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72.  Dynamics of a Relativistic Particle in an Electromagnetic Field

Jackson Chapter 12.1:  I’ve used Gaussian units in this lecture, to connect better with Jackson.

Michael Fowler UVa

The Facts

It is well-established experimentally that the equations of motion for a charged particle in an electromagnetic field are (following Jackson, so Gaussian (cgs) units here, hence the $c$ )

$$\begin{array}{l}\frac{d\overrightarrow{p}}{dt}=e\left(\overrightarrow{E}+\frac{\overrightarrow{v}}{c}\times \overrightarrow{B}\right),\\ \frac{dE}{dt}=e\overrightarrow{v}\cdot \overrightarrow{E}.\end{array}$$

These can be written in standard relativistic notation

$$\frac{d{U}^{\alpha }}{d\tau }=\frac{e}{mc}{F}^{\alpha \beta }{U}_{\beta }$$

where $\tau$ is the proper time, $dt=\gamma d\tau ,\gamma =1/\sqrt{1-v^2/c^2},$ $U^{\alpha }$ is the four-velocity $U^{\alpha }=\left(\gamma c,\gamma \overrightarrow{v}\right)=p^{\alpha }/m,$ and $F^{\alpha \beta }$ is the field tensor (for its definition see lecture 65).

Exercise:  The equivalence of these two sets of equations was given as an exercise at the end of lecture 65. If you didn’t do it then, it would be a good idea to do it now.  

The Theory: The Principle of Least Action

We first briefly review the formalism for a general dynamical system, degrees of freedom labeled by coordinates $q_i.$ The state at a given moment is specified completely by the set of positions and velocities $\left(q_i,{\dot{q}}_i\right),$ a point in configuration space.   The system’s evolution in time, determined by Lagrange’s equations, traces a path in this configuration space from a given initial state to a final state.

The action corresponding to this path is the integral               

$$S={\displaystyle \underset{t_1}{\overset{t_2}{\int }}L\left(q_i,{\dot{q}}_i\right)dt}$$

 where the usual Lagrangian $L\left(q_i,{\dot{q}}_i\right)=T-V,$ the difference of kinetic and potential energies (but we’ll find a rather different expression for a charged particle in a magnetic field, see below).

The actual physical path in configuration space followed by the evolving system has the least action of all the possible paths between the given initial and final states. Minimization of the action using the calculus of variations, i.e. varying the path a little, yields Lagrange’s equations of motion.

This is standard classical mechanics, but here we want to include relativistic motion of a particle in a field.  As we’ve discussed elsewhere, the Principle of Least Action comes ultimately from the wave nature of matter, an argument that naturally extends to the relativistic regime. We just have to figure out what relativistic action yields the dynamics we already know from experiment.

A Free Particle

We start with a free particle and no potential.  The action cannot depend on the coordinate system chosen$—$it must be Lorentz invariant. But all we have is a point particle and a path, the action being an integration over path increments. There is just one Lorentz invariant for a path increment: its interval, meaning its proper incremental time $d\tau .$ The obvious candidate for the action, then, has to be

$$S={\displaystyle \underset{t_1}{\overset{t_2}{\int }}Ldt}=-\alpha {\displaystyle \underset{a}{\overset{b}{\int }}cd\tau ,}$$

 the particle’s (Lorentz invariant) proper time, and $\alpha$ some as yet undetermined constant.  Why have we put in a minus sign?   Because we know moving clocks run slow, so if we take all possible paths beginning at the origin, then ending at the origin one minute later (as measured by a clock stationed at the origin), the trivial path of just staying at the origin takes the longest measured elapsed time (any other path needs some movement, and hence clock slowing).  The motionless path is the extremum, and, of course, it satisfies the equation of motion in zero field.

We can determine the overall constant $\alpha$ in the action by matching to the known result in the nonrelativistic limit, where the Lagrangian $L=\frac{1}{2}m{v}^2.$

Putting $dt=\gamma d\tau ,$ from the above action integral $Ldt=-\alpha cd\tau ,$ so

$$L=\alpha c\frac{d\tau }{dt}=-\alpha c\sqrt{1-\frac{v^2}{c^2}}\approx -\alpha c+\frac{\alpha v^2}{2c}.$$

The constant term is irrelevant to the minimization, matching the second term gives $\alpha =mc,$ so the action is

$$S=-m{c}^2{\displaystyle \underset{a}{\overset{b}{\int }}d\tau ,}$$

and

$$L=-m{c}^2\sqrt{1-\frac{v^2}{c^2}}.$$

Having found the Lagrangian, we can derive momentum and energy in the standard fashion:

The momentum

$$\overrightarrow{p}=\frac{\partial L}{\partial \overrightarrow{v}}=\frac{m\overrightarrow{v}}{\sqrt{1-v^2/c^2}},$$

and the energy (the Hamiltonian)

$$E=\overrightarrow{p}\cdot \overrightarrow{v}-L=\frac{m{c}^2}{\sqrt{1-v^2/c^2}}.$$

*Making Coordinates Explicit

The idea here (Jackson 12.1.B) is to escape the nonrelativistic perspective$—$here we’ll treat time as just another coordinate, like the position coordinates, and introducing a further parameter $s,$ which labels position along the particle’s path in spacetime, $x^{\mu }\left(s\right),$ and is strictly increasing. (Of course, labeling the path using proper time, as in the previous section, also treats ordinary time and space equivalently in the relativistic context, so the present section doesn’t really add much further insight, we’ve put it in to connect better with Jackson.)

Notation warning: we are following Jackson (p 584, 3^rd edition) in introducing this parameter $s.$ Unfortunately, Landau (Vol 2, Ch 2) uses $s$ for the proper time (as do some others), which we label $\tau .$

The usual nonrelativistic derivation of the equations of motion of a system by minimizing action takes an arbitrary small coordinate deviation $\delta \left(x_i\right)$ from the path and requires that the action not change to first order. We’ll do the same, but obviously to proceed it is necessary that the $x^{\mu }$ appear in the Lagrangian. With our limited options, that means $d\tau$ must be written as a function of the $d{x}^{\mu }\text{'s}\text{.}$

Now $d\tau$ is the incremental invariant interval (and to make the equations coincide with Jackson’s, here we’ll take metric (+ - - -) )

$$d\tau =\sqrt{g^{\alpha \beta }d{x}_{\alpha }d{x}_{\beta }},$$

so the action, bringing in our path label parameter $s,$

$$S=-mc{\displaystyle \underset{s_1}{\overset{s_2}{\int }}\frac{d\tau }{ds}ds}=-mc{\displaystyle \underset{s_1}{\overset{s_2}{\int }}\sqrt{g^{\alpha \beta }\frac{d{x}_{\alpha }}{ds}\frac{d{x}_{\beta }}{ds}}ds}.$$ positions $x_{\alpha }$ and “velocities” $d{x}_{\alpha }/ds.$ Following the exact variational procedure that gives Lagrange’s equations we find

$$mc\frac{d}{ds}\left[\frac{d{x}^{\alpha }/ds}{\sqrt{\frac{d{x}_{\beta }}{ds}\frac{d{x}^{\beta }}{ds}}}\right]=0.$$

At this point we note that since $s$ is a strictly increasing smooth function of $\tau ,$ the expression in the square brackets is equal to the same with $ds$ replaced by $d\tau$ everywhere:

$$\left[\frac{d{x}^{\alpha }/ds}{\sqrt{\frac{d{x}_{\beta }}{ds}\frac{d{x}^{\beta }}{ds}}}\right]=\left[\frac{d{x}^{\alpha }/d\tau }{\sqrt{\frac{d{x}_{\beta }}{d\tau }\frac{d{x}^{\beta }}{d\tau }}}\right].$$

In the second expression, the denominator is the (constant) norm of the four-velocity, $\sqrt{U_{\beta }{U}^{\beta }},$ so the variational equation becomes

$$mc\frac{d}{ds}\left[\frac{d{x}^{\alpha }/d\tau }{c}\right]=0.$$

That is, the bracketed expression is constant as a function of $s,$ therefore also as a function of $\tau ,$ and we conclude that

$$m\frac{d^2{x}^{\alpha }}{d{\tau }^2}=0.$$

Again we’ve proved that the physical path between two points in space time (with no potentials present) is one at constant velocity, that being the path of maximum proper time. Surprise.

Note:  Comparing the path in four-dimensional spacetime labeled by the parameter $s\left(x_{\alpha }\right)$ with the path of a nonrelativistic particle in three-dimensional space labeled with time $t\left(\overrightarrow{r}\right),$ it looks as if we have an extra degree of freedom. But we don’t. The path label function $s$ is arbitrary, provided only that it is increasing in proper time: all such functions give the same identical physics, there isn’t really a degree of freedom, it’s more like choosing a different gauge.

Particle in an Electromagnetic Field

Recall that for a free particle, we derived the motion from an action integral along the path, and the only Lorentz-invariant term for an increment of path was the proper time increment (dimensionally adjusted) $m{c}^{2}d\tau .$ Introducing an electromagnetic field, with four potential $A^{\mu }=\left(\phi ,\overrightarrow{A}\right),$ a possible action is

$$S={\displaystyle \underset{a}{\overset{b}{\int }}Ldt}={\displaystyle \underset{a}{\overset{b}{\int }}\left(-m{c}^{2}d\tau -e{A}_{\mu }d{x}^{\mu }\right),}$$ 

$$L=-m{c}^2\sqrt{1-v^2/c^2}+e\frac{\overrightarrow{v}}{c}\cdot \overrightarrow{A}\left(\overrightarrow{r},t\right)-e\phi \left(\overrightarrow{r},t\right).$$

This is confirmed experimentally: that is, it leads to the observed Lorentz force law, $\overrightarrow{F}=e\left(\overrightarrow{E}+\frac{\overrightarrow{v}}{c}\times \overrightarrow{B}\right),$ as we’ll now demonstrate.

Canonical and Kinetic Momenta

The canonical momentum (using upper-case $\overrightarrow{P},$ following Landau and Jackson)

$$\overrightarrow{P}=\frac{\partial L}{\partial \overrightarrow{v}}=\frac{m\overrightarrow{v}}{\sqrt{1-v^2/c^2}}+\frac{e}{c}\overrightarrow{A}\left(\overrightarrow{r},t\right).$$

We’ll reserve lower-case $\overrightarrow{p}$ for the kinetic momentum, meaning relativistic mass times velocity (the first term on the right-hand side) so the canonical momentum

$$\overrightarrow{P}=\overrightarrow{p}+\frac{e}{c}\overrightarrow{A}\left(\overrightarrow{r},t\right).$$

Writing this in terms of the spatial components of the four-velocity $U^{\mu }=\left(\gamma c,\gamma \overrightarrow{v}\right)$ and the four-potential:

$$P^i=m{U}^i+\frac{e}{c}{A}^i,i=1,2,3.$$

The energy (the Hamiltonian)

$$E=\overrightarrow{P}\cdot \overrightarrow{v}-L=\frac{m{c}^2}{\sqrt{1-v^2/c^2}}+e\phi \left(\overrightarrow{r},t\right).$$

This can also be written $P^0=m{U}^0+\left(e/c\right)A^0,$ (remember $A^0=\phi ,P^0=E/c,$ Jackson p 582) matching the expression for the spatial components $P^i=m{U}^i+\frac{e}{c}{A}^i,$ to give the four-vector

$$P^{\mu }=m{U}^{\mu }+\frac{e}{c}{A}^{\mu }.$$

Note the energy $E=c{P}^0$ is just the mass energy (relativistic mass), plus the electrostatic potential energy.  The magnetic field does no work on the particle, so doesn't appear in the energy. This is just the particle’s Hamiltonian expressed as a function of position and velocity.  But if you want to use this Hamiltonian to derive Hamilton's equations of motion, it must be written in terms of the canonical momenta, and then the vector potential reappears$—$as it obviously must, since the magnetic field affects the motion.

The equation of motion from Lagrange’s equations:

$$\frac{d}{dt}\left(\frac{\partial L}{\partial \overrightarrow{v}}\right)=\frac{\partial L}{\partial \overrightarrow{r}},$$

with $L=-m{c}^2\sqrt{1-v^2/c^2}+e\overrightarrow{v}\cdot \frac{\overrightarrow{A}\left(\overrightarrow{r},t\right)}{c}-e\phi \left(\overrightarrow{r},t\right),$

is (following Landau)

$$\frac{d}{dt}\left(\overrightarrow{p}+\frac{e}{c}\overrightarrow{A}\left(\overrightarrow{r},t\right)\right)=\frac{e}{c}\overrightarrow{\nabla }\left(\overrightarrow{A}\cdot \overrightarrow{v}\right)-e\overrightarrow{\nabla }\phi .$$

The tricky point here is that the vector potential $\overrightarrow{A}\left(\overrightarrow{r},t\right)$ is (of course) at the position of the particle, so differentiating it with respect to time must also track the particle’s position, that is,

$$\frac{d\overrightarrow{A}\left(\overrightarrow{r},t\right)}{dt}=\frac{\partial \overrightarrow{A}\left(\overrightarrow{r},t\right)}{\partial t}+\left(\overrightarrow{v}\cdot \overrightarrow{\nabla }\right)\overrightarrow{A}\left(\overrightarrow{r},t\right),$$

sometimes called the convective derivative. 

The equation of motion becomes

$$\frac{d\overrightarrow{p}}{dt}=\frac{e}{c}\overrightarrow{\nabla }\left(\overrightarrow{A}\cdot \overrightarrow{v}\right)-\frac{e}{c}\left(\overrightarrow{v}\cdot \overrightarrow{\nabla }\right)\overrightarrow{A}\left(\overrightarrow{r},t\right)-e\overrightarrow{\nabla }\phi -\frac{e}{c}\frac{\partial \overrightarrow{A}\left(\overrightarrow{r},t\right)}{\partial t}.$$

Here the $\overrightarrow{\nabla }$ is from the $\partial /\partial \overrightarrow{r}$ in Lagrange’s equations, where $\overrightarrow{r},\overrightarrow{v}$ are independent variables, so $\overrightarrow{\nabla }$ doesn’t operate on $\overrightarrow{v},$ only on $\overrightarrow{A},$ and from a well-known vector identity

$$\overrightarrow{\nabla }\left(\overrightarrow{A}\cdot \overrightarrow{v}\right)-\left(\overrightarrow{v}\cdot \overrightarrow{\nabla }\right)\overrightarrow{A}=\overrightarrow{v}\times \left(\overrightarrow{\nabla }\times \overrightarrow{A}\right)=\overrightarrow{v}\times \overrightarrow{B}$$

we have

$$\frac{d\overrightarrow{p}}{dt}=e\left(\overrightarrow{E}+\frac{\overrightarrow{v}\times \overrightarrow{B}}{c}\right),$$

as observed experimentally, and therefore confirmation of our conjectured action for a charged particle in an electromagnetic field.

Equation of Motion Directly from Least Action

(following Landau, p 64, his $ds=cd\tau .$  It’s worth doing the minimization again, staying now in four dimensions, to see how the field tensor emerges naturally.)

Now

$$\delta S=\delta {\displaystyle \underset{a}{\overset{b}{\int }}\left(-m{c}^{2}d\tau -\frac{e}{c}{A}_{\mu }d{x}^{\mu }\right)=0.}$$

That is,

$$\delta S=-{\displaystyle \underset{a}{\overset{b}{\int }}\left(mc\frac{d{x}_{\mu }d\delta x^{\mu }}{d\tau }+\frac{e}{c}{A}_{\mu }d\delta x^{\mu }+\frac{e}{c}\delta A_{\mu }d{x}^{\mu }\right)=0.}$$

Integrating the first two terms by parts, writing $d{x}_{\mu }/d\tau =U_{\mu }$ and requiring $\delta x_i\left(a\right)=\delta x_i\left(b\right)=0$,

$${\displaystyle \underset{a}{\overset{b}{\int }}\left(mcd{U}_{\mu }\delta x^{\mu }+\frac{e}{c}d{A}_{\mu }\delta x^{\mu }-\frac{e}{c}\delta A_{\mu }d{x}^{\mu }\right)=0.}$$

Now

$$\delta A_{\mu }=\frac{\partial A_{\mu }}{\partial x^{\nu }}\delta x^{\nu },d{A}_{\mu }=\frac{\partial A_{\mu }}{\partial x^{\nu }}d{x}^{\nu },$$

giving

$${\displaystyle \underset{a}{\overset{b}{\int }}\left(mcd{U}_{\mu }\delta x^{\mu }+\frac{e}{c}\frac{\partial A_{\mu }}{\partial x^{\nu }}\delta x^{\mu }d{x}^{\nu }-\frac{e}{c}\frac{\partial A_{\mu }}{\partial x^{\nu }}d{x}^{\mu }\delta x^{\nu }\right)=0.}$$

Putting the infinitesimals in terms of the four-velocity and proper time,

$$d{x}^{\mu }=U^{\mu }d\tau ,d{U}_{\mu }=\left(d{U}_{\mu }/d\tau \right)d\tau ,$$

we find (switching dummy suffixes in the third term)

$${\displaystyle \underset{a}{\overset{b}{\int }}\left(mc\frac{d{U}_{\mu }}{d\tau }-\frac{e}{c}\left(\frac{\partial A_{\nu }}{\partial x^{\mu }}-\frac{\partial A_{\mu }}{\partial x^{\nu }}\right)U^{\nu }\right)\delta x^{\mu }d\tau =0.}$$

That is, in terms of the electromagnetic field tensor,

$$mc\frac{d{U}^{\mu }}{d\tau }=\frac{e}{c}{F}^{\mu \nu }{U}_{\nu }.$$

This is of course just the equation previously derived, now written in four-dimensional form.

What’s the Hamiltonian?

Jackson (page 585) introduces a Hamiltonian $\widehat{H}=P_{\mu }{U}^{\mu }+L,$ which is Lorentz invariant (following Barut).  This is not standard practice: the Hamiltonian is conventionally thought of as the time component of an energy-momentum four-vector, as in Landau (page 49), and that will be our approach in these notes. We will not be discussing Lorentz invariant Hamiltonians.  

*An Exercise: Checking Hamilton’s Equations

We’ll check that (using the standard non-Lorentz-invariant Hamiltonian) Hamilton’s approach yields the correct equation of motion:

$$H=\overrightarrow{P}\cdot \overrightarrow{v}-L=\frac{m{c}^2}{\sqrt{1-v^2/c^2}}+e\phi \left(\overrightarrow{r},t\right).$$

(From $L=-m{c}^2\sqrt{1-v^2/c^2}+e\overrightarrow{v}\cdot \frac{\overrightarrow{A}\left(\overrightarrow{r},t\right)}{c}-e\phi \left(\overrightarrow{r},t\right),$  $\overrightarrow{P}=m\frac{\overrightarrow{v}}{\sqrt{1-v^2/c^2}}+\frac{e}{c}\overrightarrow{A}.$ )

This must be expressed in the canonical variables, in other words using $\overrightarrow{P},$ but not $\overrightarrow{v},$ leaving $\overrightarrow{r}$ the same.  Recall $P^{\mu }=m{U}^{\mu }+\frac{e}{c}{A}^{\mu }$ and in particular $P^0=m{U}^0+\frac{e}{c}{A}^0$ where $A^0=\phi ,P^0=E/c=H/c.$

The four-velocity $U^{\mu }=\left(\gamma c,\gamma \overrightarrow{v}\right)$ normalizes to $U^{\mu }{U}_{\mu }=-c^2,$ or

$${\left(H-e\phi \right)}^2-c^2{\left(\overrightarrow{P}-\frac{e}{c}\overrightarrow{A}\right)}^2=m^2{c}^4,$$

so the Hamiltonian

$$H=c\sqrt{{\left(\overrightarrow{P}-e\overrightarrow{A}/c\right)}^2+m^2{c}^2}+e\phi .$$

Hamilton’s equations are $\dot{\overrightarrow{r}}=\partial H/\partial \overrightarrow{P},\dot{\overrightarrow{P}}=-\partial H/\partial \overrightarrow{r}.$ 

That is, first,

$$\dot{\overrightarrow{r}}=\frac{c\left(\overrightarrow{P}-e\overrightarrow{A}/c\right)}{\sqrt{{\left(\overrightarrow{P}-e\overrightarrow{A}/c\right)}^2+m^2{c}^2}}=\frac{c\overrightarrow{p}}{\sqrt{{\overrightarrow{p}}^2+m^2{c}^2}}=\frac{c^2\overrightarrow{p}}{E}.$$

(Recall $E^2=m^2{c}^4+c^2{\overrightarrow{p}}^2.$ )

Notice that in  the nonrelativistic limit the denominator becomes $m{c}^2,$ the numerator is $c^2\overrightarrow{p},$ so the equation is just $\dot{\overrightarrow{r}}=\left(\overrightarrow{p}/m\right)=\overrightarrow{v}.$ $$ Away from this limit, of course, the kinetic momentum is $\overrightarrow{p}=m\overrightarrow{v}/\sqrt{1-{\left(v/c\right)}^2},$ the total energy $E=m{c}^2/\sqrt{1-{\left(v/c\right)}^2},$ so our equation can be written $\overrightarrow{p}=m_{\text{rel}}\overrightarrow{v}$ with $m_{\text{rel}}=m/\sqrt{1-{\left(v/c\right)}^2}.$

The second Hamiltonian equation is  $\dot{\overrightarrow{P}}=-\partial H/\partial \overrightarrow{r}.$ 

Now

$$\dot{\overrightarrow{P}}=\dot{\overrightarrow{p}}+e\dot{\overrightarrow{A}}\left(\overrightarrow{r}\right)/c=\dot{\overrightarrow{p}}-e\left(\overrightarrow{E}+\overrightarrow{\nabla }\phi \right)+e\left(\overrightarrow{v}\cdot \overrightarrow{\nabla }\right)\overrightarrow{A}/c$$

using $\overrightarrow{E}=-{\left(\dot{\overrightarrow{A}}/c\right)}_{\text{in place}}-\overrightarrow{\nabla }\phi$ and the last term above is from the convective derivative, since the time derivative $\dot{\overrightarrow{A}}\left(\overrightarrow{r}\right)$ includes variation from the particle moving, and

$$H=c\sqrt{{\left(\overrightarrow{P}-e\overrightarrow{A}/c\right)}^2+m^2{c}^2}+e\phi ,$$

 so

$$\begin{array}{c}-\frac{\partial H}{\partial r_i}=\frac{c{\left(\overrightarrow{P}-e\overrightarrow{A}/c\right)}_j{\nabla }_{i}e{A}_j/c}{\sqrt{{\left(\overrightarrow{P}-e\overrightarrow{A}/c\right)}^2+m^2{c}^2}}-e{\nabla }_i\phi \\ =\frac{c{p}_j{\nabla }_{i}e{A}_j/c}{\sqrt{p^2+m^2{c}^2}}-e{\nabla }_i\phi ,\end{array}$$

so Hamilton’s second equation $\dot{\overrightarrow{P}}=-\partial H/\partial \overrightarrow{r}$ is

$$\dot{\overrightarrow{p}}-e\left(\overrightarrow{E}+\overrightarrow{\nabla }\phi \right)+e\left(\overrightarrow{v}\cdot \overrightarrow{\nabla }\right)\overrightarrow{A}/c=\frac{c{p}_j{\nabla }_{i}e{A}_j/c}{\sqrt{p^2+m^2{c}^2}}-e{\nabla }_i\phi .$$

Now $\frac{c\overrightarrow{p}}{\sqrt{p^2+m^2{c}^2}}=\overrightarrow{v}$ (check this!) so

$$\dot{\overrightarrow{p}}-e\overrightarrow{E}=e{v}_j{\nabla }_i{A}_j/c-e{v}_j{\nabla }_j{A}_i/c,$$

that is, with $v_j{\nabla }_i{A}_j-v_j{\nabla }_{j}A={\left(\overrightarrow{v}\times \left(\overrightarrow{\nabla }\times \overrightarrow{A}\right)\right)}_i={\left(\overrightarrow{v}\times \overrightarrow{B}\right)}_i,$ we find

$$\dot{\overrightarrow{p}}=e\overrightarrow{E}+e\overrightarrow{v}\times \overrightarrow{B}/c.$$

*The Hamilton-Jacobi Equation

In classical dynamics, there are three standard approaches to the equations of motion: the Lagrangian, the Hamiltonian, and the (somewhat less common) Hamiltonian-Jacobi equation.

This last approach is discussed in my Classical Mechanics lecture on the topic. Briefly, the action is regarded as a function of the endpoint parameters (including time) for fixed initial conditions, and the system follows the classical path for the given initial and final variables. The formalism is a classical analogy of the Schrödinger equation.

This approach is not discussed in Jackson, but Landau uses it to give an elegant analysis of charged particle motion in a Coulomb field, and also to understand how the geometric optics limit of wave equations corresponds to the classical mechanics limit.

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