The h Bar

Slereah

00:09

Heh

0celouvskyopoulo7

yay

Slereah

The original paper on Schwartz's impossibility theorem is in french

lucky break

0celouvskyopoulo7

on the internet at my new place

Slereah

Hurray

You know it's quite lucky that out of all the weird untranslated papers there are out there, most of them are in french

0celouvskyopoulo7

@Slereah well, I would like to learn how to program a GR solver

Slereah

00:10

if they were in German or Russian that might be an issue

0celouvskyopoulo7

I would like to evolve a cubic star

Slereah

Although that could be the case and it's just that not a lot of people quote them because they don't read russian

@0celouvskyopoulo7 sounds challenging

The question I wonder is

If you pick some more or less arbitrary initial metric on the Cauchy surface

With a static distribution of matter

Will it evolve towards a stable metric?

0celouvskyopoulo7

hmm, yeah

so if you take the distribution at $t=\infty$, then plug it in for $t=0$, is it static?

Slereah

Well if that works, my guess is

Put Minkowski space or something at $t = 0$

Evolve it until it is stable

Then take that stable solution as the initial condition and check if it's stable

Also as said I'm mostly sure the metric will be $$ds^2 = -dt^2 + f(x,y,z) (dx^2 + dy^2 + dz^2)$$

Wait no, not quite

$$ds^2 = - f(x,y,z)dt^2 + g(x,y,z) (dx^2 + dy^2 + dz^2)$$

there we go

so you just gotta solve it for some induced metric on the Cauchy surface $g(x,y,z) \delta_{ij}$

With the lapse function $f$

Slereah

00:28

Apparently the proof of the impossibility theorem is just

Take the distribution defined by $x$ and $x^{-1}$

Then $x x^{-1} = 1$, but $(x \delta)[f] = 0$

So this means that $x^{-1} (x \delta) = 0$ but $(x^{-1} x) \delta = \delta$

Hence it's not associative

the proof is a bit longer because he really wants to show that a wide variety of algebras are impossible but at its core that's it

0celouvskyopoulo7

Impossibility theorem?

Slereah

That there's no algebra of distributions that's associative, obeys the Leibniz rule, contains the distribution $1$ as identity and works the usual way on smooth functions

(oh and contains $\delta$)

If you just take the algebra of distributions that are smooth functions it's fairly easy

0celouvskyopoulo7

How does the proof go?

Slereah

Basically showing that you can define an inverse function and that this implies that there are no elements $\delta$ of the algebra such that $x \delta = 0$

And that in $\mathscr D$ multiplication is always possible but not always derivation, and in $\mathscr D'$ it's the other way around

I think there's another similar proof that involves the Heaviside function and how the product of its derivation doesn't obey the Leibniz rule but I forget

Like $(H * H)' = H' = \delta$ but $(H * H)' = 2 H H' = 2 \delta$

0celouvskyopoulo7

00:51

Interesting.

Slereah

so basically you can only define algebras on distributions by either restricting it or expanding it

that's what the wavefront set and generalized functions do

wavefront set is fairly easy to understand, it's just that the product is well defined if it's well defined in the Fourier transform

$fg = \hat f \star \hat g$

which still lets you multiply rather singular distributions together

Like $\delta(x, 0) * \delta(0,y)$

0celouvskyopoulo7

01:24

@Slereah I tried to read Hormander about the wavefront set and it was unintelligible

Slereah

Read "a smooth introduction to the wavefront set" maybe

dmckee

@nbro You may have misunderstood. I wasn't suggesting that you've been waving any inappropriate part of your metaphorical anatomy, but saying that we don't put a lot of emphasis on level of formal education as a marker for expertise.

So by asking explicitly for that kind of thing you are bucking a social norm, which will, as usual with people, meet with a little friction.

To make the point @0celouvskyopoulo7 is very much my junior in formal education, but very much more accomplished in a wide variety of abstract pure maths—exactly the sort of stuff you need to master coordinate-free physics.

So suggesting that his opinion isn't valuable merely because he doesn't have enough post-nominal letters is a bit of a logical fallacy.

0celouvskyopoulo7

02:32

@dmckee thanks

loltospoon

02:45

Hello all

Could someone explain to me the general idea of how to compute 200 eigenvalues of the Schrodinger equation all at once using code?

Of course, matrices are involved here

GPhys

QM II final is over!!!!

4

I don't know how it went, because I spent the entire thing in the ER

so many emails to send

AmagicalFishy

03:30

I think I figured it out... but, the reason the proper time is a time interval is because the otherwise-length is implicitly multiplied by c... right?

David Z

@dmckee I must admit seeing "phallus waving" in the transcript threw me a bit

Avantgarde

quite the euphemism

user228700

06:20

Morning! :-)

John Rennie

Morning

What's this new Harry Potter web site?

user228700

06:39

Ah, that!

user228700

Hang on, let me show u:

John Rennie

:: John starts tapping his fingers and fidgeting impatiently ::

user228700

Sheesh, that took far longer than I anticipated, sorry.

user228700

Image incoming...

user228700

John Rennie

06:47

Is that a chat server?

user228700

@JohnRennie YEP! On Discord.

John Rennie

Discord?

user228700

It's the same platform that is hosting the other server (for the scavenger hunt) as well.

user228700

> All-in-one voice and text chat for gamers that's free, secure, and works on both your desktop and phone.

John Rennie

06:50

Hmm, Googling Hogwartaria finds nothing relevant ...

user228700

We're not gamers is all.

user228700

@JohnRennie I suspect it lay buried quite deep.

John Rennie

Did you know there is a species of dinosaur named after Hogwarts?

user228700

(Hogwartaria from Tuataria from the Tuatara on JG's wall that started the scavenger hunt)

John Rennie

Dracorex

Dracorex is a dinosaur genus of the family Pachycephalosauridae, from the Late Cretaceous of North America. The type (and only known) species is Dracorex hogwartsia, meaning "dragon king of Hogwarts". It is known from one nearly complete skull (the holotype TCMI 2004.17.1), as well as four cervical vertebrae: the atlas, third, eighth and ninth. These were discovered in the Hell Creek Formation in South Dakota by three amateur paleontologists from Sioux City, Iowa. The skull was subsequently donated to the Children's Museum of Indianapolis for study in 2004, and was formally described by Bob Bakker...

Slereah

06:52

mornin

John Rennie

The things you accidentally find while Googling ...

user228700

@JohnRennie Wow! I didn't know. Thanks!

Balarka Sen

@Slereah m

Slereah

Hey @BalarkaSen

user228700

@JohnRennie I can give u the link to join it if u want but I highly suspect that u'd feel out of place among hundreds of teenagers :-P

John Rennie

06:54

@Kaumudi.H yes, I don't think it's really my scene :-)

Slereah

Do you happen to know if there's a simple proof that if a subbundle is the retract of a bundle, then if there's a global section of the subbundle, there's a global section of the bundle?

Or is that something that requires an entire book to prove

user228700

It's so fun though! Once sorted, one can have access only to their respective common room (channel) and everything!

user228700

And we have Peeves as well! (A bot)

Balarka Sen

@Slereah Ah I have heard of this result, but I can't give you a proof. It's really interesting because the retract may have nothing to do with the bundle structure; it can distort the fibers very much, let alone being fiberwise.

I suspect any proof uses obstruction theory heavily.

Slereah

Yeah I think I'm just gonna say "See Steenrod"

I did more proof of the existence of the metric tensor on a manifold than any other GR book in history already

I'm having troubles even finding a clear enunciation of the theorem in Steenrod

user228700

06:59

@JohnR: How cool is this-

Slereah

There's plenty of theorems that look like it but it's blablabla cohomology blablabla retract over simplexes

user228700

Slereah

I don't know enough algebraic topology to cope

John Rennie

@Kaumudi.H Erm ... yes, cool - very cool ... :-)

Slereah

@BalarkaSen do you happen to know the name of that theorem

user228700

07:02

Ya know, @JohnR: sometimes I forget that you're 56 years old and might not relate to the same things as me. I have realised far too late that this is one such occasion :-P

Slereah

Having a hard time finding details

John Rennie

@Kaumudi.H by participating in a site like that you are in effect joining in a communally agreed story. In effect you're writing the story as you go along, and the suspension of disbelief and immersion is all part of the fun.

In that sense it's not so different from the DnD games that my friends used to play forty years (!!) ago and long before the invention of chat rooms.

And that can be a lot of fun, or so I'm told by apparently sane people :-)

user228700

Wow, was that worded well! :-)

Balarka Sen

@Slereah What does subbundle mean here? If it's an embedded bundle over the same base, then a global section of the subbundle immediately gives a section to the ambient bundle.

user228700

I agree. I had no idea that DnD has been around for more than 40 years, wow!

Slereah

07:08

@BalarkaSen In the case I'm interested in, the space of all Lorentzian metrics is $\approx O(n) \times Gr(n,1)$, so that the space of Lorentzian metrics can be contracted on the space of all non-vanishing vector fields

Which would imply that a metric is equivalent to the existence of a non vanishing vector field

John Rennie

@Kaumudi.H There are some who regard organised religion as merely an indulgence in a mutually agreed fiction, and that goes back a lot farther than 40 years :-) I'm not going to comment since i don't wish to be nailed to any pieces of wood!

user228700

Anyway, that server is nothing like DnD. It contains many channels to discuss HP but all that about McGonagall was just us goofing around for a minute. The bot has been programmed to say certain things when activated & that's what it was.

user228700

@JohnRennie Hehe, OK :-)

Balarka Sen

@Slereah It doesn't seem like the retract information is relevant at all. If a subbundle has a global section, so does the ambient bundle

Slereah

Does it?

Well that is good to know

John Rennie

07:13

Wow, there's a Pink Floyd exhibition at the Victoria and Albert museum

Slereah

I guess if that is true I might be able to write the full proof

Balarka Sen

Sure, a E' $\subset$ E be the subbundle, a section is $B \to E'$ taking point to each point in fiber. compose with the inclusion $E' \to E$

Slereah

True I suppose

I guess the retraction is only useful to say that this bundle is a subbundle of the metric bundle

Balarka Sen

I'm a little dull today, maybe I did not wake up properly yet

Slereah

I guess the simplest way to do it here is to take the section of $Gr(n,1)$ and take the zero section of $O(n)$?

Or whatever section is easy to do from it, I dunno

Although I'm not sure how that would work for the zero section, since the metric is $g = \sigma \alpha$, with $\sigma \in O(n)$, $\alpha \in Gr(n,1)$

Wouldn't the zero section just give the zero metric

which is not of the correct signature

In class right now so can't check Steenrod

Wait what's the zero vector in $O(n)$

It's a vector space no

I forget what the retraction retracts onto

Wait no I guess $O(n)$ doesn't form a vector space

It probably retracts on the identity I'm guessing

So just $\{ I \} \times Gr(n,1)$

What does a Grassmanian matrix look like, anyway

Slereah

07:52

I think it might be like $$\begin{pmatrix}
1 & 0 \\
0 & 0
\end{pmatrix}$$

or somesuch

No I guess not

Slereah

08:05

Ah, there we go

$$\begin{pmatrix} a & c \\ 0 & b \end{pmatrix}$$

Where the matrix has to be invertible

So $ab \neq 0$

And it has to be of rank $1$ I think so...

$$\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}$$

probably a correct Grassmannian matrix

So yeah I'm guessing it would make for a correct Lorentzian metric with a $O(n)$ matrix

Slereah

08:29

"Let $g : E' \to E$ be a $C^p$ vector bundle map between $C^p$ vector bundles over a $C^p$ premanifold with corners $X$"

I never asked for this ;w;

Shing

09:00

I am reading Feynman's Lecture Vol II, lecture 19, the least action principle.

it is just so damn good

I am always not convinced Newton Mechanics = Lagrange Mechanics

and indeed they are not same! (as Feynman said in his lecture)

Slereah

It is not, no

Lagrangian mechanics doesn't allow for non-conservative forces

Shing

We also derive F = ma from Lagrangian mechanics by 1st order approximation

I am left wondering how we get the Lagrangian for a system in the first place

the only way is by trial and error?

(as Feynman said in his lecture)

Kenshin

yes guess and check

just like how the laws of physics are created

guess and check

Slereah

Usually there are some clues as to what the lagrangian is

by symmetries and other such arguments

Kenshin

indeed, not entirely a brute force search

Slereah

09:09

for instance the lagrangian of a point particle is informed by the fact that it should be symmetric with respect to rotations, translations, and that it should be at most using second derivatives

First derivatives, even

Kenshin

@Slereah just because the laws of physics are symmetric with those things, how do we know the lagrangian must also be?

Slereah

Second in the EoM

This gives you the lagrangian $$L = f(v^2)$$

Some function of the square of the speed

@Kenshin You can show that this implies symmetry of the equation of motion wrt those symmetries

and hence measurable quantities

Kenshin

oh k ty

Slereah

The simplest Lagrangian for that is $f(v^2) = a + b v^2$

$a$ is a total derivative so it won't affect the motion

And by measurements, you can show that $b = m/2$

I'm not sure if there's an abstract argument for disallowing terms like $v^4$

or if that's just by experiment

Shing

thanks for the explaining...

although it sounds easy, I am not sure if I understand it... I am thinking of how to guess a free particle's Lagrangian... and I still can't quite get it

Slereah

09:24

Well consider it thusly

Kenshin

@Shing try guessing and checking many cases

with experience you'll learn to see what works

sure Slereah can tell you the patterns

but if you guess enough tmes, you'll see the patterns and truly recognise and understaand

Slereah

The Lagrangian will be (if we only allow up to first derivatives) a function $f(\vec x, \dot {\vec x})$

Shing

okay, thanks.

I guess the only way is to practice.

Kenshin

it's like chess, why do you move A to B? Iit is because you have already looked like 3moves ahead

The only way one can figure out why A is moved to B, is to think about what happens when A is moved to B

Slereah

Now perform a translation of this function

Avantgarde

09:26

@Kenshin that quote is going down in history

nbro

@dmckee from my experience, it's possible that people without a formal background education may also be good enough to answer my question, but I've found that in these chats people feel too confident about their knowledge and skills regarding certain topics, and most of the times they are actually incompetent. This doesn't mean that whoever has a formal background is able to answer every possible answer, but at least I'm sure it has a certain degree of experience in the field.

Shing

@Slereah I think I start to seeing it... so the Lagrangian of a free particle should be independent of $\ddot x$?

Slereah

$f(\vec x + \vec a, \vec v) \approx f(\vec x, \vec v) + \vec a f'(\vec x, \vec v)$

This is not symmetric

Kenshin

@Avantgarde thanks bro

Slereah

the same argument applies to rotations and functions of $\vec v$ that aren't $v^2$

Kenshin

09:28

@nbro what's your question bro i'll answer

Avantgarde

@Shing in many cases, yes. Though you can have second derivatives too

Slereah

Fundamentally, the lagrangian is only up to first derivatives. But I think for some effective lagrangians you can use higher derivatives

Shing

ohh, I think I get the correct impression now. thanks guys.

Avantgarde

perhaps the most important example of a lagrangian which has 2 derivatives, is general relativity

Kenshin

and in very special circumstaqnces 3 derivatives

Slereah

09:31

true

Kenshin

and sometimes ...

infinite derivatives

i made that up but it would be cool

Slereah

I think you get a lot of derivatives for like

Wave mechanics

for instance for modelling rogue waves

@Kenshin actually some lagrangians do have infinite derivatives

For instance the sine-gordon model

$$\mathscr L = g^{\mu\nu} \partial_\mu \phi \partial_\nu \phi + \lambda \cos(\phi)$$

Kenshin

nice find bro

Sha Vuklia

Kenshin

sup girl

ty for that remark

Sha Vuklia

09:35

Guys, does anyone know how they came up with this time scale factor $1/\gamma$?

@Kenshin lol you're talking to me?:P what remark?

Kenshin

ur second picture

it says "remark..."

Sha Vuklia

lol

yw

Slereah

It's the solution of the equation for the damped harmonic oscillator?

Sha Vuklia

yes

Slereah

I mean you can just check it by hand

Sha Vuklia

09:37

yea but i don't understand how they came up with it in the first place

Kenshin

everything in physics is guess and check

Slereah

Just input $Ae^{(a+ib)t}$ in the equation

Kenshin

we were just talking about that

how do you think the schrodinger equation was discovered?

Sha Vuklia

huh why? @Slereah

Slereah

I mean, I'm sure there's a 100% rigorous way of coming up with it

Kenshin

09:37

guess and check

Sha Vuklia

no but, I understand the equation of motion

Slereah

But just trying a few values is much simpler

Sha Vuklia

I just don't see why the oscillation will damp on a time scale of order $1/\gamma$

Avantgarde

@ShaVuklia when you need damping, you add a e^{-t} factor. But the exponential must be dimensionless. So the general way to damp is ee^{- \gamma t}, where \gamma has dimensions of inverse time

sorry for typo, im typing with my left hand

Sha Vuklia

hm, ok it helps that $\gamma$ has dimensions inverse time :P

like, i didn't realise that at first

Avantgarde

09:41

you want to know why it's there?

Kenshin

if you have decay of e^(kt) then the mean life time is 1/k

Slereah

$$\frac{d^2f}{dt^2} + \gamma \frac{df}{dt} = \alpha f \to \frac{d^2f}{f} + \gamma \frac{df}{f} dt = \alpha dt^2$$

Avantgarde

the reason is dimensional analysis

Slereah

Hm, I'm not sure this works for second order ODEs

Kenshin

the reason is also integration

Slereah

09:41

The old physicist trick

Sha Vuklia

yea well I know that $2\gamma=b/m$

so if I know the dimension of the drag constant, I guess inverse time will pop out

but I think what Kenshin says is what I'm looking for

about the mean life time

Avantgarde

yes

Sha Vuklia

oh, but $F=-bv$, so we get

$[b]=M/T$

Avantgarde

yeah so b/m = gamma = 1/t

Sha Vuklia

yes

Avantgarde

09:44

b/m is gamma, right? i forgot ... i did this years ago

ok

Sha Vuklia

$2\gamma=b/m$

the factor two makes the solution a bit cleaner

Avantgarde

right, we all want that

where are you studying waves from?

Sha Vuklia

well, i was supposed to use our horribly set up syllabus

but i found a different book

Avantgarde

which is?

Sha Vuklia

Introduction to Classical Mechanics by David Morin

lol

not moron :P

Avantgarde

09:48

wait what, moron? lol

Sha Vuklia

hahahhaha

sorry

:P

Avantgarde

lol i misread that

Sha Vuklia

i mistyped it too

Avantgarde

oh so you deleted it? didn't see that

Sha Vuklia

yea i edited it :P

Avantgarde

09:49

ive never heard of this book. how is it?

Sha Vuklia

ohhhh the book is a-ma-zing

Slereah

Is $Gr(n,1)$ even a subbundle of $O(n) \times Gr(n,1)$

Sha Vuklia

i hated physics this year

Slereah

I don't think so since I don't think either is a vector space

Sha Vuklia

i study both physics and math, and a month ago, i was considering quitting physics

Slereah

09:50

I think I do need the retract theorem

Sha Vuklia

because it was taught sooo badly

but this book saved me :P

Avantgarde

wow. how do you feel about it now? yeah, bad teaching can be detrimental to crucial motivation

well, good :)

Sha Vuklia

oh, i looove physics now :P

Kenshin

r u at uni @ShaVuklia or high school

Sha Vuklia

i even explained a lot of it to my dad last weekend, which says a lot

because

he literally thought there was nog gravitation on the moon

Kenshin

09:51

lol

Sha Vuklia

but eventually he could even keep up with the ideas of special relativity :P

Kenshin

my mum told me gravity was due to the earth's rotation

Sha Vuklia

@Kenshin 1-st year uni

lol my dad said: there is no gravity, because people float on the moon :P

Kenshin

lol

Sha Vuklia

luckily he could accept the fact that the moon has 1/6 mass of the earth

Avantgarde

09:52

lol

Kenshin

yteah it always looks like ppl are floating on it

Avantgarde

That's impressive. It can be hard explaining science, let alone physics, to people who are nowhere near it

Kenshin

@ShaVuklia what uni course r u doing that has physics in it?

@Avantgarde na physics is simplz

just ppls have misconceptions that's it

Sha Vuklia

i had classical mechanics and special relativity, then i had astro physics, then vibrations and waves, and now electromagnetism and intro to quantum

Kenshin

na

i mean

what degree r u doing

what type of degree has physics in it?

but cool courses btw

Avantgarde

09:54

Well, it's subjective. Every one has a different way of looking at things and understanding them.

Sha Vuklia

well it's physics?

Kenshin

oh a physics degree?

Sha Vuklia

yea

Slereah

why would you do special relativity before wave mechanics and EM

Kenshin

nice

Slereah

09:54

that is a poor idea

Sha Vuklia

physics and mathematics

Kenshin

i did physics too

and mathematics

Sha Vuklia

@Slereah it is. it traumatised me

Avantgarde

I wish I had some math too

Kenshin

y u copying me 4

what do you wnat to be @ShaVuklia?

Sha Vuklia

09:56

i'm not sure what, but i want to do mathematical physics, maybe combined with theoretical physics, as my graduate.

Avantgarde

Special relativity in the first year is a bit too much, given that you have other elementary courses and math to learn

Kenshin

yeah sounds good bro

@Avantgarde no way dude

I learnt special relativity in the first 6 weeks of my course

Avantgarde

congratulations!

Kenshin

yeah i reckon learn it asap

u know there are 14 year olds who know it so why would you delay it to 2nd year

yuggib

studying both physics and mathematics seems a good strategy to understand none of the two

Sha Vuklia

09:58

it's nice if at least you know how matrices work, so that you can understand what transformations even are

Kenshin

nice, but not necessary

Avantgarde

lol yuggi, I agree to some extent

Kenshin

matrices aren't required to understand the physics of special realtivity

Sha Vuklia

why tho? @yuggib

Kenshin

@yuggib lol nice troll

they are two strongly entwined subjects

yuggib

09:59

I am not trolling. I actually think that studying both would imply learning less things on both than you would focusing on just one

Sha Vuklia

i think studying math is great for physics. you get the real mathematical explanations for stuff like taylor approximations and linear transformations and the like. but you only need like 20% of the math you learn for physics. but it's still a good 20%

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