## Spin Angular Momentum in Acoustic Field Theory

Justin Dressel

Work with: Lucas Burns, Konstantin Bliokh, Franco Nori

Schmid College of Science and Technology
Institute for Quantum Studies
Chapman University

Brown Theoretical Physics Center Seminar
November 20, 2020

## Acoustic Spin?

• Linear acoustic waves are longitudinal
• Usually described by a scalar potential field
• Scalar potential is dynamical in Lagrangian
• Scalar fields can't have intrinsic spin

• In 2019 acoustic spin was measured!
• Acoustic velocity fields have a vector structure not in the scalar potential

• How do we understand this structure within Lagrangian field theory?

## Acoustic Spin!

• Microscopically, the origin of spin is clear
• Superimposing two orthogonal longitudinal oscillations that are out of phase produces orbital molecular motion

• Coarse-graining the molecular motion
produces smooth field with intrinsic spin

• Pressure and Velocity fields have enough structure to describe the spin
• Strong analogies to electromagnetic fields
• But what is the dynamical potential field?

## Electromagnetic

\rho\,\partial_t \vec{v} = -\vec{\nabla} p
\beta\,\partial_t p = -\vec{\nabla}\cdot\vec{v}
\vec{\nabla}\times\vec{v} = \vec{0}

(Longitudinality)

\begin{aligned} &\vec{v} : \text{velocity} \\ &p : \text{pressure}\\ &\rho : \text{mass density} \\ &\beta : \text{compressibility} \\ &c = \frac{1}{\sqrt{\rho\beta}} : \text{speed of sound} \end{aligned}
\epsilon_0\,\partial_t \vec{E} = \vec{\nabla} \times \vec{H}
\mu_0\,\partial_t \vec{H} = -\vec{\nabla}\times \vec{E}
\vec{\nabla}\cdot\vec{E} = \vec{\nabla}\cdot\vec{H} = 0

(Transversality)

\begin{aligned} &\vec{E} : \text{electric field}\\ &\vec{H} : \text{magnetic field} \\ &\epsilon_0 : \text{permittivity} \\ &\mu_0 : \text{permeability} \\ &c = \frac{1}{\sqrt{\epsilon_0\mu_0}} : \text{speed of light} \end{aligned}

## Electromagnetic

L \propto \frac{1}{2} \left[ \epsilon_0\,\vec{E}\cdot\vec{E} - \mu_0\,\vec{H}\cdot\vec{H} \right]
L \propto \frac{1}{2} \left[ \rho\,\vec{v}\cdot\vec{v} - \beta\,p^2 \right]
\sqrt{\epsilon_0}\,\vec{E} = -\partial_{ct}\vec{A}_{e} - \vec{\nabla}\phi_{e}
\sqrt{\mu_0}\,\vec{H} = \vec{\nabla}\times\vec{A}_{e}
\sqrt{\rho}\,\vec{v} = \vec{\nabla}\phi
\sqrt{\beta}\,p = -\partial_{ct}\phi

(Lagrangian Density)

(Lagrangian Density)

(Dynamical Scalar Potential)

(Dynamical Potentials)

Does not reproduce vector degrees of freedom needed to describe spin!

Correctly reproduces electric and magnetic equations of motion

\rho\,\partial_t \vec{v} = -\vec{\nabla} p
\beta\,\partial_t p = -\vec{\nabla}\cdot\vec{v}
\vec{\nabla}\times\vec{v} = \vec{0}

(Longitudinality)

\epsilon_0\,\partial_t \vec{E} = \vec{\nabla} \times \vec{H}
\mu_0\,\partial_t \vec{H} = -\vec{\nabla}\times \vec{E}
\vec{\nabla}\cdot\vec{E} = \vec{\nabla}\cdot\vec{H} = 0

(Transversality)

## Acoustic

Same basic structure as electromagnetism

But:

• $$(p,\,\vec{v})$$ is 1+3 dimensions
• involved in stress-energy tensor
• not the dynamical field in Lagrangian
• $$(\vec{E},\,\vec{H})$$ is 3+3 dimensions
• involved in stress-energy tensor
• not the dynamical field in Lagrangian

• $$(p,\, \vec{v})$$ are longitudinal
• $$(\vec{E},\,\vec{H})$$ are transverse

• $$c$$ is invariant speed of sound, not light

How do we fix the potential?

\rho\,\partial_t \vec{v} = -\vec{\nabla} p
\beta\,\partial_t p = -\vec{\nabla}\cdot\vec{v}
\vec{\nabla}\times\vec{v} = \vec{0}

(Longitudinality)

L \propto \frac{1}{2} \left[ \rho\,\vec{v}\cdot\vec{v} - \beta\,p^2 \right]
\sqrt{\rho}\,\vec{v} = \vec{\nabla}\phi
\sqrt{\beta}\,p = -\partial_{ct}\phi

(Lagrangian Density)

(Dynamical Scalar Potential)

## Acoustic

Same basic structure as electromagnetism

But:

• $$(p,\,\vec{v})$$ is 1+3 dimensions
• transforms as a "spacetime" 4-vector!
• $$(\vec{E},\,\vec{H})$$ is 3+3 dimensions
• transforms as a spacetime bivector!

• Motivated to construct "Acoustic Spacetime" by replacing speed of light with speed of sound

• Has Lorentz metric creating tangent algebra
Signature: $$(+---)$$       Algebra: $$\text{Cl}_{1,3}(\mathbb{R})$$

• Structure of potentials are manifest in 4D

How do we fix the potential?

\rho\,\partial_t \vec{v} = -\vec{\nabla} p
\beta\,\partial_t p = -\vec{\nabla}\cdot\vec{v}
\vec{\nabla}\times\vec{v} = \vec{0}

(Longitudinality)

L \propto \frac{1}{2} \left[ \rho\,\vec{v}\cdot\vec{v} - \beta\,p^2 \right]
\sqrt{\rho}\,\vec{v} = \vec{\nabla}\phi
\sqrt{\beta}\,p = -\partial_{ct}\phi

(Lagrangian Density)

(Dynamical Potentials)

### Slight Diversion

(algebra of "spacetime" 101)

## Constructing "Acoustic Spacetime" Algebra

\text{Basis: } \gamma_0,\,\gamma_1,\,\gamma_2,\,\gamma_3

Minkowski Space $$\mathcal{M}_{1,3}(\mathbb{R})$$ :

\displaystyle \begin{aligned}\eta(a+b,\,a+b) &= (a+b)^2 = a^2 + b^2 + ab + ba \\ &= \eta(a,a) + \eta(b,b) + 2\eta(a,b) \\ \\ \implies& \eta(a,b) \equiv \frac{ab + ba}{2} \equiv a\cdot b \end{aligned}
\gamma_0 \cdot \gamma_0 = 1
\gamma_k \cdot \gamma_k = -1
\gamma_\mu \cdot \gamma_\nu = \eta_{\mu\nu} = \eta(\gamma_\mu,\,\gamma_\nu)

Minkowski metric
(symmetric bilinear form)

Algebraic Axioms:   $$a,b,c\in\mathcal{M}_{1,3}(\mathbb{R})$$

1. Associativity
$$a(bc) = (ab)c$$

2. Left Distributivity
$$a(b+c) = ab + ac$$

3. Right Distributivity
$$(b+c)a = ba + ca$$

4. Contraction
$$a^2 = aa = \eta(a,\,a)$$

Axioms constructively define Clifford algebra $$\text{Cl}_{1,3}(\mathbb{R})$$

\displaystyle ab = \frac{ab + ba}{2} + \frac{ab - ba}{2} \equiv a\cdot b + a\wedge b

Symmetric part of product encodes metric:

Anti-symmetric part encodes Grassman wedge product:

(others zero)

(same as for differential forms)

## Graded Clifford Basis (Dirac Algebra)

• (contravariant) basis:
\gamma_0\gamma_1\gamma_2\gamma_3
1

0

1

2

3

\gamma_\mu\gamma_\nu = \begin{cases}-\gamma_\nu\gamma_\mu = \gamma_\mu\wedge\gamma_\nu, & \mu\neq\nu \\\quad\,\, \eta_{\mu\nu} = \gamma_\mu\cdot\gamma_\nu, & \mu = \nu \end{cases}

4

2^4 = 16\;\text{dimensions}

Bivector Planes (of rotation):

3 hyperbolic | 3 elliptic

Vector Lines (of translation):

1 timelike | 3 spacelike

Scalar Points

Pseudovector Volumes:

1 spacelike | 3 timelike

Pseudoscalar 4-Volumes

Multivector:    $$M = \langle M \rangle_0 + \langle M \rangle_1 + \langle M \rangle_2 + \langle M \rangle_3 + \langle M \rangle_4$$

• Reciprocal/dual/covariant basis:
\gamma^\mu \equiv (\gamma_\mu)^{-1} = \gamma_\mu / (\gamma_\mu)^2

## Relative Frame Simplification (Pauli Algebra)

I
1

0

1

2

3

I \equiv \gamma_0\gamma_1\gamma_2\gamma_3 \equiv \vec{\sigma}_1\vec{\sigma}_2\vec{\sigma}_3

(Spacetime-aware "imaginary unit" is the unit pseudoscalar: $$I^2 = -1$$)

4

Relative 3D space embedded as even-graded subspace:
$$\text{Cl}_{3,0}(\mathbb{R})$$

Equivalent to complexified
3-vector (Pauli) algebra since $$I$$ commutes with even grades

\vec{\sigma}_k \equiv \gamma_k\gamma_0 = \gamma_k \wedge \gamma_0

(Observed spatial directions in reference frame)
(Space axis dragged along temporal worldline)

Note duality transformation of left multiplication by $$-I$$ : Hodge star operation

Multivector:    $$M = [\alpha + (v_0 + \vec{v})\gamma_0 + \vec{A}] + I[\vec{B} + (w_0 + \vec{w})\gamma_0 + \beta]$$

## Complex Structure & Spacetime Splits

Proper Form of Multivector:

$$M = \alpha + v + \mathbf{F} + Iw + I\beta = (\alpha + I\beta) + (v + Iw) + \mathbf{F}$$

Relative Frame Form of Multivector:

$$M = \alpha + (v_0 + \vec{v})\gamma_0 + (\vec{A} + I\vec{B}) + I(w_0 + \vec{w})\gamma_0 + I\beta$$

Components:

complex scalar

complex vector

bivector

scalar

polar paravector

polar vector

axial vector

axial paravector

pseudoscalar

4-vector

scalar

bivector

pseudovector

pseudoscalar

v = \sum_\mu v^\mu\gamma_\mu = \sum_\mu v_\mu \gamma^\mu = (v^0 + \sum_k v^k\vec{\sigma}_k)\gamma_0
\mathbf{F} = \sum_{\mu,\nu} F^{\mu\nu}\gamma_\mu\gamma_\nu = \sum_{\mu,\nu} F_{\mu\nu} \gamma^\mu\gamma^\nu = \sum_k A^k\vec{\sigma}_k + I\sum_k B^k\vec{\sigma}_k

4-vector components, or a scalar and a relative 3-vector

rank-2 antisymmetric tensor components, or a pair of 3-vectors (one polar, one axial)

## Fields and Derivatives

Multivector field:   (in tangent algebra at point $$x = (ct + \vec{x})\gamma_0$$ of manifold)

$$M(x) = \alpha(x) + v(x) + \mathbf{F}(x) + Iw(x) + I\beta(x)$$

Vector derivative (Dirac differential operator):

3-Vector derivative (3-Gradient, Pauli differential operator):

$$\displaystyle \vec{\nabla} \equiv \gamma^0\wedge\nabla = \sum_k \vec{\sigma}^k \frac{\partial}{\partial x^k}$$

$$\vec{\nabla}\vec{v} = \vec{\nabla}\cdot\vec{v} + \vec{\nabla} \wedge \vec{v} = \vec{\nabla}\cdot\vec{v} + (\vec{\nabla}\times \vec{v})I$$

\displaystyle \nabla \equiv \sum_\mu \gamma^\mu \frac{\partial}{\partial x^\mu} = \gamma_0 \left[\frac{\partial}{\partial (ct)} + \vec{\nabla} \right] = \left[ \frac{\partial}{\partial (ct)} - \vec{\nabla} \right] \gamma_0
\displaystyle \nabla v \equiv \nabla \cdot v + \nabla \wedge v \Longleftrightarrow \delta v + d v

(equivalent to co-differential plus exterior derivative for forms)

(contains divergence and curl)

\nabla^2 = \nabla\gamma_0^2\nabla = (\partial_{ct} - \vec{\nabla})(\partial_{ct} + \vec{\nabla}) = \partial^2_{ct} - \vec{\nabla}\cdot\vec{\nabla} \Longleftrightarrow d\delta + \delta d

(squares to Laplacian/d'Alembertian)

### End Slight Diversion

(algebra of "spacetime" 101)

### Maxwell's Equation (in concise spacetime algebra)

\mathbf{F}(x) = \sqrt{\epsilon_0}\sum_{k=1}^3E^k(x)\gamma_k\gamma_0 + \sqrt{\mu_0}I\sum_{k=1}^3 H^k(x) \gamma_k\gamma_0 = \sqrt{\epsilon_0}\vec{E}(x)+I\sqrt{\mu_0}\vec{H}(x)
\nabla \mathbf{F} = j
j = \sum_\mu j^\mu_e \gamma_\mu + \sum_\mu j^{\mu}_{m} \gamma_\mu I = j_e + j_m I = \sqrt{\epsilon_0}(\rho_e + \vec{J}_e/c)\gamma_0 + \sqrt{\mu_0}(\rho_m + \vec{J}_m/c)\gamma_0 I

electric charge
density and current

This is Maxwell's Equation

\begin{aligned} \nabla \mathbf{F} &= \gamma_0[(\vec{\nabla}\cdot\vec{E}) + (\partial_{ct}\vec{E}- \vec{\nabla}\times\vec{H})] \\ &+ \gamma_0[(\vec{\nabla}\cdot\vec{H}) + (\partial_{ct}\vec{H} + \vec{\nabla}\times\vec{E})]I \\ &= \gamma_0(\rho_e - \vec{J}_e) + \gamma_0 (\rho_m - \vec{J}_m)I = j \end{aligned}

Electromagnetic field bivector:

Same form as complex (Riemann-Silberstein) 3-vector

Electromagnetic source complex vector:

magnetic charge
density and current

Contains all of the usual 3D Maxwell's Equations, and admits both electric and magnetic sources

\mathbf{G} = \mathbf{F}I^{-1} = \sqrt{\mu_0}\vec{H} - I\sqrt{\epsilon_0}\vec{E}

dual field

\delta \mathbf{F} = j_e, \quad \delta \mathbf{G} = j_m \\ \mathbf{G} = \star\mathbf{F}, \quad \delta = \star^{-1}d\star

differential forms equivalent

### Acoustic Equation (in concise "spacetime" algebra)

w(x) = \sqrt{\beta}\,p(x)\gamma_0 + \sqrt{\rho}\,\sum_{k=1}^3 v^k(x) \gamma_k = [\sqrt{\beta}\,p(x)+\sqrt{\rho}\,\vec{v}(x)]\gamma_0
\nabla w = \Lambda
\Lambda = \sqrt{\rho}\,g_0 + \sqrt{\beta}\,\mathbf{G}_a = \sqrt{\rho}\,g_0 + (\sqrt{\beta}\,\vec{g} - I\sqrt{\rho}\,\vec{h})

pressure
source

This is the Acoustic Equation

\begin{aligned} \nabla w &= \nabla \cdot w + \nabla \wedge w \\ &= (\partial_{ct}\,p + \vec{\nabla}\cdot\vec{v}) - (\partial_{ct}\,\vec{v} + \vec{\nabla}p) + \vec{\nabla}\times\vec{v}I \\ &= g_0 + \vec{g} - I \vec{h} = \Lambda \end{aligned}

Velocity field 4-vector:

Acoustic source spinor:

velocity
source

Contains all of the pressure and velocity equations, as well as the longitudinality condition (when $$\vec{h}=0)$$

\delta w = \sqrt{\rho}\, g_0, \quad d w = \sqrt{\beta}\,\mathbf{G}_a

differential forms equivalent

transverse
source

equation admits a spinor source (scalar plus bivector)

Main difference from EM is that the acoustic velocity field is one grade lower

## Poincare Lemma(s): Two Possible EM Potentials

\nabla \cdot \nabla \cdot (a_m I) = 0
\nabla \wedge \nabla \wedge a_e = 0
\begin{aligned} &\text{if } \nabla \wedge \mathbf{F} = 0, \\ &\exists a_e: \mathbf{F} = \nabla \wedge a_e \end{aligned}

"All exact forms are closed"

Poincare Lemma:

"All closed forms are exact."

(dda_e=0)
(d\mathbf{F} = 0 \implies \mathbf{F} = da_e)
(\delta\delta (\star^{-1} a_m)=0)

"All co-exact forms are co-closed"

\begin{aligned} &\text{if } \nabla \cdot \mathbf{F} = 0, \\ &\exists (a_m I): \mathbf{F} = \nabla \cdot (a_m I) = (\nabla \wedge a_m)I \end{aligned}

Dual Poincare Lemma:

"All co-closed forms are co-exact."

\begin{aligned} (\delta \mathbf{F} = 0 \implies \mathbf{F} &= \delta (\star^{-1} a_m) \\ &= (\star^{-1} d \star)(\star^{-1} a_m) = \star^{-1} d a_m) \end{aligned}

Electric Vector Potential Exists

Magnetic Pseudovector Potential Exists

(couples to electric sources)

(couples to magnetic sources)

## Poincare Lemma(s): Two Possible Acoustic Potentials

\nabla \cdot \nabla \cdot \mathbf{A} = 0, \quad \mathbf{A} = \vec{a} + \vec{b}I
\nabla \wedge \nabla \phi = 0
\begin{aligned} &\text{if } \nabla \wedge w = 0, \\ &\exists \phi: w = \nabla \phi \end{aligned}

"All exact forms are closed"

Poincare Lemma:

"All closed forms are exact."

(dd\phi=0)
(dw = 0 \implies w = d\phi)
(\delta\delta (\star^{-1} \mathbf{A})=0)

"All co-exact forms are co-closed"

\begin{aligned} &\text{if } \nabla \cdot w = 0, \\ &\exists \mathbf{A}: w = \nabla \cdot \mathbf{A} \end{aligned}

Dual Poincare Lemma:

"All co-closed forms are co-exact."

\begin{aligned} (\delta w = 0 \implies w &= \delta \mathbf{A}) \end{aligned}

Scalar Potential Exists

Bivector Potential Exists

(couples to pressure sources)

(couples to velocity sources)

## Gauge Freedom and Radiation Far-field

Acoustic Bivector Potential

\begin{aligned} \mathbf{F} = \nabla \wedge a_e \end{aligned}

Electric Vector Potential

\begin{aligned} a_e \mapsto a_e + \nabla \chi \end{aligned}
\begin{aligned} \nabla\mathbf{F} &= \nabla[\nabla \wedge (a_e + \nabla \chi)] \\ &= \nabla\cdot(\nabla \wedge a_e) = j_e \end{aligned}

In a particular reference frame, can fully fix (Coulomb) gauge by enforcing transversality far from sources (radiation far-field)

Lorenz-FitzGerald condition yields wave equation

\begin{aligned} \nabla \cdot a_e = 0 \implies \nabla^2 a_e = j_e \end{aligned}
\begin{aligned} a_e = (\phi_e + \vec{A}_e)\gamma_0, \quad \vec{\nabla}\cdot\vec{A}_e = \phi_e = 0 \end{aligned}
\begin{aligned} \nabla \cdot a_e = 0 \implies \nabla^2 a_e = j_e \end{aligned}
\begin{aligned} w = \nabla \cdot \mathbf{A} &= \nabla \cdot (\vec{a} + \vec{b} I) \\ &= \gamma_0[(\vec{\nabla}\cdot\vec{a}) + (\partial_{ct}\vec{a}- \vec{\nabla}\times\vec{b})] \\ &= \gamma_0(\sqrt{\beta}\,p - \sqrt{\rho}\,\vec{v}) \end{aligned}

Gauge freedom of a curl

\begin{aligned} \mathbf{A} \mapsto \mathbf{A} + \nabla\wedge \xi \end{aligned}
\begin{aligned} \nabla w &= \nabla[\nabla \cdot (\mathbf{A} + \nabla \wedge \xi)] \\ &= \nabla\wedge(\nabla \cdot \mathbf{A}) = \Lambda \end{aligned}

Lorenz-FitzGerald-like condition yields wave equation

\begin{aligned} \nabla \wedge \mathbf{A} = 0 \implies \nabla^2 \mathbf{A} = \mathbf{G}_a \end{aligned}

In a particular reference frame, can fully fix (Coulomb-like) gauge by enforcing longitudinality far from sources

\begin{aligned} \vec{\nabla}\times\vec{a} = \vec{b} = 0 \end{aligned}

Acoustic Bivector Potential

\begin{aligned} \mathbf{F} = \nabla \wedge a_e \end{aligned}

Electric Vector Potential

\begin{aligned} \vec{\nabla}\cdot\vec{A}_e = \phi_e = 0 \end{aligned}
\begin{aligned} \nabla \cdot a_e = 0 \implies \mathbf{F} = \nabla a_e \end{aligned}

Lorenz-FitzGerald-like condition yields:

\begin{aligned} \nabla \wedge \mathbf{A} = 0 \implies w = \nabla \mathbf{A} \end{aligned}

The bivector potential has been fixed to a longitudinal 3-vector potential $$\vec{x} = \sqrt{\beta}\,\vec{a}$$ with units of mean molecular displacement

\begin{aligned} \vec{\nabla}\times\vec{a} = \vec{b} = 0 \end{aligned}

Lorenz-FitzGerald condition yields:

Coulomb gauge with no sources yields:

\vec{E} = - \partial_t\vec{A}_e
\vec{H} = \vec{\nabla}\times\vec{A}_e

3-vector potential $$\vec{A}_e$$ has units of field-momentum per unit charge

In this fixed gauge, the electric field is the rate at which momentum is removed from the field, while the magnetic field describes the circulating character of that momentum

\begin{aligned} w = \nabla \cdot \mathbf{A} \end{aligned}

Coulomb-like gauge with no sources yields:

\begin{aligned} p &= \vec{\nabla}\cdot(\vec{x}/\beta) \\ \vec{v} &= -\partial_t\vec{x} \end{aligned}

In this fixed gauge, the velocity field is the rate opposing the change in mean displacement field, while the pressure field describes the expanding and contracting character of that mean displacement

Acoustic Bivector Potential

Electric Vector Potential

\begin{aligned} \vec{\nabla}\cdot\vec{A}_e = \phi_e = 0 \end{aligned}
\begin{aligned} \vec{\nabla}\times\vec{a} = \vec{b} = 0 \end{aligned}

Coulomb gauge with no sources:

\vec{E} = - \partial_t\vec{A}_e
\vec{H} = \vec{\nabla}\times\vec{A}_e

Derived spin-vector has the asymmetric form:

Coulomb-like gauge with no sources yields:

\begin{aligned} p &= \vec{\nabla}\cdot(\vec{x}/\beta) \\ \vec{v} &= -\partial_t \vec{x} \end{aligned}
\vec{S} = \vec{E}\times\vec{A}_e

Magnetic Pseudovector Potential

\begin{aligned} \vec{\nabla}\cdot\vec{C}_m = \phi_m = 0 \end{aligned}
\vec{E} = - \vec{\nabla}\times\vec{C}_m
\vec{H} = - \partial_t \vec{C}_m

Spin-vector has other asymmetric form:

\vec{S} = \vec{B}\times\vec{C}_m

Derived spin-vector has the asymmetric form:

\vec{S} = \rho \vec{v}\times\vec{x}
\vec{x} = \sqrt{\beta}\,\vec{a}
\vec{v} = \vec{\nabla}(\phi/\sqrt{\rho})
p = -\partial_t(\sqrt{\rho}\,\phi)

Acoustic Scalar Potential

Spin-vector has other asymmetric form:

\vec{S} = \vec{0}

## None of the Derived Spin Vectors are Correct!

Acoustic Bivector Potential

Electric Vector Potential

\vec{S} = \vec{E}\times\vec{A}_e

Magnetic Pseudovector Potential

\vec{S} = \vec{B}\times\vec{C}_m
\vec{S} = \rho \vec{v}\times\vec{x}
\vec{x} = \sqrt{\beta}\,\vec{a}

Acoustic Scalar Potential

\vec{S} = \vec{0}

Correct (Measured) Spin Vector

\vec{S} = \frac{1}{2}\left[ \vec{E}\times\vec{A}_e + \vec{B}\times\vec{C}_m \right]

Correct (Measured) Spin Vector

\vec{S} = \frac{1}{2} \rho \vec{v}\times\vec{x}
\vec{x} = \sqrt{\beta}\,\vec{a}

In both EM and Acoustics, the correct (measured) spin vector is the average of those derived from the two possible dynamical potentials!

How can we derive this from the field theory framework correctly?

[Toftul, Bliokh, Petrov, Nori, PRL. 123, 183901 (2019)]

[Bliokh et al. New J Phys 15, 033026 (2013)]

### Dual-symmetry : Electric-Magnetic Egalitarianism

\mathbf{F}(x) = \mathbf{f}(x)\,e^{I\theta(x)/2} = \sqrt{\epsilon_0}\vec{E}(x)+I\sqrt{\mu_0}\vec{H}(x)
\nabla \mathbf{F} = j
j = j_e + j_m I = \sqrt{\epsilon_0}(\rho_e + \vec{J}_e/c)\gamma_0 + \sqrt{\mu_0}(\rho_m + \vec{J}_m/c)\gamma_0 I

# $$\{$$

Complete EM dynamics:

A hidden gauge symmetry is now evident:

e^{-I\phi/2}\nabla \mathbf{F}e^{I\phi/2} = e^{-I\phi/2}je^{I\phi/2}

The pseudoscalar $$I$$ generates phase rotations in the usual way:

this has the effect of swapping the electromagnetic fields and source charges

e^{-I\phi/2}\nabla \mathbf{F}e^{I\phi/2} = \nabla\left[e^{I\phi/2}\mathbf{F}e^{I\phi/2}\right] = \nabla \left[\mathbf{F}e^{I\phi}\right] = \nabla \left[ (\cos\phi\,\vec{E} - \sin\phi\,\vec{B}) + (\sin\phi\,\vec{E} + \cos\phi\,\vec{B})I\right] = \nabla(\vec{E}' + \vec{B}'I)
e^{-I\phi/2}je^{I\phi/2} = je^{I\phi} = (\cos\phi\,j_e - \sin\phi\,j_m) + (\sin\phi\,j_e + \cos\phi\,j_m)I = j_e' + j_m'I

Physics is the same provided that both source and field definitions are changed in tandem

Can always choose $$\phi$$ to make all charges electric ($$j'_m = 0$$)

Vacuum fields far from sources ($$j = 0$$) always have this field-exchange symmetry

### Electromagnetic Complex Vector Potential

\mathbf{F}(x) = \nabla z(x)

Postulate vector potential:

z(x) = \displaystyle \frac{a_e(x) + a_m(x)I}{\sqrt{2}}

Must be complex potential to satisfy dual symmetry

\mathbf{F} = \displaystyle \frac{\mathbf{F}_e + \mathbf{G}_m I}{\sqrt{2}} = \frac{\nabla \wedge a_e + \nabla \cdot (a_mI)}{\sqrt{2}}
0 = \nabla \cdot a_e = \nabla\cdot a_m

Lorenz-Fitzgerald conditions assumed

\mathbf{F} = \vec{E} + \vec{H}I \qquad a_e = (\phi_e + \vec{A}_e)\gamma_0 \qquad a_m = (\phi_m + \vec{C}_m)\gamma_0
\vec{E} = \displaystyle \frac{-\vec{\nabla}\phi_e - \partial_{ct}\vec{A}_e - \vec{\nabla}\times\vec{C}_m}{\sqrt{2}}
\vec{H} = \displaystyle \frac{-\vec{\nabla}\phi_m - \partial_{ct}\vec{C}_m + \vec{\nabla}\times\vec{A}_e}{\sqrt{2}}

Relative EM fields regain symmetry in definitions

\implies \nabla\mathbf{F}(x) = \nabla^2 z(x) = j(x)

Obvious wave equation

Electric and Magnetic potentials

contribute equally
(special case of Hodge Decomposition)

### Dual-Symmetric Electromagnetic Lagrangian

L[z,\tilde{z}] = \displaystyle \frac{(\nabla z)\cdot(\nabla \tilde{z})}{2} = \frac{\mathbf{F} \cdot \mathbf{F}_{\rm conj}}{2} \neq \frac{\mathbf{F}\cdot\mathbf{F}}{2}

Must consider complex vector potential and its conjugate

z = \displaystyle \frac{a_e + a_m I}{\sqrt{2}}
\tilde{z} = \displaystyle \frac{a_e - a_m I}{\sqrt{2}}

Much like done with complex scalar potentials,
the Lagrangian must be varied with respect to
both $$z$$ and $$\tilde{z}$$ to get the correct result

\nabla^2 z = \displaystyle \frac{\nabla^2 a_e + \nabla^2 (a_m I)}{\sqrt{2}} = \nabla\mathbf{F} = 0
\nabla^2 \tilde{z} = \displaystyle \frac{\nabla(\nabla a_e - \nabla (a_m I))}{\sqrt{2}} = \nabla \mathbf{F}_{\rm conj} = 0

The first equation yields Maxwell's equation,
while the second forces each potential contribution to the total measured fields to be exactly equal

In the presence of sources, the conjugate field acquires nontrivial equations of motion that steer the measured field into a biased potential representation. This mechanism automatically locks the potential representation to match the source.

### Noether's Theorem and EM Dual Symmetry

e^{-I\phi/2}\nabla \mathbf{F}e^{I\phi/2} = e^{-I\phi/2}je^{I\phi/2}

The global phase gauge symmetry leads to an independently conserved quantity

\underline{X}(I) = [\chi I + \vec{S}I^{-1}]\gamma_0

Helicity pseudocurrent is conserved Noether charge

\chi = \displaystyle \frac{\vec{B} \cdot \vec{A}_e - \vec{E} \cdot \vec{C}_m}{2}
\vec{S} = \displaystyle \frac{\vec{E}\times\vec{A}_e + \vec{B}\times \vec{C}_m}{2}

Correct spin-density pseudovector is the helicity flux density

Correct helicity-density pseudoscalar is also obtained

These definitions only make sense when properly dual-symmetric fields are used

They match measurements only with full electric-magnetic potential symmetry

### Acoustic Spinor Potential

w(x) = \nabla \psi(x)

Postulate spinor potential:

\psi(x) = \displaystyle \frac{\phi'(x) + \mathbf{A}(x)}{\sqrt{2}}

Combines both scalar and bivector potentials in similar manner to the complex vector EM potential

w = \displaystyle \frac{w_\phi + w_\mathbf{A}}{\sqrt{2}} = \frac{\nabla \wedge \phi' + \nabla \cdot \mathbf{A}}{\sqrt{2}}
0 = \nabla \wedge \mathbf{A}

Lorenz-Fitzgerald-like condition assumed

w = (\sqrt{\beta}\,p + \sqrt{\rho}\,\vec{v})\gamma_0 \quad \mathbf{A} = \vec{x}/\sqrt{\beta} + I\,\vec{b}/\sqrt{\rho} \quad \phi' = \sqrt{\rho}\,\phi
p = \displaystyle \frac{- \partial_t\phi' + \vec{\nabla}\cdot(\vec{x}/\beta)}{\sqrt{2}}
\vec{v} = \displaystyle \frac{\vec{\nabla}(\phi'/\rho) - \partial_t\vec{x} + \vec{\nabla}\times(\vec{b}/\rho)}{\sqrt{2}}

Pressure and velocity
definitions become
symmetric

\implies \nabla v(x) = \nabla^2 \psi(x) = \Lambda(x)

Obvious spinor wave equation

Scalar and Bivector potentials

contribute equally
(special case of Hodge decomposition)

### Acoustic Spinor Lagrangian

L[\psi,\tilde{\psi}] = \displaystyle \frac{(\nabla \psi)\cdot(\nabla \tilde{\psi})}{2} = \frac{\mathbf{w} \cdot \mathbf{w}_{\rm conj}}{2} \neq \frac{\mathbf{w}\cdot\mathbf{w}}{2}

Must consider spinor potential and its conjugate

\psi = \displaystyle \frac{\phi + \mathbf{A}}{\sqrt{2}}
\tilde{\psi} = \displaystyle \frac{\phi - \mathbf{A}}{\sqrt{2}}

Must be varied with respect to
both $$\psi$$ and $$\tilde{\psi}$$ to get the correct result

\nabla^2 \psi = \displaystyle \frac{\nabla^2 \phi + \nabla^2 \mathbf{A}}{\sqrt{2}} = \nabla w = 0
\nabla^2 \tilde{\psi} = \displaystyle \frac{\nabla(\nabla \phi - \nabla \mathbf{A})}{\sqrt{2}} = \nabla w_{\rm conj} = 0

The first equation yields the acoustic equation,
while the second forces each potential contribution to the total measured fields to be exactly equal

In the presence of sources, the conjugate field acquires nontrivial equations of motion that steer the measured field into a biased potential representation. This mechanism automatically locks the potential representation to match the source.

### Finally, the Acoustic Field Spins Correctly!

\vec{S} = \displaystyle \frac{\rho\vec{v}\times\vec{x}}{2}

Correct spin-density pseudovector

The correctly measured value of the spin-density is only obtained with symmetric potentials

\bar{\vec{S}} \to \displaystyle \frac{\rho\,\text{Im}(\vec{v}^*\times\vec{v})}{4\omega}

Monochromatic field cycle-average

## Conclusions

• Acoustic fields can carry intrinsic spin, and we can now derive it!

• Spacetime Clifford algebra is a very convenient mathematical language

• The dual-symmetric structure of the electromagnetic theory becomes manifest when expressed in geometrically invariant forms

• The conserved quantity from Noether's theorem resulting from the continuous dual-symmetry in vacuum field propagation is the helicity

• A similarly symmetric spinor potential is needed in acoustics to correctly derive the spin density as part of a conserved Noether current

Thank you!

JD, Bliokh, Nori, Physics Reports 589 1-71 (2015)

Burns, Bliokh, Nori, JD, New J Physics 22 053050 (2020)

#### Spin Angular Momentum in Acoustic Field Theory

By Justin Dressel

# Spin Angular Momentum in Acoustic Field Theory

Brown Theoretical Physics Seminar, November 20, 2020. Using an acoustic analog of Minkowski geometry, we construct a Lagrangian representation of acoustic field theory that accounts for the recently measured nonzero spin-angular-momentum density of sound fields in fluids or gases. While the traditional acoustic Lagrangian representation of the measured pressure and velocity fields using a dynamical scalar potential is unable to describe the vector character of the spin, we show that the pressure-velocity 4-vector additionally admits a dynamical bivector potential that correctly accounts for the spin. In the equilibrium frame this bivector potential reduces to a displacement field with amplitude equal to the scalar potential, such that its gauge freedom is equivalent to the arbitrary choice of displacement origin. The two potentials combine into an even-graded spinor potential with dynamics that recover the observed local radiation forces and torques on small probe particles, as well as the correct canonically conserved momentum and spin densities as proper Noether currents. This twin-potential construction for acoustics closely mirrors a formulation of vacuum electromagnetism that combines both electric- and magnetic-potential representations into a manifestly dual-symmetric odd-graded potential.

• 307