Chiral Crap!

the first thing I wonder is... since $\xi$ is a symmetry of my theory and the tensor $K$ (I just gave the expression, $F$ are the standard field strengths for EM) should therefore be invariant, don't we have $\mathcal L_\xi K$ is zero?

Emilio Pisanty

Maybe this works? chat.stackexchange.com/rooms/32905/…

Danu

^lol right, that's what I thought

Emilio Pisanty

@Danu OK, reload the page and it'll work

ACuriousMind

@Danu Yeah!

Emilio Pisanty

Since we're here already, shall we keep on this one then?

Danu

10:12 PM

Err sure?

So, there are two issues for now

1. What is $K$ in terms of $Z$? It seems like $Z$ is almost the derivative of $K$

(maybe it is, I didn't check)

2. What is this current, and how do we get it from $K$?

Also @ACuriousMind $Z$ is given by $Z^\mu_{\nu\rho}=*F^{\mu\sigma}F_{\sigma\nu,\rho} - F^{\mu\sigma}*F_{\sigma\nu,\rho}$

Emilio Pisanty

Hang on, what's $K$ here?

Danu

I defined it above

the same thing as $Z$ but without the derivatives (so one less index)

Emilio Pisanty

oh, yeah, I see it

hang on, I need paper to test stuff out

Danu

Yeah me too

I'm taking the derivative and the Lie derivative of $K$

ACuriousMind

@Danu If $K$ is not the conserved quantity associated to $\xi$, then the derivative isn't expected to vanish - e.g. four-momentum is conserved, but it changes its value under boosts (because it is the conserved quantity of translations, not boosts)

Danu

10:20 PM

Right, right. So that solves one thingy

Assuming the expression is actually the Lie derivative of $K$ this does give something nontrivial

Now, why do we care about the Lie derivative of $K$ under this trafo?

Also ACM are you interested in having the (most) relevant links to papers?

There's one by Kibble (from the 60s) and one newer one (more mathematical)

ACuriousMind

@Danu Are there more than the ones in the question?

Emilio Pisanty

@ACuriousMind Not really

There's a few more but I think the question's got all the relevant ones

Danu

@ACuriousMind The question has more than what you need

the Kibble and Anco papers are the only ones I found useful so far.

So...
$$ 2\partial_\rho K^\mu_\nu=\big[\epsilon^{\mu\sigma\alpha\beta}F_{\nu\sigma}-\epsilon_{\nu\sigma\alpha\beta}F^{\mu\sigma}\big]\partial_\rho F_{\alpha\beta}+\big[\epsilon^{\mu\sigma\alpha\beta}\partial_\rho F_{\nu\sigma}-\epsilon_{\nu\sigma\alpha\beta}\partial_\rho F^{\mu\sigma}\big] F_{\alpha\beta}$$

Any cancellations, anyone?

Emilio Pisanty

No, don't unpack the $\ast F$

Danu

Why not; I thought it'd help me see cancellations faster?

"packing it back up" is simple anyways

Emilio Pisanty

10:29 PM

You can get $\partial_\rho K^\mu_{\ \nu}$ in terms of $Z$ and a transpose

$\partial_\rho K^\mu_{\ \ \nu}=Z^\mu_{\ \ \nu\rho}-Z_{\nu\ \ \rho}^{\ \ \mu}$

Danu

Also whoops I typo'd on the definition of $Z$ earlier ACM, watch out

@EmilioPisanty Right

Emilio Pisanty

There might be other symmetries to make this easier

Yeah, the two terms should cancel

Danu

also

I think we should work with -1 times the equation 2.14 in Anco

because Kibble defines his contraction of the $F$'s the other way around

so contracted index first on the second factor i.e. $F^{\mu\sigma}*F_{\sigma\nu}$ etc

you okay with that convention?

Emilio Pisanty

Yeah, that's a good point

Danu

So let's stick to these definitions:

$$Z^\mu_{\nu\rho}:= *F^{\mu\sigma}F_{\sigma\nu,\rho}-F^{\mu\sigma}*F_{\sigma\nu,\rho}$$

Emilio Pisanty

10:37 PM

Yes

Danu

and $$K^\mu_\nu:=*F^{\mu\sigma}F_{\sigma\nu}- F^{\mu\sigma}*F_{\sigma\nu}$$

Emilio Pisanty

Yeah, that's what I'm working with

good spot on the sign difference w.r.t. Anco

Danu

I also have verified now that $$\partial_\rho K^\mu_\nu=Z^\mu_{\ \ \nu\rho}-Z_{\nu\ \ \ \rho}^{\ \ \mu}$$

now, what did you mean when you said something should cancel?

Emilio Pisanty

Not sure anymore :|

Kibble reckons that $Z^{\mu\nu\rho}=Z^{\nu\mu\rho}$, eq. 6

Danu

Yes, that follows from his identity (2)

hmm yes, I see... that'd be not-so-nice huh

So the derivative of $K$ should actually vanish

Emilio Pisanty

10:49 PM

Yeah, that's pretty strange. So $K$ should be constant in space.

Danu

No, not necessarily constant in space (unless you already know it's constant in time)

Is your $E\cdot B$ constant in time?

It's conserved but not necessarily constant

Emilio Pisanty

Yeah, but all the partial derivatives of $K$ vanish. I'm afraid it's just zero by definition or something.

Danu

Oh, I'm retarded, sorry

Conserved is when you contract the indices lol

So you're right, $K$ is somehow trivial

Emilio Pisanty

$Z_{000}$ is something like $\mathbf E\cdot\dot{\mathbf B}-\mathbf B\cdot\dot{\mathbf E}$, and you've taken out the derivatives

Danu

you can't concatenate styles for letters without braces

@EmilioPisanty Right, right.

Okay, so then it's back to scratch; find some way to relate 2.14 to what we know

I still want to write it as ($\xi$ times) the Lie derivative of something

Emilio Pisanty

10:58 PM

@Danu Yeah, I'm more worried in terms of how does 2.14 emerge from Noether's theorem

Danu

Yeah, maybe it's not directly through a Lie derivative after all

In the paper you linked in the other room, we do find the following:

> The dual symmetry is associated with optical helicity so the connection between special case of zilches, optical chirality, with helicity becomes more clear

So I guess we have to perform an infinitesimal duality transformation

Emilio Pisanty

Yeah, there's definitely going to be a duality symmetry involved

Danu

and then use the usual Noether's theorem method of deriving the current

Emilio Pisanty

But I think you also need to couple it with some conformal transformation

Danu

Right, because how else are we going to get that $\xi$ in there

Emilio Pisanty

11:03 PM

So maybe the zilch is the Noether current for duality plus a Lorentz transform?

Danu

you mean a conformal transformation

what do you think @ACuriousMind?

Emilio Pisanty

Yeah, in the general case, but I suspect that zilch as in Lipkin and Kibble only needs the special case of Lorentz transformations

so $\xi^\mu=a^{\mu\nu}x_\nu$.

It would have the huge advantage of being relatively easy to implement, so one can just check

Danu

What is your reason for believing that?

Now, what does an infinitesimal duality trafo look like? :P

I've got this nice-n-general formula for the Noether current associated to a combined field-and-coordinate transformation

Emilio Pisanty

@Danu Mostly the fact that you need to get rotations involved somehow

So you can tap into the pseudoscalarness of the thing

Danu

If $\delta \mathcal L=\partial_\mu K^\mu$ for some $K^\mu$ that vanishes "at infinity" then the general current is:

$$ J^\mu:=\frac{\partial \mathcal L}{\partial \partial_\mu\phi^i} \delta\phi^i - \frac{\partial \mathcal L}{\partial (\partial_\mu \phi^i)}\partial_\nu \phi^i\delta x^\nu +\mathcal L\delta x^\mu -K^\mu $$

So... what is the infinitesimal duality transformation like? :P

Emilio Pisanty

11:13 PM

So in terms of the vector fields, it's $\mathbf E'=\cos\theta \mathbf E-\sin\theta \mathbf B$, $\mathbf B=\sin\theta \mathbf E+\cos\theta \mathbf B$

Danu

In terms of the vector potential, by any chance?

Emilio Pisanty

So I imagine something like $F'^{\mu\nu}=\cos\theta F^{\mu\nu}\pm \sin\theta {}^\ast F^{\mu\mu}$

no, it's on the force fields

At $\theta=\pi/2$ you need to completely swap them

Danu

journals.aps.org/pra/abstract/10.1103/PhysRevA.86.013845

Emilio Pisanty

@Danu Yeah, I saw that one late last night, it's on the queue

Danu

@EmilioPisanty I guess in terms of either $F$ or $A$ is fine

Emilio Pisanty

11:16 PM

@Danu Yeah, that one is just a guess, though

Or actually

should work

Danu

journals.aps.org.emedien.ub.uni-muenchen.de/pra/abstract/…

Emilio Pisanty

That's a funky link

ACuriousMind

That looks as if the duality transformation really "wants" to be a $\mathrm{SO}(2)$ transformation acting on $F \choose \ast F$ as rotation by $\theta$

Danu

It is, sorry; I'm logged into my uni stuff

@ACuriousMind Yeah, it does want to be that

Emilio Pisanty

@ACuriousMind Yeah, exactly

ACuriousMind

11:22 PM

So the infinitesimal version is just $F + \theta (\ast F)$ for small $\theta$.

Emilio Pisanty

Bizarre but that's what it says, I guess

@ACuriousMind Yeah

Danu

In fact, however, Maxwell's equations are invariant under $GL(2)$ transformations on the $F, *F$ vector

not just under rotations

Emilio Pisanty

Yeah, that's true. So does the bigger group get you anything?

Danu

@ACuriousMind Should be minus, no?

(depends on the direction of the rotation, I guess; I'm taking counterclockwise to be standard)

Emilio Pisanty

The scaling has to give you fairly trivial stuff

ACuriousMind

11:24 PM

@Danu There's room for a couple of sign conventions here ;)

Emilio Pisanty

@Danu counterclockwise is an odd term for $\begin{pmatrix}F\\\ast F\end{pmatrix}$ =P

Danu

@EmilioPisanty The rotations (first entry of vector $=: x$, duhhhh ;) )

So let's just stupidly apply this:

$$ F\mapsto F-\theta *F_{\mu\nu}$$ and then the Lorentz transformation (forgot infinitesimal version, sigh... will re-derive again)

oh yeah lol it's the antisymmetric thing

Emilio Pisanty

@Danu ??

Danu

@EmilioPisanty Infinitesimal Lorentz transformation thingy

@EmilioPisanty The paper by Barnett et al. introduces an auxiliary field btw, which I don't like.

Emilio Pisanty

Oh, yeah. $x^\mu\mapsto x^\mu+\theta a^{\mu\nu}x_\nu$, $a^{\nu\mu}=-a^{\mu\nu}$.

@Danu The $\mathbf C$ thing?

That feels pretty inevitable if you want to do the duality transformation on the potentials

whether it's actually necessary is a good question though

Danu

11:39 PM

So erm now we should just check out da Noether current right; the thing is that this is not so nice since we don't have the transformation in terms of $A_\mu$

Am I being stupid here?

Emilio Pisanty

You can always put it in directly

$\ast F$ is an explicit function of $A$

Danu

Yeah, but our transformation is in terms of $F$ and it doesn't directly (or even uniquely?) define one for $A$

Emilio Pisanty

Oh, as in, you want $A\mapsto A'=\cdots$?

Danu

mhm

Emilio Pisanty

oooh, good question

Danu

11:45 PM

I don't understand how to get my Noether current otherwise :P

If you look at my expression for $J^\mu$ you'll see that it involves a bunch of derivatives w.r.t. the derivatives of fields; there are clearly no derivatives of $F$

Emilio Pisanty

Oh, so the $\phi^i$ are just the $A^\mu$?

got it

Danu

The $\phi^i$ are the "fundamental fields" appearing in $\mathcal L$

Emilio Pisanty

ok, yeah

Danu

(they'll always have a kinetic term, hence a derivative somewhere)

Transcript for