Unit systems in electrodynamics


I learned electrodynamics, like most other undergraduate students of my generation, in SI units. They seemed like the natural choice, because we use SI units for everything else. But then I took PHY450, “Relativistic Electrodynamics”, where we use cgs-Gaussian units. At first I found this unit system strange and unnatural, but then I realized it does have some advantages over the SI system. For example, the E and B fields have the same units, as do the scalar and vector potentials. This is attractive because the scalar and vector potentials together constitute the four-current, A^i = (\phi, \mathbf{A}), and the electric and magnetic fields together form the electromagnetic field tensor,

\displaystyle F^{ij} = \begin{pmatrix} 0 & -E_x & -E_y & -E_z \\ E_x & 0 & -B_z & B_y \\ E_y & B_z & 0 & -B_x \\ E_z & -B_y & B_x & 0 \end{pmatrix}

Still, I find it very strange that cgs units don’t have a separate unit for electric charge or electric current. I think it is unintuitive to measure electric charge in terms of length, mass, and time, because electric charge isn’t defined in that way. We define speed to be distance over time, acceleration to be change in velocity over time, force to be mass times acceleration, pressure to be force over area, energy to be force times distance, power to be energy over time. We don’t define electric charge to be anything times anything or anything over anything; it’s as much a fundamental quantity as any other. For what it’s worth, there are some contexts in which it makes sense not to treat charge as a separate unit. For example, in quantum electrodynamics, the charge of the electron is often expressed in Planck units, where its value is the square root of the fine structure constant. But then again, we also measure distance in units of time sometimes (light years) and time in units of distance sometimes (think x^0 in relativity). I think that it makes as much sense to have a separate unit for charge or current as it does to have separate units for distance and time, or mass, momentum, and energy—most of the time it’s better, with some exceptions.

I think that in the nicest possible system of units, then, the equations of electrodynamics would look like this. k denotes the reciprocal of the electric constant \epsilon_0, and is 4\pi times greater than Coulomb’s constant. The magnetic field and magnetic vector potentials are c times their values in SI units. The metric convention is (+,-,-,-).

Maxwell’s equations:
\displaystyle \nabla \cdot \mathbf{E} = k\rho
\displaystyle \nabla \cdot \mathbf{B} = 0
\displaystyle \nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial(ct)}
\displaystyle \nabla \times \mathbf{B} = k\frac{\mathbf{J}}{c} + \frac{\partial\mathbf{E}}{\partial(ct)}

Lorentz force law:
\displaystyle \mathbf{F} = q\left(\mathbf{E} + \frac{\mathbf{v}}{c} \times \mathbf{B}\right)

Coulomb’s law:
\displaystyle \mathbf{F} = \frac{k q_1 q_2}{4\pi r^2} \hat{\mathbf{r}}
\displaystyle \mathbf{E} = \frac{k q}{4\pi r^2} \hat{\mathbf{r}} = \iiint \frac{k \rho}{4\pi r^2} \hat{\mathbf{r}} \, dV

Biot–Savart law:
\displaystyle \mathbf{B} = \int \frac{k}{4\pi r^2} \frac{I \, d\ell \times \hat{\mathbf{r}}}{c} = \iiint \frac{k}{4\pi r^2} \frac{\mathbf{J} }{c} \times \hat{\mathbf{r}} \, dV

Scalar and vector potentials
\displaystyle \mathbf{E} = -\nabla V - \frac{\partial \mathbf{A}}{\partial(ct)}
\displaystyle \mathbf{B} = \nabla \times \mathbf{A}
\displaystyle A^\mu = (V, \mathbf{A})

Field tensor
\displaystyle F^{\mu\nu} = \begin{pmatrix} 0 & -E_x & -E_y & -E_z \\ E_x & 0 & -B_z & B_y \\ E_y & B_z & 0 & -B_x \\ E_z & -B_y & B_x & 0 \end{pmatrix}

Potentials in electrostatics and magnetostatics (Coulomb gauge)
\displaystyle V = \frac{k}{4\pi r}q = \iiint \frac{k}{4\pi r} \rho \, dV
\displaystyle \mathbf{A} = \int \frac{k}{4\pi r} \frac{I \, d\ell}{c} = \iiint \frac{k}{4\pi r} \frac{\mathbf{J}}{c} \, dV

Retarded potentials, Lorenz gauge
\displaystyle A^\mu(\mathbf{r}, t) = \iiint \frac{k}{4\pi r} \frac{J^\mu(\mathbf{r}', t - \|\mathbf{r} - \mathbf{r}'\|/c)}{c} \, d^3\mathbf{r}'
\displaystyle A^\mu(x^\nu) = \iiiint \frac{k}{4\pi} \frac{J^\mu(x^{\nu\prime})}{c} \, 2\delta((x^\nu - x^{\nu\prime})^2) \, \theta(x^0 - x^{0\prime}) \, d^4 x^{\nu\prime}

Liénard–Wiechert potentials
\displaystyle V = \frac{kq}{4\pi r} \frac{1}{1-\vec{\beta} \cdot \hat{\mathbf{r}}}
\displaystyle \mathbf{A} = \frac{kq}{4\pi r} \frac{\vec{\beta}}{1 - \vec{\beta} \cdot \hat{\mathbf{r}}}
(\mathbf{r} is the vector from the retarded position of the charge to the observation point, r is the magnitude of \mathbf{r}, and \vec\beta is 1/c times the velocity of the charge at the retarded time.)

Poynting vector; Maxwell stress tensor; electromagnetic stress-energy tensor
\displaystyle T^{\alpha\beta} = -\frac{1}{k}F^{\alpha \gamma} F^\beta{}_\gamma + \frac{\eta^{\alpha\beta}}{4k} F^{\gamma\delta}F_{\gamma\delta}
\displaystyle T^{00} = \frac{1}{2k}(E^2 + B^2)
\displaystyle T^{0i} = T^{i 0} = \frac{1}{k} \epsilon_{ijk} E_j B_k = \frac{1}{c} S_i
\displaystyle T^{ij} = -\frac{1}{k}\left[E_i E_j + B_i B_j - \frac{1}{2}\delta_{ij} (E^2 + B^2)\right] = -\sigma_{ij}
\displaystyle \mathbf{S} = \frac{c}{k} \mathbf{E} \times \mathbf{B}
Poynting’s theorem and the statement of the conservation of momentum in electrodynamics take their usual form and are the same in all unit systems:
\displaystyle \nabla \cdot \mathbf{S} = -\mathbf{J} \cdot \mathbf{E} - \frac{\partial}{\partial t} \left[\frac{1}{2k}(E^2 + B^2)\right]
\displaystyle \partial_i \sigma_{ij} = f_j + \frac{1}{c^2} \frac{\partial S_j}{\partial t}

Electromagnetic Lagrangian density
\displaystyle \mathcal{L} = -\frac{1}{4k}F^{\alpha\beta}F_{\alpha\beta} - \frac{1}{c}J^\alpha A_\alpha

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About Brian

Hi! I'm Brian Bi. As of November 2014 I live in Sunnyvale, California, USA and I'm a software engineer at Google. Besides code, I also like math, physics, chemistry, and some other miscellaneous things.
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