text{and}\quad \text{grad}\,f = \sum g^{ij}\frac{\partial f}{\partial x_j} \dfrac{\partial}{\partial x_i}.$$

Akiva Weinberger

meaning things like $\{\{\}\}$

and not things like $\{1\}$

Ted Shifrin

A covariant one, not a contravariant one, @SubhasisBiswas. See what I just typed.

Akiva Weinberger

unless $1$ is encoded as a set

Subhasis Biswas

@TedShifrin I recognize the left one.

Ted Shifrin

That's a $1$-form, not a vector field.

Rithaniel

6:40 PM

So, you could define "a box with nothing in it" in terms of what could be in the box?

Akiva Weinberger

say again?

Adeek

I will just put it in my brain as existence of empty set is philosophically the same as just looking at an empty grass field. You know nothing exist in the grass field, but you can build things on it.

Akiva Weinberger

A pure set is something that contains only pure sets :P

Subhasis Biswas

@TedShifrin are you generalizing the $\mathbb{grad}$ to Riemannian spaces?

Adeek

I am more satisfied with this philosophically.

Ted Shifrin

6:41 PM

What I say is needed in Euclidean space as well, if you change coordinates.

Adeek

@Rithaniel I guess in your example it is the same as box being empty, but you can always put things in the empty box.

Ted Shifrin

Look at the formulas for grad in polar or spherical coordinates, for example.

Akiva Weinberger

What's $g^{ij}$ in polar coordinates

as a concrete example

$(r,\theta)$

Ted Shifrin

What's $g_{ij}$?

Adeek

This is more philosophically satisfying.

Rithaniel

6:42 PM

That is about what I was thinking, Adeek, yes.

Akiva Weinberger

$dr\cdot d\theta=0$?

Is that the right notation?

Ted Shifrin

$g_{11} = 1$, $g_{12}=0$, $g_{22}=r^2$. So what's the inverse matrix?

Adeek

At least to me.

Subhasis Biswas

@TedShifrin I will try.

Akiva Weinberger

@TedShifrin $g^{11}=1$, $~g^{12}=0$, and $g^{22}=r^{-2}$

Ted Shifrin

6:43 PM

Right.

Rithaniel

If you're talking about sets as "things in a box," it makes sense that you would need to be able to define a box with nothing in it.

Ted Shifrin

But is the box a thing itself?

Never mind.

Akiva Weinberger

OK so say we have $f(r,\theta)=\theta$

Adeek

yeah and then building other things by putting stuff in the box.

Ted Shifrin

Can you put the box in itself?

Akiva Weinberger

6:44 PM

This is a map from $\Bbb R^2\to\Bbb R/\langle2\pi\rangle$ say

Subhasis Biswas

$\displaystyle\frac{\partial \vec{r}}{\partial u^i}.\frac{\partial \vec{r}}{\partial u^j}$

Adeek

@TedShifrin I am not gonna think about that :D

@TedShifrin I am excited for my workshop trip to Taiwan.

Akiva Weinberger

so that $f(r,\theta)=\theta$ is continuous everywhere

What's $df$ and what's ${\rm grad}~f$

Adeek

it is next month 2 weeks trips of intense math finishing Toric geometry in 2 weeks

Akiva Weinberger

What's Toric geometry??

Adeek

6:45 PM

@TedShifrin My wife is worried about me going to Taiwan. I told her Taiwan is actually one of safest countries in the world.

@AkivaWeinberger It is a watered down algebraic geometry.

Akiva Weinberger

Is that when maths votes conservative?

Adeek

You build varieties out of combintorial objects.

Semiclassical

@AkivaWeinberger ow

Ted Shifrin

Ugh, DogAteMy, let's just do it locally and not worry about mods.

Adeek

But it is good testing ground for many theories @AkivaWeinberger

Akiva Weinberger

6:46 PM

@TedShifrin Fine, whatever

Rithaniel

Alternatively, you can motivate the empty set as "the set of things which don't exist." Like, all things that are both green and not green.

Adeek

@AkivaWeinberger haha

Subhasis Biswas

@TedShifrin is it the beginning of Russel's Paradox?

Ted Shifrin

Adeek, Taiwan just legalized gay marriage, so they're way ahead of most of the US now.

Akiva Weinberger

@Semiclassical I knew at least one person would appreciate it

Ted Shifrin

6:46 PM

Oh, it's a Semiclassic.

Akiva Weinberger

I'm notmisreading your reaction, am I?

Adeek

@TedShifrin I am really happy with that !

@TedShifrin I have a gay friend from Egypt. He was gonna end his life, but he went to Taiwan and is living there.

Ted Shifrin

Wow, good for him!

Semiclassical

appreciating may be the wrong word.

Adeek

Yeah I am happy for him.

Semiclassical

6:48 PM

(It’s a good bad pun)

Ted Shifrin

So what is the gradient, DogAteMy?

Akiva Weinberger

The Gay Pride parade in Jerusalem was a few days ago

Adeek

@Rithaniel Yeah that is good motivation.

Akiva Weinberger

@TedShifrin $(0,r)$?

Subhasis Biswas

gay chat

Ted Shifrin

6:49 PM

Hmm, how'd you get that?

I agree it's a multiple of $\partial/\partial\theta$.

Semiclassical

@SubhasisBiswas what?

Akiva Weinberger

It's more steep the further away from the origin you are

I wonder when Apple changed their logo from the rainbow thing it used to be

1998, according to Google

Ted Shifrin

So what's the correct gradient?

Akiva Weinberger

Not $r\frac\partial{\partial\theta}$?

Ted Shifrin

Nope.

Akiva Weinberger

6:54 PM

Or is it just $\frac\partial{\partial\theta}$

Semiclassical

Nope

Ted Shifrin

What happened to our $g^{ij}$?

Akiva Weinberger

$r^{-2}\partial/\partial\theta$?

Ted Shifrin

Much better. Now understand it :)

Akiva Weinberger

I'm having a hard time visualizing that

Semiclassical

6:57 PM

is that right? I expected only 1/r

Ted Shifrin

Well, maybe my formula was wrong. Let's see if $df(v) = \text{grad}\, f\cdot v$ works.

What's $\partial/\partial\theta$ dot itself?

Akiva Weinberger

$r^2$

Ryan Unger

Gradient should have units 1/L so it should be 1/r

Semiclassical

Hah, that was (part of) my thinking as well.

Akiva Weinberger

meaning $\partial/\partial\theta$ should be a vector of length $r$ perpendicular to the position vector

Yeah?

Ted Shifrin

6:59 PM

Well, no, the coordinate vector fields aren't unitless.

The metric becomes $dr\otimes dr + r^2\,d\theta\otimes d\theta$, so $r\,d\theta$ is unit length.

Akiva Weinberger

Meaning $r^{-2}\partial/\partial\theta$ should be a vector of length $1/r$

Yeah?

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$Mathematics$ Mathematics