The h Bar

Bernard Meurer

12:00 AM

why did my table show up in the totally wrong place?

wtf LaTeX

user218912

how does it work when like 3 or more people co-author a paper?

user218912

how do you know who wrote which part

Bernard Meurer

@0celo7 You didn't answer my question still :p

0celo7

@ACuriousMind I'm 99% sure we proved that theorem in my algebra class.

What am I thinking of if not that?

Bernard Meurer

@0celo7 @ACuriousMind github.com/bemeurer/boolean-vector-space/blob/master/…

Does Section 3 look good?

0celo7

12:02 AM

Yes

ACuriousMind

@GPhys Well, it doesn't suffice that it's constant along "a" helix, it would have to be constant along all such helices

That'd be a very weird potential indeed

@0celo7 100% sure you didn't because there are fields of order $p^k$ for arbitrary $k\in\mathbb{N}$.

@BernardMeurer Define "good"

0celo7

wtf did we prove then

Bernard Meurer

@ACuriousMind How can I improve it?

(Update your page, changes made)

also, what do i put where the $?$ is?

it won't be $\neg x$ right?

GPhys

@ACuriousMind see question 3: cosmo.nyu.edu/roman/courses/dynamics_2016/homework2.pdf (note, due date passed, I'm just asking questions our TA was not sure about when we discussed it)

Bernard Meurer

is it $(\neg x \vee 1)$?

halp

is lost

GPhys

12:09 AM

I had thought initially the movement along where the equipotential was constant would always be a symmetry, but not I realize that is not true in general

ACuriousMind

@BernardMeurer I'm a bit confused by your explanation of the Boolean algebra to begin with. First you talk about a generic set, and then suddenly you talk about truth values, and say "a" and "b" are now "0" and "1". Don't you mean the universal bounds should be 0 and 1?

GPhys

the first question wasn't straightforward either, I guess (a particle in a constant magnetic field). Translation along the axes orthogonal to the direction of the magnetic field is a symmetry even though that transformation changes the lagrangian since it only changes the lagrangian by a total time derivative

Bernard Meurer

@ACuriousMind the truth values should be x, y I guess

for x,y in B

ACuriousMind

what?

Bernard Meurer

But the thing is that Axiom 8 proves that B only has two distinct elements

0celo7

12:12 AM

@ACuriousMind Would it kill you to say "I don't want to look at your proof"

ACuriousMind

@GPhys Exactly those transformations which map a given equipotential surface to itself are symmetries.

@0celo7 Yes.

0celo7

@ACuriousMind Why?

Bernard Meurer

@ACuriousMind Axiom 8 defines the universal bounds, and makes it clear that only 2 elements can exist

Right?

ACuriousMind

@BernardMeurer You didn't prove that. If you insist on showing the algebra is a field explicitly, I don't see why you just assume that

Bernard Meurer

@ACuriousMind Good idea :)

ACuriousMind

12:15 AM

It's certainly not evident from anything you've written that a Boolean algebra has only two elements

GPhys

@ACuriousMind Okay that makes sense

ACuriousMind

@GPhys I find that "helical" and "toroidal" are both rather vague descriptions of how exactly the equipotential surfaces look like

Bernard Meurer

@ACuriousMind Doesn't that follow from Axiom 8?

Or am I wrong to state so?

I get that I need to prove it, but It makes sense right?

0celo7

@BernardMeurer no

GPhys

@ACuriousMind I had a similar thought at the time

0celo7

12:16 AM

you write $a,b,\dotsc$ in the first line

implying there are more than two

Bernard Meurer

@0celo7 If there exists an operator $a\rightarrow a'$ such that $a\cdot a'=\emptyset$ and $a+a'=I$

GPhys

Our TA grading the homeworks initially claimed that x and y translation were not symmetries in the constant magnetic field problem (assuming the magnetic field is in the z direction)

but I convinced him during lecture that it is, in fact, a symmetry

Bernard Meurer

Dang it I had the logic for this in my head a second ago

Sigh

GPhys

@ACuriousMind if and only if?

ACuriousMind

@BernardMeurer I think you can't prove that. Unless your definition deviates from the standard, a Boolean algebra has $2^n$ elements

Bernard Meurer

12:20 AM

$n$?

GPhys

as in, if a transformation doesn't map a potential to itself, is it always not a symmetry?

ACuriousMind

@BernardMeurer Some integer

0celo7

@ACuriousMind Positive, no?

GPhys

right now my working definition of a symmetry is a transformation that doesn't change the Lagrangian by more than a total time derivative

ACuriousMind

sure, 1/2 elements doesn't make any sense

0celo7

12:21 AM

@ACuriousMind You never know.

Bernard Meurer

@ACuriousMind I'll ask on MathSE

0celo7

I've seen papers that explicitly state, let $M$ be an $n$-dimensional manifold, $n$ positive.

Bernard Meurer

I don't know how to prrove this

0celo7

I want to know what these negative dimensional manifolds are

Bernard Meurer

I had the most wonderful proof in my head a second ago

0celo7

12:22 AM

@BernardMeurer yeah yeah

Fermat tried that shit 400 years ago and we didn't buy it then.

3

Q: All finite boolean algebras have an even number of elements?

This seems obvious but I wanted to check, since I don't see it mentioned anywhere. If we define a boolean algebra as having at least two elements, then that algebra has a minimal element (0) and a maximal element (1). Since each element has a unique complement, 0* = 1 and 1* = 0. If I add a thir...

boolean-algebra

@BernardMeurer It seems highly nontrivial, and you don't get 2

ACuriousMind

@GPhys Sure, that's correct. I'm reluctant to say "if and only if" because you are allowed to transform $t$ too, after all, and so there might be strange symmetries that don't preserve the equipotentials because they compensate for it by strangely transforming $t$.

0celo7

You get $2^n$ like Mr. grouchy grouch said

@BernardMeurer Also a quick wiki search tells me that a Boolean algebra forms a ring, not a field

GPhys

@ACuriousMind I would be interested in an example of such a potential, but that does answer my question in the way I intended, thanks

0celo7

so it won't be a vector space over itself

it's just a module

Bernard Meurer

@0celo7 Well but does it forming a ring imply it not forming a field?

0celo7

12:26 AM

If it formed a field, they would say that instead of saying it's a ring

a field is more specific

ACuriousMind

Well, the two-element Boolean algebra is clearly a field

Bernard Meurer

@ACuriousMind Exaclty

ACuriousMind

Just the general ones might not be

0celo7

"clearly" -- only if you have a PhD in algebra!

I still don't know what a boolean algebra is

ACuriousMind

wait

Bernard Meurer

12:37 AM

0

Q: Proving that a Boolean algebra only has 2 elements

We define a Boolean algebra as a set of $B$ elements $a,b,\dots$ which satisfies the following axioms$^{[1]}$: $B$ has two binary operators $\wedge$ or $\cdot$ (logical AND) and $\vee$ or $+$ (logical OR) Idempotence $a\wedge a = a \vee a = a$ Commutative law $a \wedge b = b \wedge a$ $a \...

abstract-algebra proof-writing boolean-algebra

GPhys

@ACuriousMind this is only for a potential $V(q_1,q_2,\dotsc ,q_i)$, right? (as opposed to depending on, say, $\dot q$)

I was thinking with, e.g., the constant magnetic field problem (say, in the $z$ direction), translation in $x$ or $y$ does change the potential but is still a symmetry

but in that case the potential depends on $\dot x$ and $\dot y$

otherwise that's a counterexample to the if and only if, I guess

GPhys

1:13 AM

@ACuriousMind So, to recap, I should be able to prove this: Suppose $V=V(q_1,\dotsc ,q_i)$ is a potential that depends only on the generalized coordinates. Then, suppose $\zeta[\epsilon]$ is a coordinate transformation parametrized by $\epsilon$ that does not include a time transformation. Then, $\zeta[\epsilon]$ corresponds to a symmetry if and only if it transforms the potential like $V\mapsto V$. That is, such that the entire potential maps to itself.

(I just wrote that out so I can work on it later when I have more time)

ACuriousMind

@GPhys I don't think that's necessarily true

Consider the symmetry that leads to the conservation of the Lenz-Runge vector as an example

0celo7

@ACuriousMind Getting out my algebra notes

AHA!

I was wrong!

Dammit!

GPhys

@ACuriousMind Is there a modification of the statement to what you were saying earlier to make it true (that's my translation of what I thought you were saying before)

0celo7

The prime field of a finite field is $\Bbb F_p$

GPhys

looking at the Lenz-Runge vector now

ACuriousMind

1:22 AM

@GPhys Well...that was what I was saying before, then I thought a bit more about it and tried to find an example of a symmetry that's a) highly dependent on the potential but b) not really related to a spatial symmetry of that potential, and I remembered the 1/r potential/the Lenz-Runge vector.

GPhys

ah

@ACuriousMind Oh my

@ACuriousMind But it does work in one direction though, right? That is, if $V\mapsto V$ then it corresponds to a symmetry?

ACuriousMind

1:39 AM

Well, yeah, kinda tautologically. If you transform only the $q$ and keep the $p$ fixed, then the standard Hamiltonian of $p^2 + V(q)$ is invariant if $V\mapsto V$ under the transformation.

GPhys

Sure

err

yes

@ACuriousMind I was thinking it wasn't true, but $p^2$ can be changed by at most a total time derivative I guess

user218912

@0celo7 this guy in my class showed me that you don't even need to lower all the indices/ use dummy indices to do the lagrangian vector field derivatives. because the metrics cancel out.

user218912

so you can just take the derivative normally and get the same answer.

0celo7

@IceLord Correct, but that's not rigorous.

You first have to justify being able to do that.

user218912

well mine is rigorous af.

user218912

1:52 AM

like 7 lines of steps for each derivative.

0celo7

Well, no one actually does the derivatives like that

user218912

lol

0celo7

but now that you understand how to do them like that, you can do the shortcut

user218912

but in the process I learned index notation so it was good for me anyway.

user218912

brb

GPhys

1:59 AM

@ACuriousMind Or, maybe it's not so clear. The transformation can still depend on the $q_i$s of each other (like in rotations). Under this it is not so clear that $p^2$ only differs by a total time derivative?

0celo7

@IceLord Dammit I need hardcover Lee

ask more questions

wtf is "one-to-one into"

user218912

@0celo7 I understand everything so far.

user218912

problem set 2 in 1 day, I'll be asking a lot of questions then.

Sir Cumference

CRAP

need help

Anyone know how to import a vector from a pdf paper into microsoft word?

0celo7

a vector?

what does that even mean

Sir Cumference

2:08 AM

Basically an image whose quality doesn't get worse no matter how much you zoom in

As opposed to a raster (JPEG, PNG, etc.), which gets more pixelated when you zoom in

0celo7

does copy paste work

Sir Cumference

Nope

user218912

@SirCumference google is your friend.

Sir Cumference

@IceLord Hasn't helped

I need figure 1 from here

arxiv.org/pdf/0911.4872v1.pdf

0celo7

LINK TO THE ABSTRACT

user218912

2:09 AM

@SirCumference superuser.com/questions/302045/…

Sir Cumference

Why tho

0celo7

@IceLord nice

you destroyed him once again +1

user218912

lol

Sir Cumference

No, I'm afraid you haven't

I already got the vector graph out

I need to figure out how to put it in word

0celo7

control V

noob

Sir Cumference

2:11 AM

Doesn't work...

0celo7

Sure.

user218912

@SirCumference is it eps or svg or emf?

Sir Cumference

I've never figured out how to get vectors in word before. LaTeX equations (which retain their quality when you zoom) won't import, and pdfs won't either

@IceLord Pdf

0celo7

@IceLord I'm making a reading list for you.

user218912

for physics?

0celo7

2:13 AM

No

Well, GR books will be on there

Does that make it physics?

user218912

@0celo7 yes

0celo7

ok start reading Zee

user218912

I'm reading like 5 books right now.

user218912

@SirCumference does this help

0celo7

My shelf is looking good.

I removed the second to last physics book.

HE is now no longer on my shelf of Holy books.

user218912

2:15 AM

what's the last?

0celo7

Straumann GR

user218912

that has a lot of differential geometry in it.

0celo7

still the best reference for formulas in basic Riemannian geometry

user218912

I am still developing a book shelf.

0celo7

Well, I have three shelves + a pile

my "main" shelf is right next to my desk

by shelf I mean a single shelf, not a shelving unit.

user218912

2:20 AM

okay.

user218912

I enrolled in a grad stat mech course. that makes 3 grad courses this semester.

user218912

qft = okay, condensed matter = bird, stat mech = difficult af.

user218912

I already missed 3 weeks of lectures and I'm so lost.

0celo7

lol

there's this dude in my analysis class

he just showed up on monday

transferred in from single variable

prof's like "who are you"

he says "oh I transferred from 341"

Prof says "ok" ::turns around:: "so we were talking about the closure of point separating function algebras on compact metric spaces, right?"

that dude is screwed

user218912

That's like me in stat mech.

0celo7

2:27 AM

@MAFIA36790 hi

want to see my new books :P

user218912

@0celo7 how can he switch in this late?

0celo7

I don't know. But the class is really hard

user116211

@0celo7 Where? Where? Show it, man!

user116211

@0celo7 o/

Sir Cumference

@IceLord Forget that

I imported the pdf into Illustrator and cropped out the graph into an svg

Now how do I put it into word?

0celo7

2:29 AM

so pretty

user218912

@SirCumference ...

Sir Cumference

Yeah. Just found that...

Thanks tho

Now I realize I can just drag and drop eps files...

user116211

@0celo7 AMS Chelsea is awesome!!

0celo7

2:45 AM

@MAFIA36790 yes it is pretty good

the print quality is very nice too.

user116211

I'm very much liking the brown front-cover with those golden writings.

0celo7

yeah

but they're only old books, basically

AMS Chelsea reprints classical texts that are out of print

user116211

@0celo7 Like Dover?

0celo7

Yes

But these are much nicer ;)

user116211

Yeh, seeing that.

user116211

2:49 AM

Anyways, JD has pulled off his nomination. Now, it's four: Alfred, tpg, Jim-the king, ACM.

0celo7

hmm maybe ACM won't win after all

this tpg fellow seems to have it

@IceLord I need Lee :o

user116211

Frankly, I don't know very much about him.

user116211

@0celo7 NOOO!!

user116211

Most of our first choice was ACM.

user116211

ACM and Jim - that must be it.

0celo7

2:55 AM

I don't like this flimsy paperback

it's basically an embarrassment

user116211

Well, not bad though but nothing to AMS; I have Callahan from Springer; they look pretty same.

user116211

Only Bourbaki: Set Theory looks better in the Springer cover.

0celo7

oh I have some springer books

like 20

but this one is special

user116211

I have three.

0celo7

from this angle it looks fine:

user116211

2:59 AM

The Wileys are the worst.

0celo7

user116211

Nay; dull looking ;/

0celo7

cat tooth marks:

user116211

@0celo7 ;P

0celo7

and then the glorious binding:

user116211

3:01 AM

I have many dog teeth/foot marks too but not in the front cover.

user116211

@0celo7 Seems to be an old library book.

0celo7

the book is basically falling apart :/

@MAFIA36790 no it's mine

my 40 year old Spivak is in better shape than this 2 yo springer book

user116211

@0celo7 I know; but since the front cover was coming out, that's why the old, library feeling came.

0celo7

I didn't even abuse it

user116211

@0celo7 2yr? ohh!

0celo7

3:02 AM

the front just detached

my Shankar is holding up pretty nicely though

user116211

@0celo7 Wasn't there a 1 or 2 months guarantee?

0celo7

it's even older

@MAFIA36790 It lasted about 6 months until it did that

user116211

ohh.

0celo7

when I get rich I'll buy a new one

user116211

sure ;))

user116211

3:04 AM

Have you anything in mind @0celo, that might make you rich faster?

0celo7

lots of money?

user116211

~~apart from the stripper plan~~

0celo7

small million dollar loan perhaps

user116211

Ah!

user116211

Anyways, I studies Numerical Analysis yesterday and learned some pretty cool things:

user116211

3:09 AM

Newton's Forward Formula for equal width interpolation; Lagrange's Interpolation Formula, Numerical Integration - Quadrature Formula, Trapezoid Formula and Simpson's One Third Formula.

0celo7

I learned three of those some time ago

my calculus prof was a numerical analyst

user116211

@0celo7 They are pretty easy though.

user116211

@0celo7 ohh.

0celo7

@MAFIA36790 Bernard's fav. cover

@BernardMeurer ^

user116211

Holy hell! Can't say anything about that ;P I'm now very much fond of the brown cover; so can't expect a dark dull green cover.

0celo7

3:22 AM

hey that's my favorite mathematics text of all time...

user116211

@0celo7 Sure, but I'm talking about the cover; not the content inside.

0celo7

I happen to like the cover too

user116211

oops.

user116211

AMS has created a standard now; so ....

0celo7

the regular AMS books are not that nice

user116211

3:23 AM

ooh!!

GPhys

@MAFIA36790 Gaussian Quadrature?

If not, you certainly missed the most important one!

0celo7

for example

GPhys

although in practice you probably want some combination of trapezoid, romberg, and gaussian quadrature

0celo7

very good print quality but the cover is meh

GPhys

since trapezoid is probably the best for noisy integrals

and computers are fast these days anyway

0celo7

3:26 AM

AMS GSM books stick out like a sore thumb on math shelves

user116211

@GPhys: $$I=h\left[ny_0 +\frac{n^2}{2}\Delta y_0 + \frac{\frac{n^3}{3}-\frac{n^2}{2}}{2}\Delta^2 y_0 + \frac{\frac{n^4}{4}-n^3 +n^2}{6}\Delta^3 y_0 +\ldots\right]$$

user116211

General Quadrature formula.

user116211

@GPhys I have not read Romberg yet.

GPhys

@MAFIA36790 Look up Romberg integration for probably the simplest version of what you can do with normal expansions and error estimates using evenly spaced points

user116211

@GPhys Trapezoid is easier to compute than Simpson.

user116211

3:28 AM

@GPhys sure.

GPhys

@MAFIA36790 Gaussian Quadrature is generally the best you can do but you sacrifice evenly spaced points

user116211

@0celo7 Not that bad.

GPhys

also in practice you want to do things like trapezoid integration adaptively

user116211

Actually I have Kelley's topology which looks very much the same.

0celo7

WHAT

GPhys

3:30 AM

when you double the number of points in the trapezoid rule you can write it as the previous integral estimate + a function of half the points

0celo7

I NEED THAT BOOK

GIVE ME

user116211

@0celo7 But I have one!

GPhys

so like this you can adaptively calculate the error when doing trapezoid integration without adding extra point calculations

0celo7

How did you get it

It's out of print

user116211

@0celo7 It is pretty cheap here at Amazon.in.

user116211

3:30 AM

@0celo7 Noo!!

GPhys

@MAFIA36790 Do you know Python?

0celo7

@MAFIA36790 how about this

user116211

@GPhys no; I'm just a first year learner ;P

GPhys

oh

user116211

@0celo7 I have the pdf but haven't read.

GPhys

3:31 AM

@MAFIA36790 gaussian quadrature is amazing

0celo7

You can get Kelley topology for like $20, but it doesn't look anything like Helgason...

And it's a crappy reprinted paperback

user116211

@0celo7 Kelley is actually great; especially the introductory order-theory in the first chapter.

GPhys

@MAFIA36790 using $n$ sample points, you can get an exactly correct integral value for any polynomial up to order $2n-1$

with gaussian quadrature

user116211

@0celo7 Yes, now published by maybe Wiley or so. Wait...

GPhys

so, say, by sampling 5 points you can exactly integrate any 9th order polynomial

user116211

3:33 AM

@GPhys Holy hell!!

GPhys

so, for any function well approximated by polynomials (so, any smooth function), gaussian quadrature works extremely well

0celo7

@MAFIA36790 lol you like saying that

GPhys

@MAFIA36790 it gets better: gaussian quadrature is as simple as riemann sums computationally

(provided you have pre-stored "weight" values. but there are libraries of these weights widely available)

user116211

@GPhys This can be done, I know that, thanks to Weierstress Theorem; we want just a function to be continuous over the concerned interval.

user116211

amazon.in/gp/product/8181283287/…

user116211

3:36 AM

Ah! Springer!!

GPhys

well, the functions not "well approximated by polynomials" are e.g. functions with asymptotes in their derivatives like sqrt(x) integrated from 0 to 1, but actually gaussian quadrature still works better for these functions than other methods due to the way it samples

user116211

The original front cover was blue.

GPhys

the other functions where you don't want to use gaussian quadrature are extremely noisy functions

user116211

@GPhys I would definitely read that!

GPhys

where the polynomial-like assumption is clearly invalid and you should just be using the trapezoid rule

the correct way to deal with e.g. integrating sqrt(x) is to just do an integral transformation to get rid of the derivative asymptote

0celo7

3:37 AM

@MAFIA36790 lies

what is this

GPhys

but for noisy functions you should just use the trapezoid rule (the fitting of simpson's rule is bad in very noisy functions too)

user116211

@0celo7 Kelley Topology :(

user116211

@GPhys How can I use trapezoid rule if the polynomial assumption is invalid? It comes from the Quadrature formula with the assumption that the degree of the polynomial is one. I didn't get that.

GPhys

@MAFIA36790 Another way to interpret the trapezoid rule is just connecting the dots of your function evaluatoin points and calculating the integral under the shape you just traced out

indeed, this is where "trapezoid" comes from

it's a trapezoid between any two points

user116211

@GPhys ahh!

GPhys

3:45 AM

so that works pretty much just in general

user116211

What book did you follow @gphys?

GPhys

I learned it long enough ago to not know where I learned it originally

and if I was curious again it's easy enough to rederive

user116211

I'm now following Scarborough, Freeman and a bit of Whittaker.

user218912

I realized the only way I can learn stuff is to do exercises.

user116211

@IceLord Nay ;/

user116211

3:48 AM

You first learn the stuff and then apply it and furnish it doing the exercises.

user218912

nah

user218912

from what I've been experimenting with, I just go straight to the exercises and look up what I need to solve it.

user218912

then I remember it forever.

user218912

if I just read it I'll forget it in 5 mins.

0celo7

read harder

user116211

3:49 AM

Anyways, @gphys, I find Newton's General Divided Difference Formula easier to compute than the Lagrange's Interpolation Formula. Do you know what advantages I get using the latter?

user116211

@IceLord WTH!!

0celo7

@IceLord there are books w/o exercises

I have some

user116211

Bourbaki

user116211

No exercises.

user116211

But still a great book!!

user218912

3:50 AM

well my reading memory is bad.

0celo7

you don't need exercises if you prove all of math, dude

user218912

or that^

user116211

^^^

user116211

@0celo7 That's my point!!

0celo7

exercises have two purposes:

they're exercises

and they let you state things the reader should know, but you don't have the time/space/patience to prove them

for Bourbaki option two was not an issue so they don't have exercises

user116211

3:52 AM

@0celo7 They prove everything.

0celo7

Books like HE or Li or KN don't have exercises, but you'll write pages and pages trying to decipher what they're saying

user116211

They even taught how to write formulas.

user116211

Anyways, I'm leaving for the college.

0celo7

Bourbaki taught many modern mathematicians how to write mathematics

user116211

@0celo @IceLord o/

0celo7

3:53 AM

bye

user116211

@0celo7 nods

user218912

bye.

0celo7

writing birthday cards is the worst

@IceLord what page of gourgoulhon?

user218912

25

0celo7

highly nonstandard

burn the book

I'm trying to set mine on fire rn

user218912

3:56 AM

lol

user218912

how is that nonstandard?

user218912

ofc you would know better but

user218912

it seems reasonable to me

0celo7

I do know better

let me define spacetime

getting out the standard reference for Lorentzian manifolds

user218912

oh boy

0celo7

3:58 AM

A spacetime is a time-oriented Lorentzian manifold.

user218912

okay then what is minkowski space?

0celo7

$\Bbb R^4$ with the Minkowski metric and a time orientation

both orientations are equivalent

well, maybe not a time orientation

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