SI Units and Gauss Units

It happened many times that I cannot figure out which unit system I’m using in my derivation. Should I put a $c$ hare, or should I include a $4\pi$ there? It seems that I need to make a conclusion for these two unit systems, this article is based on the appendix of Jackson’s Classical Electrodynamics.

Units and Dimensions

Physics is an experiment science. What we can read directly from a measure tool for a physical quantity is just a number, which is the multiple of a standard quantity. This standard quantity can be a defined basic unit or a derived unit defined in both magnitude and dimension through theory and experiment in terms of the basic units.

In fact, we have some freedom to choose which one to be basic unit or derived unit, that’s the origin of various unit system.

Both SI units and Gauss units system choose mass, length and time to be basic quantities. But SI system uses the meter, kilogram, and second as basic units, while Gauss system uses centimeter, gram, and second.

In classical mechanics, there isn’t a much difference between MKS and CGS systems, except for a power of 10. When it comes to electrodynamics, we will encounter some new physical quantities, and need to define new basic units(SI) or choose the coefficients of some physics laws(Gauss). Thus, quite a difference between SI units and Gauss units emerged.

SI Units and Gauss Units

The most important quantities we will encounter in electromagnetic include charge $q$ , electric field $\mathbf{E}$ , current $I$ , and magnetic field $\mathbf{B}$ . We should find out the relationships between them in order to define their units, that is, The three basic phenomenons in electromagnetic field, the Coulomb’s law, Ampere’s force law, and Faraday’s law, will help us to confirm these relationships.

The Coulomb’s law states the electrostatics force induced on a particle of charge $q'$ by another charge $q$

(1) $\begin{equation*} F_1=k_1\frac{qq'}{r^2} \end{equation*}$

The electric field has the same definition in both SI and Gauss units system

(2) $\begin{equation*} E=\frac{F}{q'} \end{equation*}$

According to Ampere’s force law, the force per unit length between two infinitely long, parallel wires separated bt a distance $d$ and carrying currents $I$ and $I'$ is

(3) $\begin{equation*} \frac{dF_2}{dl}=2k_2\frac{II'}{d} \end{equation*}$

And the magnetic field at the wire $I'$ is defined as

(4) $\begin{equation*} B=\alpha \frac{\frac{dF_2}{dl}}{I'} \end{equation*}$

where $\alpha$ is different in SI and Gauss systems.

Look at Eq.1, we can ensure the constant $k_1$ if we defined the unit of $q$ , then we can find out the $k_1$ by experiment. However, we always define the current as

(5) $\begin{equation*} I=\frac{dq}{dt} \end{equation*}$

thus, we can fix $k_2$ in Eq.3 and $k_1$ at the same time as long as we defined the unit of current, this is the way we taking in SI units, where $1A$ “is that constant current which, if maintained in two straight parallel conductors of infinite length, of negligible circular cross section, and placed one meter apart in vacuum, would produce between these conductors a force equal to $2\cdot10^{-7} N/m$ “. Then $k_2=10^{-7} N/A^2$ , and the experiment results indicate $k_1=8.99\times 10^9 N\cdot m^2/c^2$ . Usually, we denote $k_1=\frac{1}{4\pi\epsilon_0}, k_2=\frac{\mu_0}{4\pi}$ . In fact, we can find out that $\frac{k_1}{k_2}=\frac{1}{\epsilon_0\mu_0}=c^2$ .

In Gauss units system, we choose $k_1=1$ , thus the unit of charge is a derived unit from Eq.1. In turn, we can get the unit of current from Eq.5, and ensure $k_2=\frac{1}{4\pi c^2}$ in Eq.3.

The unit of $E$ is also a derived unit in both of the units system from Eq.2. The only thing left is the unit of $B$ right now. In fact, we choose $\alpha=1$ in Eq. 4 in SI unit system, and $\alpha=\frac{1}{c}$ in Gauss unit system. Thus, we have the derived unit for $B$ though Eq.4.