The h Bar

ZeroTheHero

00:03

For those in a demonstrative mood: anti-powerpoint-party.com/en

ACuriousMind

@ZeroTheHero ...last update 7 years ago?

ZeroTheHero

02:46

@ACuriousMind according to wiki: As of February 2021, the party had 4,632 members, making it the eighth largest party in Switzerland. (see en.m.wikipedia.org/wiki/Anti-PowerPoint_Party)

…so it’s still alive and still promoting flip charts.

Semiclassical

04:38

so, here's a change from second edition Sakurai to third edition that i don't quite see the reasoning for:

second edition: "Show that the differential cross section for the elastic scattering of a fast electron by the ground state of the hydrogen atom is given by $$\frac{d\sigma}{d\Omega}=\left(\frac{4m^2e^4}{\hbar^4 q^4}\right)\left(1-\frac{16}{[4+(qa_0)^2]^2}\right)^2.$$ (Ignore the effect of identity.)"

third edition: "Show that the differential cross section for the elastic scattering of a fast positron by the ground state of the hydrogen atom is given by [same expression]."

there's the deletion of "effect of identity", but i didn't understand what that meant in the first place

oh. maybe it's just to avoid any questions of the Pauli principle while the scatterer interacts with the bound electron in the hydrogen atom?

Nihar Karve

07:28

@Semiclassical yeah, with identical particles you'd also have to include the exchange amplitude, which alters the cross-section

bolbteppa

@Slereah I've been looking at the Dirac paper again, it's just simply brilliant, if you want to iron out any points/'sentences in the first few pages this is a good time, e.g. the real motivation for going to 6D etc

Nihar Karve

@Semiclassical in particular, you'd have to use the Mott scattering formula (with the form factors etc) to get the right $\theta$-dependence

I don't have Sakurai, but I'm pretty sure this problem would be before introducing RQM, so he probably wants to avoid that discussion

Slereah

They should put a skip button on papers

Skip all the thanks preface, introduction and basics

Slereah

07:52

Strangely a lot of Clifford algebra related articles on conformal geometry

apparently there are things you may do with the extra dimensional vectors

bolbteppa

Well the relation to $SO(2,D)$ and $SO$ having spinors is probably why

Slereah

Probably

Apparently one vector they enjoy naming $e_0$ and one $e_\infty$

bolbteppa

Point at infinity?

Slereah

since one of them will help define the infinity

yeah

Many such cases

I think the $e_0$ here is the famed homogeneous coordinate such that $(x, y) \sim (x : y : 1)$

bolbteppa

Same idea to what Dirac does

The question is, why are we even going from $R^n$ to $R^{n+1,1}$, when you do it the usual way of deriving the conformal algebra from the conformal Killing equation you can see it at the end, but this 'global' way does it immediately but why

If you ignore the signature of the metric (as Dirac does), then $SO(D)$ is the 'conformal group' (associated to the $SO(D-2)$ rotation group)

Slereah

08:07

That's why I think the stereographic example is probably a bit easier to check

Since it's a little more visual

bolbteppa

I don't know why we're even doing stereographic projections either

Slereah

It is apparently a fairly common process?

Get the conformal geometry of something from the projective model

bolbteppa

'S is the projective (or Möbius) model of conformal geometry'

Conformal geometry

In mathematics, conformal geometry is the study of the set of angle-preserving (conformal) transformations on a space. In a real two dimensional space, conformal geometry is precisely the geometry of Riemann surfaces. In space higher than two dimensions, conformal geometry may refer either to the study of conformal transformations of what are called "flat spaces" (such as Euclidean spaces or spheres), or to the study of conformal manifolds which are Riemannian or pseudo-Riemannian manifolds with a class of metrics that are defined up to scale. Study of the flat structures is sometimes termed Möbius...

I'd like to appreciate this as a starting point a bit more basically

Slereah

I think it's just that if you take $\mathbb{R}^{n+1, 1+1}$, the projectivized version of the null cone has the shape of the appropriate space?

So $\mathbb{R}^{n+1, 1}$ has the sphere $S^n$ if you take the projection of the con

Not 100% sure where the conformal transformation pops up in this mess, but I assume that one of the subgroup of $PO(n+1, 1)$ has it in somewhere

bolbteppa

08:31

There's just a lot of hand-waving with this perspective, the usual CFT approach at least avoids any doubt

Slereah

Yeah but I like to have also the global perspective on geometry

bolbteppa

The real goal is to describe a geometry where you can measure angles, but distances are ignored right, so in $x^2 + y^2 + z^2 - t^2 = R^2$ we see the radius is now allowed to vary so should be added as a new variable, which brings you to $\mathbb{R}^5$, these coordinates live on a sphere, so using projective coordinates brings you to $\mathbb{R}^6$,

but how do we add a new coordinate to $x^2 + y^2 + z^2 - t^2 - R^2 = 0$, hand-waving we view this as the $w = 0$ case of $x^2 + y^2 + z^2 - t^2 - R^2 = w^2$, but I could have added a $-w^2$

Slereah

I'm not sure that's particularly less hand-waving 🤔

bolbteppa

Dirac talks about ignoring the signature so okay, if we do that then the conformal group of $SO(D-2)$ is $SO(D)$, but if we don't then I could get that the conformal group of $SO(1,3)$ is $SO(2,4)$ by adding $-w^2$ but $SO(3,3)$ if I added a $+w^2$...

So why does the usual infinitesimal method pick out $SO(2,4)$...

Slereah

I think the basic idea is that if you add $(1+1)$ dimension, the shape of the light cone is gonna be the same shape as the original quadric?

Ironically for once it's nlab that doesn't have much to say : ncatlab.org/nlab/show/M%C3%B6bius+space

I'm not 100% sure if the whole extra dimension thing is totally warranted or if it's just a weird historical leftover because that's how people dealt with projective spaces in the 19th century

Slereah

08:54

Also apparently that model is also used to make the model homogeneous???

bolbteppa

A 2-D unit sphere in $(2+1)D$ Euclidean space $\mathbb{R}^{2+1}$ is $x^2 + y^2 + z^2 = 1$. By writing this as
$$0 = (z^2 - 1) + x^2 + y^2 = - 2\frac{(1-z)}{\sqrt{2}} \frac{(1+z)}{\sqrt{2}} + x^2 + y^2 = - 2 w_0 w_3 + x_1^2 + x_2^2$$
we are now in Minkowski space $\mathbb{R}^{2+1,1}$, so what's the analogue for 4D Minkowski space of all this

Slereah

Because if you have say $\mathbb{R}^n$, the $0$ is a special point, but that's not the case in the horosphere

which is the projectivized null cone

@bolbteppa The Minkowkski equivalent is apparently the Einstein static universe

Which is $(S^1 \times S^3 / \mathbb{Z}_2, -dt^2 + g_{S^3})$

bolbteppa

If we have $(x,y)$ coordinates in $\mathbb{R}^2$, and we want a 2D unit sphere, we need to add a third coordinate and write $z^2 + x^2 + y^2 = 1$. So with $(t,x,y,z)$ we'd have to add a $w$ and form $w^2 + x^2 + y^2 + z^2 - t^2 = 1$ in 5D, then writing $w^2 - 1 = - 2 x_{-1} x_5$ we now have 6 coordinates

Slereah

which you can embed into $\mathbb{R}^{3+1, 1+1}$ via... miraculous means?

bolbteppa

I think that's probably the logic, it's basically what they do here

Slereah

08:58

arxiv.org/pdf/0706.3055.pdf

It's in there I think

ACuriousMind

@Slereah the embedding is exactly the same independent of signature (modulo replacing $x^2$ by $g(x,x)$)

it's just that you get different shapes from the projective null cone depending on signature

bolbteppa

It looks the same as working with a sphere of radius $R$ and adding a $w$ to end up with $w^2 + x^2 + y^2 + z^2 - t^2 - R^2 = 0$ where the radius is now a variable and we preserve the sphere under transformations $\lambda \eta_{\mu \nu}$...

Slereah

Question is, if all of this is true, does this mean that hidden in regular special relativity, there is a hidden theory of conformal spheres in there?

or circles, anyway

ACuriousMind

Careful: Normal relativity is in 4d, so the corresponding "conformal space" would be 2d and for 2d the conformal compactification is pretty useless because it doesn't reflect all conformal transformation

bolbteppa

In the 2D Euclidean plane with $ds^2 = dx^2 + dy^2$, to get 2D sphere's we need to go to 3D and work with $z^2 + x^2 + y^2 = R^2$ or $(z,x,y,R)$ satisfying $z^2+x^2+y^2-R^2=0$ and $\eta_{\mu \nu}' = \lambda \eta_{\mu \nu}$. So in 4D SR we need to look at a 4D sphere which means going to 5 and 6 D

So SR is hidden in the conformal theory, not the other way around

Slereah

09:07

@ACuriousMind alas

ACuriousMind

alas what

Slereah

The models one can visualize tend to not be the best

ACuriousMind

I still feel you're fixating far too much on this silly trick to show that conformal trafos are $\mathrm{SO}(p+1,q+1)$ :P

Slereah

I try to understand its various aspects :p

ACuriousMind

@Slereah what I mean is that a 2d starting space isn't even a model

Slereah

09:09

Yeah I vaguely remember that it doesn't have an effective conformal structure or something

Like $\mathrm{Conf}(2) = \mathrm{Iso}(2)$ or something

Still technically a conformal structure, just not a very interesting one :p

but you know me, I believe in understanding models by every crazy possible formalisms

ACuriousMind

the whole reason 2d CFT is so special is because conformal trafos in 2d are very different from any other dimension

so in this case the "simplest" model is an extremely special case that doesn't teach you anything about the general case

Slereah

I noticed yeah

They are very adamant about $n > 2$

IIRC the 2D model is boring and the $(1+1)D$ model is a nightmare

Infinite dimensional conformal group or something

ACuriousMind

that's in Schottenloher, there isn't really "a" conformal group

you get the infinite dimensional Witt algebra as (one half of) the algebra of transformations, but it doesn't "integrate" into a nice group

ah, no, wait

there is a conformal group and its infinite-dimensional, yeah

the group that doesn't exist is a "Virasoro group"

Slereah

How can a group not exist if it's a group 🤔

ACuriousMind

I mean that you have the Virasoro algebra and the question "what Lie group is this the algebra of?" doesn't have a good answer (because there's no unique notion of the Lie correspondence for the infinite-dimensional case)

anyway, yes, the conformal group of $\mathbb{R}^{1,1}$ is infinite-dimensional - it's two copies of all orientation-preserving diffeomorphisms of the circle

Slereah

09:20

Does this have any bearing on SR?

ACuriousMind

I have absolutely no idea

Slereah

09:39

bivector.net/…

This paper seems to give the conformal transformations in terms of the Clifford algebra of the homogeneous model

bolbteppa

09:56

They begin by talking about going from 3D to 5D space as being useful and then introduce extra basis vectors which square to some choice of signatures all out of thin air like everything else

Slereah

Apparently if you consider the cone defined by the quadric for the Einstein universe, the metric tensor induced on that cone is a degenerate metric $(0, -, +, +, ...)$

And the kernel is generated by the null curves

I think hence the projectivized version is just the appropriate Einstein universe

Slereah

10:11

"The set $\pi(C^{2,n}$ is a smooth hypersurface $\Sigma$ of $RP^{n+1}$. This hypersurface turns out to be endowed with a natural Lorentzian conformal structure. Indeed, for any $x \in C^{2,n}$,$\hat{q}_x^{2,n}$ the restriction of $q^{2,n}$ to the tangent space $T_xC^{2,n}$, that we call $\hat{q}_x^{2,n}$, is degenerate. Its kernel is just the kernel of the tangent map $d_x\pi$.

Thus, pushing $\hat{q}_x^{2,n}$ by $d_x\pi$, we get a well defined Lorentzian metric on $T_{\pi(x)}\Sigma$. If $\pi(x) = \pi(y)$, the two Lorentzian metrics on $T_{\pi(x)}\Sigma$ obtained by pushing $\hat{q}_x^{2,n}…

Slereah

10:28

The thing I remain unsure of is why this projective trick gives out the appropriate conformal structure

I haven't really seen any justification as to why that would work

John Rennie

11:06

Apparently there are no known primes of the form:

1 2 3 4 5 6 7 8 9 10 11 ... n

i.e. you construct the number by concatenating the digits of all numbers up to n. This has allegedly been checked for n ≤ 1000000 (though I haven't confirmed this for myself).

Yet another of the strange and probably completely useless problems in number theory.

Slereah

I think the scaling is done by considering the bundle of the null cone over the quadric $\pi : C \to Q$, and then the scaling seems to be done by a section of that bundle

Or something similar, not quite sure

Slereah

11:32

hal.archives-ouvertes.fr/tel-03195071/document

Sometimes I am so glad to be French

Flight of the Conchords Ep 8 'Foux Da Fa Fa'

►

Slereah

11:55

"Given a point $p$ in $\mathrm{Ein}(n)$, the lightcone with vertex $p$, denoted by $C(p)$, is the set of lightlight geodesics containing $p$. In the projective model, if $p = \pi(u)$, with $u$ some isotropic vector of $\mathbb{R}^{2,n}$, then $C(p) = (P \cap C^{2,n})$, where $P$ is the degenerate hyperplane $P = u^\top$"

bolbteppa

12:14

In 2D if you go from $x^2 - t^2$ to $z^2 + x^2 - t^2 = R^2$ you then want to preserve $z^2 + x^2 - t^2 - R^2 = 0$, where is the complex analysis :\

Slereah

Hestenes seems like maybe a cool book to look into for such things

bolbteppa

I don't think there's a single thing that GA can do that something else can't

Slereah

12:34

GA?

bolbteppa

Geometric Algebra

Slereah

Oh

I mean it's just the evil twin of Clifford algebras

Conformal geometry is a bit of a tough thing because there's apparently like 5 or 6 different formalisms for it

Which is true for most things, certainly, but there isn't a lot of unified treatments

I've skimmed a few so far and I still haven't seen what the hell an ambient model is

and few of them even bother actually to give examples of specific conformal transformations

bolbteppa

The direct derivation of the global SCT from the $SO(4,2)$ formulation is good, every other derivation of it I've seen is just a mess/guess

Samyak Marathe

is anyone up?

PM 2Ring

@Slereah Greg Champion - The French Song.

Samyak Marathe

12:44

I need help about mechanics

its urgent, if anyone can help

PM 2Ring

Nataly Dawn, (wife of Patreon founder, Jack Conte), likes to sing in French, from time to time. Here's a nice arrangement of Charles Trenet's Que reste-t-il de nos amours?

Slereah

Why do we talk about the SCT for the conformal group and not inversions?

Are inversions not a conformal transformation?

"The circle inversion map is anticonformal, which means that at every point it preserves angles and reverses orientation (a map is called conformal if it preserves oriented angles)."

Close

I guess that's why the SCT has two inversions

PM 2Ring

Inversions certainly preserve angles.

bolbteppa

An inversion is a discrete transformation, thus it has no infinitesimal parameter associated to it either and so no Lie algebra generator, but an inversion followed by a translation followed by another inversion is a continuous transformation which has a continuous parameter parametrizing it (the translation) and so it has a Lie algebra generator on doing an infinitesimal translation, this is a SCT

PM 2Ring

I have a lovely paper I found years ago that explains how to do hyperbolic geometry on the Poincaré disc, using old-school compass & straight-edge constuctions, via circle inversion.

Slereah

12:58

Does that mean you could extend the conformal group with discrete transformations?

same way you can have them in the orthogonal group

bolbteppa

They are encoded into the SCT, an inversion is the special case where you do no translation in the SCT

Slereah

Aren't you doing two inversions, though?

bolbteppa

From the point of $SO(4,2)$, a SCT is trivially just a 'Lorentz' transformation

Oh right yeah

But they are still built into it

Slereah

If $b = 0$ that's just the identity, no?

So that you can't really do an inversion by itself

ACuriousMind

@Slereah you posted a screenshot not too far up that does exactly that when it talks about the special conformal group

that's the same "special" as in "special orthogonal group"

bolbteppa

13:01

An inversion preserves $ds^2$ up to a scale factor, a sequence of them similarly preserves $ds^2$ up to a scale factor, the combined effect of them (with a translation in between) is called a SCT, but in reality it's a sequence of 'inversion+translation+inversion'

Slereah

The past is a foreign country my friend

Let me see if I can find it

ACuriousMind

i.e. that source does not inherently define a "conformal" trafo to be orientation-preserving, and so has to call the orientation-preserving trafos "special conformal"

bolbteppa

So you don't need to append discrete transformations onto it

ACuriousMind

e.g. Schottenloher also doesn't require conformal transformations to be inherently orientation-preserving, but then calls "the conformal group" the identity component of the group of conformal transformations of the conf. comp.

(orientation-reversing transformations can never be in the identity component)

bolbteppa

If you take $ds^2 = \eta_{\mu \nu} dx^{\mu} dx^{\nu}$ and plug $x_{\mu} = \frac{x_{\mu}'}{x'^2}$ into it, this is a discrete inversion, it preserves $ds^2$ up to an overall factor $x'^{-4}$. The SCT is just
$$x_{\mu} \to \frac{x_{\mu}}{x^2} \to \frac{x_{\mu}}{x^2} + a_{\mu} \to \frac{\frac{x_{\mu}}{x^2} + a_{\mu}}{(\frac{x_{\mu}}{x^2} + a_{\mu})^2}$$

Slereah

13:05

I see

ACuriousMind

so it's not as if there was universal agreement that "conformal" means "orientation-preserving", just people agree that the thing we're usually interested in is the orientation-preserving ones, much like we are usually interested in the identity component of e.g. the Lorentz group

Slereah

Is that total group related to the Möbius group, by any chance

It seems to pop up a lot on the topic

ACuriousMind

uhhh

the Möbius group is just the conformal group of $\mathbb{C}$ (or rather its comp. the Riemann sphere), no?

Slereah

I think so?

But that one seems to explicitely admit the inversions

For some reasons

ACuriousMind

so?

Slereah

13:08

idk, it seems strange that this one does but not the conformal group?

ACuriousMind

it's like people talking about "the Poincaré group" and they sometimes really do mean the full group and sometimes just the identity component

language is sloppy

Slereah

it is true that I think basically any subset of spacetime transformation has a "generic" version and one that preserves orientation

Just didn't think of the SCT as being one of them

ACuriousMind

I also don't think that there is somehow universal agreement that "the Möbius group" explicitly includes inversions

Slereah

I have certainly seen it included more often than for the conformal group

bolbteppa

From the point of view of $SO(4,2)$ you define your $6$-vector $x_{A}$ and let it transform under $x_A' = \Lambda_A^B x_B = (e^{\frac{i}{2} \omega^{CD} J_{CD}})_A^B x_B$ and define the SCT as being generated by $K_{\mu} = J_{5 \mu} - J_{6 \mu}$ and figure out the $x_A'$ explicitly using $(J_{AB})_C^D = i (\eta_{AC} \delta_B^D - \delta_A^D \eta_{BC})$

and then just use this in the $4$-vector $y_{\mu} = \frac{x_{\mu}}{x_5 + i x_6}$ for $\Lambda = e^{i \beta K}$ i.e. work out $y_{\mu}'$, it is a direct derivation which is very difficult the Lie algebra way, usually books resort to guessing to find the explicit formula

ACuriousMind

13:11

e.g. wiki seems to think Möbius transformations are orientation-preserving, too

Slereah

But then here's a terrible and dumb question

If the various discrete symmetries of the Poincaré group have consequences for particle physics, what happens for inversions in CFTs?

Do CFT theories have an inversion number 🤔

Although apparently inversions are connected to the part of the group connected to spatial reversal

ACuriousMind

it's literally the difference between O and SO

Slereah

I see

ACuriousMind

the orientation preserving trafos are $\mathrm{SO}(p+1,q+1)$, the general ones are $\mathrm{O}(p+1,q+1)$

Slereah

Is it related to the fact that you can perform rotations as series of reflections?

ACuriousMind

13:17

is what related to that fact?

Slereah

I don't know, you have two continuous symmetries that can be performed as the action of two discrete symmetries?

bolbteppa

This discusses them as $\det(I) = - 1$ transformations on page 5

Slereah

Thanks

bolbteppa

On pages 3-4 that derives the SCT in an ugly way from the infinitesimal result directly, but at least it doesn't guess the SCT

ACuriousMind

@Slereah what?

the claim is that "inversion" and "reflection" are part of the same connected component - but a lot of the other elements in that component are "reflection + orientation-preserving trafo" or "inversion + orientation-preserving trafo", not "discrete" reflections/inversion

Slereah

13:26

Fair enough

ACuriousMind

Qmechanic shows here when exactly reflection and inversion are in the same component

@Slereah as for the significance of inversion as a symmetry of a field theory, see projecteuclid.org/download/pdf_1/euclid.cmp/1103860419

Slereah

Thanks

Hm

Semiclassical

something a student just asked me which I'm forgetting the 'right' answer to

whenever we do scattering theory, we have to use momentum states and deal with the fact that they're not normalizable

formally you avoid that by quantizing scattering states in a box of size L^3

and then you find at the end that the size of the box doesn't actually matter, so the actual value of L is irrelevant

nevertheless: is there a good physical meaning to L? even if it's just at the level of "your box had better be at least "this big" to justify scattering theory"

bolbteppa

*(some people do scattering that way)

I am not afraid of the continuous spectrum :p

Semiclassical

lol

Slereah

13:40

Conformal group has 15 dimensions, that's like... 6 spatial rotations, 1 timelike rotations, 8 translations

bolbteppa

6 Lorentz transformations, 4 translations, 4 SCT's, and a dilation

Slereah

The numbers match up, yes

Semiclassical

this is one place where i might be able to justify my fascination with non-relativistic pilot wave theory

Slereah

I just need to figure out how they fit with each other

Semiclassical

e.g. arxiv.org/abs/1210.7265

Slereah

13:41

the Lorentz transformations are applied to the projectivized null cone

Semiclassical

(though that's 1D not 3D)

bolbteppa

The SCT is a vector because you snuck a translation in there sandwiched between the inversions

Slereah

Yeah I guess the translations + SCTs match up with the 8 translations

bolbteppa

'(made using unphysical plane-wave states)'...

Slereah

Is the dilation corresponding to the O(1) symmetry?

I think it might, considering that the scaling is related to the point on the null cone you consider

bolbteppa

13:43

In terms of $SO(4,2)$ the dilation is $J_{56}$, the translation is $J_{5 \mu} + J_{6 \mu}$ and the SCT is $J_{5 \mu} - J_{6 \mu}$ and Lorentz is $J_{\mu \nu}$

Semiclassical

@bolbteppa unphysical in the sense that you can't actually prepare a true plane wave

Slereah

ah yeah, they don't perfectly match up I guess

But close enough

Semiclassical

usually you deal with that by appealing to Gaussian wavepackets

this is in the same spirit, but just a different set of scattering states

bolbteppa

We can't prepare free particles (plane waves), but on the other hand all we can do in QFT is measure free particles (using scattering experiments), it does not make sense to try to bypass using free particles

Slereah

I probably need to draw a little cone and projecting surface to convince myself it works

Semiclassical

13:46

@bolbteppa i mean, the point is that "free particles" strictly speaking don't exist. it's just that one usually gets past this by not worrying about the distinction

none of this requires pilot wave theory, though. ultimately it's just doing scattering theory with different scattering states

e.g. arxiv.org/abs/0910.1513 doesn't invoke pilot wave theory at all

bolbteppa

Strictly speaking, all we can measure is free particles, so strictly speaking they are the only things that exist on a formal level

(From a qft perspective, which at the end of the day is the most fundamental experimental thing we have)

Slereah

No particle is truly free

Due to interaction with the universe and capitalism

Semiclassical

i mean, you can't and don't prepare states with infinite coherence length. you just prepare it to have a coherence length which is long enough that the distinction doesn't matter :P

that said i haven't seen a version of the above story which is directed at the QFT context, so i can't really speak to that

i also don't actually know how you do QFT scattering theory with finite coherence lengths

bolbteppa

Preparing something on this level means something like interacting with what was initially a free particle and ensuring it scatters into an asymptotic free particle state where you have some idea what it is

In reality we are burdened with approximations, but theoretically we can always just set up a free particle and blame ourselves for not being able to describe it properly, we do the same thing in classical physics all the time too

Daniel Underwood

14:08

@Slereah seize the means of pair production?

Slereah

How often do you read the preface/introduction of a book

"It states that any smooth conformal mapping on a domain of Rn, where n > 2, can be expressed as a composition of translations, similarities, orthogonal transformations and inversions: they are Möbius transformations (in n dimensions)."

Liouville got my back on the issue

Hm

There's two compactifications of Euclidian space

The conformal one, which is the one point compactification $\mathbb{R}^n \cup \{ \infty \}$, and the projective compactification, $\mathbb{R}^n \cup P_\infty(\mathbb{R}^n)$

The Minkowski equivalent of the first is the Einstein universe, but what would be the second one?

As far as I can guess, the Penrose diagram is the injection of Minkowski space into the Einstein universe, and therefore probably the conformal compactification thereof

But is there one for the projective case?

I know Weyl talked about it a lot but I'm not 100% sure he ever went anywhere specific with it

I guess a priori it should be the same boundary as Euclidan space, but it would be separated into a variety of regions

ACuriousMind

14:30

Jan 21, 2021 at 11:22, by Charlie

To write a Marxist QFT book one would first have to seize the means of pair production

Slereah

The conformal coordinate nonsense is apparently also related to polyspherical coordinates :

encyclopediaofmath.org/wiki/Pentaspherical_coordinates

Semiclassical

@bolbteppa amusingly, another student question just now was actually in precisely the vein i was speaking to

when we do time-dependent perturbation theory with a harmonic perturbation, you need to assume that the perturbation lasts a long time

Daniel Underwood

@ACuriousMind do you have this chat indexed in your brain? :P

Semiclassical

but how can that be justified for a particle scattering? "surely" that interaction should only be happening while the particle is in the vicinity of the atom, and therefore only interacting for a short period of time

Slereah

Check out this suggestive form

ouh la la

Semiclassical

14:39

the out is that what you're sending in is a plane wave, which has an infinite ("long enough") coherence length

and therefore spends an infinite amount of time interacting with the atom

ACuriousMind

@DanielUnderwood maybe?

bolbteppa

15:00

@Semiclassical I think my mind would break if I tried to interpret e.g. qed scattering in that way :p

@Slereah tractors

Semiclassical

i mean, the point is that one should just not think too literally about scattering theory :P

or at least not borrow the intution of "scattering events are brief"

bolbteppa

I just can't see how to rationalize 2D holomorphic transformations as conformal transformations by going to 4D and analyzing $x^2 + y^2 + z^2 - t^2 = 0$

Semiclassical

otherewise the use of time-dependent perturbation theory in scattering theory is entirely unintelligible

bolbteppa

Maybe saying the point at infinity occurs when $z^2 - t^2 = 0$ holds so we have to preserve $(x + iy)(x - i y) = z \overline{z} = 0$ in the 2D plane which for $dz d \overline{z}$ gives $\frac{dz}{dz'} \frac{d \overline{z}}{d \overline{z}'} dz' d \overline{z}'$

ACuriousMind

@Semiclassical not so sure about the "surely" :P

Semiclassical

15:10

well that's why they were in scare quotes!

i do find scattering theory very hard to follow tbh

ACuriousMind

Sure, you need to "switch off" the potential completely in order to return to free states, but that's why we talk about the free input/output states as the asymptotic past/future

everything interacts with everything else all the time, however neglegibly

we're actually making an error when we pretend we can switch off the interaction completely

Semiclassical

right, but the point is also that you shouldn't imagine this as "a wave packet comes in, interacts with the particle, and then comes out"

or, at least

you shouldn't think of said wave packet being narrow

ACuriousMind

yeah, no, you shouldn't imagine some sort of sharp or even vaguely well-defined boundary there

bolbteppa

Technically in the Hydrogen atom or something the interaction is there forever right

ACuriousMind

the "switch off the interaction" part in scattering theory is part of our model, not part of the world

this is very different as opposed to cases where you literally have an experiment where you can turn the perturbation on an off

Semiclassical

15:13

@ACuriousMind well, you could imagine a wavepacket being something like this:

oops

ACuriousMind

what I'm saying doesn't relate to the shape of the in/out states at all

Semiclassical

i.e., a wavepacket whose spatial extent is reasonably well-defined

ACuriousMind

not so sure about step potentials, but potentials like $\frac{1}{r}$ "interact" with your wave packet everywhere and forever regardless of how localized they are

Semiclassical

hmm, true

you do have to make some assumptions about the potential being local in scattering theory, tho

ACuriousMind

sure, because otherwise not even the asymptotic future will guarantee that you reach a regime where the interaction becomes negligible

Semiclassical

15:17

i'm not sure what a good 'generic' incident scattering state would be for something like a 1/r potential tho

ACuriousMind

iirc you do a partial wave expansion in terms of Coulomb waves for that

Semiclassical

gross

bolbteppa

$\psi = (1 + \frac{1}{i k^3 r(1 - \cos \theta)})e^{ikz + \frac{i}{k} \ln(kr - kr \cos \theta)} + \frac{f(\theta)}{r}e^{ikr - \frac{i}{k} \ln(2kr)}$

So it's like $e^{ikz}$ with messy corrections

Semiclassical

that first one looks weird

would've expected $1+i k^3 r(1-\cos \theta)$ or some such

bolbteppa

In terms of scattering cross sections, the extra terms turn out to be irrelevant

Semiclassical

15:27

@ACuriousMind i'd also expect the Yukawa potential to be what's "really" relevant in most cases

@bolbteppa well. so long as the leading terms don't vanish :P

Slereah

15:57

Is there a version of projective space where directions matter?

the boundary is just a sphere instead of / Z2

What if I only wished to quotient by $\mathbb{R}^+$

Because sometimes directions are important

Slereah

17:22

It really saddens me sometimes that in the end, there are maybe like 8 math questions totals, but each of them have a billion nuance

Semiclassical

17:49

something i'm forgetting: in what sense does the Coulomb potential have a valid Fourier transform? obviously one can pass to the Yukawa potential, take the FT, and then pass to the Coulomb limit, but i'm forgetting if that's the only meaningful way to interpret the Coulomb FT

(obviously one can regularize in different ways)

bolbteppa

18:14

4

A: Fourier transform of the Coulomb potential

This is not really a complete answer, but more of an interesting observation which might help to understand the regularized result better. First observe that \begin{equation} \int_V d^3 r \frac{e^{i\mathbf q \cdot \mathbf r}}{r} = 2\pi \int_0^R dr \int_{-1}^1 du \, r e^{iqru} = \frac{4\pi}{q} \in...

That's an interesting hand-waving one that's easy to remember

You can also do IBP and use $\nabla^2 \frac{1}{r} = 4 \pi \delta(r)$ which gives it in a line or two

Semiclassical

yeah, i noticed that answer too

eigenvalue

23:56

Can anyone here confirm my calculation (tensor product rules)? I have this pseudo-Riemannian metric given:
$g=-(\cos\alpha dx+\sin\alpha dy)\otimes(-\sin\alpha dx+\cos\alpha dy)$

I want to know how it looks like in the "conventional metric representation". Is the following calculation correct?

$g=-(\cos\alpha dx+\sin\alpha dy)\otimes(-\sin\alpha dx+\cos\alpha dy)=(-\cos\alpha dx)\otimes(-\sin\alpha dx)+(-\cos\alpha dx)\otimes(\cos\alpha dy)+(-\sin\alpha dy)\otimes(-\sin\alpha dx)+(-\sin\alpha dy)\otimes(\cos\alpha dy)=(\cos\alpha\sin\alpha)dx\otimes dx-(\cos^{2}\alpha)dx\otimes dy+(\sin^{…

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