hmm , I think I need something clearer lol

I am cursed. . I can't understand it lol

@Danu I know

I had this notion that it might also refer to penrose notation for some reason tho

hi guys can i delete an old post of mine that I feel is useless for this website (physics.stackexchange.com/questions/172360/…) , it drew almost no attention, etc

it's a homework type of post

@no_choice99 If you see the "delete" text at the lower left of the post you can delete it.

I believe the system only prevents owners from deleting material that has upvoted answers or is locked by a moderator.

ok thank you very much

ok finally figured it out . . . physics.ohio-state.edu/~mathur/italy1.pdf

was not that hard either, just a lot of substitution, then some regular plots accompanied by new vocabulary

exciting stuff

Just got panda express . . . . yeeehaaaa!!

So what I have learned so far from trying to do path integrals- - - I never really learned calculus well enough

----- QFT is a big integration orgy

QFT has a shitload of integrals

especially once you start doing loops

It's amazing people actually sit down and compute these things. I used to think at least some computer thing could understand and help in analytically computing these things but neh, gotta do it by hand lol

I am going to go for a run and then come back , and try to see if today would be the day I understand some of the gr things that have been eluding me for years. Hopefully I can find a gr book that is written in the spirit of anthony zee. Zee reduced the path qft to a regular integral, and introduced feynman diagrams at that point already, then upped the anti gradually . . . . . till the end. It was quite an enjoyable excursion.

@0celo7: Something to chuckle for you: Are Bajorans of human origin?

@kevinTahN. No one learns calculus well enough until they have to really apply it. Except Feynman, of course.

Hey does anybody have a copy of library.seg.org/doi/abs/10.1190/1.1441417 ?

My school's library apparently doesn't have it...

This hat is just great

Feynman creates calculus for his own understanding @dmckee

According to his quote, what I cannot create, I do not understand

yeah, but he also boasted a lot

That's his personality.

@skullpetrol He could've created a well-defined path integral then :P

:D

Susskind said Feynman always looked for the simplest way to approach problems.

Feynman's scientific style was always to look for the simplest, most elementary solution to a problem that was possible. If it wasn't possible, you had to use something fancier. But no doubt part of this was his great joy and pleasure in showing people that he could think more simply than they could. But he also deeply believed, he truly believed, that if you couldn't explain something simply you didn't understand it.

His creations that he did understand must have been made as simple as possible.

I would argue that the fields have grown more complex, since his time, and that today a simple explanation is not feasible.

For some things, yes.

But the danger of not being able to see the forest for the trees will always exist.

@dmckee Feynman is the man. I recently read "Quantum man" by Krauss. There are all sorts of achievements attributed to him that I was not even aware of. Also yes, I am only beginning to appreciate how far calculus can go. I need to work hard to be able to begin applying it. The good all days of just calculating areas, volumes, velocities etc are long gone. I have to be an adult now lol

Feynman is the man.

@kevinTahN. functional integration, I highly doubt, but there is SymPy for doing some symbolic calculations in Python.

(and incidentally half the point of Wolfram Alpha is to show step-by-step solutions for things)

well automatic integration exists, but its probably harder compared to get a mathematica document to integrate for you...

well that statics are different...

Fuck, I have a large project due in 10 hours and I can't do the derivation.

Help me o'atheist god

@Mikhail Dawkins can't help you physics

@kevinTahN. What else do you do with calculus that is not a fancy version of the things you just listed

@ACuriousMind I already know the answer Mr. Ayy Lmao

Also fancy hat...

@Slereah dear god

What do you guys think of my answer for this: physics.stackexchange.com/q/223788/6049

@Slereah we should write a GR book in only penrose notation

@Slereah just scan in our drawings

There is a group theory book all in penrose notation

I know

@Slereah apparently $\Lambda^kT^*$ is a functor on $\mathsf{Diff}$

@Slereah hmm apparently the dimension of a manifold can be nonconstant across the manifold

wot

What manifold has non-constant dimensions

a non-pure manifold

wat

stop trol statements

fu

apparently topology only says that $n$ is locally constant

what's an example tho

also what's the transition function between atlases gonna be

no clue

just lies

empty statements

so bad

example: $S^1\sqcup\mathbb{R}^2$?

Hm, I guess

So only disconnected manifolds, tho, mb

we could ask the math chat

who on earth would want to define manifolds to allow such things

well according to this book the definition of manifold does not forbid it

well yeah, if you define them differently than usually

usually you define an $n$-manifold as something that is Hausdorff, 2nd countably and locally homeomorphic to an open set in $R^n$

never seen anyone define a "manifold" as some object such that for each neighbourhood there exists a natural number $n$ such that ...

he fixes $n$ so he's contradicting himself

what book is that

"Natural Operations in Differential Geometry"

what are you reading it for

GR

wtf is a germ, anyway

equivalence class

I can read wiki too

what does that mean

ok then read wiki

no

then prove that the different definitions of tangent vectors coincide

I've done that before

it's quite boring

ok

then why don't you know what germs are

I did

This is a germ

I just forgot

can you prove that the usual basis vectors for the tangent space are linearly independent with the germ definition

The proof of that is in Wald. Or Straumann.

so Wald and Straumann can

yah

Straumann, Straumann

Can do whatever a Strau can

@0celo7 is that just a DG book or something else?

@Huy GR

ah ok

are there any somewhat common applications of $RP^n$ in physics?

string theory

your favourite field right

no

:(

I would be more interested in it if 1) there were good books on it 2) SUSY didn't suck

3) physicists could explain group theory

what's SUSY

there's an intro course on it next semester at my uni

wondering if I should attend it to see what it's about

@Huy is that what you wanted

SUSY is supersymmetry

It is the maximal extension of the Poincarré group or some shit I forget

With group elements to switch between representations

that's $\{0\}$

@0celo7 I know some group theory

can you calculate tensor products in $E_8\times E_8$

No this is it

maybe if I cared about it

@Huy that's the easy shit

well yeah it's an intro course

not superstring theory

it says intro to string theory

not superstring theory

so ELI5

ok well that's fine

You can also just read non-string SUSY

did he get banned

no

Did I?

@Huy so is that what you wanted me to prove??

no but nvm

ELI5 what I'd learn in that course

I wanna see if it's at all interesting for me if I don't do anything more than that intro

do you want some lecture notes that cover that stuff?

no I want an ELI5 from you tl;dr

you'll see that the string contains the graviton

if you ask me "what do I learn in functional analysis" I would try to summarize some of it and not give you a book

then the prof will go ape shit and say string theory needs more funding

what's graviton

quantum of the gravitational field

we have enough funding

over here profs don't need to beg for more funding

Gravitons are closed strings

Well one mode of a closed string, IIRC

basically you will show that the closed string has several modes

From what I vaguely recall closed strings pull double duty as graviton and dilaton

graviton, Kalb-Ramond and dilaton

What is Kalb Ramond

two-form gauge field

gross

is it true that nothing in String theory is yet verifyable because we can't do such measurements yet

the dilaton gives you the expansion parameter in the string perturbation series

also some shit with the Euler number of the worldsheet

it's not very interesting tbh

@Huy well it reproduces everything we know of, sorta

but I'm not sure what the new predictions are...and we certainly can't measure them

what's the 7 or 8 dimension fuzz about

it's 26, 11 or 10

ah ok

for what? for the world we live in?

26 in baby string theory

11 in super string theory

10 in M theory, which is not a string theory

but somehow string theorists study it...not sure what that's about

I quit trying to string theory when there were stupid things like 246 dimensional reps of $E_8\times E_8$ floating around

heheh

and then they threw in some dynkin diagrams for fun

for what? for the world we live in?

I know some dynkin diagrams

for $E_8\times E_8$

and what does E_8 \times E_8 model

or $\mathrm{SO}(32)\times\mathrm{SO}(32)$

much simpler

at least I know what the fuck $\mathrm{SO}(32)$ is

What is it

32-dimensional special orthogonal matrices

Is it 5 bits of rotation

Well yes but why 32

because Witten

what is it rotating

Flavors?

meat

ok so not interesting

waits for @ACuriousMind to see the horror

More importantly

Do they have a diagram for SO(32)

Maybe some Young tableaux

no

Or maybe

Penrose notation for SO(32)

one of my profs hated penrose

he might have had a good reason

he said that some scientists are really good at research and some are really good at selling themselves well, and that penrose belonged to the second group

String theory ^

Help

I need somebody

Isn't $B_{\mu\nu}$ the torsion field

I seem to recall something like that

I think it is...

I don't really care anymore.

@Huy for?

did you ever complete that calculation?

maybe after the pedagogy "exam"

I am "studying" for it atm

so why do you need somebody

cuz Paul said so

Paul?

..

dunno who Paul is

@Huy ??

@0celo7 !!

this is such a waste of time

@ChrisWhite I think the statics tags are fine, the -fields can go

@Huy what

@0celo7: pedagogy shit

20mins till exam

@ACuriousMind you should give @Huy the tl;dr on string theory

I'm too confused to do it

Haven't slept in a while

I don't think there is a tl;dr on string theory

@ACuriousMind he wanted one

chat session!

@0celo7 Uh...it starts in 2 hours

@ACuriousMind Being in your time zone is crazy

@ACuriousMind how does one live 6 hours ahead of civilization

My barbarian mind does not understand this question.

@ACuriousMind Ooga booga frooga doo?

SUSY, where are you?!

Don't leave us!

For those who missed it: The CMS announcement just revealed that there are no SUSY signals in their current data.

@0celo7 define civilization

@yuggib people who live in [GMT-5].

@Danu That'll make a lot of people quite sad

@0537 Then I prefer to live in uncivilized countries where you can find something e.g. like this:

I'm too dumb to do the ODE $f''-\frac 2gf'g'=0$, $f$ and $g$ being both variables in $x$. I tried to divide by $f'$, getting $f''/f'=2 g'/g$. How to continue? Both sides could be any function of $x$, can't they?

@Bass Hint: Logarithms.

ba dum tss! thanks!

it's always logarithms

Anyone know what the numbers under your little avatar in chat mean?

For example, @yuggib has 4507 underneath.

I think it's supposed to be total network-wide reputation

oddly enough I'm not seeing those numbers now

Chat session?

Half an hour-ish

as in the Dirac delta function?

@DavidZ Yes.

Hm. Well, if you want mathematical rigor, I have no idea, but in practice I consider it to be an integral over the subspace on which $f(x,y) = 0$

oh wait, that'd be if it's $\int \mathrm{d}x\, \mathrm{d}y\, \delta(f(x,y))$

hmm...

@yuggib U.S. of A.

@DavidZ Yes, indeed.

@DanielSank For "nice" $f$, you have $\delta(f(x)) = \sum_i \frac{\delta(x-x_i)}{\lvert f'(x_i)}$ for the real roots of $f$.

@ACuriousMind Indeed, but what happens when I want to integrate over one variable?

Treat $y$ as a constant, I suppose

@DavidZ Right, if we were integrating over all space we could e.g. parametrize the manifold on which the function is zero and then just do the integral.

@DavidZ I'm having trouble understanding what that means.

>manifold

Treat it like 5 :-P

I mean, can you elaborate?

it could be some plane curve which has self-intersections, no?

So, you get $\int \delta(f(x,y))\mathrm{d}x = \sum_i \int \frac{\delta(x-x_i)}{\lvert f'(x_i,y)\rvert}\mathrm{d}x = \sum_i \frac{1}{\lvert f'(x_i,y)\rvert}$ since $\int \delta(x)\mathrm{d}x = 1$.

Hmmm

@DavidZ Thinking.

I see your problem now, if I write another $y$-integral the results are completely different depending on how one expands the $\delta$.

@ACuriousMind huh?

@ACuriousMind This seems wrong.

what is $f'(x_i,y)$ even

Ah, the $x_i$ don't make sense!

@0celo7 It's probably $\partial f / \partial x$.

@ACuriousMind exactly

$f(x,y)$ doesn't have roots "in one variable"

^^

that's a middle school mistake, mister

So...I'd guess $\delta(f(x,y)) = \sum_i \frac{\delta(x-x_i)\delta(y-y_i)}{\lvert \nabla f(x_i,y_i) \rvert}$ for $(x_i,y_i)$ the roots of $f$.

^^

But it is probably wrong, the denominator especially could be different

@ACuriousMind Interesting.

@ACuriousMind how does one prove it in 1 dimension again?

@ACuriousMind Actually I bet that's right and can be proved in the same way as for one variable.

I'll try to derive it.

linear expansion and change of variables?

@0celo7 It's easy:

$f(x) \approx f(0) + x \, (df/dx)(0) + \cdots$

so...what I just said

@0celo7 Yes, jeezus I was typing.

@DanielSank Ah, yes, you are right, the Taylor expansion just gives the gradient.

@ACuriousMind yeah

Write the delta distribution as the limit of a mollifier

substitute in the mollifier argument the function $f(x,y)$

and compute the integral in $x$

then take the limit and you should get the distribution corresponding to $\int dx \delta(f(x,y))$

of course it should only be true for regular enough $f$s

Assuming $x$ and $y$ are independent, it should work out that $$\int\mathrm{d}x\,\delta(f(x,y)) = \sum_{i: f(x_i,y)=0} \frac{\delta(x - x_i)}{\partial_x[f](x_i, y)}$$

I think

@DavidZ the condition $f(x_i,y)=0$ doesn't really make sense unless you expect that $f$ is zero on an entire line

@yuggib Yeah that works too, but I was hoping to see it another way.

@0celo7 : Hm.

@0celo7 Wrong way around, 11 in M theory, 10 for superstring, I think.

@ACuriousMind Thanks for the push. The important thing, in some sense, is that the first order terms in a multi-variable Taylor expansion only involve one variable at a time.

@ACuriousMind I don't see what you mean... $f(x_i, y) = 0$ seems like a perfectly sensible equation, and given $y$ you can find values of $x_i$ for which it holds

@DavidZ Oh you mean the sum itself to be dependent on $y$!

Hmmmm

Yeah, of course

In physics-world, at least, $\int \mathrm{d}x \delta(f(x, y))$ implicitly requires that $y$ have some kind of externally-given value

or so is my understanding

@DavidZ so the indices may be an uncountable number, and therefore the sum is an integral...

@DavidZ Doesn't it just yield a function of $y$?

@DanielSank Yep

@DavidZ Why does $x_i$ have a subscript but $y$ doesn't? Is $y$ standing for some arbitrary number of dimensions here?

@yuggib sure, but you can have that in 1D delta-function integrals too

@DanielSank In my opinion the only hope is to get a distribution acting on one-variable functions

@yuggib Well, let me explain why I asked this. I have a probability distribution $P(x,y) = (1/r) \delta ( x^2 + y^2 - r^2)$. I want to know the marginal probability of, say, just $y$.

@DanielSank because $x_i$ represents a root - a value of $x$ (as a function of $y$) which makes $f(x_i, y) = 0$.

@DavidZ Your l.h.s has an integral over $x$ and the r.h.s still depends on $x$, something has gone awry there

@DavidZ Oh, $x_i$ is a root... ok.

@ACuriousMind oh right, forgot an integral

@DavidZ Okay

$$\int\mathrm{d}x\,\delta(f(x,y)) = \int\mathrm{d}x\sum_{i: f(x_i,y)=0} \frac{\delta(x - x_i)}{\partial_x[f](x_i, y)}$$