Convert formula from CGS to SI

Question

I'd like to convert this formula \begin{equation} l^2 =\frac{c\hbar}{eH} \end{equation} where $l$ is a length, and $H$ is in oersted, to SI units. I am pretty sure it uses CGS, since Oe is mentioned in the text, and its from a theory paper (Kawabata1980, eq.3).

Answer 1

As a plasma physicist I use the NRL Plasma Formulary to convert between CGS and SI units. It can be downloaded from here.

On page 18 it gives you a prescription on how to convert any formula. Remember to convert both sides of the equation. For your problem I get

$$ l^2 = \frac{\varepsilon_0 c^2 \hbar}{eH} $$

Step by step instruction:

1. Identify all the quantities in your equation (with $\alpha=10^2\mathrm{cm\;m}^{-1}$ and $\beta=10^7\mathrm{erg\;J}^{-1}$)
   • $l$ length, factor $\alpha$
   • $c$ velocity, factor $\alpha$
   • $\hbar$ action = energy $\times$ time, factor $\beta \times 1$
   • $e$ charge, factor $(\alpha \beta / 4 \pi \varepsilon_0)^{1/2}$
   • $H$ magnetic intensity, factor $(4 \pi \mu_0\beta/\alpha^3)^{1/2}$
2. Replace all quantities in the equation $$ \alpha^2 l^2 = \frac{\alpha c \; \beta \hbar}{(\alpha \beta / 4 \pi \varepsilon_0)^{1/2}e\;(4 \pi \mu_0\beta/\alpha^3)^{1/2}H} $$
3. Simplify $$ l^2 = \frac{c \; \hbar}{(1 / \varepsilon_0)^{1/2}e\;\mu_0^{1/2}H} = \frac{\varepsilon_0 c \; \hbar}{(\varepsilon_0 \mu_0)^{1/2}eH} $$
4. Use $c = 1/\sqrt{\varepsilon_0 \mu_0}$ $$ l^2 = \frac{\varepsilon_0 c^2 \hbar}{eH} = \frac{\hbar}{e\mu_0 H}$$

Answer 2

It's not completely clear to me what's going on there, but the authors provide a useful foothold by stating (under equation 4) that the cyclotron frequency of the problem is $$ \omega=\frac{eH}{mc}, $$ and this needs to correspond directly with the SI expression, $$ \omega_\mathrm{SI}=\frac{eB}{m}. $$ From here you can get the correspondence $$ \frac{eH}{c} \leftrightarrow (eB)_\mathrm{SI}. $$ Taking this and running with it, you get the SI version $$ l^2=\frac{\hbar}{eB}. $$ This checks out dimensionally: $eB$ is a force per unit velocity, with dimensions $[F/v]=[M\,T^{-1}]$, which matches $[\hbar/l^2]$.

Moreover, this coincides with Holger's black magic (which is probably the way to go), which reduces to $l^2=\hbar/e\mu_0H$, and this coincides with $l^2=\hbar/eB$ in vacuum.