Heaviside-Lorentz system of units

$\DeclareMathOperator{\Div}{div}
\DeclareMathOperator{\Rot}{rot}
\newcommand{\parder}[2]{\frac{\partial {#1}}{\partial {#2}}}$$\setCounter{0}$

I will want to construct the theory of light propagating through weird media to eventually make predictions about results of ellipsometric experiments on such. Ellipsometry is a geometrical method and gives the results in terms of angles or ratios. This allows us to choose convenient system of electro-magnetic units which reduces clutter in the formulas.

In this post I want to start with electro-magnetism written in the Heaviside-Lorentz system of units. This is naturally well covered in many books and for quick reference the Wikipedia page is very useful. Here, I will take bit different route and show how to make the transition from SI (International system of units) to HLU (Heaviside–Lorentz units), as to show the motivation for such steps.

Maxwell’s equations in SI

First we write the macroscopic Maxwell’s equations in some material medium. Let the $\mathcal{D}$, $\mathcal{E}$, $\mathcal{H}$ and $\mathcal{B}$ be the SI electric and magnetic vector fields, $\rho$ the scalar charge density field and $\mathbf{j}$ the vector current density field.

$\begin{align}
\begin{aligned}
\Div\mathcal{D} &=\rho,
&\Rot\mathcal{H} &=\parder{\mathcal{D}}{t}+\mathbf{j},
\\
\Rot\mathcal{E} &=-\parder{\mathcal{B}}{t},
&\Div\mathcal{B} &=0.
\end{aligned}
\label{eqMaxSI}
\end{align}$

The Maxwell’s equations written in SI are nice and clean. The basic fields are the $\mathcal{E}$ and $\mathcal{B}$, which appear in the two homogeneous equations, while the induced fields are the $\mathcal{D}$ and $\mathcal{H}$, which appear in the inhomogeneous two equations – those containing the external charges and currents. The polarization charges and currents induced in the material are contained in the material relations.

The usual situation in optics is that there are no external charges, $\rho = 0$, or currents, $\mathbf{j} = 0$, and the material relations are formulated in a way that the $\mathcal{E}$ and $\mathcal{H}$ fields are considered as bare field strengths while $\mathcal{D}$ and $\mathcal{B}$ are induced. This choice is alright as long as we don’t deal with moving media and relativistic effects.

Simple material relations in SI

In a simple dielectric medium, the electric displacement field $\mathcal{D}$ is a combination of the electric field $\mathcal{E}$ and the polarization $\mathcal{P}$, which contains the response of the material to the electric field, $\mathcal{P} = \varepsilon_0 \chi_e \mathcal{E}$. Here $\chi_e$ is dimensionless dielectric susceptibility, and the $\varepsilon_0$ is vacuum permittivity – a constant we need to get the units in order.¹

$\mathcal{D} = \varepsilon_0 \mathcal{E} + \mathcal{P} = \varepsilon_0 \mathcal{E} + \varepsilon_0 \chi_e \mathcal{E} = \varepsilon_0 (1+\chi_e )\mathcal{E} = \varepsilon_0 \varepsilon \mathcal{E}$,

where the $\varepsilon$ is the dimensionless relative permittivity – dielectric constant. Note that $\mathcal{D}$ and $\mathcal{P}$ are measured in the same units, i.e. volume density of dipole moment.

Almost analogical formulas appear for the magnetic quantities. We consider the magnetic induction $\mathcal{B}$ as a combination of the magnetic field $\mathcal{H}$ and the magnetization of the material $\mathcal{M}$. Magnetization is again the response of the material to the field, $\mathcal{M} =\chi_m \mathcal{H}$. Here $\chi_m$ is dimensionless magnetic susceptibility.

$\mathcal{B} = \mu_0 ( \mathcal{H} + \mathcal{M} ) = \mu_0 ( \mathcal{H} + \chi_m \mathcal{H}) = \mu_0 (1+\chi_m )\mathcal{H} = \mu_0 \mu \mathcal{H}$,

with the relative permeability $\mu$, and vacuum permeability $\mu_0$.² From the perspective of optics around visible range, it is customary to state here that materials do not show magnetic response at optical frequencies, so $\mu = 1$ and $\mathcal{B} = \mu_0 \mathcal{H}$. However, we want to have more general theory and include also terahertz frequencies, where we might observe magnetic, $\mu$, resonances. Worth noting is the position of the $\mu_0$ constant in front of the bracket and accordingly that $\mathcal{H}$ and $\mathcal{M}$ have the same unit – as a volume density of magnetic dipole moments.

General material relations and transition to HLU

We can imagine complex material with magneto-electric and electro-magnetic coupling, adding terms with dimensionless proportionality constants $\alpha$ and $\alpha’$:

$\begin{align}
\begin{aligned}
\mathcal{D} &=\varepsilon\varepsilon_0\mathcal{E} + \alpha\sqrt{\varepsilon_0\mu_0}\mathcal{H},
\\
\mathcal{B} &=\mu\mu_0\mathcal{H} + \alpha’\sqrt{\varepsilon_0\mu_0}\mathcal{E}.
\end{aligned}
\end{align}$

Here we eventually see all the pesky constants which are needed to keep the SI units right. But we will deal with them now – the shape of the relations can give us the idea how we should redefine the fields. Let’s divide the first equation by $\sqrt{\varepsilon_0}$ and the second by $\sqrt{\mu_0}$. Then define the new fields $\mathbf{D}$, $\mathbf{E}$, $\mathbf{B}$ and $\mathbf{H}$, now measured in the some new units.

$\begin{align}
\begin{aligned}
\mathbf{D} &= \frac{\mathcal{D}}{\sqrt{\varepsilon_0}},
&\mathbf{E} &= \sqrt{\varepsilon_0}\mathcal{E},
\\
\mathbf{B} &= \frac{\mathcal{B}}{\sqrt{\mu_0}},
&\mathbf{H} &= \sqrt{\mu_0}\mathcal{H}.
\end{aligned}
\label{eqFieldsConv}
\end{align}$

With the new fields the constitutive relations clear up and we already see that all four fields are now measured in the same units:

$\begin{align}
\begin{aligned}
\mathbf{D} &=\varepsilon\mathbf{E} + \alpha\mathbf{H},
\\
\mathbf{B} &=\mu\mathbf{H} + \alpha’\mathbf{E}.
\end{aligned}
\end{align}$

Transforming Maxwell’s equations to HLU

Let’s plug the $\mathbf{D}$, $\mathbf{E}$, $\mathbf{B}$ and $\mathbf{H}$ fields \eqref{eqFieldsConv} to the SI Maxwell’s equations \eqref{eqMaxSI} and see what happens. Actually, what happens is that again we have all the equations littered with the square roots:

$\begin{align}
\begin{aligned}
\sqrt{\varepsilon_0}\Div\mathbf{D} &=\rho,
&\frac{1}{\sqrt{\mu_0}}\Rot\mathbf{H} &=\sqrt{\varepsilon_0}\parder{\mathbf{D}}{t}+\mathbf{j},
\\
\frac{1}{\sqrt{\varepsilon_0}}\Rot\mathbf{E} &=-\sqrt{\mu_0}\parder{\mathbf{B}}{t},
&\sqrt{\mu_0}\Div\mathbf{B} &=0.
\end{aligned}
\end{align}$

From the first equation already, we can get a hint that we might also measure the charge in some new units and accordingly the charge and current densities and also the polarization. When we are at it, let’s redefine the magnetization field as well.

$\begin{align}
q’ = \dfrac{q}{\sqrt{\varepsilon_0}},\quad
\rho’ = \dfrac{\rho}{\sqrt{\varepsilon_0}},\quad
\mathbf{j’} = \dfrac{\mathbf{j}}{\sqrt{\varepsilon_0}},\quad
\mathbf{P} = \dfrac{\mathcal{P}}{\sqrt{\varepsilon_0}}, \quad
\mathbf{M} = \sqrt{\mu_0}\mathcal{M}.
\end{align}$

Using relation $ 1/c = \sqrt{\varepsilon_0\mu_0} $ we get the Maxwells back in shape:

$
\begin{align}
\begin{aligned}
\Div\mathbf{D} &=\rho’,
&\Rot\mathbf{H} &=\frac{1}{c}\parder{\mathbf{D}}{t}+\frac{1}{c}\mathbf{j’},
\\
\Rot\mathbf{E} &=-\frac{1}{c}\parder{\mathbf{B}}{t},
&\Div\mathbf{B} &=0.
\end{aligned}
\end{align}$

This set of equations is decorated by $c$ to balance the spatial and temporal derivatives of the HLU fields, because those all have same unit. In the following we will see what is actually the Heaviside-Lorentz unit for field.

Understanding the units

Let us have a closer look on the dimensions of the SI fields. From the Gauss’s law – the very first equation of \eqref{eqMaxSI} – we can extract that unit of $\mathcal{D}$ is coulomb per square meter, $[\mathcal{D}] = \text{C}/\text{m}^{2}$, which is the area charge density or equivalently the volume density of dipole moment, $\text{C}\cdot\text{m}/\text{m}^{3}$. From the second equation we get that $[\mathcal{H}] = \text{C/m}\cdot\text{s}$, that is the same as area charge density multiplied by velocity, or volume density of magnetic moments, $\text{A}\cdot\text{m}^{2}/\text{m}^{3}$.

The units of $\mathcal{E}$ and $\mathcal{B}$ we simply deduce from the Lorentz force, $[\mathcal{E}] = \text{N/C}$, but also volt per meter, and $[\mathcal{B}] = \text{N/C}\cdot\text{m}\cdot\text{s}^{-1}$, that is force per coulomb and velocity unit, which is defined as tesla. It can be easily checked that this works together in the Maxwell’s equations – as it must.

This seems to be fundamental as we reduced the units of the fields to four intuitively “irreducible” quantities: mass, charge, length and time.³ It also makes sense that the bare “force” fields $\mathcal{E}$ and $\mathcal{B}$ describe the effect on charged mass, while the induced fields $\mathcal{D}$ and $\mathcal{H}$ are written in the form of configuration of charges and currents. I should remind you that for historical reasons the nomenclature as well as the form of the constitutive relations is bit misleading.

In the Heaviside-Lorentz form, we get the SI unit of all the fields $\mathbf{E}$, $\mathbf{D}$, $\mathbf{B}$, $\mathbf{H}$, $\mathbf{P}$, $\mathbf{M}$ in the same dimension, $\sqrt{\text{N}}/\text{m}$. On top of that, the HLU charge $q’$ is measured in $\sqrt{\text{N}}\cdot\text{m}$. This is easily shown after expressing the units of $\varepsilon_{0}$ and $\mu_{0}$.

Surprisingly, the coulomb completely disappeared from the HLU measurement system. It seems as the intuitive understanding is different here. We suddenly deal with some abstract force fields – abstract in a way that they carry only square root of force, so we need to combine them with the charge to get back to physically meaningful quantity in newton units. For example, we can write the scalar Coulomb’s law:

$F=\dfrac{q’_{1}q’_{2}}{4\pi r^2}$,

where we cancel the meters squared and the $4\pi$ constant has geometrical meaning, as the spherically symmetric radial field from a point charge is spreading over surface area of the sphere, that is $A = 4\pi r^2$. The Lorentz force in HLU has following form

$\mathbf{F} = q’\mathbf{E} + q’\left( \dfrac{\mathbf{v}}{c} \times \mathbf{B}\right) $,

where we see the hint of measuring velocities in units of $c$, but we will not go that way, and keep the speed of light as a constant measured in meters per second.

In summary, we did not really introduced new units here. We redefined the electromagnetic fields and electric charge, in terms of different SI units. As a result we have clear formulas and practically, we will be dealing with numbers free from high positive or negative decimal exponents. The outcomes of theory or experiment can be always presented in terms of the dimensionless quantities $\varepsilon$, $\mu$, $\alpha$ and $\alpha’$.

¹
Permittivity of vacuum, $\varepsilon_0$, sounds like some “material property” of free space, but is in fact system-of-unit constant. It is needed in places where we are connecting electromagnetic units derived from charge in coulombs, with mechanical units of kilogram, meter, second and particularly newton. Unit of $\varepsilon_0$ is given as $\text{F}/\text{m}$, where farad as an unit of capacitance is coulomb per volt, $\text{C}/\text{V}$, volt is an electric potential unit, i.e. joule per coulomb, $\text{J}/\text{C}$, and joule we can write as $\text{N}\cdot\text{m}$. Together:

$[\varepsilon_0] = \dfrac{\text{F}}{\text{m}} = \dfrac{\text{C}^2}{\text{N}\text{m}^2}$,

which also matches the units in Coulomb’s force law. We will also handle $\sqrt{\varepsilon_0}$, which has the unit of $\text{C}/\sqrt{\text{N}}\cdot\text{m}$. Here we might be worried about the square root of newton, but as we will see, this comes out directly for the other HL quantities as well. One might even speculate that interaction-force unit of newton might be actually square of something more basic, in the same sense as we understand that “area” is measured in squared length.

²
As for the permittivity, the permeability of vacuum, $\mu_0$, is a constant needed by the system of units, not a meaningful material property. Since recent redefinition of SI in 2019, it has the approximate value of $4\pi\cdot10^{-7}$ $\text{H}/\text{m}$ and has to be determined by experiment (before 2019 it was a defined exact value). The unit henry can be understood from the Faraday’s law of induction, or coil self-inductance, as voltage induced by change of current, that is $\text{H}=\text{V}\cdot\text{s}/\text{A}$, with ampere being coulomb per second, and, as above, expressing volt as newton-meter per coulomb, we get

$[\mu_0] = \dfrac{\text{H}}{\text{m}} = \dfrac{\text{N}\text{s}^2}{\text{C}^2}$.

And again we want also the square-root, $\sqrt{\mu_0}$, for which we get the unit of $\sqrt{\text{N}}\cdot\text{s}/\text{C}$. From here we can also see that $\sqrt{\varepsilon_0\mu_0}$ has the unit of inverse velocity, since $ 1/c = \sqrt{\varepsilon_0\mu_0} $.

³
Note that coulomb is not a base SI unit, but is very close, since coulomb is ampere times second, $\text{C} = \text{A}\cdot\text{s}$. As well, newton is not a base SI unit, but is constructed from kilogram, meter, and second, $\text{N} = \text{kg}\cdot\text{m}\cdot\text{s}^{-2}$.

Elusive Minus

World behind the mirror.