The h Bar

12:00 AM

I don't mean to sound pretentious, but I'd probably be bored by that.

@ACuriousMind What do gauge transformations do for us? Like so what if I multiply the field by a symmetry group. Is this a useful property to know about?

0celo7

@bolbteppa They spend 20+ pages on GR review.

No graduate ST texts do that.

ACuriousMind

@StanShunpike Oh, the gauge symmetries tell you that your description of the system contains superfluous parameters - two field (or coordinate) configurations that are related by a gauge transformation describe the exact same physical configuration/state

0celo7

@bolbteppa Also it's 170 pages.

Stan Shunpike

@0celo7 True. That's why I got confused reading them when I tried initially.

0celo7

12:03 AM

@StanShunpike You have to know your stuff for BBS. I don't care what it says on the cover, it's not really introductory.

Hey all

Hi

yo

Is it fair to say that fusion is the most important technological achievement that will happen in our lifetime, excluding the interwebs?

@StanShunpike If you have Scribd, you can get Polchinski legally.

Stan Shunpike

@ACuriousMind So would that be like saying R^3 has 6 DOFs because you have 3 translations and 3 rotations? I'm just trying to understand what you mean by superfluous parameters...

@0celo7 I'm lucky. My library has everything. It's enormous.

0celo7

12:05 AM

@StanShunpike R^3 has three degrees of freedom.

Hence the 3.

@StanShunpike Then get Polchinski! Or maybe Zwiebach if you're not confident.

Stan Shunpike

@0celo7 I know that. I was giving an example of saying having extra parameters that you don't need.

Asking if this was like

bolbteppa

I think they included it to make it easier to indoctrinate young people into their propaganda lol but you can always just skim read the review chapters if you need or just jump into chapter 3 on constrained Hamiltonian systems (the stuff we've been discussing for 2 days!) it's always a good idea to browse the 'dumb' books on a subject to make sure you don't miss anything

Stan Shunpike

an example where a gauge symmetry has extra paramter

ACuriousMind

@StanShunpike No, it's like, well you took this four-potential $A^\mu$ to describe your vector field, and it has four real d.o.f., but...well, there are just two physical polarisations.

Or, like, you took a phase space with two $q$ and two $p$, but there's a relation that says $p_1=p_2$

Essentially, what you call "independent" variables...aren't independent variables

Danu

@0celo7 ...if it happens :D

0celo7

12:08 AM

@Danu I'm writing a scholarship app telling them I'm gonna change the world and usher in an amazing technological achievement. It'll happen.

Hopefully.

Maybe.

:/

Danu

Maybe mention world domination

0celo7

@Danu Suitcase H bombs.

@Danu You'd better watch out for my Nobel speech. I'll be mentioning strangled chickens.

Danu

Also, I'm not completely convinced it'll bring as much good as one may (naively?) imagine

0celo7

Also that one crazy German who thinks the disjoint union of two intervals is a pedagogically relevant example.

Danu

It won't solve what is in my view the more important problem: Distributing stuff (e.g. energy) in a 'good' way (fair or whatever)

0celo7

12:13 AM

@Danu We can put fusion reactors on planes though.

And spaceships.

Unlike fission reactors, they can be shut off in a heartbeat.

And the fuel is not radioactive.

Danu

I know about the obviously very attractive advantages

0celo7

@Danu You have to define fair first.

Danu

I've just grown very wary of forgetting about logistics

@0celo7 Note the 'whatever': I can't and won't make it precise

0celo7

@Danu I think the idea is that the reactors can be made small enough that we can do away with these monstrous 500-mile long power lines we have in America.

Danu

My point mainly being that perhaps the problems we face are not so much due to lack of resources, but due to failure to efficiently allocate them.

0celo7

12:16 AM

We can have reactors more spread out.

@Danu Partially, yes.

Power lines, for one, are terrible. Very inefficient.

@Danu Do you know of any good LQG resources? I'm tempted to make a thread, I didn't see a good one when I searched.

Stan Shunpike

@ACuriousMind I don't think I follow. Maybe I need to read more or try some problems out.

Danu

@0celo7 No. I think it's a pretty fringy subject so there are probably not (m)any good textbooks on it

0celo7

@Danu Fringy like crackpot or fringy like no funding?

@StanShunpike What don't you get exactly?

Danu

@0celo7 A bit of both?

I didn't study it so I don't know.

Sofia

12:38 AM

@ACuriousMind I see that you voted for closing this question. Initially I took it in hand, but I didn't have any wish to stay and understand an article in which I am not interested. However, look what the OP replies to you, (and I don't understand): (I continue)

@ACuriousMind Can someone do a derivation of the Lagrangian as an IVP problem instead of the traditional boundary value method? What for God's sake is IVP?

ACuriousMind

@Sofia IVP = Initial value problem

Hm, I didn't actually vote for closing yet because I was out of close votes until now

OP's reply that they really are just asking about someone explaining the PDF in more detail doesn't make me any more sympathetic, though.

Sofia

@ACuriousMind It is somehow stupid, isn't it? The minimal action principal is finding the path that minimize the action between two points. How can it be transformed into an initial values problem?

ACuriousMind

@Sofia Well, the Euler-Lagrange equations are differential equations that are solved with initial data alone - specfying start-point and start-velocity determines the trajectory completely for non-ugly systems

Sofia

@ACuriousMind But he is not at the step of the E_L equation, he is at the building the Lagrangian, i.e. even before putting the minimum condition.

ACuriousMind

@Sofia I've skimmed the paper now. OP hasn't even understood what it's about.

Sofia

12:47 AM

@ACuriousMind but, what bothers me is what he wrote in continuation. Please see: Here is a supporting document that gives you almost all the steps but can you fill in the pieces and make it easier to follow. *

ACuriousMind

It's not about formulating the action principle as an initial value problems, it's about formulating an action principle for an initial value problem

Which is just the inverse Lagrangian problem all over again, and this paper does some weird trick to construct an "action" that can be used for general dissipative systems

The question is definitely too broad, there are five pages of explanation right there. If that's not sufficient, no SE answer can be.

Sean

@Danu does LQG have significantly less of a following than ST?

Sofia

@ACuriousMind no, no, it bothers me that in his mind is that we should somehow re-write the paper in a clearer form for him. Don't you understand his words in the same way?

bolbteppa

That is a potentially genius question, idk why you guys don't like it

ACuriousMind

@Sofia Of course, that's why I voted to close

Stan Shunpike

12:51 AM

@0celo7 Well, okay I understand that $A^\mu$ constitutes a 4-vector. And as @ACuriousMind says, we use this to describe our vector field and it has 4 real degrees of freedom. But then he said something about polarizations and I got lost there. I forget, what are the ways light can be polarized? there's circular polarization...and

bolbteppa

I've never thought about that before, but I know it's a humongous issue in the literature that people fuck up over this non-conservative question all the time

Stan Shunpike

@0celo7 And I didn't really follow how the polarization related to the DOFs....

ACuriousMind

@bolbteppa It's not a genius question. The paper contents are interesting, but the question is not. The question is "Please explain this paper to me".

Stan Shunpike

@ACuriousMind Exactly.

Sofia

@ACuriousMind I am not against him, but don't we have to explain him that this site doesn't do such things as he expects? His protests seem very vehement.

@ACuriousMind I wanted to tell him this, but I needed additional opinions. He seems very offended.

Stan Shunpike

12:55 AM

I remember the first question I posted got downvoted. I posted it like a homework problem. And I didn't intend it that way.

I felt discouraged initially, but I kept at it.

And a few months later I posted that same question but formated differently, and it got 3 upvotes.

bolbteppa

I would love it if someone would explain the jist of that paper and whether it falls foul of this famous paper physics.gatech.edu/files/u24/publications/0007.pdf

ACuriousMind

@StanShunpike This one? It has no downvote, and doesn't look look like homework to me. Did you delete your question?

(Not that it's bad to delete one's badly received questions, just, well, curious)

@Sofia: I left a second comment saying why this question is too broad. I can't (and won't) do anything else with it.

Sean

Holy cow @StanShunpike you ask a lot of questions. Which certainly isn't a bad thing

Stan Shunpike

I also get a lot of good answers :)

@ACuriousMind It was this one math.stackexchange.com/questions/1137420/…

ACuriousMind

@bolbteppa The paper doesn't take a Lagrange multiplier approach, so the paper you link has nothing to say about it.

Stan Shunpike

1:02 AM

@ACuriousMind I originally posted it on Physics SE I think. But the problem was, I just didn't do a good job writing the question in a way that would be useful for other SE members.

I basically copy and pasted my LaTex file directly into the question box

And I think people didn't take kindly to some of the unconventional formatting I had used.

But I still wanted an answer, so I tried again and this time I got a nice answer.

It was an exercise from MTW

0celo7

@StanShunpike You know that photons have two degrees of freedom, right? Classically, we have two polarizations.

Stan Shunpike

Yeah, I just can never remember which ones. Aren't there multiple kinds of polarization?

0celo7

Quantumly, the quantum field that describes the photon can only have two degrees of freedom.

@StanShunpike There are different kinds, but always two directions.

ACuriousMind

@0celo7 BTW, Wiktionary says quantumly is a word ;)

0celo7

Quantumly, we eliminate one degree of freedom by the gauge condition. I don't remember where the other one goes. @ACuriousMind, does the EOM take care of the second?

David

1:08 AM

hello can someone tell me quickly what δ means in thermodynamics? Is any different from the symbol for an inexact differential? I don't this is worth making a post in a question.

0celo7

@ACuriousMind $20 you did that.

ACuriousMind

@0celo7 The page was created on 5th July 2010

0celo7

"SemperBlotto" screams crazy German scientist.

ACuriousMind

No, that wasn't me

Sofia

@bolbteppa what means holonomic?

0celo7

1:09 AM

@Sofia It's a type of constraint of the form $f=0$.

Stan Shunpike

@0celo7 in 3space, we choose one of the directions as the direction the photon is traveling. that leaves 2 directions left. Are those the two polarizations you mean?

0celo7

@David Actually not 100% sure about that, don't listen to me.

@StanShunpike EM waves are transversal. That means two degrees of freedom about the direction of motion.

@ACuriousMind Quantumly, where do the photon degrees of freedom go? One is the gauge condition, is the other the EOM?

ACuriousMind

@0celo7 For massless vector fields, both d.o.f. are killed by gauge freedom, for massive vector fields, it is indeed the e.o.m that kills one

0celo7

Ok.

Sofia

@bolbteppa good idea to suggest an easier article (so I hope) and not let him go empty-handed, but are you aware that he was looking for some Lagrangian that would address also dissipation?

Stan Shunpike

1:12 AM

@ACuriousMind @0celo7 What does that mean "killed" by gauge freedom?

0celo7

@ACuriousMind For example, the Lorentz gauge only kills one. I know radiation + $A^0=0$ kills two though.

ACuriousMind

@StanShunpike Hah, that's just metaphoric speak for "They are irrelevant for a physical description"

bolbteppa

@ACuriousMind it does matter, Lagrange multipliers are a shortcut (as the paper says), it's just a way of ensuring you can start from $F = ma$ and derive the Euler-Lagrange equations, it's a way of ensuring those conditions without assuming $F = ma$, there are apparently hundreds of papers that make mistakes on this simple point, including Goldstein's book, this paper apparently disproved Goldstein it's that subtle, and I'm sure the paper the guy linked to is making some silly mistake also

0celo7

@StanShunpike The gauge constraints determine components of $A^\mu$.

ACuriousMind

@0celo7 Yeah Lorentz gauges are incomplete. In Gupta-Bleuler quantization, one d.o.f. is killed by the Lorentz/Feynman gauge condition and the other because you can show that the "spurious" states decouple

The BRST formalism shows generally that gauge fields in D dimensions have D-2 d.o.f.

0celo7

1:15 AM

@ACuriousMind Spurious being the longitudinal mode?

bolbteppa

Lanczos' Variational Principles of Mechanics has a derivation of Euler-Lagrange from $F = ma$ and I don't see how this guy can magically change initial and boundary conditions, it's amazing if it's a way to avoid issues with velocity-dependent potentials (but more than likely wrong)

0celo7

@ACuriousMind The idea that the little group of massless particles is $SO(D-2)$ is crucial to string theory, as you know.

ACuriousMind

@0celo7 mhh...I think the spurious direction is dependent on the specfic gauge implemented, and usually a mixture of longitudinal and time-directed

Stan Shunpike

@ACuriousMind So does gauge fixing relate to the symmetry groups $g(x)$ you were talking about earlier?

Where you had $g(x)\phi(x)$

ACuriousMind

@StanShunpike Gauge fixing means carrying out a specific $g(x)$ and then trying your best to forget that you ever had gauge freedom, essentially

0celo7

1:18 AM

@StanShunpike The idea behind gauge fixing is that the theory is invariant under a gauge transformation. Then you perform a particular gauge transformation to constrain your fields a certain way.

Stan Shunpike

@0celo7 Why? What does that achieve?

ACuriousMind

For example, for the $A^\mu$, you impose $\partial_\mu A^\mu = 0$, and always implicitly carry out the gauge trafo that produces an $A^\mu$ fulfilling that.

0celo7

Since the fixing is irrelevant to the theory as a whole, this is permissible and makes life much easier.

@StanShunpike Another one is $\nabla\cdot \vec A=0=A^0$.

ACuriousMind

@0celo7 Ah, but you have to show that your results do not depend on the gauge that is chosen, which is very annoying

Stan Shunpike

Ah, so it's a way to specify constraints on the 4-potential itself? To narrow down the properties of it?

ACuriousMind

1:19 AM

It should be irrelevant, but you can't assume that

Stan Shunpike

or rather values it can take

0celo7

@ACuriousMind Are there any gauge anomalies?

Classically I wouldn't expect any problems.

ACuriousMind

@0celo7 The problem is - even classically, how do you know that what you derived there isn't just valid in this specific gauge? Have you proven that your gauge fixing condition really only intersects each gauge orbit once? Is it even possible to choose a global gauge condition? (Even classically, it is not, these things are known as Gribov ambiguities)

@StanShunpike Yes, you choose a constraint on the potential to make it "lose the superfluous d.o.f", since every equation effectively eliminates some d.o.f.

0celo7

@ACuriousMind Ha, Zeidler discusses Gribov three volumes in.

Stan Shunpike

@ACuriousMind Is superfluous dof handwaving? I don't understand the quotes.

0celo7

1:26 AM

@ACuriousMind What's the tl;dr on the Gribov ambiguity? How is it resolved?

ACuriousMind

@0celo7 It isn't. Stay with the constrained Hamiltonian system and do BRST quantization :P

Stan Shunpike

What does tl;dr stand for?

ACuriousMind

@StanShunpike It's not really handwaving...I did the quotes because it is not a technical term

0celo7

@ACuriousMind I remembered that Weinberg mentions it. He just says "it won't bother us here because we use axial gauge".

Stan Shunpike

Ah, okay. So does it reduce from 4 to 2 degrees of freedom in this case? or does the number of dofs reduced depend on choice of gauge

?

0celo7

1:28 AM

@StanShunpike The 2 is physical.

ACuriousMind

@0celo7 But also: In almost all applications you can ignore Gribov, he's a harmless old man ;)

0celo7

Haha

@StanShunpike too long; didn't read

Stan Shunpike

LOL

@0celo7 Aren't the 4 dofs of the 4 potential physical?

0celo7

@StanShunpike No, because the photon only has two polarization directions.

The 4 d.o.f. of the 4-potential is an unfortunate fact of group/representation theory.

Stan Shunpike

What do you mean?

0celo7

1:31 AM

Massless spin-1 particles simply don't have 4 dof.

I don't think massive ones have 4 dof either.

ACuriousMind

@0celo7 Nope, they have three

0celo7

Yeah, 4 doesn't even make sense.

ACuriousMind

The fourth would be a spin-0 part, essentially

0celo7

@ACuriousMind You mean a second one?

ACuriousMind

@0celo7 Ah, a true spin-0 part, not a spin-1 part with 0 projection

0celo7

1:32 AM

The vector rep would have to split into a triplet and singlet?

ACuriousMind

Yep

0celo7

@ACuriousMind That's what I meant.

ACuriousMind

Okay then :)

0celo7

@ACuriousMind What's the rep theory proof that that doesn't happen?

ACuriousMind

I do have some difficulty sometimes to tell whether people are just a bit sloppy/different in their expression or if they mean something actually different from what I mean

Stan Shunpike

1:33 AM

I couldn't figure out what representation theory is. From what I read, it sounds like a way to describe properties about groups using "representations" of them with matrices so we can talk about their properties using stuff we know from linear algebra. Is that correct or am I just completely wrong?

0celo7

@ACuriousMind I phrased that poorly, you just found the correct words.

So the former I guess.

@ACuriousMind Is it just that we have an $SO(3)$ 3-vector rep for spin one and we have to complete it into a 4-vector?

But this of course does not add an extra spin mode.

bolbteppa

@StanShunpike this post of mine physics.stackexchange.com/questions/164245/… and Landau's QM chapter on representation theory are phenomenal to understand the idea of representation theory.

Stan Shunpike

Which book is Landau again?

bolbteppa

Quantum Mechanics

0celo7

@StanShunpike There's 10 (?) of them.

ACuriousMind

1:37 AM

@StanShunpike Basically the right idea, just that we aren't really after learning something about the groups but rather about what kind of objects they can "act" (i.e. be represented) on. (this e.g. shows you that there are only things with half-integer and integer spins, because rotations can't act any other way)

Stan Shunpike

OH!

That's so cool!

@0celo7 10 of what?

@bolbteppa thx for the info. I will check both out.

ACuriousMind

@0celo7 Well, in QFT, you have to look at $\mathrm{SO}(1,3)$, and Wigner's classification tells you how the allowed irreducible reps look there. Talking about $\mathrm{SO}(3)$ is a bit of a shortcut.

0celo7

@StanShunpike Landau is a 10 volume set IIRC.

Stan Shunpike

@bolbteppa btw that book has 3 volumes apparently, did you have a specific one in mind?

oh wow!

okay nvm

thanks for correcting me. 10 volumes. jeeez

0celo7

Course of Theoretical Physics

The Course of Theoretical Physics is a ten-volume series of books covering theoretical physics that was initiated by Lev Landau and written in collaboration with his student Evgeny Lifshitz starting in the late 1930s. It is said that Landau composed much of the series in his head while in an NKVD prison in 1938-39. However, almost all of the actual writing of the early volumes was done by Lifshitz, giving rise to the often repeated witticism, "not a word of Landau and not a thought of Lifshitz". The first eight volumes were finished in the 1950s, written in the Russian language, and translated...

ACuriousMind

1:40 AM

@0celo7 $A^\mu$ has four components, initially, because it is a 1-form on the spacetime, and it's transform under the Lorentz group is essentially already given by that.

bolbteppa

There are 10 Landau's, to quote myself quoting G.K. Ustinova yesterday
"If you intend to become a good physicist, don’t waste your time. Study the complete Course of Theoretical Physics by Landau and Lifshits"
www.researchgate.net/post/Any_recommended_reading_for_physics_undergraduate_student

0celo7

@bolbteppa Lifshits

bolbteppa

I guess it's a possible Russian transliteration for a native speaker

Stan Shunpike

So it is just volume 3 tho for the QM

0celo7

Yes

ACuriousMind

1:42 AM

@0celo7 It gets very confusing in QFT because there are two kinds of representations, the unitary ones, which are Wigner's and act as $A^\mu\mapsto UA^\mu U^\dagger$, and the ordinary ones, which act as $A^\mu\mapsto \Lambda^\mu_\nu A^\nu$.

0celo7

@ACuriousMind Remind me how that works. I know the general results (scalars are spin 0, vectors are spin 1, tensors spin 2, etc.).

It's not something I've remembered in much detail or know where to relearn.

@ACuriousMind I remember this from Weinberg.

One acts on the Hilbert space and the other on Minkowski space, right?

ACuriousMind

@0celo7 I'm not sure I can give a quick reminder of this, it's quite involved

bolbteppa

@ACuriousMind pretty sure the first is basically a spinor representation (change of basis) and the second is an orthogonal representation

ACuriousMind

@0celo7 Yeah

@bolbteppa Nah, the first is an infinite-dimensional rep because there are no unitary finite-dimensional reps of the Lorentz group. The second is just the usual finite-dim trafo as a vector field.

The spinors are also finite-dim reps of $\mathrm{SO}(1,3)$, which are linear reps of $\mathrm{SU}(2)\times\mathrm{SU}(2)$ (up to complexification/compactification/some other detail I might not remember). The group theory goes a bit crazy here. Just be satisfied that there are spinors ;)

The QFT representation theory is a bit confusing because you always switch between looking at the Hilbert spaces and the fields. It's weird, but it works even if you don't think too hard about it

bolbteppa

Yeah I know but I think what I'm saying is correct though it would take me ages to justify it (and I'm probably wrong because of the dimensionality basis stuff :( )

There is a really nice reason why it splits into $SU(2) \times SU(2)$

ACuriousMind

1:49 AM

Here's Qmechanic playing with this:

17

A: How do I construct the $SU(2)$ representation of the Lorentz Group using $SU(2)\times SU(2)\sim SO(3,1)$ ?

Here is a mathematical derivation. We use the sign convention $(+,-,-,-)$ for the Minkowski metric $\eta_{\mu\nu}$. I) First recall the fact that $SL(2,\mathbb{C})$ is (the double cover of) the restricted Lorentz group $SO^+(1,3;\mathbb{R})$. This follows partly because: There is a bi...

Also, there's this unitarian trick or whatnot, the Wikipedia article is quite comprehensive, but difficult to follow if you don't already know what you are looking for

Also, the fact is it doesn't split into $\mathrm{SU}(2)\times\mathrm{SU}(2)$, because that'd be $\mathrm{SO}(4)$ (compactness and all).

0celo7

@ACuriousMind Busy for a moment, will take a look at this in a bit.

Stan Shunpike

@ACuriousMind I like how he says for a laugh. I didn't know QMechanic joked.

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