# Electromagnetic hamiltonian factor of 1/c question

by copernicus1
Tags: 1 or c, electromagnetic, factor, hamiltonian
 P: 83 I often see the EM Hamiltonian written as $$H=\frac1{2m}\left(\vec p-\frac ec\vec A\right)^2+e\phi,$$ but this confuses me because it doesn't seem to have the right units. Shouldn't it just be $$H=\frac1{2m}\left(\vec p-e\vec A\right)^2+e\phi,$$ since the vector potential has units of momentum per unit charge? And if so, why do so many authors put in the factor of 1/c?
 PF Gold P: 1,165 Electromagnetic hamiltonian factor of 1/c question This is just a convenient choice of definition of ##\mathbf A##, introduced already in classical electromagnetic theory. In this definition, the electromagnetic force in field ##\mathbf E,\mathbf B = \nabla \times \mathbf A## is given by $$\mathbf F = q\mathbf E + q \frac{\mathbf v}{c}\times \mathbf B.$$ Then E and B have the same units and the velocity v appears always in the companionship of speed of light ##c## as the ratio ##\frac{\mathbf v}{c}##. This has its practical advantages in relativistic theory. For example, it is very convenient to use in description of the motion of a particle in an external EM wave. In this convention, E and B have usually magnitudes of the same order of magnitude. The magnitude of the magnetic force term is then easily estimated from the value of v/c. Also approximate low-velocity approximations are best formulated in terms of v/c.
 Sci Advisor Thanks P: 2,471 Unfortunately, in electromagnetism there are still (at least) three systems of units in use. The oldest are the Gaussian units, where the Lagrangian reads $$\mathcal{L}=-\frac{1}{16 \pi} F_{\mu \nu} F^{\mu \nu} - \frac{1}{c} j_{\mu} A^{\mu},$$ where $(A^{\mu})=(c \Phi,\vec{A})$ is the four-vector potential of the electromagnetic field $F_{\mu \nu} =\partial_{\mu} A_{\nu}-\partial_{\nu} A_{\mu}$, and $j^{\mu}=(c \rho,\vec{j})$ the four-dimensional current density. I've used the west-coast convention $(\eta_{\mu \nu})=\text{diag}(1,-1,-1,-1)$ and the four-vector $(x^{\mu})=(c t,\vec{x})$, keeping all factors of $c$. Because of the "irrational" factor $1/(4 \pi)$ in front of the kinetic term of the gauge field, this system of units is called irrational CGS system (CGS standing for centimeters, grams, seconds, which form the basic units in this system). Another CGS system just differs by this factor $1/(4 \pi)$. This is the rationalized Heaviside-Lorentz system of units and usually used in relativistic quantum field theory and thus theoretical high-energy physics. It has the advantage to put the factors $1/(4 \pi)$ where they belong and to reflect the physical dimensions of the quantities best. Of course, electromagnetics is relativistic and thus this system of units is the most natural one. Here, the Lagrangian reads $$\mathcal{L}=-\frac{1}{4} F_{\mu \nu} F^{\mu \nu}-\frac{1}{c} j_{\mu} A^{\mu}.$$ From this Lagrangian it follows that the force on a point charge without magnetic moment is given by $$\vec{F}=q \left (\vec{E}+\frac{\vec{v}}{c} \times \vec{B} \right ).$$ The official system of units, called SI (for Systeme International), in use in experimental physics and engineering everywhere on the world is taylormade for practictal purposes and provides welldefined accurate realizations of the units. In theoretical electromagnetics it's on the other hand a disease, if you ask me, because the beautiful Lorentz symmetry of the relativistic theory is spoiled. Of course, there is no principle problem to use it also in theory, but then you always get questions like, what are $\epsilon_0$ and $\mu_0$? The answer is they are conversion factors to transform from the SI units to the more natural Gauß or Heaviside-Lorentz units. Also the SI adds a fourth basis unit to the three mechanical units (the SI concerning the mechanics is an MKS system, using metre, kilogram, second), the Ampere for the electric current. The Lorentz force in this units reads $$\vec{F}=q \left (\vec{E}+\vec{v} \times \vec{B} \right).$$ The only physical universal constant in electromagnetics is the velocity of light, $c$, and it's related to the conversion factors of the SI by $$c=\frac{1}{\sqrt{\mu_0 \epsilon_0}}.$$ The Lagrangian reads $$\mathcal{L}=-\frac{1}{4 \mu_0} F_{\mu \nu} F^{\mu \nu} -j_{\mu} A^{\mu}.$$