The h Bar

12:02 AM

@ACuriousMind Srednicki defines it like Dine...I think he's saying the constant is irrelevant, which it is. I'll give this more thought later.

user54412

12:13 AM

@StanShunpike I hesitate to recommend Wald as a starting place for anything basic. Carroll usually covers such things well.

Stan Shunpike

I have Carroll too. I like Wald a lot. Perhaps I should go back and forth between them.

Carroll's notes online are fantastic.

user54412

Wald is good -- one of only 3 GR texts I regularly look at -- but if you jump right into the middle of the book you'll see everything cast in terms of $\mathcal{I}^+(J^-(S))^\circ$ and such, and it's quite an endeavor to backtrace through his definitions.

Stan Shunpike

Although not as bad as MTW, who seemingly decide to make up their own notation.

0celo7

12:33 AM

@StanShunpike Wald's notation is standard I believe, you've just never seen the topics before that he treats with that "unusual notation".

I was under the impression you had already read Carroll, my bad.

@ChrisWhite I'm assuming MTW is another. What is the third?

@StanShunpike : @ChrisWhite is right about Wald being a poor starter. If you haven't already, give Zee's black hole chapter a serious read and then read the relevant chapters in Carroll. They use similar notation IIRC so you should be fine. If you want to read Wald, you can't go back any further than chapter 8, perhaps even chapter 7. All of chapter 9 builds off of chapter 8. Chapter 11 builds off of chapters 9 and 8. Chapter 12 (black holes) builds off of chapters 9 and 11.

ACuriousMind

Have you memorized the chapter contents or are you actually looking at the book right now?

0celo7

@ACuriousMind I just finished it a week ago, and am looking at it.

@ACuriousMind I have Zee pretty much memorized at this point. It's pretty abused after only 2 years of ownership, because I've read it so much.

ACuriousMind

I see

0celo7

@ACuriousMind In hindsight, his treatment of differential geometry is abhorrent. The diffeomorphism is mentioned twice: once in the introduction where he says it is useless and the second time in an appendix where he defines it and says once more that it is useless.

ACuriousMind

@0celo7 Uh...what? That's an...unusual thing to say, at least.

0celo7

12:47 AM

@ACuriousMind :: hunts for specific quote ::

@ACuriousMind While I realize the need for and the benefit of precise definition, for the most part I simply plead membership in the Feynman “Shut up and calculate” school of physics. Thus, I won’t trouble your sleep with assertions such as “A bijective differentiable map of a manifold, whose inverse is also differentiable, is called a diffeomorphism.”

Regarding statements like this, I think that another Einstein quote may be apropos: “We should take care not to make the intellect our god; it has, of course, powerful muscles, but no personality.”

@ACuriousMind In the aforementioned appendix: "After I wrote this appendix during final revision of this book, I sent it to a colleague, one of those distinguished physicists listed in the preface. He sent back a terse email, complaining that the inclusion of diffeomorphism is “like a scratch on a record playing the sublime music of gravity.” Then he told me, if I must include this kind of stuff, to find a better place to hide it."

ACuriousMind

So, he writes "bijective differentiable map whose inverse is also differentiable" every time? Or does he simply not discuss such requirements and just tacitly assumes them?

0celo7

@ACuriousMind The latter.

ACuriousMind

I'm not sure which of these I find more horrible

0celo7

@ACuriousMind XD

Note that he views the wedge as superfluous notation.

Changing the title to A. Zee - Handwaving would not be that far off.

However, it is quite excellent for what it is.

It has around 80 pages or so on dS, AdS and KK theory.

ACuriousMind

@0celo7 Gnnnnnnh. My precious wedge! Superfluous!

0celo7

1:00 AM

It also squeezes in a good SR overview, an introduction to the action principle and quite a bit of phenomenology.

ACuriousMind

He's got a point though, you can go your entire life as a physicist without ever seeing a wedge

0celo7

@ACuriousMind He's a field theorist at heart though.

I've read some of his papers on TQFT...no wedges.

It's strange.

ACuriousMind

I believe Zee stands firmly with the index notation, then?

Stan Shunpike

I have Carroll, MTW, Zee, Hartle, Weinberg and now Wald

I like Wald and Weinberg the best.

ACuriousMind

It's equivalent, you can essentially do everything in indices. It's so cluttered though, to my eyes.

0celo7

1:03 AM

@ACuriousMind He writes $X=X^i\partial_i$ once in an appendix, calls it a "rich man's tool", and discards it.

ACuriousMind

lol

0celo7

@ACuriousMind You'd get a kick out of some of the handwaving he does. It's unfollowable at times.

ACuriousMind

@0celo7 Well, good handwaving is an art in itself.

0celo7

@ACuriousMind He handwaves the Hawking temperature and Bekenstein entropy in the prologue :D

ACuriousMind

I can enjoy it, just don't demand of me to do it myself - my handwavy arguments tend to produce utterly wrong results ;)

0celo7

1:08 AM

He sets $c=\hbar=k=1$ and does dimensional analysis :D

ACuriousMind

Haha. That sounds good

0celo7

@ACuriousMind Set $c=\hbar=k=1$. We see that $GM$ is a length and hence an inverse mass. Since $k=1$, temperature has dimensions of energy, or equivalently dimensions of mass. Hence $$T_H\sim\frac{1}{GM}$$ From thermo we have $dE=TdS$. Here $E$ is just $M$. Integrating $dS/dM=1/T_H\sim GM$, we obtain $$S\sim GM^2$$ Now we use the fact that $R\sim GM$ and hence $A\sim R^2$, and conclude that $$S\sim\frac{R^2}{G}$$

@ACuriousMind Art.

ACuriousMind

Heh, pretty

0celo7

1:25 AM

@ACuriousMind :: sigh :: I messed up and have to read 80 pages of psychology for tomorrow.

Cheerio.

Stan Shunpike

Question: on the wikipedia page on the Lagrangian for classical electrodynamics

They have a Lagrangian with two terms

the field term is just the one with the faraday tensor

and the interaction term involves A and J

They then say

In the interaction term, the four-current should be understood as an abbreviation of many terms expressing the electric currents of other charged fields in terms of their variables; the four-current is not itself a fundamental field.

What does that statement mean?

ACuriousMind

@0celo7 Hehehe...ah, the joys of general education. My condolences, although psychology is one the fields I actually find interesting.

Stan Shunpike

@ACuriousMind best psych book i've ever read is The Science of Acting

ACuriousMind

@StanShunpike Well, it means what it says ;) $A_\mu J^\mu$ is simply the catch-all interaction term, and $J^\mu$ is not a dynamical variable, i.e. you do not derive/use an Euler-Lagrangian equation for it, only for its constituents which are dynamical variables - or you even treat it as a given, e.g. if you just have a wire in the theory that carries a current and you're not interesting in the specifics of that current, only in the field it creates

gonenc

As I learned from @ACuriousMind I am not allowed to post convention questions on SE hence I deleted my question but there is nothing that stops me from asking this question in chat :) Why do we flip the coordinate axis while drawing a minkowski diagram. ie. why does one draw time in the y- and position in the x-axis?

Stan Shunpike

1:39 AM

@ACuriousMind So the four-potential is a dynamical variable? Why? Why not J? Maybe I'm not thinking about this correctly. In classical electrodynamics, the electromagnetic field is a fundamental field correct? Is the electromagnetic four potential another form of this field?

ACuriousMind

@StanShunpike You get the EM field as the derivative of the four-potential as $F= \mathrm{d}A$. And yes, $A$ is a dynamical variable - after all, Maxwell's equations are the e.o.m. derived from this action

@gonenc I, personally, have not the slightest idea. But note that the Minkowski diagrams are not like diagrams for functions where the x-axis would be an independent and the y-axis the dependent variable - time and space are both independent variables.

gonenc

@ACuriousMind Indeed they are both independent variables and that is probably why they flip the axis so that you can know that this diagram is different. But it just disturbs me when I see time in the y axis and I had to pay really good attention when using Minkowski Diagrams since I always drew it incorrectly the other way around!

0celo7

@ACuriousMind Each test I study less and less. This time I'm only going to read the "learning outcomes" and do the vocab. I'm not going to read all this.

I have a 97 right now.

@ACuriousMind So my calc teacher had no problem with me writing "no poles $\in\mathbb{R}$" (she probably ignored it), but I did get a 95 because on the optimization problems I didn't do a second derivative test.

ACuriousMind

Well, the second derivatives are relevant, it's the difference between stationary and extremal points, right?

0celo7

@ACuriousMind It wasn't relevant to these problems.

ACuriousMind

1:51 AM

Also, I don't think I ever had to read 80 pages for a test. American schools are weird.

0celo7

Just a dumb formality.

@ACuriousMind AP U.S. History was more than 80.

Depending on the teacher, U.S. schools can be quite rigorous.

@ACuriousMind In my notes I defined "Déjà vu" as topology.

ACuriousMind

@0celo7 Huh? I don't get it.

0celo7

@ACuriousMind I ask you the same topology question every week

ACuriousMind

Ah :D

0celo7

By the way, how the heck is lattice QCD Lorentz-invariant? The definition of the Wilson loops certainly isn't invariant...

@ACuriousMind • Retroactive interference – the disruptive effect of new learning on the recall of old information

The bane of self-study. (doing psych vocab)

ACuriousMind

2:00 AM

@0celo7 Um, it isn't as far as I can tell. Since spacetime is discrete here, the underlying "spacetime symmetry" is the symmetry of the lattice rather than the Lorentz symmetry, I think.

0celo7

@ACuriousMind Dine says it should be Lorentz invariant as the lattice spacing goes to zero. By the way, how small is the lattice spacing? Are we talking proton radius or more Planck length?

I have not read the full chapter, however, and can't do it now.

ACuriousMind

Yeah, in the continuum limit, Lorentz invariance is obtained/restored

And the lattice spacing is as small as your computer you're simulating with can handle ;)

I seem to recall that you can get pretty good QCD results with only 128 points or so

0celo7

128 points means nothing without an overall scale.

user54412

That sounds like a computational solution represents an equivalence class of physical solutions

ACuriousMind

Well, it depends on which effects you're interested in, but I think the total volume of the spacetime cancels out of most things, so it isn't really relevant

Hm, no, wait, it doesn't cancel

But, again, it depends - you need to know what you are looking for to choose the spacing and volume sensibly

0celo7

2:07 AM

@ACuriousMind It's about time someone solved QCD analytically.

ACuriousMind

I refer you to @ChrisWhite:

2 days ago, by Chris White

Ah, analytic solutions! I believed in those once, in my youth.

It's about time that someone solves the damn millenium problem and shows 4D Yang-Mills is well-defined, first, though.

user54412

I'm actually sitting here trying to coax mathematica to give me some analytic solutions

0celo7

@ACuriousMind When you are old and decrepit, yelling at kids on your lawn, you can also yell about their "fancy analytic QCD solutions" and go on about having to learn Wilson loops and do month-long calculations that use more electricity than a small African nation

BTW the underline fixer tool thingie works in chat again

@ACuriousMind I still don't get why the mass gap is such a big deal.

ACuriousMind

@0celo7 Because we don't see any massless glueballs.

@0celo7 Also, lol

0celo7

@ACuriousMind They're hanging out with the sparticles, man

Right there with the pentaquarks and gravitons

Mfw we find the gravitino before the graviton

user54412

2:15 AM

::sigh:: Yes, mathematica, $-1 + 4Mr / (a^2 + 2r^2 + a^2 \cos2\theta)$ really does equal $-1 + 2Mr / (r^2 + a^2 \cos^2\!\theta)$

0celo7

@ACuriousMind • Confirmation bias -- Superstring theorists

Is superstring phenomenologist an oxymoron?

ACuriousMind

@0celo7 Then I've been taught by an oxymoron this semester :P

@ChrisWhite It doesn't believe you? ;)

0celo7

@ChrisWhite Mathematica needs to brush up on trig. I'm sure there's a plug-in.

Stan Shunpike

on page 37 in wald, he suggests rather than measuring the change in a vector field as we go around a patch on a manifold, instead we multiply the vector field by a dual vector field and compute the change in the scalar. Something to that effect. I can quote him if necessary.

Anyways, why can we just multiply by an arbitrary dual vector field?

0celo7

@StanShunpike one moment, have to get it

Stan Shunpike

2:20 AM

under figure 3.3

0celo7

@StanShunpike We measure the change of the scalar product of the two.

In the end we hope $\omega_a$ just falls out.

Stan Shunpike

Oh, and that we don't need it? That's the hope?

0celo7

Yes, and that's what happens on the next page.

Stan Shunpike

Okay that makes a lot more sense. It seemed arbitrary but this explanation clears up something I've been misunderstanding for a while. Great. Thanks for clarifying.

Another question. I saw Wald simple upgrade the standard partial derivatives in the KG equation to covariant ones.

This got me thinking: gravity and the SM are usually pitted against each other as though they aren't yet compatible. But do we currently accept that QFT works for curved spacetime? Or is that still something that's problematic?

That is, does curved spacetime make QFT more difficult and does the SM work in curved spacetime?

0celo7

@StanShunpike He's using the KG field as a simple model. See chap 14 for QFT in curved spacetime.

Yes and no/maybe.

Definitely yes for the first one, I don't think so for the second one.

When he uses the KG field in the first few chapters, it's not a quantum field, just a classical scalar one.

@ACuriousMind Has anyone been able to write the full SM in curved spacetime, without quantizing gravity?

I'd image the technical difficulties are immense.

ACuriousMind

2:32 AM

@0celo7 Well, writing ain't the problem, I think, just make all the derivatives covariant and use the spin-connection thingy for the spinors. I'm not sure if there are any predictions made from this except for Hawking radiation/Unruh effect. I imagine @Danu knows more about this than me.

0celo7

@ACuriousMind Those effects are generic QFT in curved spacetime effects, not SM in curved spacetime.

ACuriousMind

Well, I don't know any others. My QFT in curved spacetime is practically non-existent.

0celo7

@ACuriousMind I'd learn it if I had more time.

Currently reading SUSY, or at least the SM review in the first few chapters of Dine's book.

0celo7

2:55 AM

I'm off. Gotta sleep early tonight.

ACuriousMind

3:16 AM