Why does the formula for magnetic field force include the speed of light in the denominator in cgs units? Where does the extra $c$ go in SI units?


2 Answers 2


The natural way to write it in this notation is

$$F = q(E + \beta \times B)$$

where $\beta$ is the velocity measured in natural units - the velocity as a fraction of the speed of light.

In the CGS system, we instead write $\beta = \frac{v}{c}$ and the equation becomes

$$F = q(E + \frac{v}{c} \times B)$$

That's not silly enough, though, so we go really crazy and invent two new constants and give them obscure names: "permittivity of free space" ($\epsilon_0$) and "permeability of free space" ($\mu_0$) and relate them so $c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$.

If that's where it ended the equation would be

$$F = q(E + v \sqrt{\mu_0 \epsilon_0} \times B)$$

But that would still not be silly enough. Instead, we then decided to invent a new magnetic field given by $B' = \sqrt{\mu_0/4\pi} B$ and a new electric field given by $E' = \frac{E}{\sqrt{4\pi \epsilon_0}}$. This would give us the new formula

$$F = q(\sqrt{4\pi \epsilon_0}E' + v \times \sqrt{4\pi \epsilon_0} B')$$

but we need to be more sillier, so we define a new measure of charge by $q' = q\sqrt{4\pi\epsilon_0}$. At last we get

$$F = q'(E' + v \times B')$$

and there you have the SI Lorentz force law. So the answer is that the speed of light appears only because of complicated ways of changing the units.

It's a bit instructive to look at the energy densities of the fields. In the original units they're just $\frac{E^2}{8\pi}$ and $\frac{B^2}{8\pi}$. But we made a bunch of units changes, so these become $\frac{4\pi \epsilon_0 E'^2}{8\pi} = \frac{\epsilon_0 E'^2}{2}$ and $\frac{4\pi B'^2}{8 \pi\mu_0} = \frac{B'^2}{2\mu_0}$

So the cost of making the speed of light go away in the Lorentz force law is that we pick up these strange $\mu_0$ and $\epsilon_0$ constants that hide the speed of light and flop around through all the subsequent formulas.


There's a pretty sweet and complete discussion of this in this set of notes by Littlejohn. Basically, the fields $\vec{E}, \vec{B},\vec{H}$ and $\vec{D}$ have the same units in the cgs system. Simple dimensional analysis show that the usual SI equation $$ \vec F=q(\vec E+\vec v \times \vec B) \qquad \stackrel{\hbox{cgs}}{\Rightarrow} \qquad \vec F=q\left(\vec E+\frac{\vec v}{c} \times \vec B\right) \tag{1} $$ if the equation on the right (in cgs units) is to remain dimensionally consistent. In principle, the quantity $c$ with units $M T^{-1}$ that must appear in (1) for dimensional homogeneity could be any speed although it is related to the SI quantities $\epsilon_0$ and $\mu_0$ in vacuum by $$ c=k/\sqrt{\epsilon_0\mu_0}\, . $$ The choice $k=1$ follows from the experimental observation that electromagnetic waves in vacuum travel at a that speed.

The choice for the units of the fields - well in fact for the units of electric charge and magnetic poles - is discussed at length in this rather old but still relevant

R.T. Birge, On electric and magnetic units and dimensions, American Journal of Physics, May 1934 pp. 41-48.

(Birge actually has a series of articles on this topic in the early issues of Am.J.Phys.)

Birge shows that one can consistently define the electric force using $$ F=\frac{q_1q_2}{r^2} $$ which gives the units of $q$ as $M^{1/2}L^{3/2}T^{-1}$, without reference to the Ampère as an independent unit. In addition, one can also consistently define the magnetic force as $$ F=\frac{m_1m_2}{r^2} $$ where $m_1$ and $m_2$ are magnetic poles, also with units in $M^{1/2}L^{3/2}T^{-1}$.

[On an unrelated note: I'm disappointed to see that the batman isn't popular as a unit of weight.]


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.