This quantum world/Feynman route/Lorentz force law

Lorentz force law

To incorporate effects on the motion of a particle (regardless of their causes), we must modify the action differential $dS=-m{c}^{2}dt\sqrt{1-v^2/c^2}$ that a free particle associates with a path segment $d𝒞.$ In doing so we must take care that the modified $dS$ (i) remains homogeneous in the differentials and (ii) remains a 4-scalar. The most straightforward way to do this is to add a term that is not just homogeneous but linear in the coordinate differentials:

${(}^{*})dS=-m{c}^{2}dt\sqrt{1-v^2/c^2}-qV(t,𝐫)dt+(q/c)𝐀(t,𝐫)\cdot d𝐫.$

Believe it or not, all classical electromagnetic effects (as against their causes) are accounted for by this expression. $V(t,𝐫)$ is a scalar field (that is, a function of time and space coordinates that is invariant under rotations of the space coordinates), $𝐀(t,𝐫)$ is a 3-vector field, and $(V,𝐀)$ is a 4-vector field. We call $V$ and $𝐀$ the scalar potential and the vector potential, respectively. The particle-specific constant $q$ is the electric charge, which determines how strongly a particle of a given species is affected by influences of the electromagnetic kind.

If a point mass is not free, the expressions at the end of the previous section give its kinetic energy $E_k$ and its kinetic momentum ${𝐩}_k.$ Casting (*) into the form

$dS=-(E_k+qV)dt+[{𝐩}_k+(q/c)𝐀]\cdot d𝐫$

and plugging it into the definitions

${(}^{*}{}^{*})E=-\frac{\partial dS}{\partial dt},𝐩=\frac{\partial dS}{\partial d𝐫},$

we obtain

$E=E_k+qV,𝐩={𝐩}_k+(q/c)𝐀.$

$qV$ and $(q/c)𝐀$ are the particle's potential energy and potential momentum, respectively.

Now we plug (**) into the geodesic equation

$\frac{\partial dS}{\partial 𝐫}=d\frac{\partial dS}{\partial d𝐫}.$

For the right-hand side we obtain

$d{𝐩}_k+\frac{q}{c}d𝐀=d{𝐩}_k+\frac{q}{c}[dt\frac{\partial 𝐀}{\partial t}+(d𝐫\cdot \frac{\partial }{\partial 𝐫})𝐀],$

while the left-hand side works out at

$-q\frac{\partial V}{\partial 𝐫}dt+\frac{q}{c}\frac{\partial (𝐀\cdot d𝐫)}{\partial 𝐫}=-q\frac{\partial V}{\partial 𝐫}dt+\frac{q}{c}[(d𝐫\cdot \frac{\partial }{\partial 𝐫})𝐀+d𝐫\times (\frac{\partial }{\partial 𝐫}\times 𝐀)].$

Two terms cancel out, and the final result is

$d{𝐩}_k=q\underset{\equiv 𝐄}{\underbrace{(-\frac{\partial V}{\partial 𝐫}-\frac{1}{c}\frac{\partial 𝐀}{\partial t})}}dt+d𝐫\times \frac{q}{c}\underset{\equiv 𝐁}{\underbrace{(\frac{\partial }{\partial 𝐫}\times 𝐀)}}=q𝐄dt+d𝐫\times \frac{q}{c}𝐁.$

As a classical object travels along the segment $d𝒢$ of a geodesic, its kinetic momentum changes by the sum of two terms, one linear in the temporal component $dt$ of $d𝒢$ and one linear in the spatial component $d𝐫.$ How much $dt$ contributes to the change of ${𝐩}_k$ depends on the electric field $𝐄,$ and how much $d𝐫$ contributes depends on the magnetic field $𝐁.$ The last equation is usually written in the form

$\frac{d{𝐩}_k}{dt}=q𝐄+\frac{q}{c}𝐯\times 𝐁,$

called the Lorentz force law, and accompanied by the following story: there is a physical entity known as the electromagnetic field, which is present everywhere, and which exerts on a charge $q$ an electric force $q𝐄$ and a magnetic force $(q/c)𝐯\times 𝐁.$

(Note: This form of the Lorentz force law holds in the Gaussian system of units. In the MKSA system of units the $c$ is missing.)

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