The h Bar

ACuriousMind

12:00 AM

@0celo7 Probably, I didn't read all of it, but that's all that happens, really.

0celo7

@ACuriousMind Is the second theorem the one with Lie derivatives?

ACuriousMind

@0celo7 Noether's second theorem concerns the "conserved" currents for local symmetries instead of global symmetries. They are not that often that interesting because they are seldomly gauge invariant.

0celo7

@ACuriousMind Do we need Noether's second theorem here because conformal trafos depend on position?

ACuriousMind

Just like the $j=\epsilon T$ aren't "true" conserved currents, since they are not invariant under conformal tranformations

But yes, the conformal currents are really from the second, not the first theorem

@NeuroFuzzy Oh, I held a talk about elementary WKB once, but the accompanying hand-out is in German and quite dense

0celo7

@ACuriousMind They're the currents of a specific trafo, right?

No reason to expect them to be invariant.

bolbteppa

12:06 AM

In the Hamiltonian case I can write $(1 + iH)\psi(1 - iH) = \psi + i[H,\psi] = \psi + i\int [T_{00},\psi]dt$ yeah? I checked Peskin and Schroder and $H = \int T_{00} dV$ is in there, so $i[H,\psi] = i[\int T_{00}dV,\psi] = i\int [T_{00},\psi]dV$ right? Almost?

ACuriousMind

@bolbteppa Almost, $H = L_0 + \bar L_0$, as far as I recall

But essentially right

@0celo7 Yeah, it's unreasonable to expect the current of specific local trafos to be invariant, but that makes them not all that useful.

0celo7

@ACuriousMind $T_0$?

ACuriousMind

Argh

$L_0$

0celo7

:P

ACuriousMind

that's right, right?

bolbteppa

12:12 AM

@ACuriousMind yeah $H = L_0 + \bar{L}_0$, and in complex coordinate $H = z \partial_z + \bar{z} \partial_{\bar{z}}$ right?

0celo7

@ACuriousMind There's a factor of 2 in (closed) string theory. But no one cares.

ACuriousMind

@bolbteppa Yep, seems right, the expressions in $z$ are always one degree higher than the $L_i$.

@0celo7 You're right, no one cares about constants ;)

0celo7

@ACuriousMind That's the line between physicists and engineers.

Or good engineers and bad engineers.

ACuriousMind

@0celo7 Yep. I'm happy on my side of the line :)

bolbteppa

So where does $T_{\varepsilon} = \dfrac{1}{2\pi i}\oint \varepsilon( z) T(z) dz$ come from? I'm fine about the $\dfrac{1}{2 \pi i}$ for now and the constant time thing, but why do they have a $\varepsilon (z)$ AND the EM tensor in that integral? If we agree to define $e:=\oint dz\epsilon T$ in my calculation $(1 + ie) \psi (1 - ie) = \psi + ie \psi - i\psi e = \psi + [e,\psi] = \psi(0) + i(d\psi /de)de$ it makes sense, BUT where did $e:=\oint dz\epsilon T$ come from?

0celo7

12:18 AM

@bolbteppa $j=\epsilon T$

I've said this countless times...

@ACuriousMind $g=2$ torus out of an octagon is pretty neat, I must say.

Danu

Did you guys see that Sofia apparently decided to give up on participating on this site?

Sofia

3.8k 2 4 23

ACuriousMind

@0celo7 Gluing surfaces out of polygons is, like, the best thing ever

0celo7

I see.

bolbteppa

Compare:
$(1 + ie)A(1-ie) = A + i[e,A] = A + i[\int \varepsilon TdV,A] = \int [\varepsilon T,A]dV$
$(1 + iH)A(1 - iH) = A + i[H,A] = A + i[\int_V T_{00} dV,A] = A + i \int[T_{00},A]dV$
One calculation has a $\varepsilon$, the other doesn't. One time we define $H = \int TdV$, the next we define it $H = \int j dV = \int \varepsilon T dV$, I'm so confused...

0celo7

@ACuriousMind What the link between a $k$-gon and a $g$-torus?

ACuriousMind

12:23 AM

My bachelor's thesis essentially worked by just thinking about surfaces as polygons

Danu

@0celo7 Quite, IMO

ACuriousMind

@Danu It was inevitable, in a sense. I'm not surprised

Danu

@ACuriousMind I feel slightly bad

0celo7

@bolbteppa Note that you are missing you trafo parameter $t$ in your Hamiltonian calc.

ACuriousMind

@0celo7 You can glue a surface without boundary of genus $g$ out of a $4g$-gon, I think

0celo7

12:27 AM

Cool.

ACuriousMind

the funnier way is gluing everything together out of pairs of pants, which are glued from a nonagon, though

Yep, boundary-less surfaces of genus $g$ are just $4g$-gons.

0celo7

@ACuriousMind Is the proof nontrivial?

ACuriousMind

@Danu I...don't. She isn't a bad person, but she simply didn't understand how this site works

0celo7

Does anyone know the full story of what happened?

bolbteppa

@0celo7 woah... What is it, $t$, $\varepsilon$, $z$? Ahhh

Danu

12:29 AM

If you followed chat, I don't think you missed much @0celo7

I'm a little bit saddened by the fact---or at least I think it is---that this was mostly due to language/cultural differences

0celo7

@Danu I know that she posted a wrong answer that got accepted.

ACuriousMind

@0celo7 If you know how to take the gluing recipe $x_1 x_2 x_1^{-1} x_2^{-1} x_3 \dots x_{2g}x^{-1}_{2g}$ and know that the genus is a topological invariant, then it is trivial, because the gluing recipe produces a surface of genus $g$ ;)

0celo7

@ACuriousMind You a French mathematician now?

Kyle Kanos

@Danu Yeah, she was commenting above about being too frustrated with downvotes w.r.t. FTL communication

0celo7

@KyleKanos Wait, I thought she was wrong? Why would she be upset if she's wrong?

Danu

12:33 AM

It was not an obvious right/wrong thing

Language issues, again, were at the root of it all in my view

alarge

@ACuriousMind Well, I think it boils down to inexperience of how internet, and internet communities in particular, work. Getting all worked up about how some of the thousands of people visiting the site might, (pseudo)anonymously, disagree with you is not good for your health.

Kyle Kanos

@0celo7 My understanding is that she advocates against it and is taking the downvotes on her answers about the lack of FTL as being an implicit "for FTL comm."

ACuriousMind

@0celo7 I strive to be ;) It's really easy once you've become familiar with the notion of "gluing", and once you feel comfortable with seeing the sphere and the Möbius strip and the Klein bottle as different gluings of a square

bolbteppa

@0celo7 I think it's supposed to be written $U\psi U^* = (1 + iH\varepsilon)\psi(1 - iH\varepsilon) = \psi + i[H \varepsilon ,\psi] = \psi + i[\int T_{00} \varepsilon dV,\psi]dV$!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

0celo7

@ACuriousMind I'm plenty comfortable with squares, $4g$-gon is just cray-cray.

bolbteppa

12:34 AM

OMG!!!

WAHOO!!!

0celo7

@bolbteppa Lol.

Let it be known that I totally figured it out.

bolbteppa

That is insane...

Danu

what the heck are you on about?

bolbteppa

Yeah man! Thanks for the point about the parameter!

0celo7

@Danu The most nontrivial trivial calculation ever.

ACuriousMind

12:37 AM

@alarge Yep, correct in every respect.

bolbteppa

lolwut, so because time generates symmetries of the hamiltonian we can write $U = e^{-iHt}$, but when conformal transformations generate symmetries of the Hamiltonian we can write $U = e^{-iH \varepsilon}$, so if something is a symmetry of the Hamiltonian it goes in the exponent for some reason

0celo7

@bolbteppa The exponential is how you go algebra -> group.

Danu

@bolbteppa This doesn't seem surprising at all

bolbteppa

Yeah but why does the parameter change depending on whether it gives a symmetry of the Hamiltonian?

0celo7

@bolbteppa What?

bolbteppa

12:43 AM

It's fine, wahoo!!!

0celo7

@Danu Have you looked at Nakahara? It seems quite good.

Danu

nope

I have this idea that it's a math for physicists book

...but I never actually tested it

0celo7

@Danu I'm interested in the sections on bundles, Kahler geometry and Chern classes.

bolbteppa

Nakahara + Frankel seem like a pretty good combo for that stuff

Danu

what book by Frankel

ACuriousMind

12:47 AM

@bolbteppa I...don't think it's $\mathrm{i}H\epsilon$ there, it has to be $\mathrm{i}Q_\epsilon \epsilon$, where $Q_\epsilon$ is the Noether charge corresponding to the infinitesimal transformation $\epsilon$.

bolbteppa

I'd like to give you all a secret, chapter 12 of Nakahara is literally just generalizing the theory of integral equations to manifolds, that is where a lot of the ideas are coming from if you want to be ready for it, otherwise it'll all look complicated and the big words will scare ;)

ACuriousMind

And the Noether charge is, as 0celo7 said, just the integration of $T\epsilon$ along a circle.

@Danu: Have you tried reading Quantization of gauge systems yet? You asked about it some weeks ago.

bolbteppa

@ACuriousMind so you're saying $H = Q$ if that's true

ACuriousMind

@bolbteppa No, $Q = \oint T\epsilon$.

I don't get what the Hamiltonian would have to do with a transformation on an arbitary field $\psi$ at all.

Danu

Wow, I just found an amazing treasure trove: topology.org/tex/conc/differential_geometry_books.html#physics

@ACuriousMind Naw, I'm busy studying my elementary abstract algebra for now

bolbteppa

12:51 AM

@ACuriousMind so how does $Q$ fit in to this calculation $U\psi U^* = (1 - iH\varepsilon)\psi(1 + iH\varepsilon) = \psi - i[H \varepsilon ,\psi] = \psi + i[\int T_{00} \varepsilon dV,\psi]dV$ Either $H\varepsilon = \int T_{00} \varepsilon dV$ or $H\varepsilon = Q\varepsilon = \int T_{00} \varepsilon dV$ right?

ACuriousMind

@Danu Better that way. Hennaux assumes you are quite comfortable with differential complexes and other advanced algebra

Danu

In any case, I should at least study Riemannian geometry first

ACuriousMind

@bolbteppa How is your $U$ defined?

bolbteppa

@Danu the Frankel book he mentions on that page

ACuriousMind

And what's $\psi$.

Danu

12:53 AM

@bolbteppa That's how I found it

bolbteppa

@ACuriousMind if we had Schrodinger's equation and $U = e^{-iHt}$ then to first order @ACuriousMind if we had Schrodinger's equation and $U = e^{-iHt}$ then to first order $U\psi U^* = (1 - iHt)\psi(1 + iHt) = \psi - i[H t ,\psi] = \psi + i[\int T_{00}t dV,\psi]dV$, if I change $t$ to $\varepsilon$ and only worry about $U$ to first order I can write $U\psi U^* = (1 - iH\varepsilon)\psi(1 + iH\varepsilon) = \psi - i[H \varepsilon ,\psi] = \psi + i[\int T_{00} \varepsilon dV,\psi]dV$.

Either $H\varepsilon = \int T_{00} \varepsilon dV$ or $H\varepsilon = Q\varepsilon = \int T_{00} \varepsilon dV$ right? My book uses $Q$ so I wonder...

That silly silly thing cost me like 2 days, ahh... you have no idea how helpful it is!

ACuriousMind

@bolbteppa: 1. Not every unitary transformation $U$ has to be an exponential of the Hamiltonian. 2. For example, The infinitesimal rotation about the $z$-axis is given by the $\mathrm{SO}(3)$ generator $L_z$ by $\psi \mapsto \psi + [L_z,\psi]$. The finite trafo is given by $\psi \mapsto \mathrm{e}^{L_z\theta} \psi \mathrm{e}^{-L_z\theta}$. The Hamiltonian has nothing to do with trafo as such.

bolbteppa

@ACuriousMind sure but in this case it has to be because conformal transformations generate symmetries of the Hamiltonian (hand-wavey, potentially wrong)

ACuriousMind

@bolbteppa It doesn't matter what they are symmetries of, the thing in the exponential is the Noether charge, not the Hamiltonian.

bolbteppa

@ACuriousMind so if that's true then you're saying the Noether charge is equal to the Hamiltonian in my example?

ACuriousMind

1:05 AM

@bolbteppa Only if your transformation is a time translation, i.e. the transformation is generated by $L_0$. If it is not, the thing in the exponential is not the Hamiltonian.

I don't get how you arrived at your conclusion that it has to be that Hamiltonian at all

Stan Shunpike

@ACuriousMind did you say yesterday a $k$-tensor doesn't act upon forms?

ACuriousMind

@StanShunpike Yes.

Observe that you can expand to a notion of a $(k,l)$-tensor though, which eats $k$ vectors and $l$ covectors.

Stan Shunpike

We built our $k$-tensors out of $1$-tensors right?

bolbteppa

@ACuriousMind how do you derive that $U = e^{iQ}$ is a conformal transformation?

ACuriousMind

Since, for example, a $1$-form is at every point just a covector, you can feed forms to tensors in that sense, then.

Stan Shunpike

1:13 AM

Huh?

bolbteppa

@ACuriousMind yeah you're right! physics.stackexchange.com/questions/137499/…

ACuriousMind

@bolbteppa There's no need to derive it, that's how you get the finite transformation out of the infinitesimal one - the exponential map!

bolbteppa

That is mental...

ACuriousMind

@bolbteppa Nah, just Lie theory ;)

@StanShunpike Okay, how about you ask what you originally wanted to ask before I began to blather? ;)

bolbteppa

So basically I had two huge issues, A & B in $e^{iAB}$ lol

Awesome!

0celo7

1:20 AM

@ACuriousMind Obligatory prayer to Urs

ACuriousMind

@0celo7 You're never gonna live me down that I called him my spirit animal, are you? :D

0celo7

@ACuriousMind Urskin

That might be too reddit/4chan for you.

ACuriousMind

His name even means the same as mine. I just realized that, it's spooky

bolbteppa

You guys rock

One last thing, I don't really see why $\int j_0 \varepsilon dx^1 = \int T_{00}\varepsilon^0 + T_{01}\varepsilon^1 \mapsto \dfrac{1}{2\pi i}\oint T(z) \varepsilon dz$ I just wave my hands :( Stupidly I think $\oint j_0 d(\dfrac{z-\bar{z}}{2i})$ which already doesn't lead to what we want :(

ACuriousMind

@bolbteppa The actual time/space coordinates live on a cylinder that's conformally mapped to the complex plane with the time coordinate becoming the radial coordinate and the (compact) spatial coordinate becoming the angle - so integrating in space on the worldsheet/2D spacetime is the same as integrating along a circle in the plane

bolbteppa

1:31 AM

Yeah it makes sense hand-wavily so but if you try to work it out from the integral explicitly it looks like I wrote it which makes no sense, is it even possible?

ACuriousMind

@bolbteppa Oh, that's a good question. I never tried to work that out explicitly oO

bolbteppa

The cylinder map is $w = (t,x) \mapsto z = e^w = e^t,e^{ix}$ so you integrate along a circle, but fuck when you try it computationally, all the sources, all, just wave hands!

0celo7

I think there is a bit of hand waving involved.

The Ward identities, of course, are the stronger statement.

First of all, I don't think $j$ is defined in the cylinder coordinates.

You derive it in the plane.

ACuriousMind

@0celo7 One should be able to reverse the transformation (except at temporal infinity) and recover an expression for the current on the cylinder, no?

0celo7

@ACuriousMind Sure, but that's not very productive, is it?

ACuriousMind

1:37 AM

@0celo7 No, I don't think so^^

bolbteppa

So does it fall out of the Ward identities unquestionably?

Stan Shunpike

@ACuriousMind Back from cena. So, I read that $k$-tensor is short for covariant tensor

cena = dinner in spanish

0celo7

@bolbteppa Yes.

You can work backwards and derive the charge from the Ward identities.

Stan Shunpike

@KyleKanos Norah Jones's father is a famous sitar player. He's amazing

I study his stuff for some of the sitar pieces I have written

ACuriousMind

"Ward identities" = "quantum Noether"

Stan Shunpike

1:40 AM

This came to mind from the Chex Mix track

bolbteppa

It looks to me like they use this to go from the normal ward to the 'conformal ward'

0celo7

@bolbteppa You need the complex Stokes theorem to go from normal to conformal.

Stan Shunpike

@ACuriousMind In other words, when you said $k$-tensors don't act on forms, do you mean contravariant?

Or is that a useless distinction?

ACuriousMind

@StanShunpike Well, contravariant tensors - multilinear maps on the dual space instead of the space itself - act on covectors instead of vectors. If you take a tensor field - a map that associates to every point a tensor - then it can eat a covector at every point, and since forms associate covectors to every point, tensor fields can indeed act on forms.

Stan Shunpike

Oh really, I thought covariant acted on covectors...

ACuriousMind

1:47 AM

@StanShunpike Oh. Well. I mix the two words up every time :)

bolbteppa

@0celo7 Okay cool, how was 4.29 in Blu's String Theory $T(z)T(w) = ...$ so obvious using the (incorrect) $[\delta,\delta]$, Blu's CFT book let me down by defining this fucker and showing it works :( He claims it's obvious in the ST book!?

ACuriousMind

If you have a metric, though, the distinction between vectors and covectors is indeed not that useful, because you can always raise/lower indices to get vectors from covectors and vice versa.

Stan Shunpike

LOL

OH

0celo7

@bolbteppa I have that specific calculation in my notebook, hold up.

Stan Shunpike

Is that only possible with a metric?

@ACuriousMind why?

0celo7

1:48 AM

@bolbteppa I will use the sign conventions of Blu ST.

ACuriousMind

@StanShunpike Well you define the covector to a vector $v$ as the map on the tangent space $g(v,-)$, where $g$ is the metrix.

bolbteppa

@StanShunpike if you have a metric, i.e. a symmetric bilinear form, then there is a 1-1 correspondence between linear functionals and inner products, by the Riesz Representation Theorem I think

Stan Shunpike

The minus....just means empty slot?

ACuriousMind

If you haven't got a metric, you can't canonically associate a covector to a vector in that way.

Stan Shunpike

OH

ACuriousMind

1:49 AM

@StanShunpike Yes, $-$ is empty slot

Stan Shunpike

That's so helpful!

I was so confused by that

0celo7

@bolbteppa Note first that, due to the Jacobi identity, $[\delta_{a},\delta_{b}]\phi(z)=-[[T_a,T_b],\phi(z)]$. Hence it suffices to analyze $[T_a,T_b]$.

I'll use Latin letters for the trafos because reasons.

bolbteppa

@0celo7 woah, how is that obvious!?

Kyle Kanos

@StanShunpike If he's so famous, how come I've never heard of him?

Stan Shunpike

He's old as hell.

He died a while ago.

0celo7

1:50 AM

@bolbteppa Work it out using the Jacobi identity.

Stan Shunpike

I think he's famous in India.

0celo7

Use $\delta\phi=-[T,\phi]$.

Stan Shunpike

My aunt says he's really big in Malaysia, that's where she's from

Kyle Kanos

Google-fu says he was born in 1920 & died in 2012. Fathered Norah at 59 y/o

0celo7

@bolbteppa You can work that out later. I'll continue.

Stan Shunpike

1:52 AM

@KyleKanos Yeah, I learned some neat sitar tricks from his stuff that I uses in my songs sometimes.

bolbteppa

@0celo7 okay cool, yeah I think I see it

Stan Shunpike

I came across him because I wanted to write indian music

Just for fun and wanted some authentic sounding melodies

But he's unusual because he blends classical music with indian music

So a lot of his works are symphonic in the Western sense

But with intense indian components

0celo7

@bolbteppa We have, for the commutator of charges $$[T_a,T_b]=\oint dz\oint dw\,a(z)b(w)[T(z),T(w)]=\oint_{\{0\}}dw\,b(w)\oint_{\{w\}}dz\,a(z)\mathcal{R}[T( z )T(w)]$$

ACuriousMind

@StanShunpike: Let me get this straight: You are a composer, who is an econ major, who also has a big interest in physics?

0celo7

No $2\pi i$ because reasons.

Stan Shunpike

1:54 AM

Yes....

My mother almost turned into a pro musician but became a psychologist. My dad is an economist / lawyer. And I guess the physics is my own since no one else in my family seems to like that

ACuriousMind

@StanShunpike That's the most polymath-like person I've ever known :D

Stan Shunpike

LOL

0celo7

Here $\mathcal{R}$ is denotes radial ordering.

bolbteppa

yeah

Stan Shunpike

My artist name is Lepton.

0celo7

1:56 AM

Now we plug in the ansatz $$T(z)T(w)\sim \frac{c/2}{(z-w)^4}+\frac{2T(w)}{(z-w)^2}+\frac{\partial T(w)}{(z-w)}$$

Stan Shunpike

I do not actively promote my songs on SE since that's not its purpose, but you might like this one: soundcloud.com/gotlepton/sono-tuo-per-sempre

0celo7

@bolbteppa And we check that it gives the consistent trafo.

You want me to continue?

Stan Shunpike

When I make a new one, I just put in on my profile page

bolbteppa

Yeah definitely!

0celo7

$$[T_a,T_b]=\oint_{\{0\}}dw\,b(w)\oint_{\{w\}}dz\,a(z)\left[\frac{c/2}{(z-w)^4}+ \frac{2T(w)}{(z-w)^2}+\frac{\partial T(w)}{(z-w)}\right]$$

bolbteppa

1:59 AM

Oh woah, where did that ansatz come from

0celo7

Now we use the Cauchy theorem. I got $$\oint_{\{w\}}dz\,a(z)\left[\frac{c/2}{(z-w)^4}+ \frac{2T(w)}{(z-w)^2}+\frac{\partial T(w)}{(z-w)}\right]=\frac{c\partial^3 a(w)}{3!\cdot 2}+2\partial a(w)T(w)+a(w)\partial T(w)$$

for the inner integral

bolbteppa

yeah

0celo7

The ansatz is just the OPE guessed cleverly.

We good so far?

bolbteppa

Yeah

Stan Shunpike

@ACuriousMind I don't understand something you said earlier

Why do we need the tensor product to define $k$-tensors?

I mean, I saw that we use it to define them

But can't we just state it is a multilinear map?

We don't need tensor products then right?

That's why I asked if there was some advantage to defining them this way.

0celo7

2:02 AM

So we have $$[T_a,T_b]=\oint_{\{0\}}dw\,b(w)\left[\frac{c\partial^3 a(w)}{3!\cdot 2}+2\partial a(w)T(w)+a(w)\partial T(w)\right]$$

ACuriousMind

@StanShunpike It sounds good :) I guess I'm a bit one-dimensional because my parents are both chemists (not with a university degree). I'm the first one in my extended family to really pursue an academic degree and have a realistic chance of completing it.

bolbteppa

@Danu the dude writing that book just came in here

geometry.org/tex/conc/dgchaps.html

0celo7

@bolbteppa The first term vanishes because it is analytic ($\oint dz\,f(z)=0$)

Stan Shunpike

You read fiction :)

@ACuriousMind that's not one dimensional

that's at least 2

lol

0celo7

Integrate the second term by parts.

bolbteppa

2:03 AM

okay

ACuriousMind

@StanShunpike Heh, alright, because my father is a massive nerd :D

Stan Shunpike

omg really? Does he have a large library? I read sooooo much. My room is overflowing with books

I can't tell if that's a trait nerds share or just smart people

0celo7

You end up with $$[T_a,T_b]=\oint_{\{0\}}dw\, [b(w)\partial a(w)-a(w)\partial b(w)]T(w)=\delta_{a\partial b-b\partial a}$$

$\Box$

Stan Shunpike

@0celo7 what does $T_a$ stand for?

ACuriousMind

@StanShunpike We (well, now, only my parents) live in quite a small flat. It's filled with books, and the cellar is full of old (=80's pulp fiction and such) books I read through as a young'un

0celo7

2:05 AM

@StanShunpike Infinitesimal conformal trafo by $a(z)$.

Stan Shunpike

and that's a commutator?

0celo7

@StanShunpike Yes.

Stan Shunpike

What are you integrating with respect to?

like what is $dw$

0celo7

@ACuriousMind That's the calculation I was freaking out about over the other day because of a minus ;)

@StanShunpike You don't know complex analysis?

bolbteppa

lol

woah

Stan Shunpike

2:06 AM

LOL no but I'm looking for books on it for sure. I haven't had a decent recommendation yet.

I could probably learn a bunch quickly if I had one

0celo7

@StanShunpike I learned mine from Cahill and it served me decently.

Nothing crazy.

ACuriousMind

@StanShunpike Ah, yes, you define the tensors as multilinear maps. But it's...pretty to see that they're all essentially built out of $1$-tensors.

So you can just define and check stuff on the $1$-tensors and "linearly extend" it to all tensors

Stan Shunpike

@0celo7 Physical Mathematics by Cahill?

0celo7

Yes

Chapter 5

Stan Shunpike

@ChrisWhite Do you have a favorite book on complex analysis?

0celo7

2:09 AM

He doesn't actually tell you how to do a basic contour integral though.

The wiki article is actually good for that.

Stan Shunpike

en.wikipedia.org/wiki/Methods_of_contour_integration this one?

0celo7

@StanShunpike No, but that article is good too.

Line integral

In mathematics, a line integral is an integral where the function to be integrated is evaluated along a curve. The terms path integral, curve integral, and curvilinear integral are also used; contour integral as well, although that is typically reserved for line integrals in the complex plane. The function to be integrated may be a scalar field or a vector field. The value of the line integral is the sum of values of the field at all points on the curve, weighted by some scalar function on the curve (commonly arc length or, for a vector field, the scalar product of the vector field with a d...

Stan Shunpike

Oh, I didn't realize by contour you meant line integrals

0celo7

You can always come in here and bug me and the skull.

@StanShunpike complex line integral = contour integral

Stan Shunpike

AH

okay, that's good to know

I thought I had to learn some new complex integral involving measure theory or something

ACuriousMind

2:13 AM

@StanShunpike: When I moved to Heidelberg, I took two large boxes full of my favourite books with me. Almost everyone looked at me as if I was crazy when I told them that. For me and my parents, it was the most natural thing in the world.

Stan Shunpike

I'm still not great at that

0celo7

@ACuriousMind What's the value of a contour integral around Western Europe?

Stan Shunpike

@ACuriousMind I know right! Me too. Why not? They are like treasure troves to me. It's where I find all my information.

@0celo7 waiting for punchline....

0celo7

@StanShunpike You won't get it :P

Zero, because all the Poles are in Eastern Europe.

ACuriousMind

@0celo7 Uh...the residue at the Mont Blanc? ;)

0celo7

2:15 AM

Lol what

Stan Shunpike

@KyleKanos Do you know any good complex analysis books?

ACuriousMind

It's the highest mountain in Western Europe, seemed natural to consider it as a pole :D

0celo7

@StanShunpike Complex analysis is the result of Cauchy and Riemann sitting down and deciding calculus was not awesome enough already.

3

@ACuriousMind I know it's the highest mountain...

Stan Shunpike

hahahahha I love that way of phrasing it

That's awesome

ACuriousMind

@StanShunpike Well, I don't have that much books giving me information, I actually took my favourite fiction books with me

bolbteppa

2:16 AM

@0celo7 thanks man i'll write it up tomorow!

Stan Shunpike

Well, those are fun too. All my fiction is audio entertainment though. I collect and listen to radioshows from 1935-1955

Kyle Kanos

@StanShunpike Don't know any complex analysis book

ACuriousMind

And I wept recently when I heard that Terry Pratchett had died. The only death of a person I didn't really know that has ever touched me.

Stan Shunpike

wow that was 4 days ago

Kyle Kanos

Still don't know who this guy is

Stan Shunpike

2:18 AM

(me neither but i have mad google fu)

ACuriousMind

@KyleKanos You should. He liked to say that being funny doesn't mean that you're not also being serious, and most of his books are testament to that

0celo7

I have no time/desire to read fiction right now.

ACuriousMind

@0celo7 That's alright. I like to be immersed in other worlds, to explore what our imagination is really capable of, and I gladly sacrifice time I could have spent more "usefully" for that.

0celo7

@ACuriousMind Pop quiz: solid sphere moment of inertia

ACuriousMind

@0celo7 lol, ask Wiki. I don't memorize stuff that's always at my fingertips :P

0celo7

2:25 AM

@ACuriousMind (2/5)MR^2

I hope.

I have a rotational motion test tomorrow.

ACuriousMind

@0celo7 Correct, according to Wiki ;)

user54412

@StanShunpike never found a good one

ACuriousMind

@0celo7 I so do not miss school tests.

Stan Shunpike

@ChrisWhite I'll ask my uncle.

He's a mathematician. maybe he knows one

user54412

It also depends what you're going for

Kyle Kanos

2:28 AM

3

Q: Which book on complex analysis is good for self study?

Which book on complex analysis is good for self study? I am an average student and have just a very basic knowledge of this subject.I want to cover up to Runge's Theorem. I heard about few books- Gamelin's Complex Analysis; a text by Churchill and another Ahlfors' Complex analysis. Thank you.

complex-analysis complex-numbers several-complex-variables

25

Q: Complex Analysis Book

I want a really good book on Complex Analysis, for a good understanding of theory. There are many complex variable books that are only a list of identities and integrals and I hate it. For example, I found Munkres to be a very good book for learning topology, and "Curso de Análise vol I" by Elon ...

complex-analysis reference-request

user54412

a bare minimum is the Cauchy(-Goursat) theorem, Cauchy's integral formula, and the Cauchy Residue theorem

user54412

also the analytic-holomorphic equivalence

ACuriousMind

@ChrisWhite That one's like a dream come true. Isn't there some quote saying that complex analysis is where all the stuff you wished were true, is true?

Kyle Kanos

@ChrisWhite: Do you know when astronomers stopped using "nebula" to mean galaxy? I'm reading/skimming Baade & Zwicky's 1934 paper & they use nebula in place of galaxy in a few places (e.g., A similar frequency has been found by Hubble in the well-known Andromeda nebula.)

user54412

if you're into pure math you want to cover things like harmonic functions, whereas more applied courses would focus on things like conformal mapping

0celo7

2:32 AM

@ChrisWhite I just kinda assume this at this point. What are the original definitions that are not immediately related?

Kyle Kanos

Or was it that they actually thought it was a nebula and later found that to not be the case...hmm

ACuriousMind

@0celo7 analytic = expandable as a power series that converges everywhere. holomorphic = complex differentiable.

0celo7

Ah yes.

Stan Shunpike

@ChrisWhite Conformal mappings is applied mathematics????

Why?

user54412

@KyleKanos "In 1924, Edwin Hubble established the distance to classical Cepheid variables in the Andromeda Galaxy, until then known as the Andromeda Nebula, and showed that the variables were not members of the Milky Way."

Stan Shunpike

2:34 AM

@ACuriousMind The more physics I learn, the more exciting I find it.

user54412

wikipedia obviously things Baade and Zwicky were old-fashioned

Kyle Kanos

@ChrisWhite Hmm. Seems strange that Hubble wrote that in 1924 and B&Z chose to perpetuate nebula

Stan Shunpike

@ACuriousMind When my sister asks why I like it, I often say because it's like a really cool story with magic that actually is real!

ACuriousMind

@StanShunpike <3

You speak my mind.

user54412

@StanShunpike solving differential equations is applied math, and there are all sorts of tricks for solving them on 2D domains by mapping the domain to something else

Stan Shunpike

2:37 AM

like what?

rings? fields? manifolds?

what kind of thing do you map to?

user54412

@KyleKanos I've seen "nebula" at least through the 30s -- might be a question for HSM as to when the switch was made (or when the term "galaxy" even came into its present meaning)

0celo7

I'm derping majorly. "A narrow but solid spool of thread has radius $R$ and mass $M$. If you pull up on the thread so that the CM of the spool remains suspended in the air at the same place as it unwinds, what force must you exert on the tread?"

user54412

Old astro papers are cute in a way -- for one, almost every paper is equal parts theory, observation, and instrumentation

0celo7

No smithereen of an idea what to do.

Is the upward force just $Mg$?

$Mg-F_{up}=0$

@ACuriousMind :3 help

Stan Shunpike

@0celo7 :3 looks like a handlebar mustache

user54412

2:45 AM

@0celo7 Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better

user54412

;)

0celo7

@ChrisWhite D,:

ACuriousMind

@0celo7 I think you have to factor in that part of the force goes into just unwinding the spool, i..e is not actually applied to the centre. It's a yo-yo, essentially, isn't it?

0celo7

@ACuriousMind I think so, but I'm not sure.

And I don't know how to do that.

user54412

My wild guess is there's a factor of 2 that comes out

ACuriousMind

2:47 AM

But then, I think that the question is a bit ill-posed - is the spool suspended in the air, and they are asking what you need to do so that it unwinds so that it doesn't fall down?

Or does the spool lie on a table?

Stan Shunpike

Okay, so David Z has set me straight about flagging and that flagging is a good thing to do when you do it intelligently.

But what about downvoting.

Like, this may be silly, but I like to keep my rep points and so I don't downvote

0celo7

The one question I have no idea how to do is even numbered.

Stan Shunpike

is that the wrong strategy

?

ACuriousMind

@0celo7 Look at the "mathematics" section here

They derive an equation for $y(t)$, and you seem to need to solve it for $y(t) = 0$ identically, yes?

@StanShunpike Heh. First, realize that downvoting questions doesn't cost rep

0celo7

@ACuriousMind I'm going to pull the high school card and say that's too complicated for this homework.

Stan Shunpike

2:52 AM

It costs -1?

@ACuriousMind

ACuriousMind

@StanShunpike No, only on answers

Downvoting questions is free

Stan Shunpike

OH

Really?

Wow

ACuriousMind

Yep, only answers cost rep

user54412

and I believe points lost downvoting answers are refunded if the answer is deleted

Stan Shunpike

That's a different ball game.

NeuroFuzzy

2:54 AM

you could conspire to have someone in chat upvote you every time you downvote ten times

Stan Shunpike

Glad I asked!

I frequently see hw questions that I have always been inclined to downvote but reluctant because I thought it would cost me rep.

ACuriousMind

@NeuroFuzzy That's...not the purpose of voting, and I think the moderators would be on to that quite quickly

NeuroFuzzy

How would they know? It's not like this room is being recorded

ACuriousMind

@NeuroFuzzy It is

There's a transcript that goes back into the beginning of time

NeuroFuzzy

@ACuriousMind whaaaaaaaaaat

ACuriousMind

2:56 AM

Also, the mods are in here quite frequently

You don't really think Qmechanic would miss anything, would you? ;)

NeuroFuzzy

I would still urge @StanShunpike to go for it.

this is... sarcasm, by the way.

Stan Shunpike

I don't even know what we're talking about.

Lolol

I didn't follow this at all.

lolol

i just typed in

ACuriousMind

@NeuroFuzzy Totally randomly, here's a post from you more than a year ago.

Stan Shunpike

virasoro generators and it wanted to change it to bizzaro generators

user54412

more votes is better -- too few votes and the S/N ratio becomes too large to reliably differentiate the good from the bad

user54412

2:59 AM

Also, I see @KyleKanos will soon knock me out of the top 5

ACuriousMind

I advise everyone to downvote and upvote more liberally. Votes are the mechanism by which things are sorted on SE, and reluctance to vote means visitors are not very well able to distinguish between good and bad stuff.

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