# 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.

• Sep 6, 2014 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.

• Thanks. In order to make this question findable by others with the same problem, I should probably change the title of the question. For instance "Dimensional Analysis and Electromagnetism" since Gaussian units are so often used in the history of papers on electromagnetism. Its truly horrifying that different "units" systems differ not only in their units but in their underlying dimensions. Enough damage is done simply by people failing to get their units consistent and commensurable in arithmetic expressions. Inconsistent dimensionality between different systems multiplies confusion. Sep 7, 2014 at 0:21

Here's the conversion. I will write Gaussian quantities as $$\hat{φ}, 𝐚, 𝐛, 𝐞, 𝐝, 𝐡, 𝐣, \hat{ρ}, q, \hat{ε}, \hat{μ}$$ and the corresponding SI quantities as $$φ, 𝐀, 𝐁, 𝐄, 𝐃, 𝐇, 𝐉, ρ, e, ε, μ.$$

Kinematic and mechanical quantities $$𝐯$$ (velocity), $$𝐩$$ (momentum), $$m$$ (rest mass), $$M$$ (moving mass), $$E$$ (energy), $$𝐅$$ (force), $$P$$ (power) do not differ between the two. I will also use $$𝔏$$ to denote the Maxwell-Lorentz Lagrangian density, which is also the same for the two.

Corresponding to the equations in Gaussian form: $$𝐛 = ∇×𝐚, \hspace 1 em 𝐞 = -∇\hat{φ} - \frac{1}{c} \frac{∂𝐚}{∂t}, \hspace 1em ∇·𝐛 = 0, \hspace 1em ∇×𝐞 + \frac{1}{c}\frac{∂𝐛}{∂t} = 𝟎, \\ ∇·𝐝 = 4π\hat{ρ}, \hspace 1em ∇×𝐡 - \frac{1}{c}\frac{∂𝐝}{∂t} = \frac{4π𝐣}{c}, \hspace 1em ∇·𝐣 + \frac{∂\hat{ρ}}{∂t} = 0, \\ 𝔏 = \frac{|𝐞|^2 - |𝐛|^2}{8π}, \hspace 1em 𝐝 = \hat{ε}𝐞, \hspace 1em 𝐛 = \hat{μ}𝐡, \\ 𝐅 = q\left(𝐞 + \frac{𝐯×𝐛}{c}\right), \hspace 1em P = q𝐯·𝐞$$ are the equations in SI form: $$𝐁 = ∇×𝐀, \hspace 1 em 𝐄 = -∇φ - \frac{∂𝐀}{∂t}, \hspace 1em ∇·𝐁 = 0, \hspace 1em ∇×𝐄 + \frac{∂𝐁}{∂t} = 𝟎, \\ ∇·𝐃 = ρ, \hspace 1em ∇×𝐇 - \frac{∂𝐃}{∂t} = 𝐉, \hspace 1em ∇·𝐉 + \frac{∂ρ}{∂t} = 0, \\ 𝔏 = \frac{ε_0|𝐄|^2}{2} - \frac{|𝐁|^2}{2μ_0}, \hspace 1em 𝐃 = ε𝐄, \hspace 1em 𝐁 = μ𝐇, \\ 𝐅 = e\left(𝐄 + 𝐯×𝐁\right), \hspace 1em P = e𝐯·𝐄,$$ with $$c = 1/\sqrt{ε_0μ_0}$$.

The conversions are: $$(\hat{φ}, 𝐚, 𝐛, 𝐞) = \sqrt{4πε_0} (φ, 𝐀c, 𝐁c, 𝐄), \\ (𝐝, 𝐡, 𝐣, \hat{ρ}, q) = \frac{1}{\sqrt{4πε_0}}\left(4π𝐃, \frac{4π𝐇}{c}, 𝐉, ρ, e\right), \\ (\hat{ε}, \hat{μ}) = \left(\frac{ε}{ε_0}, \frac{μ}{μ_0}\right).$$ In the reverse direction, they are $$(φ, 𝐀, 𝐁, 𝐄) = \frac{1}{\sqrt{4πε_0}}\left(\hat{φ}, \frac{𝐚}{c}, \frac{𝐛}{c}, 𝐞\right), \\ (𝐃, 𝐇, 𝐉, ρ, e) = \sqrt{4πε_0}\left(\frac{𝐝}{4π}, \frac{c𝐡}{4π}, 𝐣, \hat{ρ}, q\right), \\ (ε, μ) = (ε_0 \hat{ε}, μ_0 \hat{μ}).$$ There is no direct mention of $$ε_0$$ or $$μ_0$$ in the Gaussian version, only of $$c$$; while in the SI version, there is no direct mention of $$c$$, only of $$ε_0$$ and $$μ_0$$ and it's confined to the (vacuum version of the) constitutive relations and Lagrangian density, nowhere else. All the other equations are $$c$$-independent and (in fact) non-metrical. They live at a deeper level of geometry where there is no concept of a metric, parallelism, orthogonality, congruence, speed, or any distinction between space-like and time-like.

If the dimensions are denoted $$Q$$, $$P$$, $$M$$, $$L$$ and $$T$$, respectively for electric charge/flux, magnetic charge/flux, mass, length and time duration, then the dimensions for the various quantities in SI will be $$[\left(φ, 𝐀, 𝐁, 𝐄, 𝐃, 𝐇, 𝐉, ρ, e, ε, μ\right)] = \left(\frac{P}{T}, \frac{P}{L}, \frac{P}{L^2}, \frac{P}{LT}, \frac{Q}{L^2}, \frac{Q}{LT}, \frac{Q}{L^2T}, \frac{Q}{L^3}, Q, \frac{QT}{PL}, \frac{PT}{QL}\right).$$ The dimensions of $$P$$ and $$Q$$ are conjugate and multiply out to those of action: $$PQ = \frac{ML^2}{T}.$$

From this, you can determine what the dimensions for the Gaussian quantities are: $$[\left(\hat{φ}, 𝐚, 𝐛, 𝐞, 𝐝, 𝐡, 𝐣, \hat{ρ}, q, \hat{ε}, \hat{μ}\right)] = \left(\frac{G}{L}, \frac{G}{L}, \frac{G}{L^2}, \frac{G}{L^2}, \frac{G}{L^2}, \frac{G}{L^2}, \frac{G}{L^2T}, \frac{G}{L^3}, G, 1, 1\right), \hspace 1em G = \sqrt{\frac{ML^3}{T^2}}.$$ They are, needless to say, awkward - square roots of dimensions involving $$M$$, $$L$$ and $$T$$. But at least they got rid of the $$P$$'s and $$Q$$'s. You might say that they paid no mind to the $$P$$'s and $$Q$$'s.

Finally, the kinematic and mechanical dimensions are independent of the system: $$[(𝐅, P, E, 𝐩, m, M, c, 𝐯, 𝔏)] = \left(\frac{ML}{T^2}, \frac{ML^2}{T^3}, \frac{ML^2}{T^2}, \frac{ML}{T}, M, M, \frac{L}{T}, \frac{L}{T}, \frac{M}{LT^2}\right).$$