Chiral Crap!

Emilio Pisanty

6:02 PM

So two candidates for the symmetry associated with the zilch

arxiv.org/abs/1408.2802

ACuriousMind

@EmilioPisanty: Found the correct thing, I think: arxiv.org/abs/1303.0687v3

Emilio Pisanty

reckons it's $$F\mapsto F+\ell^{\alpha\beta}\partial_\alpha\partial_\beta {}^*\!F$$

@ACuriousMind Oh, yeah, that's also interesting

ACuriousMind

@EmilioPisanty Yeah, I think that paper is somewhat bad because they mix Noether's theorem with wildly applying the equations of motion. What I linked about is a proper off-shell quasi-symmetry of the action, and, ironically, a lot simpler than what the other papers do

Emilio Pisanty

@ACuriousMind Yeah, part of the problem is the vast number of different systems of language that people use

so it's a fair bit of work to determine whether two people are talking about the same thing or not

This is also relevant

journals.aps.org/pr/abstract/10.1103/PhysRev.150.1251

I'm fairly sure that it's not the same symmetry as in Lashkari-Ghouchani, arxiv.org/abs/1408.2802

ACuriousMind

@EmilioPisanty The L-G paper doesn't have a symmetry, as far as I can tell. They use the equations of motion in their derivation, this always gives $\delta S = \int \mathrm{d} J$ for some current form $J$.

Emilio Pisanty

6:09 PM

@ACuriousMind So it's wrong or at least misleading?

In any case the Philbin symmetry looks different to both L-G and Dass journals.aps.org/pr/abstract/10.1103/PhysRev.150.1251

ACuriousMind

@EmilioPisanty The result and derivation is correct, I think, but they are, contrary to their claims, neither using a symmetry of the action nor Noether's theorem in its proper form. They just use that $\delta S= \int \mathrm{d} J$ on-shell and then massage this expression further to get a nice expression for a conserved quantity

I've tried to do their transformation and get that it's a symmetry, but it really doesn't look like one to me. Their generalized duality transformation is not a symmetry of the action. It becomes $\delta S = 0$ only upon use of $\ast F = \pm F$, which the EM field only fulfills on-shell in vacuum

Emilio Pisanty

Hmmm. And you're not allowed to restrict your symmetry to on-shell?

ACuriousMind

Well, as I just said in the main chat, an "on-shell symmetry" is a vacuous notion, as Qmechanic keeps saying in his answers. You can apparently derive conservation laws from such on-shell transformations, but that doesn't make them symmetries, and finding the "correct" on-shell transformation to get useful conservation laws seems complete guesswork to me

I can't access the Dass paper, but the statement that every conservation law has a symmetry transformation associated to it seems very daring in that generality. Could you tell me what he writes for that zilch symmetry?

@EmilioPisanty However, note that Philbin's symmetry does not give the zilch as the naive Noether current - you have to make the result gauge-invariant first. The other calculations do not have that issue, they use gauge-invariant transformations from the start, which could be the reason they have to work on-shell

Emilio Pisanty

6:48 PM

@ACuriousMind Oh, yeah, I see it, eq. (8) in Philbin.

(I can pop you the Dass paper by email if you want btw. No eprint that I can see.)

So Dass gives the symmetry transformation as $$\bar{\delta}A_\alpha =\lambda_{\mu\nu}G_{\mu\nu\alpha\beta\sigma\tau}A_{\beta,\sigma\tau}$$

Where $\lambda$ is symmetric and $G$ is a mess of $\epsilon$s and $\delta$s

And $\bar\delta$ is the "local variation"

This is apparently a subset of the transformations of the form $$\delta A_\mu=a_{\alpha\beta}b_{\mu\nu}A_{\nu,\alpha\beta}$$

([sic] on the repeated non-raised indices)

$b$ antisymmetric

and this definitely includes Philbin's symmetries

ACuriousMind

Yes, that seems to include Philbin's transformation

Emilio Pisanty

I don't see how it would connect with Lashkari-Ghouchani, though

ACuriousMind

Well, two papers kinda agreeing on what transformation to use is a start :D

Emilio Pisanty

@ACuriousMind Yeah, I'm not sure how full the agreement is, though

Philbin is very limited in scope, with fewer handles on the transformation (I think), which I imagine is what makes him only get a limited hold on the full $Z^{\lambda\mu\nu}$.

He seems to get $Z^{000}$ and its current but no more.

ACuriousMind

Yep. Either he wasn't able to guess the full general transformation, wasn't interested in it, or had limited space

Emilio Pisanty

6:58 PM

I suspect the second tbh.

The optical helicity - EM duality connection is interesting, though

That definitely seems pretty set.

cf Philbin and Cameron/Barnett/Yao

iopscience.iop.org/article/10.1088/1367-2630/14/5/053050/pdf and journals.aps.org/pra/abstract/10.1103/PhysRevA.86.013845

Emilio Pisanty

7:26 PM

Also if you don't mind the $\mathbf C$ pseudovector potential, iopscience.iop.org/article/10.1088/1367-2630/14/12/123019/pdf

It seems pretty complete

but starts off to the tune of $$\mathcal L=-\frac18\left(\partial_\alpha A_\beta-\partial_\beta A_\alpha\right)\left(\partial^\alpha A^\beta-\partial^\beta A^\alpha\right)-\frac18\left(\partial_\alpha C_\beta-\partial_\beta C_\alpha\right)\left(\partial^\alpha C^\beta-\partial^\beta C^\alpha\right)$$

Also, section 8 of iopscience.iop.org/article/10.1088/1367-2630/14/5/053050/pdf has some compelling arguments against zilch as a fundamental quantity

with the terrible luck of calling their replacement "ij-infra-zilch"

Danu

@EmilioPisanty Yikes! :P

So, the thing Philbin gets is not actually $Z^\mu_{\nu\rho}$, agreed?

Only stuff related to $Z^0_{00}$

ACuriousMind

@EmilioPisanty Well, since we have a gauge theory anyway, I see little harm in adding additional degrees of freedom. Since they show that one may implement now the duality transformation on the level of the potentials, it seems to me this might be the "correct" way to think about duality transformation - essentially introducing a dual for the potential

@Danu Yes, Philbins paper is limited to the $Z^0_{00}$.

Danu

So I expect it to not be very useful in looking for other interesting quantities.

ACuriousMind

However, Cameron/Barnett seems to get all the zilches with their duality trafo on $A,C$.

In particular, they present an argument that every symmetry of Maxwell's equations is equivalent to a symmetry of their extended Lagrangian, which would satisfactorily explain why Anco et al. get all this from symmetries of the equations.

Emilio Pisanty

7:41 PM

@ACuriousMind Yes, that's correct.

They've got a bunch of stuff

Danu

It's a bit curious: They just introduce some other potential for fun but it's somehow really not that different from the usual setup because its field strength (curvature tensor) is just the dual to $F$. Am I understanding it correctly?

Emilio Pisanty

@Danu Not completely sure. They get $G=*F$ from $C$ in the same way that $F$ comes from $A$

ACuriousMind

@Danu Yep. I think the basic idea is: They observe the duality transformation is not a symmetry of the standard Lagrangian, basically because $F$ and its dual are not independent. Then they make them articificially independent by introducing an additional potential, which then gives the correct transformation. However, your theory now carries the constraint that the field strength of one is the dual of the other as additional, which you must remember for consistent transformations

Emilio Pisanty

And they seem to give $A$ and $C$ equal standing

Danu

@ACuriousMind Exactly.

Emilio Pisanty

7:50 PM

@ACuriousMind Or yeah, that

Danu

And as in other cases that I've seen, you must be careful about when to impose the constraint.

Also, it's so nice and elementary that paper, you can just read it like a novel 8)

Emilio Pisanty

@Danu Yeah, this is what (good) scientific papers are meant to read like

The tone of all the others is why I was pulling my hair out

Danu

@EmilioPisanty I'm not sure :P Depending on the audience.

I mean... if you want to be at the level of deriving the Noether current explicitly you're not going to derive the AdS/CFT duality in any foreseeable amount of time :P

For these purposes, I think the level of the exposition :)

Emilio Pisanty

@Danu I don't think it's to do with the level of the material. They do a lot of pretty technical stuff, but they explain it thoroughly yet concisely

Just good writing style

Not always possible, though, some stuff is just too technical

ACuriousMind

Well...I think the issue is that Anco's goal is "Classify all possible conserved quantities in a rigorous manner", while what we were looking for was more "What symmetry of the action corresponds to zilches, and how do I say that in physicists' terms?" Note Barnett doesn't classify all currents or all possible forms of distinct symmetries, while Anco actually counts the possible conserved currents order by order

Emilio Pisanty

8:04 PM

Yeah, Barnett doesn't get a full classification, so there's the danger that they missed stuff.

As it is it feels like a much more accessible framework, though

Mostly, I think Barnett is the interesting one

The thing that worries me is that I don't see the optical spin and the ij-infra-zilches (urgh, what a horrible name) of iopscience.iop.org/article/10.1088/1367-2630/14/5/053050/pdf in the symmetry framework of the later one iopscience.iop.org/article/10.1088/1367-2630/14/12/123019/pdf

ACuriousMind

Oh god, who had the brilliant idea to name the PDF document?

Danu

@EmilioPisanty It's really easy to follow :)

@ACuriousMind Lol

ACuriousMind

@EmilioPisanty They mention it in the discussion at the end

Emilio Pisanty

@ACuriousMind Yeah, but it's pretty brief

It'd be interesting to know which symmetry gives those

I imagine a coupling with the conformal symmetry in the same way that they got the stress-energy tensor in §6.2

Or maybe it's squarely in this non-local class

Hmmmmm

But also this:

> It seems, however that such quantities [ij-infra-zilches] are not related between reference frames in a simple manner.

Danu

8:31 PM

So @EmilioPisanty the quantities 6.31 are not of use to you?

Emilio Pisanty

@Danu Not sure

Hang on

Yeah, probably

Danu

And the continuation of the chain 6.34,6.35,6.36?

Emilio Pisanty

8:48 PM

@Danu To be honest with you, the practical need has sort of vanished over today. The arguments I had that the specific field I'm interested in is locally chiral have mostly crumbled, so the contradiction w.r.t. the results of the optical helicity and the optical chirality (i.e. they vanish) is no longer there.

At this stage I want to understand what's going on (cause it's all pretty cool) and yeah, I think we're mostly there, and that the next term in the hierarchy is (6.31) to whatever the next relevant order is.

So yeah, I'm pretty ready to call it in

Do you want to write up (6.31)?

For the bounty =D

Danu

@EmilioPisanty Awww

Emilio Pisanty

@Danu Nah, it's fine.

Still a bit puzzling though.

Danu

@EmilioPisanty I don't think I could resist that...

Emilio Pisanty

Hah, get ready for it, cause that's research ready for you.

In case you're curious, it's the field in nature.com/nphoton/journal/v9/n11/full/nphoton.2015.181.html, two circularly polarized waves sent in at an angle.

The fields themselves are globally their own mirror image, so up front it seems like it can't be chirally discriminatory. However, if you put an off-axis detector in the far field then the experiment is no longer its mirror image so you've got a shot.

Danu

I guess it's easier to start by answering this question:

4

Q: What symmetry is associated with conservation of Lipkin's zilch?

The 'zilch' of an electromagnetic field is the tensor $$ Z^{\mu}_{\ \ \ \nu\rho}=^*\!\!F^{\mu\lambda}F_{\lambda\nu,\rho}-F^{\mu\lambda}\,{}^*\!F_{\lambda\nu,\rho} \tag1 $$ given in terms of the electromagnetic field tensor $F^{\mu\nu}$ (and therefore in terms of the electric and magnetic fields ...

Do you want to do that yourself?

Emilio Pisanty

8:57 PM

@Danu Up to you. Happy for you to get the rep.

Danu

Okay, I'll start writing.

Emilio Pisanty

@Danu Excellent =).

Emilio Pisanty

9:12 PM

@Danu Still pretty mysterious, though.

Just checked Barnett's first high-order one, and it also vanishes

Danu

@EmilioPisanty Okay... Interesting.

Emilio Pisanty

The experiment could be chirally sensitive

i.e. mirror enantiomers could produce (different) mirror-image far-field spectra

but it seems it just doesn't want to

It can't be enantiomerically selective unless it's locally chiral

and it just doesn't seem to be

Oh well

Plenty of cool stuff unearthed in the process so no biggie.

Danu

@EmilioPisanty enantiomerically... yes....?

What in the heck does that mean!

Emilio Pisanty

@Danu Heh, finally, one up for me (to a gazillion up for you =P). Enantiomers are mirror images of each other.

An enantiomerically selective experiment can distinguish between the two.

Danu

I see.

Emilio Pisanty

9:20 PM

i.e. distinguish between e.g. these two

So you need the molecule to 'feel' a locally chiral field, e.g. a circularly polarized pulse.

Hence the need for local measures of chirality.

Danu

Yes, right.

Emilio Pisanty

9:36 PM

@Danu Interesting discussion in iopscience.iop.org/article/10.1088/1367-2630/15/3/033026/pdf

Near eq. (2.7).

Danu

@EmilioPisanty Is this the paper Barnett mentions as also considering the same Lagrangian at some point?

Emilio Pisanty

The duality transformation respects the EOM but messes with the standard lagrangian

Danu

^yeah, exactly

That is one way (the one I'm using in my answer, too) to motivate the change of Lagrangian.

Emilio Pisanty

@Danu Yeah, their ref. [10] in this one iopscience.iop.org/article/10.1088/1367-2630/14/12/123019/meta, mentioned at the end of the introduction

Could be worth linking to as an alternative reference for that step.

Slightly more in depth as to why a change of lagrangian is really inevitable.

Cameron and Barnett just jump from 3.9 to 3.11 without a lot of elaboration.

Danu

it's natural ;D

My answer to your simpler question is already getting quite long.

Emilio Pisanty

9:42 PM

@Danu Seems pretty natural to me, to be honest. If you've got something $f(A)$ which is not symmetric but $f(C)$ will do equally well, take $(f(A)+f(C))/2$ and you're probably in the sweet spot.

Danu

Exactly :)

Emilio Pisanty

@Danu That's probably quite natural and all for the best =).

@Danu But Bliokh's 2.6 is really crying out "complementary integrals" at the top of its lungs

Danu

10:14 PM

@EmilioPisanty physics.stackexchange.com/questions/223991/…

0

A: What symmetry is associated with conservation of Lipkin's zilch?

This answer will mostly follow this excellent (and quite readable!) paper, pointed out to me by Emilio himself, in the exposition. This is another paper that contains similar considerations. For an extended discussion on this and closely related topics, see this chatroom. There are a number of p...

Emilio Pisanty

@Danu Yeah, just saw it

Danu

@EmilioPisanty Let me know if you find any errors... Lol!

@EmilioPisanty I already spotted one! :D The $\xi^{\alpha\beta}$ is symmetric, not antisymmetric!

Emilio Pisanty

@Danu Presumably your second link is meant to go to Bliokh?

Danu

@EmilioPisanty ...oops

Emilio Pisanty

Or somewhere that's not also the first link?

=P

Danu

10:19 PM

Done.

Emilio Pisanty

While you're there it's probably best to change to the DOIs

just because

you know, something

Danu

right...

Emilio Pisanty

Boom, yeah, excellently phrased.

Danu

@EmilioPisanty Thanks.

Emilio Pisanty

> perhaps-not-completely-unnatural

Love it

Danu

10:25 PM

@EmilioPisanty I mean... ;) Best I could come up with!

Emilio Pisanty

@Danu lol

It's spot on though.

Danu

So, when we look at this chain of terms from the paper (adding derivatives to the transformation), how do I actually check if the thing I will get is a time-even pseudoscalar or not?

This may be trivial, but I'm not really too comfortable with these notions :P

Time-even seems to indicate I better be taking even-order terms in the chain.

(i.e. odd number of indices on my conserved current)

Emilio Pisanty

@Danu You just look at it and see how it behaves under the transformations.

Clearest example is $$\mathbf E\cdot \dot{\mathbf B}$$

Danu

@EmilioPisanty So how do I distinguish a pseudoscalar again? :P

Given an expression in terms of $F$, its dual and its derivatives.

Emilio Pisanty

$\mathbf E$ is odd under parity and $\mathbf B$ is even under parity, so $\mathbf E\cdot \dot{\mathbf B}$ is odd under parity. Same for time-even ($\mathbf E$ odd, $\mathbf B$ even).

In terms of the $F^{\mu\nu}$... I'm not so sure.

Danu

10:30 PM

Hmm...

Emilio Pisanty

Thing is, $F^{0i}$ is odd under parity and $F^{ij}$ is even. Ditto under time reversal.

So if you've got a complicated contraction then it could in principle be both at the same time which would be terrible.

Danu

Right... So it's about the representations of $SO(3)$ instead of the Lorentz group

(as I should've expected)

Emilio Pisanty

You're looking at the $00\cdots 00$ component of something, though.

@Danu Yeah, to a point. You can still get Lorentz pseudoscalars, though.

Not sure exactly how they work, because $\mathcal M^4$ is quite different.

Danu

Right, but that should mean something along the following lines: A scalar under Lorentz trafo's that is odd under parity (which implies something for which rep of SO(3) it lives in, which is not reflected by the Lorentz index structure)

Emilio Pisanty

@Danu Yeah, it's probably not well reflected by the Lorentz index structure.

Note also that it's the reps of O(3) that matter, because under SO(3) scalars and pseudoscalars are the same.

Danu

10:35 PM

Oh, sorry, of course! Duh :P SO(3) cuts out the reflections.

Emilio Pisanty

@Danu Thing with this one is that it's only ever the 000...0 component of the thing that can be an O(3) (pseudo)scalar, because otherwise you get O(3) vector indices.

Danu

@EmilioPisanty Yes, very good.

So I only have to look at the zero component parts

Emilio Pisanty

@Danu Yes. Which means that you need to choose some timelike axis

i.e. a reference frame

You could just start with "let $\hat{\mathbf e}$ be a timelike four-vector"

Danu

@EmilioPisanty Blurgh

Emilio Pisanty

exactly

Danu

10:40 PM

I'm not so sure about htis.

So... say I've got something in this hierarchy

I'm pretty sure I need to take the even-number-of-derivatives ones for time-evenness

Now, the 0000000 component

How do I check whether it's a pseudoscalar? :P

Or should I really just reconnect to $B$ and $E$?

Emilio Pisanty

It should actually be pretty easy

Danu

Yeah, well it'll get a little annoying because I'll have a lot of contractions probably

Or no, wait haha

I will always just have one

Emilio Pisanty

You're looking at 6.31 of Cameron and Barnett, right?

Danu

So it shouldn't be a big deal.

Emilio Pisanty

$H^{\alpha\beta\cdots\zeta}$

All of the derivatives on the RHS are in the indices in $H$

So it's the same contraction throughout.

Same as for $Z$

so you're actually done

With the time-even you only need to do the time derivatives in steps of two, which you're luckily enough already doing.

You're only ever doing $G^{0i}F^{i0,00\cdots0}$ and $F^{0i}G^{i0,00\cdots0}$.

Danu

10:46 PM

Yeah, exactly

Emilio Pisanty

Both are parity-odd because exactly one ingredient is.

Danu

Makes total sense

It's simple because we keep on just contracting once

Emilio Pisanty

So actually

Danu

So no need to consider multiple combinations of indices

Emilio Pisanty

The whole thing just collapses for monochromatic fields

Danu

10:47 PM

You're pretty much answering your own question :P

Do you want me to even write up the answer? :P You could just as well do it

Emilio Pisanty

because $$H\propto \mathbf E\cdot \frac{\partial^{2n}\mathbf B}{\partial t^{2n}}-\mathbf B\cdot \frac{\partial^{2n}\mathbf E}{\partial t^{2n}}$$

which is a bummer but oh well

it is what it is.

Danu

Yeah...

Emilio Pisanty

@Danu Naw, go for it, you deserve it, you've been a champ

Danu

It's funny how the really crucial thing turns out to find a lucid-enough paper :P

Emilio Pisanty

Plus it's a bucket load of rep gone into a rep black hole otherwise ;P

Danu

10:49 PM

@EmilioPisanty Hahahahaha

BRING EM HERE 8)

Emilio Pisanty

@Danu Yeah, this is why communicating the science is every bit as important as getting it right.

Danu

11:47 PM

@EmilioPisanty Hmkay, I'm not going to make it tonight but I'm working on it. I guess tomorrow or the day after I'll manage to post my answer.

Transcript for