scitation.aip.org/content/aip/journal/jmp/16/12/10.1063/…

Oh boy

Most Backlund transformations are for garbage equations, though

Thos one is

$$f_{xt} + f_{yy} + f_{xx} + (3f^2)_{xx} + f_{xxxx} = 0$$

what are you even supposed to do with that

Is it generally true that $L_{[X,Y]}=[L_X,L_Y]$ ($L$ the Lie derivative)? I feel like this is the Jacobi identity, when acting on vector fields

@Danu It's true, but why would it be the "Jacobi identity"?

pnas.org/content/77/8/4393.full.pdf

Oh boy

It's abstractly the statement that $\text{Vector fields}\to \mathrm{End}(\text{Tensor fields}), X\mapsto \mathcal{L}_X$ is a Lie homomorphism.

@ACuriousMind Lie algebra structure on vector fields.

@ACuriousMind Yeah, I realized that, too.

@Danu Hm? The Jacobi identity would be a statement about cyclic permutations, I'm not sure what you're trying to say

@ACuriousMind Vector fields are kind of the canonical thing---in my mind, at least, that Lie algebras are supposed to model. The Jacobi identity axiom for Lie algebras corresponds to the identity above, if you consider $L$ as acting on vector fields.

Well lie algebras are just vector fields

...the Jacobi identity axiom guarantees the structures match, yes.

@Danu 1. Lie algebras are the left-invariant vector fields on a Lie group. 2. Oh, you are right, it is the Jacobi identity in that way.

@ACuriousMind All Lie algebras are realized in this way?

You can always make a group out of such an algebraic structure?

@Danu I read your recent answer on books for differential geometry for physics and I wanted to know if you've looked at the new book by Eschrig?

@Danu Being the vector fields on a Lie group/being the tangent space at the identity of a Lie group is an equivalent definition of a Lie algebra, yes. "Lie integration" associates to every Lie algebra a simply connected Lie group.

How does one have a notion of exponential or anything like that without a diff. geo. construction a priori?

@Danu amazon.com/Topology-Geometry-Physics-Lecture-Notes/dp/…

Hm

Perhaps

@Danu It's a very non-trivial result, and it's called Lie's third theorem

@ACuriousMind Interesting. Google doesn't seem to return things.

Ah, okay.

So I wasn't being dumb ;D

Perhaps a solution to the Sine Gordon equation on a time cylinder would be

An infinite Backlund superposition

Hmm, that's such a beautiful result.

Separated by an adequate distance

To be similar to the geodesic on the cylinder

I mean I'm not sure that would be useful for anything

But it would be a solution

@3075 I'll have a short look.

Also I'm not sure how one would do an infinite Backlund transform

@Danu It is, and that's why you "esssentially never" need to study Lie groups - just study the algebras instead ;)

@Danu Thank you.

Because together with knowing that every Lie group comes from its universal cover, this means that all information about the group is encoded in the algebra+the fundamental group, which is both purely algebraic data generally perceived as easier to deal with

I have no idea what I'm doing really

Although I guess technically a static soliton would be a solution here

But that's the most boring soliton

well that and $\varphi = 0$

@ACuriousMind Hmm... I still don't see how that follows.

At least some topological information must be lost.

@3075 For my taste, it's not enough of the definition-theorem-proof style. It seems to cover a lot of topics.

@Danu I'll read it then, thanks.

I wonder if $\phi^3$ theory is doable to solve

@3075 Didn't like any of my other references? ;)

Surely $\Box \phi = \phi^2$ should be a bit more manageable

@Danu The marvel of Lie theory is that there isn't.

What's a function with the second derivative equal to its square

@Danu I looked at nakahara and naber, both were good, but well, this book is blue so...

I'm sure it's some tangent related function

@ACuriousMind Well... As you just said, you're only going to get simply connected things.

@3075 I'm not a fan of either :\ But I care more about rigor and depth than needed for physics.

@Danu Yeah, but you also know that every Lie group has a universal cover which is again a Lie group

@Danu I can see that. I don't really care about that right now though.

So to determine a property for an arbitrary Lie group (say, the possible representations), you determine it for the universal cover and then look which survive the projection by the covering map.

@ACuriousMind Yeah, but I'd like to understand $SO(n)$, too, please ;)

@ACuriousMind Fair enough.

And since the covering map has finite and discrete fiber that step is usually not hard

What would be hard is finding out things about the universal cover - but there you can just look at the algebra

Question: Why is it that when you go to an infinite amount of dimensions you no longer need your orthonormal basis vectors to have a norm of 1? Why is there a special normalization when u go to an infinite amount of dimensions

@Danu Bah, that's just $\mathrm{SU}(n)/\mathbb{Z}_n$. Go for $\mathrm{SO}(p,q)$, things get a bit more difficult there

Also how can you have an infinite amount of orthogonal vectors?

@GavinN. I think you've fallen prey to the usual trick physicists pull: The position eigenstates are not a basis.

@GavinN. Think about the energy eigenstates of the harmonic oscillators.

I don't know about quantum I'm just in Shankar's math chapter

I haven't learned that stuff yet

@GavinN. By having a vector space of infinite dimension? (I know that sounds tautological, but that's all there is to it)

@ACuriousMind Those filthy physicists

@ACuriousMind What? How does that work by dimension?

@Danu Ehhhhhh

You're right, that's nonsense

Dunno why I wrote that

That sounded way too good to be true, haha.

WAIT

@GavinN. What do you mean by an orthonormal basis that no longer has norm 1, then? Orthonormal bases do have norm 1, even in the infinite-dimensional case.

Sine gordon has multiple vacua

Does it have domain walls?

How do the solitons interact with the domain walls

@ACuriousMind WHAT

Like in an infinite amount of dimensions your dirac delta function isn't 1 but some other number

Then how does the "expansion in |x>" bullshit work

@Danu It's still fine because all $\mathrm{SO}(n)$ are just doubly-connected - they are a $\mathbb{Z}_2$ quotient of their universal cover

Yeah, Spin(n)

I knew that.

@0celo7 We have been over this at least 3 times :P

@ACuriousMind link???

@Danu See, and we don't need to know anything about Spin(n) except that it's the "integration" of $\mathfrak{so}(n)$.

SO(2) is a circle, though

(which is why nobody does know anything, haha)

It's a Z quotient of R

@Danu ...that might be true :D

@Slereah I FORGOT $n > 2$, OKAY?

:D

Never forget n = 2

n = 2 is the magic dimension

@Slereah @ACuriousMind clarified here

Seriously why aren't position eigensates a basis

You need $n\geq 2$ to the the second homotopy group to vanish.

@0celo7 they aren't rays

@0celo7 Hmmm...It appears while we've talked about rigged Hilbert spaces, you were never there when I said something like that. So I retract my statement

@ACuriousMind yeah I know they're in the rigged Hilbert space

@0celo7 Ah, and the point is that they are in the rigged space and not in the actual space

Ok ok

Aren't they a basis in compact spaces, though?

So they are not a basis because a basis should lie within the space it's a basis of

I need to sharpen my question.

Like QFT on S

monkeys pls help me. Why does your orthogonality condition change when you go to an infinite dimensions (going from dirac delta that is 1 at I=j to a dirac delta that is not necessarily 1 at I=j)?

@ACuriousMind Why does bullshit physicist math work for QM?

Which then implies that your basis vectors are not of norm 1

@Slereah No, the position operator is still continuous on that

Beautiful, this long exact homotopy sequence. I always wonder why homotopy seems more important than homology in physics.

What about the momentum basis?

There are only a countable amount of them, no?

@Slereah But you get half of what you want - the momentum states become actual states

Hurray

@0celo7 It's telling you functional analysis sucks ;)

@0celo7 Well, because physicists just have discarded all parts of the bullshit math that don't lead to meaningful results

In that basis btw, $\langle p \vert q \rangle = \delta_{pq}$

It's all good

@Danu what

It's not so much that the "physicists' math works" as that they've had decades to cleverly circumvent the parts where it doesn't

@ACuriousMind but why can we expand in position eigensates

Oh yeah

You should have seen early QFT

It was a disaster

@ACuriousMind can you give an example?

@0celo7 That's a non-trivial result of functional analysis (either something about kernels (in the sense of functional analysis not of linear algebra) on $L^2(\mathbb{R}^n,\mathrm{d}x)$ or about rigged spaces, depending on your preference for the formulation)

I'm going to be taking physicist quantum mechanics next two semesters, I need to be prepared

Well for instance early QFT, relativistic quantum mechanics, used to have negative probabilities

Also you might notice, when you do QM, that one of the most basic exercize of non-relativistic QM is the barrier potential

@0celo7 You can derive all sorts of non-sensical statements when you try to take expectation values with respect to position or momentum eigenstates, a favourite thing is to derive $-\mathrm{i}\hbar = 0$.

But it is rarely done in QFT

And there is a good reason for that

(it used to be all sorts of fucked up)

@ACuriousMind I am well aware of that trick, no need to link it

@ACuriousMind Can we view a smooth curve $c:I\to M$ as a smooth homotopy between the maps $f_0:\{*\}\to c(0)$ and $f_1:\{*\}\to c(1)$?

here $*$ is...something.

So it's a homotopy $F:\{*\}\times I\to M$?

"We construct a stable domain wall ring with lump beads on it in a baby Skyrme model"

What the fuck is even that abstract

Skyrim?

baby skyrim

@ACuriousMind Is that German Skyrim hack out yet,

@0celo7 Yes!

@0celo7 No! :(

Apparently a nice way to do QFT on a CTC spacetime is

Going Euclidian

BUT

It only works for a small number of spacetimes!

Apparently you can do it for Godel though

@ACuriousMind Cool.

physics.stackexchange.com/questions/89958/… Why does the guy say that the first integral is inconsistent in the first answer?

@GavinN. Because $\lvert y \rangle = 0$ means that your states are not a basis, they are just zero!

@ACuriousMind Ok, I want to, once and for all, prove that a connected manifold is smoothly arcwise connected.

what notation are you using @ACuriousMind? It looks nonsensical on my comp

> By going through the local parametrizations, it is easy to show that arcwise connected components are clopen in X, and hence if X is connected it is arcwise connected.

ARGH

It's not easy because what do you do about the chart overlaps??

@GavinN. You need to activate ChatJaX

@GavinN. ...LaTeX

looks perfectly fine to me

arxiv.org/abs/gr-qc/0511029

hm

I am intrigued

sciencedirect.com/science/article/pii/S0021782401012296

also

It is said that the objects in space will not decrease or increase their velocity, (i.e. there's no friction or something to stop it). But why is not taken to account the gravity force? If you are in a velocity against the direction of the Earth your velocity should decrease because of the force. Can someone help me in my misunderstanding?

I have too many papers open

"Manuscript received 3 September 2001"

Ahah

I wonder what happens next week

@PichiWuana You have no misunderstanding. Gravity does act on objects in space.

It's not "objects in space" that will not increase or decrease their velocity, it's "objects one which no force acts".

@ACuriousMind Let's say a rocket is launched from the surface of the Earth to the space, with a specific acceleration. The real acceleration will be the acceleration of the rocket (with the fuel) minus the acceleration of the gravity?

@PichiWuana yes

And then if the fuel is finished, the speed will decrease because of the gravity force?

@PichiWuana Whether it "decreases" depends on how exactly the rocket accelerated - if it flew such that it got into orbit, the spend will not "decrease", but the rocket will be in a stable orbit.

@ACuriousMind My case is the rocket being launched upwards and not getting into orbit

@PichiWuana then yes

@0celo7 : Where is Duffield? Writing a book. My working title is The Physics Detective. It's tough work.

Oh so it will have a similar movement to a vertical throw in Earth?

^important

@PichiWuana It will be different because it's continuously accelerating and getting lighter until it runs out of fuel, but if you're just referring to "going up then coming down", then yes, it's similar :P

Yes, that's what I meant

Thank you much for your help

@JohnDuffield Oh good

Can I have an advance copy?

I'll email you my address

@JohnDuffield So what did you get suspended for?

I got a month recently

If you fall from 9.8 meters how many seconds does it take to hit the ground?

@GavinN. Less than 2 seconds?

@GavinN. Actually, 1.41421 seconds

"Roland the Farter (known in contemporary records as Roland le Fartere, Roulandus le Fartere or Roland le Petour) was a medieval flatulist who lived in 12th century England. He held Hemingstone manor in Suffolk and 30 acres (120,000 m2) of land in return for his services as a jester for King Henry II. Each year he was obliged to perform "Unum saltum et siffletum et unum bumbulum" (one jump, one whistle, and one fart) for the King's court at Christmas."

History is pretty great

.

@Slereah wtf