# 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 $E^i=F^{0i}$ and $B^i=\tfrac12 \epsilon^{ijk}F_{jk}$) and its dual $^*\!F^{\mu\nu}=\tfrac12 \epsilon^{\mu\nu \rho\sigma} F_{\rho\sigma}$, with commas denoting partial derivatives. This tensor is conserved on-shell in vacuum, in the sense that $$\partial_\mu Z^{\mu}_{\ \ \ \nu\rho} =\partial^\nu Z^{\mu}_{\ \ \ \nu\rho} =\partial^\rho Z^{\mu}_{\ \ \ \nu\rho} =0$$ whenever $F^{\mu\nu}$ satisfies the vacuum Maxwell equations, $$\partial_\mu F^{\mu\nu}=0 \quad\text{and}\quad \partial_\mu{}^*\! F^{\mu\nu}=0.$$

This conservation law, which gives in total ten conserved charges, was found by Lipkin [J. Math. Phys. 5, 696 (1964)], though the form $(1)$ was first given by Kibble [J. Math. Phys. 6, 1022 (1965)]. Since its discovery the zilch has apparently been a bit of an odd child, with its physical interpretation a bit out in a lurch, but it is definitely an integral part of the bigger framework of the conservation laws of the EM field.

To give it a bit of a more concrete feeling, the most accessible component of the zilch is $Z^{000}$, which has been called the optical chirality: $$C=Z^0_{\ \ \ 00}=\mathbf B\cdot\frac{\partial \mathbf E}{\partial t}-\mathbf E\cdot\frac{\partial \mathbf B}{\partial t}.$$ This is a pseudoscalar (odd under parity) but otherwise quite similar to the electromagnetic energy density, so there's definitely a lot of involvement of the Lorentz group action in this quantity.

In general, conservation laws tend to have a tight association with the symmetry properties of the system. Noether's theorem provides a conservation law for every appropriate symmetry, and it has a converse which guarantees the existence of symmetries given suitable conservation laws, though it seems that the situation is more complicated for gauge theories.

I would like to know how this principle applies to the Lipkin zilch tensor. (In particular, if there is no such symmetry, I would like a clear and compelling argument of why this is the case.) The literature is not particularly clear or (for me) easy to decode, so I think it's worthwhile asking this outright, so that there's a clear answer on the record: what symmetry of the electromagnetic field is associated with the conservation of Lipkin's zilch tensor? Moreover, how exactly does this symmetry relate to the conservation law? Through a direct application of Noether's theorem, or are there more subtleties in play?

I have made some inroads into the literature and I'm happy to discuss what I have read already and what I haven't found yet but I think it's probably for the best if I just leave this question clean for now.

• Another similar paper might be this one: arxiv.org/abs/1408.2802. They claim the zilch can be related to some sort of current densities, but other than that I haven't found much of an interpretation in terms of general symmetries. – gented Dec 14 '15 at 14:20
• @Gennaro Yeah, that one's on my radar, I'm still trying to work out exactly what they're saying the symmetry is. – Emilio Pisanty Dec 14 '15 at 14:53

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 papers which make all kinds of claims about how one can (attempt to) derive the zilch tensor as a conserved quantity associated to a symmetry: Most of the relevant literature is linked either in this question or in another recent question by Emilio on the same topic.

## Electromagnetic duality and choosing an action

Browsing the literature, it becomes clear that the conservation of the zilch tensor has something to do with so-called electromagnetic duality transformations. It is well known that Maxwell's equations can be written in the following form:

$$\partial_\mu F^{\mu\nu}=0 \qquad \qquad \partial_\mu * F^{\mu\nu}=0$$

where we defined the Hodge dual of the electromagnetic field strength (or curvature) tensor $F_{\mu\nu}$ by the equation $*F^{\mu\nu}=\frac{1}{2}\epsilon^{\mu\nu\alpha\beta}F_{\alpha\beta}$. This form of the equations makes it obvious that they are symmetric under a transformation of the form

$$\begin{pmatrix} F_{\mu\nu} \\ *F_{\mu\nu}\end{pmatrix}\mapsto \begin{pmatrix} aF_{\mu\nu}+b*F_{\mu\nu} \\ c*F_{\mu\nu}+ d F_{\mu\nu}\end{pmatrix}$$

In particular, if the transformation matrix corresponds to an element of $SO(2)$ then we say that the transformation is an electromagnetic duality rotation. Now, it's nice that the equations of motion possess this symmetry but closer inspection shows that, in fact, the standard Maxwell action

$$S=-\frac{1}{4}\int\mathrm d^4 x F_{\mu\nu}F^{\mu\nu}$$

is not invariant under an infinitesimal duality rotation

$$\begin{pmatrix} F_{\mu\nu}\\*F_{\mu\nu}\end{pmatrix} \mapsto \begin{pmatrix}F_{\mu\nu}-\theta *F_{\mu\nu} \\ *F_{\mu\nu}+\theta F_{\mu\nu} \end{pmatrix}$$

This is one way to motivate a change of action to one that does manifest this symmetry of the equations of motion: One can then also hope to derive the zilch tensor as a conserved current of some duality-related symmetry.

One can make different choices for an alternative Lagrangian, but the paper I link at the beginning of this answer presents a particularly simple and clearly motivated choice: We attempt to consider the dual field strength tensor as an independent variable, call it $G_{\mu\nu}$, and form a Lagrangian that treats $F_{\mu\nu}$ and its dual in a symmetric manner. Then, if we keep in mind the constraint that $G_{\mu\nu}$ is really the dual of $F_{\mu\nu}$ when we perform transformations on the fields, we can get a consistent treatment while having a Lagrangian that is manifestly duality-invariant.

To do this, we introduce a second, dual (electric) four-potential (in addition to the usual $A^\mu$), and call it $C^\mu$. We can form the associated field strength tensor

$$G^{\mu\nu}=\partial^{[\mu}C^{\nu]}(\equiv *F^{\mu\nu})$$

A natural choice for a new Lagrangian to replace the standard Maxwell Lagrangian is then

$$\mathcal L=-\frac{1}{8}(F_{\mu\nu}F^{\mu\nu}+G_{\mu\nu}G^{\mu\nu})$$

This is clearly invariant under a duality rotation, which takes the following form:

$$A^\mu\mapsto A^\mu\cos\theta -C^\mu\sin\theta \qquad \qquad C^\mu\mapsto C^\mu\cos\theta +A^\mu\sin\theta$$

Moreover, we note that this Lagrangian also gives rise to the usual equations governing electrodynamics (after imposing the constraint $G_{\mu\nu}=*F_{\mu\nu}$ since the Euler-Lagrange equation read:

$$\partial_\mu F^{\mu\nu}=0 \qquad \qquad \partial_\mu G^{\mu\nu}=0$$

## General Noether currents (with example)

It is simple to derive (and this is also explicitly carried out in the paper that I'm following) that under a general variation of the potentials

$$A^\mu\mapsto A'^\mu=A^\mu+\delta A^\mu \qquad \qquad C^\mu\mapsto C'^\mu=C^\mu+\delta C^\mu$$

the variation of the Lagrangian is

$$\tag{\star} \delta \mathcal L = \frac{1}{2}\partial_\nu (F^{\mu\nu}\delta A_\mu+ G^{\mu\nu}\delta C_\mu)$$

Of course, we need to keep in mind our relation between $G_{\mu\nu}$ and $F_{\mu\nu}$ at all times, hence we must impose that our transformation does not leave the constraint surface, i.e. $\partial_{[\mu}C'_{\nu]}\equiv \frac{1}{2}\epsilon_{\mu\nu\alpha\beta}\partial^{[\alpha}A'^{\beta]}$. When this is satisfied, we also obtain a symmetry of the new Lagrangian (as explicitly demonstrated in the paper I linked).

Answering a first, obvious question, we use equation ($\star$) to find the conserved current associated to an infinitesimal duality rotation:

We obtain

$$\delta \mathcal L = \frac{\theta}{2}\partial_\nu(-F^{\mu\nu}C_\mu + G^{\mu\nu}A_\mu)\equiv 0$$

and hence the requirement that $\delta \mathcal L$ vanishes (as it should, for a symmetry of the theory!) yields a conserved current

$$\partial_\nu \kappa^\nu=0 \qquad \qquad \kappa^\nu=G^{\mu\nu}A_\mu - F^{\mu\nu}C_\mu$$

The conserved charge it defines,

$$Q_\kappa=\int \mathrm d^3 x \kappa^0$$

can be shown to be simply the optical helicity, but I will not dwell on the details of this. Instead, we press on toward our final goal:

## The zilch tensor as a Noether current

In order to connect to the zilch tensor, we need a slightly more obscure symmetry. Nevertheless, this transformation bears some resemblance to the duality transformation, and can be viewed as a perhaps-not-completely-unnatural variation on a theme. It is given by:

$$A_\mu\mapsto A_\mu-\xi^{\alpha\beta}\partial_\alpha G_{\beta\mu} \qquad\qquad C_\mu\mapsto C_\mu+\xi^{\alpha\beta}\partial_\alpha F_{\beta\mu}$$

To see that this indeed defines a symmetry of Maxwell's equations, we note that on the level of field strength tensors, it induces the transformation

$$F_{\mu\nu}\mapsto F_{\mu\nu}-\xi^{\alpha\beta}\partial_\alpha\partial_\beta G_{\mu\nu} \qquad\qquad G_{\mu\nu}\mapsto G_{\mu\nu}+\xi^{\alpha\beta}\partial_\alpha\partial_\beta F_{\mu\nu}$$

We see that $\xi^{\alpha\beta}$ is symmetric in its indices (this is forced by the derivatives appearing, and is therefore not an additional assumption). Taking the divergence of both transformed tensors and using the equations of motion for both $F_{\mu\nu}$ and $G_{\mu\nu}$ on each equation, we see that the new curvature tensors indeed still satisfy the equations of motion. Applying equation ($\star$) now leads us to conclude that:

\begin{align*}\delta \mathcal L &=\frac{1}{2} \partial_\nu(-F^{\mu\nu}\xi^{\alpha\beta}\partial_\alpha G_{\beta\mu}+G^{\mu\nu}\xi^{\alpha\beta}\partial_\alpha F_{\beta\mu}) =\frac{\xi^{\alpha\beta}}{2}\partial_\nu(G^{\mu\nu}\partial_\alpha F_{\beta\mu} - F^{\mu\nu}\partial_\alpha G_{\beta\mu})\\ &=\frac{\xi^{\alpha\beta}}{2}\partial_\nu Z^\nu_{\beta\alpha}\equiv 0 \end{align*}

Since $\xi^{\alpha\beta}$ does not vanish we must conclude that $\partial_\nu Z^\nu_{\beta\alpha}=0$, i.e. the zilch tensor arises as a conserved current associated to this symmetry.

• I once showed the second paper you mention to Mark Srednicki. He thinks that the two-potential formulation isn't wrong per se, but it's kind of useless, because it only applies to the source-free Maxwell's equations, which have pretty trivial dynamics - every solution's just a superposition of electromagnetic plane waves. Any free theory has an infinite number of conserved quantities (e.g. the amplitudes of the Fourier modes), so he doesn't think that finding a few more is terribly exciting. He said he'd be interested if someone found a new conserved quantity that works in the presence... – tparker Dec 22 '16 at 3:11
• ... of charges. Also, he found the separation of orbital and spin angular momenta to be artificial and unphysical since it's gauge-dependent. – tparker Dec 22 '16 at 3:11
• This is very easy to read and highly informative. Fantastic, thanks a bundle! – Selene Routley Jul 22 '17 at 11:10

There is already a good answer by Danu. In this answer, we give a few more details on how to apply Noether's theorem.

I) If we assume the lore that Lipkin's zilch is associated with EM duality ${\bf E} \leftrightarrow \pm {\bf B}$, then it is a well-known observation that the EM Lagrangian density

$$\tag{A} {\cal L}_1~=~-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}~=~\frac{1}{2}({\bf E}^2-{\bf B}^2), \qquad F_{\mu\nu} ~:=~ d_{[\mu} A_{\nu]},$$

does not possess EM duality symmetry, despite the vacuum Maxwell equations (ME) do. In fact, the Lagrangian density ${\cal L}_1\leftrightarrow -{\cal L}_1$ changes sign under EM duality. Hence the Noether's theorem is not applicable to the Lagrangian density (A).

II) Danu's answer and Ref. 2 consider instead two copies of EM,

$$\tag{B} {\cal L}_2 ~=~-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}-\frac{1}{4}G_{\mu\nu}G^{\mu\nu},\qquad F_{\mu\nu} ~:=~ d_{[\mu} A_{\nu]},\qquad G_{\mu\nu} ~:=~ d_{[\mu} C_{\nu]} ,\qquad$$

and want to apply Noether's theorem to the two-potential formulation (B) instead. We regard the two-potential formulation (B) as merely a trick to explore symmetry & conservations laws; not necessarily as a physical model in its own right.

III) Ultimately we want to implement the duality condition

$$\tag{3.7} G_{\mu\nu} ~=~ (\star F)_{\mu\nu} \qquad \Leftrightarrow\qquad -F_{\mu\nu} ~=~ (\star G)_{\mu\nu}$$

in Ref. 2, but not too soon, since this would make the Lagrangian density

$$\tag{C} \left. {\cal L}_2\right|_{G=\star F}~\equiv~0$$

vanish even off-shell (wrt. the ME) because the Hodge dual

$$\tag{D} \star^2~=~-1$$

squares to -1 in 1+3D Minkowski space. [The Minkowski sign convention $(+,-,-,-)$ follows Ref. 2.] This important point (C) is mentioned in Ref. 2 between eqs. (3.11-12).

IV) OP's definition (1) of Lipkin's zilch agrees with eq. (4) in Ref. 1. It reads

$$\tag{4} Z_{\mu\nu,\rho}~=~ F_{\mu\lambda}~ d_{\rho} G_{\nu}{}^{\lambda} -(F\leftrightarrow G)$$

after converting to the two-potential formalism.

The version of Lipkin's zilch in eq. (9) of Ref. 1 is interesting, since that version displays more symmetry. It reads

$$\tag{9} Z_{\mu\nu,\rho} ~=~ \frac{1}{2} \left[ F_{\mu\lambda}~ d_{\rho} G_{\nu}{}^{\lambda} -(F\leftrightarrow G) \right] + (\mu\leftrightarrow \nu)$$

after correcting a typo $\rho\leadsto\nu$ in Ref. 1, and converting to the two-potential formalism. To derive OP's definition (1) from eq. (9), one must use the eq. (2) in Ref. 1. It reads

$$\tag{2} F_{\mu\lambda} G^{\nu\lambda}~=~\frac{1}{4}\delta_{\mu}^{\nu} ~F_{\kappa\lambda}G^{\kappa\lambda},$$

which in turn relies on the duality condition (3.7).

V) It is straightforward to use definition (9) and the consequences

$$\tag{10}\Box F^{\mu\nu}~\approx~0, \qquad \Box G^{\mu\nu}~\approx~0,$$

of Maxwell's equations (ME)

$$\tag{3} d_{\mu}F^{\mu\nu}~\approx~0, \qquad d_{\mu}G^{\mu\nu}~\approx~0,$$

to derive the first

$$\tag{11} d_\rho Z_{\mu\nu}{}^{\rho}~\stackrel{(9)+(10)}{\approx}~0$$

of the three on-shell conservation laws for Lipkin's zilch. [Here the $\approx$ symbol means equality modulo e.o.m. The words on-shell and off-shell refer to whether e.o.m. are satisfied or not.]

VI) Ref. 2 considers the infinitesimal transformation

$$\tag{6.20} \delta A_{\alpha}~=~\xi^{\mu\nu}d_{\mu} G_{\nu\alpha}, \qquad \delta C_{\alpha}~=~-\xi^{\mu\nu}d_{\mu} F_{\nu\alpha},$$

where $\xi^{\mu\nu}$ are (not necessarily $\mu\leftrightarrow \nu$ symmetric) infinitesimal parameters. This implies

$$\tag{6.21} \delta F_{\alpha\beta} ~\stackrel{(B)+(6.20)}{=}~\Delta G_{\alpha\beta}, \qquad \delta G_{\alpha\beta} ~\stackrel{(B)+(6.20)}{=}~-\Delta F_{\alpha\beta},$$ $$\Delta~:=~\xi^{\mu\nu}d_{\mu}d_{\nu} ~=~\frac{1}{2}\xi^{\{\mu,\nu\}}d_{\mu}d_{\nu}.\qquad$$

VII) The transformation (6.20) is a quasi-symmetry of the Lagrangian density

$$\tag{E} \delta {\cal L}_2 ~\stackrel{(B)+(6.21)}{=}~ \frac{1}{2}G^{\alpha\beta} \Delta F_{\alpha\beta} -(F\leftrightarrow G) ~\stackrel{(F)}{=}~d_{\mu}f^{\mu}(\xi),$$

where

$$\tag{F} f^{\mu}(\xi)~:=~\frac{1}{4}\xi^{\{\mu,\nu\}}d_{\nu}F_{\alpha\beta} G^{\alpha\beta} -(F\leftrightarrow G)~\stackrel{(3.7)}{=}~0.$$

VIII) The full Noether current is

$$\tag{G} J^{\alpha}(\xi)~:=~j^{\alpha}(\xi) - f^{\alpha}(\xi)~\stackrel{(F)}{=}~j^{\alpha}(\xi),$$

where $j^{\alpha}(\xi)$ is the bare Noether current, see eq. (I) below. The corresponding on-shell conservation law reads

$$\tag{H} d_{\alpha} J^{\alpha}(\xi)~\approx~0.$$

The bare Noether current becomes Lipkin's zilch (4)

$$j^{\alpha}(\xi) ~:=~\frac{\partial{\cal L}_2}{\partial A_{\beta,\alpha}} \delta A_{\beta} +\frac{\partial{\cal L}_2}{\partial C_{\beta,\alpha}} \delta C_{\beta} ~\stackrel{(B)+(6.20)}{=}~G^{\alpha\beta}\xi^{\mu\nu}d_{\mu} F_{\nu\beta} -(F\leftrightarrow G)$$ $$\tag{I}~\stackrel{(4)}{=}~-\xi^{\mu\nu}Z^{\alpha}{}_{\nu,\mu}.$$

IX) In this answer, we have on purpose postponed the use of the duality condition (3.7) until after the conservation law (H) is established.

Looking at eqs. (E) & (F), the transformation (6.20) becomes an exact off-shell symmetry of ${\cal L}_2$ if we use the duality condition (3.7). Recalling eq. (C), this is not too surprising! Note that the transformation (6.20) respects the duality condition (3.7).

X) Altogether, eq. (H) becomes the second

$$\tag{J} d_{\alpha}Z^{\alpha}{}_{\nu,\mu}~\approx~0$$

of the three on-shell conservation laws for Lipkin's zilch. By the symmetry (9), it is also the third conservation law.

XI) In conclusion, it would be interesting to see eq. (11) as a Noether conservation law. Also it would be interesting to develop further a conservation law that does not rely on the duality condition (3.7). We leave that for the future.

References:

1. T.W.B. Kibble, J.Math.Phys. 6 (1965) 1022.

2. R.P. Cameron & S.M. Barnett, New Journ. Phys. 14 (2012) 123019.