Definition of units

In some old scientific papers and books, the physics units are usually written in Gaussian units. Which cause lots of troubles for the young researchers since the new publications all written in International System of Units (SI) units. Thus known the conversion method between the two units system is very import. The conversion starts from the definition of units. Actually, the units system is like kind of value normalization, where the value basis is the unit standard that we choose. Here we choose the definition of value of length as an example. As the ancient Chinese philosopher Laozi said in the "book of Taoism”: “Long and short become obvious only when we compare them” (长短相形《老子，道德经》500 B.C.E). In the definition of value for length $L$, if we choose $L_0$ as unit basis, the value of length can be measured by comparing with this basis as: $\bar{L} = \frac{L}{L_0}$. Different units system means the choice of different unit bases. In SI units, the length unit is $L_{S0} = 1$ m, while in Gaussian unit the length unit is $L_{G0} = 1$ cm. For a certain length $L$, it can be expressed by the two unit systems as:

\begin{equation} L = \bar{L}_S L_{S0} = \bar{L}_G L_{G0} \Leftrightarrow \frac{\bar{L}_S}{\bar{L}_G} = \frac{L_{G0}}{L_{S0}} \end{equation}

Thus, we can discover that the ratio of length values is inversely proportional to the ratio of its unit basis. This rule apply to other physical values as well. Thus, to make conversion between different unit systems, we only need to know the ratio between its unity bases.

Conversion table of SI and Gaussian units

In particular, for the SI and Gaussian (CGS) unit, the ratio between their unity bases is summarized in the table below. Where we take reference from the textbook of prof. ZHAO Kaihua [赵凯华, 陈熙谋. 新概念物理教程: 电磁学 [M]. 高等教育出版社, 2003.].

Notice that not all the physical values are listed in this table, other values should be able to be derived from the values in this table. And the $c$ in this table stand for pure number $c = c_G = 3\times 10^{10}$ of light velocity in Gaussian units. This is different from the value of speed of light in SI units as: $c_{SI} = 3\times 10^{8}$. It should also be careful that the symbol of length and inductance are all $L$, but only length is choose as elementary unit.

The SI units system choose 7 units (length [$L$] = m, mass [$M$] = kg, time [$T$] = s, current [$I$] = A, Amount of substance [$N$] = mol, thermodynamic temperature [$\Theta$] = K, Luminous intensity: [$J$] = cd) as elementary units to derive all the other physical units [https://www.nist.gov/pml/weights-and-measures/metric-si/si-units]. However, CGS Gaussian units system only use mechanic unit ($L$, $M$, $T$) to derive the electro-magnetic units as well as all the other physical values [赵凯华, 陈熙谋. 新概念物理教程: 电磁学 [M]. 高等教育出版社, 2003.].

Examples of SI and Gaussian units conversion

Here we use several specific examples to show the general procedure to convert physical formula under SI units to Gaussian units.

eg: Faraday’s Law:

• Wrong derivation:

Faraday’s equation in SI units:

$\nabla\times\vec{B}_{SI} = \mu_0\vec{J}_{SI} + \mu_0\epsilon_0\frac{\partial\vec{E}_{SI}}{\partial t}$

Some websites give derivation in this manner [http://www.pgccphy.net/ref/gaussian-conv.pdf] using the conversion relation:
$\frac{B_G}{B_{SI}} = \sqrt{\frac{4\pi}{\mu_0}}$,
$\frac{J_G}{J_{SI}} = \frac{1}{\sqrt{4\pi\epsilon_0}}$,
$\frac{E_G}{E_{SI}} = \sqrt{4\pi\epsilon_0}$
from wikipedia [https://en.wikipedia.org/wiki/Gaussian_units]. And the Faraday’s law in Gaussian becomes:

$\nabla\times\vec{B}_G = \frac{4\pi}{c}\vec{J}_G + \frac{1}{c}\frac{\partial\vec{E}_G}{\partial t}$

This is correct form of Faraday’s law in Gaussian units system. However, there are some hidden errors in the above derivation process. Because the length unit hidden in the nabla operator $\nabla = \frac{\partial}{\partial\vec{r}}$ is NOT converted.

• Correct derivation:

Here we will show the strict derivation of units conversion for the Faraday’s law using the conversion relation from the book of prof. ZHAO Kaihua.

Using the relations: $\frac{r_{SI}}{r_G} = 10^{-2}$, $\frac{B_{SI}}{B_G} = 10^{-4}$, $\frac{J_{SI}}{J_G} = \frac{10^5}{c_G}$ and $\frac{E_{SI}}{E_G} = (\frac{10^6}{c_G})^{-1} = \frac{c_G}{10^6}$. We convert Faraday’s law for Gaussian unit like:

\begin{equation} \begin{array}{ll} & \frac{\partial}{\partial(10^{-2}\vec{r}_G)}\times(10^{-4}\vec{B}_G) = \mu_0 \frac{10^5}{c_G}\vec{J}_G + \mu_0\epsilon_0\frac{\partial}{\partial t}(\frac{c_G}{10^6}\vec{E}_G) \\ \Rightarrow & 10^{-2}\nabla_G\times\vec{B}_G = \frac{\mu_0}{4\pi}4\pi\frac{10^5}{c_G}\vec{J}_G + \frac{1}{c_{SI}^2}\frac{c_G}{10^6} \frac{\partial\vec{E}_G}{\partial t} \\ \Rightarrow & \nabla_G\times\vec{B}_G = \frac{4\pi}{c_G}\vec{J}_G + \frac{1}{c_G}\frac{\partial\vec{E}_G}{\partial t} \\ \end{array} \end{equation}

Among the derivation, these relations are used: $\frac{\mu_0}{4\pi} = 10^{-7}$, $\mu_0\epsilon_0 = \frac{1}{c_{SI}^2}$, $\frac{1}{c_{SI}^2}c_G = \frac{c_G^2}{c_{SI}^2}\frac{1}{c_G} = \frac{10^4}{c_G}$.

eg: Columb force

Under SI units, the law of column force is given as:

\begin{equation} F_{SI} = \frac{1}{4\pi\epsilon_0}\frac{q_{1_{SI}} q_{2_{SI}}}{r_{SI}^2} \end{equation}

Then we use the relation that:$\frac{F_{SI}}{F_G} = 10^{-5}$, $\frac{q_{SI}}{q_G} = (\frac{c}{10})^{-1}$, $\frac{r_{SI}}{r_G} = 10^{-2}$ to get:

\begin{equation} \begin{array}{ll} & 10^{-5}F_G = \frac{1}{4\pi\epsilon_0}(\frac{c}{10})^{-2}\frac{q_{1_G}q_{2_G}}{10^{-4}r_G^2} \\ \Leftrightarrow & F_G = \frac{1}{4\pi\epsilon_0}\frac{10^{11}}{c^2}\frac{q_{1_G}q_{2_G}}{r_G^2} \\ \Leftrightarrow & F_G = \frac{q_{1_G}q_{2_G}}{r_G^2} \end{array} \end{equation}

Among the conversion:

\begin{equation} \frac{1}{4\pi\epsilon_0}\frac{10^{11}}{c^2} = \frac{1}{4\pi\mu_0\epsilon_0}\mu_0\frac{10^{11}}{c_G^2} = \frac{\mu_0}{4\pi}c_{SI}^2\frac{10^{11}}{c_G^2} = \frac{\mu_0}{4\pi}(3\times 10^{8})^2\frac{10^{11}}{(3\times 10^{10})^2} = \frac{\mu_0}{4\pi}10^7 = 1 \end{equation}

Notice that we used the constant value: $(\frac{\mu_0}{4\pi})_{SI} = 10^{-7}$.

eg: Reduce MHD equation

The reduced MHD equation in SI units is [G. Y. Fu, and H. L. Berk, Effects of pressure gradient on existence of Alfvén cascade modes in reversed shear tokamak plasmas, Phys. Plasmas 13, 052502 (2006)]:

\begin{equation} \rho\frac{\partial^2}{\partial t^2}\vec{\xi} = -\nabla\delta P + \delta\vec{J}\times\vec{B} + \vec{J}\times\delta\vec{B} \end{equation}

In the paper of prof. L. Chen [Liu Chen and R. B. White, M. N. Rosenbluth, Excitation of Internal Kink Modes by Trapped Energetic Beam Ions, Physical Review Letters, Vol 52, 13, 1984.], the reduced MHD equation in Gaussian units is given as:

\begin{equation} \rho\frac{\partial^2}{\partial t^2}\vec{\xi} = -\nabla\delta P + \frac{1}{c}\delta\vec{J}\times\vec{B} + \frac{1}{c}\vec{J}\times\delta\vec{B} \end{equation}

To make this conversion, we need to use these value relations: $\frac{\rho_{SI}}{\rho_G} = \frac{m_{SI}L_{SI}^3}{m_G L_G^3} = 10^3$, $\frac{\xi_{SI}}{\xi_G} = 10^{-2}$, $\frac{t_{SI}}{t_G}= 1$, $\frac{P_{SI}}{P_G} = \frac{F_{SI}/L_{SI}^2}{F_G/L_G^2} = 10^{-1}$, $\frac{J_{SI}}{J_G} = \frac{I_{SI}/L_{SI}^2}{I_G/L_G^2} = (\frac{c}{10})^{-1}10^4 = \frac{10^5}{c}$, $\frac{B_{SI}}{B_G} = 10^{-4}$.

Then the reduced MHD equation in Gaussian units shall be:

\begin{equation} \begin{array}{ll} & \rho_{SI}\frac{\partial^2}{\partial t^2}\vec{\xi}_{SI} = -\frac{\partial}{\partial\vec{r}_{SI}}\delta P_{SI} + \delta\vec{J}_{SI}\times\vec{B}_{SI} + \vec{J}_{SI}\times\delta\vec{B}_{SI} \\ \Leftrightarrow & 10^3\rho_G\frac{\partial^2}{\partial t^2}10^{-2}\vec{\xi}_G = -\frac{\partial}{10^{-2} \partial\vec{r}_G}10^{-1} \delta P_G + \frac{10^5}{c}\delta\vec{J}_G\times 10^{-4}\vec{B}_G + \frac{10^5}{c}\vec{J}_G\times 10^{-4}\delta\vec{B}_G \\ \Leftrightarrow & \rho_G\frac{\partial^2}{\partial t^2}\vec{\xi}_G = -\frac{\partial}{\partial\vec{r}_G} \delta P_G + \frac{1}{c}\delta\vec{J}_G\times \vec{B}_G + \frac{1}{c}\vec{J}_G\times\delta\vec{B}_G \\ \Leftrightarrow & \rho_G\frac{\partial^2}{\partial t^2}\vec{\xi}_G = -\nabla\delta P_G + \frac{1}{c}\delta\vec{J}_G\times \vec{B}_G + \frac{1}{c}\vec{J}_G\times\delta\vec{B}_G \\ \end{array} \end{equation}

This is the correct form of reduced MHD equation under Gaussian units.