The h Bar

ooolb

21:00

delusional people?

CooperCape

oh right

I see

So neither of them have very intelligent not asshole people?

Slereah

But

One thing is

Is the reparametrization a solution to the EoM?

Oh wait

i guess the EoM changes with the reparametrization

ooolb

@CooperCape there's no subreddit for that

because it's not entertaining

CooperCape

Oh, right

Slereah

$$\ddot x = 0 \to f' \ddot x + f'' \dot x = 0$$

ooolb

21:16

@CooperCape we should make a subreddit for those kind of people and call it r/@BalarkaSenRecentMessages

r u flattered bsen

CooperCape

Not sure if it would pass though...

rather specific

Good pr for h bar though, right?

Balarka Sen

I think I am a moderately intelligent asshole so that doesn't sound right

ooolb

i think it's best for all of us that the h bar doesn't become popular

just sayin

we need to lay low

Balarka Sen

Let's keep the SE chat top secret

ooolb

i mean the h bar chat logs are already indexed by google

that's already quite bad

Balarka Sen

21:21

truly

@ooolb Look at the bright side though. We say a lot of edgy shit that's recorded in the internet, but at least it's not as recorded as Harry Gindi

do you know who that guy is

google him. you should find a encyclopedia dramatica page

thank me later

ooolb

see also: math, autism, forums

lol the hyperlinks

that's hilarious

Anonymous

@BalarkaSen He looks like the JD of math

Balarka Sen

indeed...

ooolb

well JD didn't know any physics at all

Slereah

He doesn't need to

He has Einstein and the evidence

ooolb

21:27

was gindi a crackpot or just being edgy and also not wanting to admit he is wrong?

Balarka Sen

gindi wanted to rewrite mathematics using topos theory

he was also an ass

and wanted to fight with everyone who would say his stuff are wrong-headed, uninspiring, or perhaps purely wrong

0celo7

talk moved to thursday due to inclement weather

ooolb

where are you right now @0celo7

0celo7

why

ooolb

i don't think anything is ever cancelled here unless it's like -50 or 3+ feet of snow (overnight so it can't be plowed in time)

@0celo7 i wanted to check the weather

Slereah

21:35

@BalarkaSen is he nlab

bolbteppa

This is a nightmare

0celo7

@BalarkaSen is it obvious that homology classes have smooth embedded representatives?

Balarka Sen

It is not obvious because it is false

0celo7

when is it true

bolbteppa

Spot the error

\begin{align}
L &= -\frac{1}{2 \pi \alpha'} \sqrt{(\dot{x} \cdot x')^2 - x'^2 \dot{x}^2} \\
\frac{\partial L}{\partial \dot{x}_{\mu}} &= - \frac{1}{2 \pi \alpha'} \dfrac{(\dot{x} \cdot x') \dot{x}^{\mu} - x'^2 \dot{x}^{\mu}}{\sqrt{(\dot{x} \cdot x')^2 - x'^2 \dot{x}^2}} \\
\frac{\partial^2 L}{\partial \dot{x}_{\nu} \partial \dot{x}_{\mu}} &= -\frac{1}{2 \pi \alpha'}\{ [(\dot{x} \cdot x') \dot{x}^{\mu} - x'^2 \dot{x}^{\mu}] \frac{\partial }{\partial \dot{x}_{\nu}} \dfrac{1}{\sqrt{(\dot{x} \cdot x')^2 - x'^2 \dot{x}^2}} + \\

Slereah

21:42

The error is that the tex doesn't parse

Also you couldn't pay me to read this

bolbteppa

Last thing should be equal 0

Slereah

The error is that it's not equal to zero

Balarka Sen

@0celo7 Consider the dumb example of $2\alpha$ in $\Bbb R^2 - \{0\}$ where $\alpha$ is the generator of $H_1$.

bolbteppa

This is why no book ever includes it

0celo7

that is a stupid example

ok, let's consider $H_k(M,\Bbb Z)$, and I want a representative of a class

Balarka Sen

21:44

Yes, because you asked a stupid question. The right question should be when it can be represented as a map from a manifold

0celo7

then $\alpha$ would be the rep, which is just the unit circle

@BalarkaSen what?

Balarka Sen

@0celo7 Huh? I just said $2\alpha \in H_1(\Bbb R^2 - 0)$ can't be represented by an embedded submanifold. This is not the right question.

You should really ask if any homology class $\alpha \in H_k(M; \Bbb Z)$ can be written as $\alpha = f_*[N]$ where $f : N^k \to M$ is a map from a $k$-manifold.

(In this case, an immersion of S^1 in R^2 - 0 of winding number 2 does the trick)

bolbteppa

Holy s

If you read the beginning of this, it's like a sneaky trick to avoid this calculation

3

Q: How to find the rank of the matrix $\frac{\partial ^2\mathcal{L}}{\partial \dot{X^\mu} \partial \dot{X^\nu} }$ for the Nambu-Goto string Lagrangian?

In this case $$\mathcal{L}~=~-T\sqrt{-\dot{X^2}X'^2+(\dot{X}\cdot X')^2}.$$ I was reading some books and papers about the constraints in the Nambu-Goto action, and all of them say something like the matrix has two zero eigenvalues corresponding to the eigenvectors $\dot{X^\mu}$ and $X'^\mu$,...

With the two eigenvectors

0celo7

@BalarkaSen um, ok?

bolbteppa

That is potentially genius

0celo7

21:48

I'm just trying to make sense of the intersection form

Balarka Sen

@0celo7 then it's an interesting question. The answer is still no, but always true if $k < 7$

Oh then you're in the world of 4-manifolds

0celo7

@BalarkaSen wtf

Balarka Sen

yeah it's Thom's theorem

34

Q: When is a Homology Class Represented by a Submanifold?

Possible Duplicate: Cohomology and fundamental classes Given an oriented manifold $M$ and an oriented submanifold $\phi:N\to M$ we can obtain a homology class $\phi_*[N]\in H_*(M)$ where $[N]$ is the fundamental class of $N$. In general, it is not true that every hom...

at.algebraic-topology gn.general-topology

0celo7

christ

where does one even learn this stuff

bolbteppa

We can see $\dot{X}^{\mu}$ is a null eigenvector from
$$\frac{\partial ^2\mathcal{L}}{\partial \dot{X^\mu} \partial \dot{X^\nu} }\dot{X^\mu}=\frac{\partial }{\partial \dot{X^\nu} }\left( \frac{\partial \mathcal{L}}{\partial \dot{X^\mu} }\dot{X^\mu}\right)-\frac{\partial \mathcal{L}}{\partial \dot{X^\nu} }=
\frac{\partial }{\partial \dot{X^\nu} }\left( \frac{\partial \mathcal{L}}{\partial \dot{X^\mu} }\dot{X^\mu}-\mathcal{L}\right)=0$$
but this depends on computing the Hamiltonian for this specific NG action and seeing it's zero, (circular since we wanted to compute the Hessian and then use …

Balarka Sen

21:51

@0celo7 Oh. I am a little wrong.

You can represent $2\alpha$ in $H_1(\Bbb R^2 -0)$ by an embedded submanifold

Just consider two disjoint concentric circles centered at 0

Basically $\alpha + \alpha$

You can't represent it be a connected submanifold

Sorry

0celo7

where does one learn this

Bredon?

Balarka Sen

I have no earthly idea

0celo7

well how do you know it

Balarka Sen

I picked it up, from here and there, from books of lore and tales of old

Do you want to understand the intersection form for 4-manifolds?

bolbteppa

That procedure only gives us one constraint, still need to find another, and then using those two constraints we form the constrained Hamiltonian

0celo7

21:54

My advisor gave me a copy of conference proceedings he edited and there's stuff about 4-manifolds in it

but if you can't give me a book that explains it, I say fuck it

can't be bothered

Balarka Sen

lots of 4-manifold texts out there

0celo7

I don't want a 4-manifolds text, I wanted to read the 20 pages in this book

Balarka Sen

in which book

actually i don't want to know

i have stuff to be doing

cya

bolbteppa

Oh my god, so usually in physics your Lagrangian is not reparametrization invariant but your EOM are, but if your action is reparametrization invariant, you have $\int dt L(q,\frac{dq}{dt}) = \int d \tau \frac{dt}{d\tau} L(q,\frac{d \tau}{dt} \frac{dq}{d \tau})$ implying $\frac{dt}{d\tau} L(q,\frac{d \tau}{dt} \frac{dq}{d \tau}) = L(q,\frac{dq}{dt})$ i.e. $\frac{1}{a}L(q,a\dot{q}) = L(q,\dot{q})$ so that from $L(q,a \dot{q}) - a L(q,\dot{q}) = 0$ we have

$\frac{d}{da}[L(q,a \dot{q}) - a L(q,\dot{q})] = \frac{\partial L}{\partial \dot{q}}\frac{dq}{dt} - L = H = 0$

So the Hamiltonian is going to be zero for NG automatically

Slereah

yes

Also true of GR, since it is diffeomorphism invariant

bolbteppa

22:07

I literally remember skipping this parametrization thing in CoV because I had no feel for it

bolbteppa

22:32

Why does reparametrization invariance imply there are no secondary constraints

Slereah

It does?

Errr

Isn't there some relation for secondary constraints

like $[H, \phi] = 0$ or something

And since $H = 0$

bolbteppa

(S)he says it here between eq. 5 and 6 physics.stackexchange.com/a/184756/25851

Slereah

It will always be true

bolbteppa

Maybe it's because the Hamiltonian is zero it implies it

Slereah

or something yeah

ACuriousMind

22:33

What @Slereah said - reparametrization invariance implies zero Hamiltonian, and zero Hamiltonian implies no secondary constraints.

Slereah

yeah it's like

Every constraint will be true no matter if it's on shell?

ACuriousMind

@Slereah No, it's like - the time evolution of all constraints is trivial, because time evolution is trivial, so there are no additional relations to satisfy

CooperCape

Lol @BalarkaSen how have I only just heard about Logan Paul doing a video where the thumbnail had a dead body smh.

bolbteppa

So secondary constraints seem to arise from forming Poisson brackets of the constraints with the constrained Hamiltonian reducing to non-trivial Poisson brackets of the constraints with the canonical Hamiltonian, but in this case they are zero automatically?

ACuriousMind

The secondary constraints are what has to be satisfied so that initial data that fulfill the primary constraints still fulfill them at all later times - $[H,\phi]$ is simply the infinitesimal time evolution, and it has to be equal to zero modulo the constraints, so these are the secondary constraints.

bolbteppa

22:39

Hmm, so you take a time derivative of primary constraints, and this becomes a PB with the constrained Hamiltonian, and this should be zero (or weakly zero whatever that means), i.e. if $\psi_a(p,q) = 0$ is your constraint, then $\frac{d \psi_a}{dt} = \{ \tilde{H},\psi_a \} = 0$ should hold so the constraints are constant and hold for all times, which means $\{ H_c + u^b \psi_b,\psi_a \} = \{H_c,\psi_a \} + u^b \{\psi_b , \psi_a \} = 0$, but if $H_c$ is zero something something

ACuriousMind

@bolbteppa The second bracket also vanishes since you can assume without loss of generality that there are no second-class constraints. If there are, you can eliminate them by replacing the Poisson bracket with the Dirac bracket on their constraint surface.

0celo7

I need to rearrange my shelves

Slereah

your spice rack

0celo7

the books

Slereah

I don't have a lot of things on my spice rack

Salt

Tabasco

cinnamon

0celo7

22:50

@Slereah that shit is evil

Slereah

It's alright

I don't even have pepper

0celo7

wtf

I want to move the Reed and Simon saga to the analysis section

bolbteppa

Gah, why Dirac why

ACuriousMind

Or, wait, no, that's not quite correct

0celo7

acm doesn't know physics

he's a computer programmer now

Slereah

22:54

@bolbteppa the alternative is much worse

You have to find the actual physical configuration space

I don't even know if it's possible for most systems

ACuriousMind

On third thought, what I said was correct :P

0celo7

even worse

bolbteppa

So the story is, you have the NG action, we know it's rep invariant so that the Hamiltonian is zero automatically, so we know the Hessian must possess constraints, and further we know the constraints must be primary alone, and so using the trick posted above (or directly computing baby derivatives which is taking ages) we find that $\dot{x}^{\mu}$ and $x'^{\mu}$ are null eigenvectors, the only question is proving those two eigenvectors are the only ones, QMech's proof seems insane in that link

0celo7

@Slereah I'm reading an actually intelligible article on hyperbolic PDE

I'm already learning from that class

Slereah

link plz

What are the actual other eigenvectors of the hessian, anyway

It's 4 by 4

It has one zero eigenvector

What's the other three

(for point particles, I mean)

0celo7

22:59

spacelike probably

Slereah

I guess it might be like

bolbteppa

The point-particle action only has one eigenvector

apparently

Slereah

Initial speed?

Does it?

I thought it has one zero eigenvector, to fix with one constraint $p^2 - m^2 = 0$

ACuriousMind

The action is not an operator, it does not have eigenvectors at all :P

Slereah

And all the other eigenvectors were fine

@ACuriousMind The Hessian

0celo7

23:00

link.springer.com/chapter/10.1007/978-3-0348-8895-0_2

Slereah

Since the number of constraint depends on the rank of the hessian of the lagrangian

bolbteppa

Yeah, Blumenhagen ch. 2 says the point-particle action only has one null eigenvector, and the no. of null eigenvectors is the number of primary constraints, which is $p^2 - m^2 = 0$

Slereah

Please, don't say null for zero :p

Null means other things in SR too

bolbteppa

Haha I know

I was hoping to get away with that instead of saying 'eigenvectors of the 0 eigenvalue' each time :p

Slereah

nothing gets by me

bolbteppa

23:03

Maybe a trick in the NG case is that the Poisson brackets of the two primary constraints is closed and that's the proof there are only two...

0celo7

huh, the magnetic field on a hypersurface is the curvature of the induced connection

Slereah

@0celo7 I think my thesis advisor worked on some topic like that

Sir Cumference

@JohnRennie I mean, I've only ever heard that a negatively curved universe can collapse. Aren't such universes necessarily finite?

ACuriousMind

@bolbteppa It's not even necessary for the constraint algebra to close

Slereah

It involved the Atiyah-Singer theorem

0celo7

23:04

why does fucking physics involve so many damn algebras

in no other field, not even algebra, do so many algebras appear

2

bolbteppa

Scherk says the constraints $\phi_a$ and the canonical Hamiltonian must form a closed algebra $\{ \phi_a,\phi_b \} = c_{abc} \phi_c$ and $\{ H_c, \phi_a \} = c_{ab} \phi_b$ where the last relation holds when $H_c = 0$ automatically..,

If Dirac constraint theory wasn't so unintuitive...

Slereah

Exterior algebra, tensor algebra, Clifford algebra, Weyl algebra, Dirac algebra, Colombeaux algebra, C* algebra, ortholattice algebra, Lie algebra

ACuriousMind

@bolbteppa Ah, sorry, I was thinking of the generating set of gauge transformations. You're right, the constraint algebra has to close.

Slereah

The SUPERALGEBRA

0celo7

@ACuriousMind are you drunk?

bolbteppa

23:07

So does it close for the NG primary constraints :p

ACuriousMind

@0celo7 Why would you think so?

Slereah

because you're German

ooolb

lmao

0celo7

@ACuriousMind you've been saying nonsense

Slereah

Are there any famous magma in physics

0celo7

23:08

Slereah

or is everything a group, algebra, semigroup, etc etc

0celo7

that is an $\in$ where an $\epsilon$ should be

do I even want to know what a magma is

Slereah

Partial groupoid

In abstract algebra, a partial groupoid (also called halfgroupoid, pargoid, or partial magma) is a set endowed with a partial binary operation. A partial groupoid is a partial algebra. == Partial semigroup == A partial groupoid ( M , ∘ ) {\displaystyle (M,\circ )} is called a partial semigroup if the following associative law holds: Let x , y , z ∈ G {\displaystyle x,y,z\in G} such that x ∘ y...

they don't even list an example

of a magma that isn't something else

Magma (algebra)

In abstract algebra, a magma (or groupoid; not to be confused with groupoids in category theory) is a basic kind of algebraic structure. Specifically, a magma consists of a set equipped with a single binary operation. The binary operation must be closed by definition but no other properties are imposed. == History and terminology == The term groupoid was introduced in 1926 by Heinrich Brandt describing his Brandt groupoid (translated from the German Gruppoid). The term was then appropriated by B. A. Hausmann and Øystein Ore (1937) in the sense (of a set with a binary operation) used in this article...

oh wait

there it is

still no example, tho

48

Q: What's a groupoid? What's a good example of a groupoid?

Or more specifically, why do people get so excited about them? And what's your favorite easy example of one, which illustrates why I should care (and is not a group)?

ct.category-theory examples big-picture groupoids

bolbteppa

Ok so that Regge book computes the PB of the two NG primary constraints with themselves and one another and seems to say they are equal zero, but who knows if more constraints exist tbh apart from that insane QMech proof physics.stackexchange.com/a/184756/25851

0celo7

"for any reasonable physical system, $V\ge 0$"

why?

ACuriousMind

23:17

@bolbteppa Computing the rank of a matrix is in general a rather annoying problem, you need to perform either row reduction or compute minors for a general algorithm, so you shouldn't expect an easy proof.

It's either magic tricks or lengthy computation.

bolbteppa

Yeah this could potentially be an impossible thing to prove

I found something interesting on this

Slereah

@0celo7 potential energy?

I think it's to have the hamiltonian bounded from below

0celo7

Why should that be the case

bolbteppa

Apparently the 'second Noether theorem' implies that this will have two 0 eigenvalue eigenvectors immediately because of reparametrization invariance

and this is why the point particle action has 1

Slereah

@0celo7 runaway reaction

You'd just get particles radiated away

bolbteppa

23:26

amazon.com/Introduction-Relativistic-String-B-Barbashov/dp/…

That book has a big appendix proving it, insane

Slereah

Since it's a perfectly reasonable reaction to have the vacuum radiate

bolbteppa

Turns out magically we can skip performing Gaussian elimination on the Hessian of the Nambu-Goto action haha

Slereah

You can show it for Klein Gordon with φ^3

bolbteppa

You just need Noether's second theorem to prove the 'two parameters implies two 0 eigenvalue eigenvectors' excuse is legit

Slereah

Cubic interaction has those tadpoles diagrams that just mie the vacuum radiate

bolbteppa

23:32

Oh my god, that book literally derives both NG constraints and the two eigenvectors foolproof

You just basically derive Noether's theorem and you get everything

0celo7

23:52

@bolbteppa did you download it illegally

bolbteppa

@0celo7 books.google.com/…

enumaris

weez

bolbteppa

How do you get $\lambda_{\mu \nu}$ on that page, i.e. the Hessian, from what he says

Where $L_{\mu}(\partial x, \partial^2 x) = \dfrac{\partial}{\partial \tau} (\dfrac{\partial \mathcal{L}}{\partial \dot{x}^{\mu}}) + \dfrac{\partial}{\partial \sigma} (\dfrac{\partial \mathcal{L}}{\partial x'^{\mu}}) = 0 $

Since you're so interested

vzn

@enumaris where ya been man? something for you + all the other coderz + hackerz bet youll like it

in theory salon, 8 hours ago, by vzn

AlphaGo documentary / imdb ← outstanding! :)

enumaris

trying to find a job lol

vzn

23:58

still working with ML?

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