# Convert formula from CGS to SI

I'd like to convert this formula $$l^2 =\frac{c\hbar}{eH}$$ 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).

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 $c = 1/\sqrt{\varepsilon_0 \mu_0}$ $$l^2 = \frac{\varepsilon_0 c^2 \hbar}{eH} = \frac{\hbar}{e\mu_0 H}$$

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.

• Dimensional analysis probably isn't enough to manage unit conversions when electromagnetism is involved. Oct 4, 2016 at 14:00
• @EL_DON Indeed it isn't, which is why the bulk of this answer is based on the cyclotron frequency. It's still a useful consistency check, which is why I included it. Oct 4, 2016 at 14:09