# Dimensional Analysis in Electromagnetism (SI vs Gaussian-cgs)

Looking at Konopinski's formula for conjugate momentum (in the comment after equation 3 of "What the Vector Potential Describes"):

$$\mathbf{p}= M \mathbf{v} + q\mathbf{A} /c$$

it is plain enough that $$M \mathbf{v}$$ is the momentum, but if we naively take the usual notion of the magnetic vector potential $$\mathbf{A}$$to have dimensions of magnetic flux/length or, equivalently, momentum/charge (weber/meter = (kg $$\cdot$$ m/s)/coulomb) the following dimensional expression obtains for the right hand side of the $$+$$:

charge $$\cdot$$ ((mass $$\cdot$$ velocity)/charge)/velocity

So in a naive dimensional analysis this simplifies to mass; but this is not commensurable with momentum (mass $$\cdot$$ velocity) on the left side of the $$+$$.

Likewise, in that same comment, Konopinski describes the "interaction energy" as:

$$q[\phi-\mathbf{v}\cdot\mathbf{A}/c]$$

a naive dimensional analysis would notice a similar discrepancy -- this time in that the electric scalar potential, phi is energy/charge whereas the dimensional expression of the right hand side of the $$-$$ is:

velocity $$\cdot$$ (momentum/charge)/velocity

which, again, naively, simplifies to momentum/charge rather than the required energy/charge.

Obviously, naive dimensional analysis doesn't work here.

• – Qmechanic Sep 6 '14 at 23:39

You seem to be doing dimensional analysis in SI units. The paper seems to using Gaussian units. The magnetic field differs between these units by a factor $c$. In SI units we have $$\mathbf F = q(\mathbf E + \mathbf v \times \mathbf B \tag{SI})$$ but in Gaussian units $$\mathbf F = q(\mathbf E + \frac{\mathbf{v}}{c} \times \mathbf B \tag{G}).$$ The latter definition makes electric and magnetic fields have the same unit and is the form used in the paper you linked.