The h Bar

9:00 PM

Suppose one oscillator has variables $a$ and $a^*$ while the other has $b$ and $b^*$.

Okay, let me make the objection a different way: This is nice for practical calculations, but it is unnatural from the Hamiltonian viewpoint since $a,a^\ast$ are not canonical variables (their Poisson bracket acquires an ugly $\mathrm{i}$).

I see the point however that it is reasonable that Dirac et al. did know about this classical method

0celo7

Hmm, I've never seen this before

DanielSank

If you couple your oscillators (through a spring, or whatever) you'll find that the interaction term has the form $$a b^* e^{(\omega_a - \omega_b)t} + a^* b e^{(\omega_b - \omega_a) t} + ab e^{(\omega_a + \omega_b)t}+ a^* b^* e^{-(\omega_a + \omega_b)t}$$

The terms with $\pm(\omega_a + \omega_b)$ wildly off resonance and so contribute little to the dynamics.

They are, subsequently, dropped.

This is the classical version of the rotating wave approximation and you never would have though of it if you were thinking in terms of $x$ and $p$.

Note further that the amplitude and phase of $a$ are the action-angle variables for the problem.

@ACuriousMind No way, dude, look up action-angle variables!

The $i$ isn't ugly at all. It's standard classical mechanics.

It's how people have done serious mechanics calculations for a long time.

0celo7

I don't remember imaginary numbers showing up in action angle variables...

DanielSank

@0celo7 I'm not surprised. Physicists seem to be obsessed with taking simple ideas from classical mechanics, reinventing them in quantum textbooks and lectures, and then lying to their students by saying that those ideas are "quantum".

ACuriousMind

9:07 PM

@DanielSank Action-angle variables are reached by canonical transformations, and their Poisson bracket has the standard form

0celo7

That's what I thought. How are $\mathrm i$'s gonna pop up on your real manifold?

ACuriousMind

The point is that just like in QM the $[x,p] = \mathrm{i}$ becomes $[a,a^\dagger] = 1$, we would get $\{a,a^\ast\} = \mathrm{i}$ from $\{x,p\} = 1$.

And the non-unity Poisson bracket clearly shows that the coordinates cannot be reached by a canonical transformation, since canonical transformations leave the Poisson bracket invariant.

DanielSank

@ACuriousMind Well, if I have an action $J$ and angle $\theta$ I can surely form a phasor $J \exp(i \theta)$.

I abused language there. Apologies.

0celo7

Phasor?

DanielSank

Yes, phasor.

0celo7

9:09 PM

Huh.

ACuriousMind

@DanielSank Yes, that is correct, but the things that in the end actually appear as coordinates on the phase space are Fourier series in those phasors that are manifestly real. No $\mathrm{i}$ appears anywhere in the Poisson brackets or the actual coordinates

DanielSank

@0celo7 There is very, very little that we learn in quantum courses which is truly quantum.

It takes entanglement and Bell states, or other similarly subtle effects to find an experiment where the result cannot be explained via classical mechanics or E&M.

0celo7

@ACuriousMind Fourier series?

I have no clue what you guys are talking about any more.

DanielSank

@ACuriousMind I don't understand your point.

I use Fourier transforms/series all the time to represent real signals.

The complex phases have real meaning.

I don't claim that such representations are less valid just because I have to remember to take the real part to get the time domain signal.

Similarly, we can write all of quantum mechanics without the $i$ if we want, it's just a turbo-inconvenient thing to do.

Whether or not there's an $i$ is not some deep profound thing. It's just a question of how to represent stuff.

0celo7

...we can?

DanielSank

9:13 PM

@0celo7 Sure.

You could literally write Schrodinger's equation as a pair of real equations.

0celo7

I thought e.g. conservation of probability requires complex numbers.

DanielSank

It's dumb, and you'd just be re-inventing complex algebra, but you could do it.

My point is that it's similarly dumb to reject the use of complex representations in classical mechanics.

@ACuriousMind do you think it's somehow illegitimate to say that the impedance of an inductor is $i \omega L$?

ACuriousMind

@DanielSank Using complex phases for oscillations and such things is completely fine. Even the scheme of defining $a= x+\mathrm{i}p$ is fine on a computational level. However, the pair of phase space coordinates $(a,a^\ast)$ is uncanonical, and in a strict Hamiltonian treatment such a coordinate change is not allowed because it is not induced by a canonical transformation.

DanielSank

Where'd the $i$ come from? There aren't an imaginary numbers in my electrical circuit.

0celo7

@DanielSank Yeah I don't understand that.

DanielSank

9:16 PM

@ACuriousMind Ok. I guess I just find "that representation doesn't have the properties of representations in class XYZ" a bit unimportant. Or at least, I don't know why you bring it up.

Am I missing something?

0celo7

@ACuriousMind Can you give me a tl;dr on what "the characteristic of $F$ is not 2" means?

PhD algebra has found its way into my geometry, I don't like it

ACuriousMind

@DanielSank You probably miss the appreciation for the abstract Hamiltonian formalism, while I miss the appreciation for computational ease ;)

DanielSank

@ACuriousMind I get why Hamiltonians are nice... but I don't see how that point is important in what started this whole discussion.

17 mins ago, by ACuriousMind

Okay, let me make the objection a different way: This is nice for practical calculations, but it is unnatural from the Hamiltonian viewpoint since $a,a^\ast$ are not canonical variables (their Poisson bracket acquires an ugly $\mathrm{i}$).

ACuriousMind

@0celo7 It means: "For functions of several variables, being alternative and being antisymmetric are equivalent", which fails in characteristic 2

DanielSank

^ What are you objecting to there?

ACuriousMind

9:19 PM

@DanielSank Let me re-read the conversation. I have a feeling that I'm indeed objecting to something that isn't there.

0celo7

@ACuriousMind Hmm. What does alternative mean if not antisymmetric?

DanielSank

26 mins ago, by ACuriousMind

@DanielSank Not actually allowed in classical mechanics, since coordinates are real variables

Perhaps "not actually allowed" isn't what you meant there.

0celo7

Does alternating mean $f(w_1,\dotsc, w,\dotsc, w,\dotsc, w_n)=0$?

ACuriousMind

@DanielSank Yeah. I apologize, no idea what my actual problem here was. Everything is fine.

DanielSank

@ACuriousMind Haha, ok.

ACuriousMind

9:22 PM

I did mean "not actually allowed" but fixing that is trivial :P

DanielSank

Move along, everyone. Nothing to see here.

ACuriousMind

@0celo7 Yes

DanielSank

Thanks for the opportunity to brain dump about the RWA not being quantum.

0celo7

RWA?

@ACuriousMind So, what does the characteristic actually do?

DanielSank

Rotating Wave Approximation

Rotating wave approximation

The rotating wave approximation is an approximation used in atom optics and magnetic resonance. In this approximation, terms in a Hamiltonian which oscillate rapidly are neglected. This is a valid approximation when the applied electromagnetic radiation is near resonance with an atomic transition, and the intensity is low. Explicitly, terms in the Hamiltonians which oscillate with frequencies ω L + ω 0 {\displaystyle \omega _{L}+\omega...

ACuriousMind

9:24 PM

@0celo7 Please ask a more specific question

DanielSank

Unfortunately, the wiki article starts right off with quantum mechanics.

0celo7

@ACuriousMind If I write down the polarization $f(v+w,v+w)=0$, then I get $f(v,w)=-f(w,v)$.

DanielSank

Like I said, physicists are obsessed with pretending standard classical mechanics techniques are quantum.

0celo7

So I'm wondering what the characteristic "does", because I got that by just using multilinearity.

ACuriousMind

@0celo7 Yes, but $-1=1$ in characteristic 2.

DanielSank

9:25 PM

I think this happens because they learn these ideas in quantum courses/books and don't understand what parts of those books are just Hamiltonian mechanics and which parts are actually quantum.

0celo7

What the hell?

DanielSank

It's kinda sad.

@ACuriousMind Obviously.

@0celo7 You'd know that if you were a programmer.

Two's complement

Two's complement is a mathematical operation on binary numbers, as well as a binary signed number representation based on this operation. Its wide use in computing makes it the most important example of a radix complement. The two's complement of an N-bit number is defined as the complement with respect to 2N; in other words, it is the result of subtracting the number from 2N. This is also equivalent to taking the ones' complement and then adding one, since the sum of a number and its ones' complement is all 1 bits. The two's complement of a number behaves like the negative of the original number...

0celo7

Aha, looking up what the characteristic is on the PhD level site Wikipedia makes that clear.

@ACuriousMind So, $\Bbb R$ is a field of characteristic 0?

ACuriousMind

@0celo7 Yes. Did you ask me questions about characteristics without actually knowing what a characteristic is?

0celo7

...crap

Sorry

ACuriousMind

9:27 PM

@DanielSank Yes, that's right

DanielSank

@ACuriousMind Yeah. It's probably because all the classical mechanics books are so fantastically bad.

And all for the same reason!

They fail to distinguish between variables and functions.

This amazing confusion has made every classical mechanics book I've ever read, except for Jose and Saletan, completely un-understandable.

This confusion is so pervasive that it's also poisoned most of the thermodynamics books.

In my opinion, it's one of the biggest disasters in physics literature.

0celo7

ok @ACuriousMind I have put my algebra textbook on my desk and will consult it before asking you anything in the future

@DanielSank You've said that before, but I don't think i understood what you meant (I still don't)

DanielSank

@0celo7 Let me cook up a very simple example.

One of the first places I saw this confusion was in discussing the change of variables formula in integrals.

0celo7

@DanielSank I think the classical confusion is that the Lagrangian $L(q,\dot q)$ depends on the variables separately.

One treats $q$ and $\dot q$ as variables, but they're functions of $t$! Huh?

But as soon as you write $L:TQ\to\Bbb R$ it becomes clear.

DanielSank

@0celo7 That's a very good example. Yes, thank you.

@0celo7 Exactly yes.

So now you know what I mean.

This confusion is a virus in physics literature.

It renders countless books very hard to understand.

0celo7

9:33 PM

So, what we mean by $\int L\,\mathrm d t$ is

Hmm, I'd have to think about the abstract definition.

$\int L(c_*)$.

where $c_*$ is the tangent bundle path traced out by the differential of a curve $c:\Bbb R\to M$

DanielSank

@0celo7 I don't remember the vocabulary enough to comment, but I think you're right.

0celo7

So $q,\dot q$ are variables, but on shell they're functions

Something like that

DanielSank

I'm not sure what you mean. I just try to only think in terms of functions.

Variables, to me, are there only to help compute stuff.

For example, suppose I have a function $f$ defined by the equation $f(x) = x^2$.

Now I want to compute $g$ such that $g$ is the derivative of $f$.

I have to use variables to do that: $df/dx = 2x$.

So $g$ is defined by the equation $g(x) = 2x$.

That language "$f$ is a function defined by the equation..." first came to my attention in Munkres's book and it changed my life.

It gave me a way to unravel the messes we find in so many physics books.

0celo7

Lol, changed your life?

DanielSank

Yes.

Before that, I didn't understand the difference between functions and variables.

0celo7

9:39 PM

I don't think I had that confusion, so I'm having a hard time understanding why this was a revelation.

DanielSank

@0celo7 Good. I'm happy for you.

0celo7

Sigh, I'm trying to understand what you're saying, not showing off.

DanielSank

@0celo7 I wasn't being facetious at all. I really am emotionally uplifted by the thought that you don't have to suffer the same confusion I did!

I don't joke around much about notation or language.

They're too important.

bolbteppa

What about $S(x,y,y') = S(t,x,x') = \int_1^2 \frac{\sqrt{1+(y')^2}}{x}dx = \int_1^2 \frac{\sqrt{1+(x')^2}}{t}dt$? Here is an $L = L(t,q,\dot{q})$, is this not valid?

0celo7

What are you talking about?

@DanielSank Is there any notation/language in diff geo/riem geo/GR that you find confusing?

DanielSank

9:52 PM

@0celo7 Heh. Well... GR is one place where a substantial fraction of authors actually try to use reasonable notation.

I think this happens because if they didn't the subject would be completely impenetrable.

Wald actually does something I like: he uses Roman letters for "variables" and greek letters when he's trying to just indicate the signature of a tensor and what's contracted with what.

bolbteppa

I gave you an $L = L(t,q,q')$ that can't be written as $L = L(q,q')$ with $t$ as a parametric variable you can neglect, it's intrinsic for this lagrangian,

0celo7

@DanielSank Yes, that's abstract index notation

It's actually not good

If you want to write a basis, you need to write $e^a_\mu$ for a basis vector.

Leads to much confusion.

Because then one also has to wonder what a tetrad is in that notation...and I do not know the answer

DanielSank

@0celo7 Nah, you need better notation for the basis vectors, that's all.

Write the index over the basis set somewhere else.

I don't care where.

0celo7

@DanielSank Huh? You need the $a$ index because it's a vector, and the $\mu$ lets you know which one it is.

DanielSank

Above/below the $e$... whatever.

bolbteppa

9:55 PM

I don't think classical mechanics books are doing anything wrong

DanielSank

@bolbteppa I, obviously, disagree :)

bolbteppa

Maybe they wont state the fundamental lemma of the calculus of variations and just assume it, but I don't see any problem with the notation, would love to know what they are doing wrong that I'm missing

DanielSank

@bolbteppa For example:

If I have a function $f$, then $\dot{f}$ is the derivative of this function.

However, mechanics books write $L(q, \dot{q})$.

Here, $L$ is a function from $\mathbb{R}^2 \rightarrow \mathbb{R}$.

$L$ exists apart from any notion of variables.

AND it's completely confusing to name the variables using a notation that's supposed to indicate the derivative of a function.

Slereah

I think u mean a function from the cotangent bundle to R

DanielSank

@Slereah Well, fine.

I think physics books should do two things:

Slereah

10:01 PM

1) teach about apple brewing

DanielSank

1a) Use mathematicians' notation for functions, e.g. $f: \mathbb{R} \rightarrow \mathbb{R}$

This helps distinguish the function $f$ from it's definition via a formula, e.g. $f(x) = x^2$.

2) Don't use function notation for variables.

0celo7

what does 2) even mean?

DanielSank

@0celo7 For example, if you're going to use $\dot{q}$ as a variable name, you need to be very, very explicit that there's no relation between $q$ and $\dot{q}$.

They're just placeholders to help us operate on functions.

0celo7

ok

DanielSank

I gotta go

bolbteppa

10:04 PM

Hmm, I've never seen any mechanics book even address that tbh, the only book I seen address this was Gelfand's calculus of variations, or Elsgolts, where they define functionals $S$ on normed function spaces, I think Arnold does the same thing

0celo7

@ACuriousMind I have a set $C=\{1,\dotsc, p\}$, and I want the operator $A_p$ on $C$ to "add $p$". Should that be "add $p$, then $\mod p$"?

No, that could give me $0$

Ignore ^

Rubber ducking.

Slereah

Is @ACuriousMind the rubber duck in this scenario

0celo7

Yes.

DanielSank

hey look, it's @Mostafa

0celo7

@ACuriousMind If you have some great insight on cross sections of products of permutation groups, I'd love to hear it. I don't really get why they're useful except for the definition

bolbteppa

10:12 PM

The way to get around worrying about that is to define your action $S$ as a functional on a function space $S : \mathcal{F} \rightarrow \mathbb{R} | y \mapsto S(y) = \int_a^b F(x,y,y')dx$ and you don't worry about $F$ :p

I checked Burke, and your Lagrangian is a one form on the 'line-element contact bundle of configuration space' with variables $(x,y,y') = (t,x,x')$, same gist as considering it as a functional

ACuriousMind

@0celo7 I'm not sure what a "cross section of products of permutation groups" is.

0celo7

@ACuriousMind Have you heard of "cross section" in the context of permutation groups before?

or groups in general?

ACuriousMind

I'm also not sure how I managed to accumulate so much crap in this cupboard. At this rate. I'm throwing away more than I'm taking with me

0celo7

wtf, are you moving or something?

ACuriousMind

@0celo7 No, but you have to consider that a large part of my algebra education was in German

@0celo7 Yes, closer to the uni :P

0celo7

10:23 PM

For some reason I'm studying Grassmann algebras over general fields, and the algebra is not nice

We have the permutation group $S_{p+q}$, and $S_p\times S_q$ naturally regarded as a subgroup of it

A cross section is a subset of $S_{p+q}$ such that it contains exactly one element in each left coset of $S_p\times S_q$

ACuriousMind

Doesn't ring a bell

0celo7

@ACuriousMind Ok, but am I right that something in the cross section is of the form $\pi\circ(\sigma,\tau)$ with $\pi\in S_{p+q}, \sigma\in S_p,\tau\in S_q$?

Thanks ducky

@ACuriousMind One runs into problems when defining Grassmann algebras on vector spaces with field charactersitics that are primes $\le$ dimension of the vector space.

there's a very abstract way of doing it via cross sections that works in all cases

0celo7

11:08 PM

@DanielSank So, is there anything confusing in diff geo or not?

I'm writing a little summary of important results

ACuriousMind

@0celo7 Why on earth are you looking at Grassmann algebras over fields with non-zero characteristic?

0celo7

@ACuriousMind I'm...not sure.

user3183724

11:26 PM

@0celo7 Hello!

DanielSank

@0celo7 That's a vague question.

Some authors are confusing. Others are not.

Also, I barely studied diff geo at all, so I'm not good to ask.

Someone give me a physics question to answer.

I feel like actually contributing to the site today.

Danu

11:42 PM

lolool

WHY WOULD YOU DO THAT

physics.stackexchange.com/questions?sort=unanswered

Find an unanswered question

user3183724

@DanielSank physics.stackexchange.com/questions/283372/… seems like an excellent question.

ACuriousMind

Hey, I thought of Daniel when editing that and then forgot to ping him!

Danu

@DanielSank See the above linked post

I know that guy in real life

And I told him to ask you in chat---I guess he didn't.

He's actually a decently close friend of mine :)

user3183724

Decently close indeed. You told me to ask him about a different question though, which I didn't ask him personally but he did end up answering

Danu

hehe

DanielSank 2quantum5me

ACuriousMind

11:52 PM

@user3183724 Why do you have different user names/numbers on different sites? Took me a while to realize you asked that question :P

user3183724

@ACuriousMind I don't think I do. However it might certainly be an idea to make it into a real name, after being here for about 2 years.

Danu

Who uses real names

eww

ACuriousMind

@user3183724 You're on math/chat with 3183724 but you asked that question/are on physics with 129412

@Danu lol

Danu

YOU DON'T KNOW ME MAN

ACuriousMind

Does anyone really know anyone?

user3183724

11:55 PM

Interesting. I figured stackexchange just created some number for me and would be consistent throughout the different sites, guess that's why you don't assume

0celo7

since when are you a philosopher?

DanielSank

@Danu wat?

Danu

@DanielSank You're 2quantum5me

ACuriousMind

C'mon now, chat flag system, wtf am I supposed to do with Russian messages?

Danu

@ACuriousMind You know what's coming

ACuriousMind

11:56 PM

@Danu Bring it

0celo7

@Danu I think that 5 needs to be a 4.

Danu

Haddaway - What Is Love [Official]

►

ACuriousMind

Yay!

Danu

@0celo7 I think you shouldn't think too much

DanielSank

@user3183724 Lolwut? I answered a question you asked?

Danu

11:56 PM

Note that he has different user id's

user3183724

Yes, about the analytic expression for the Purcell effect

Ah, I see now. So in the chatroom my user ID is different than on the physics stack exchange, that is what @ACuriousMind meant. Well that's just terrible

DanielSank

@user3183724 Ah yes. So wait a minute... your name in chat differs from your name on the main site?

@user3183724 Yes, that is fantastically confusing.

user3183724

I thought you meant mathchat, but you meant math/chat.

Lets see if I can remedy this.

ACuriousMind

@user3183724 Yep. that's because your chat parent user is your math.SE account, so you appear in all chats with the number from math.SE although you might have different numbers on the individual sites

Danu

TL;DR: Don't use "userxljr2peiou-9238q-" as name

dmckee

11:58 PM

You can fix it be (1) setting a username rather that just being another number and (b) syncing your profiles.

Danu

^

user129412

Then how do I stay anonymous Danu?

ACuriousMind

Incoming username: "Another number" :P

dmckee

And apparently it is not necessary to use a consistent labeling scheme for steps in the procedure.

user129412

Although you've already spoiled that

DanielSank

11:59 PM

Ho boy... This question is perhaps asking too many things.

ACuriousMind

Unless Danu is a loner who has only one friend, I don't see what he's really spoiled here

dmckee

You stay anonymous by choosing a moniker that isn't your actual name.

0celo7

@dmckee I tried that

dmckee

Nor something that the people who know you would instantly spot as possibly being your alias.

0celo7

didn't work

DanielSank

11:59 PM

Whenever I see a post like that one, where I read for a while and find a part that says "As a final part I am also interested in..." I just stop and go do something else.

Danu

@ACuriousMind Proof: Clear

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