Relativistic Lagrangians for interactions with electromagnetic fields
For describing relativistic particles, we will use the Lagrangian formalism, from which we can derive the equations of motion by requiring the action \[\begin{equation} \require{physics} S = \int \dd{t} L(\vb{r}, \vb{v}, t) = \int \dd{\tau} \tilde{L}(x, u) \end{equation}\] to vanish due to the variations in the dynamical variables \(\vb{r}\) and \(\vb{v}\) in \(L\) and \(x^{\mu}\) and \(u^{\mu}\) in \(\tilde{L}\). If we have one of both expressions, we can change to the other one by using \[\begin{equation} L(\vb{r}, \vb{v}, t) \dd{t} = \tilde{L}(x, u) \dd{\tau} \thinspace . \end{equation}\]
The relativistic Lagrangian for a free particle
Suppose we would like to describe a free relativitic particle. We then require the action to be stationary with respect to variations in \(x^{\mu}\) and \(u^{\mu}\). The Euler-Lagrange equations of motion then become \[\begin{equation} \dv{\tau} \qty( \pdv{\tilde{L}}{u^{\mu}} ) = \pdv{\tilde{L}}{x^{\mu}} \thinspace . \end{equation}\] For a free particle, the Lagrangian (with the dynamical variables of the system the particle’s position \(x^{\mu}\) and its velocity \(u^{\mu}\)) is: \[\begin{equation} \tilde{L}(x, u) = - m c \sqrt{u_{\mu} u^{\mu} } \thinspace , \end{equation}\] with the corresponding equation of motion: \[\begin{equation} \dv{p_{\mu}}{t} = 0 \thinspace . \end{equation}\]
Its \(\mu = 0\) component describes that the total energy of the particle should be conserved: \[\begin{equation} \dv{E}{t} = 0 \end{equation}\] and the other components describe that the total relativistic momentum of the particle should be conserved: \[\begin{equation} \dv{\vb{p}}{t} = 0 % \thinspace . \end{equation}\]
The interaction between a \(4\)-current and the electromagnetic field
Suppose we want to describe an electromagnetic field characterized by the \(4\)-vector \(A^{\mu}(x)\), which has sources due to the \(4\)-current \(j^{\mu}(x)\). The only dynamic variables then are \(A^{\nu}, \partial_{\mu} A^{\nu}\): \(x^{\nu}\) and \(j^{\nu}\) are not dynamic variables of the system, simply because we are only describing the interactions of the field with itself and with a \(4\)-current whose origin is unspecified.
The total Lagrangian density \(\mathcal{L}(A^{\nu}, \partial_{\mu} A^{\nu})\) for this system is a sum of two contributions: \[\begin{equation} \mathcal{L} = \mathcal{L}_{\text{rad}} + \mathcal{L}_{\text{int}} \end{equation}\] The first contribution arises from the electromagnetic field merely existing: \[\begin{align} \mathcal{L}_{\text{rad}} &= - \frac{1}{4 \mu_0} F_{\mu \nu} F^{\mu \nu} \\ % &= \frac{1}{2 \mu_0} \qty( \norm{\vb{E}}^2 - \norm{\vb{B}}^2 ) \end{align}\] and the second contribution arises from the interaction between the electromagnetic field and the \(4\)-current: \[\begin{align} \mathcal{L}_{\text{int}} &= - A_\mu j^\mu \\ &= - \qty(\rho \phi - \thinspace \vb{j} \cdot \vb{A}) \thinspace . \end{align}\] The Lagrangian density for this interaction term is the simplest choice that ensures gauge and Lorentz invariance, and is referred to as minimal coupling.
Both contributions to the Lagrangian density are Lorentz invariant because they are Lorentz scalars. The gauge invariance of the radiation term is evident and the gauge invariance of the action of the interaction term is a direct consequence of the continuity equation.
The relativistic action for the electromagnetic field is defined by \[\begin{equation} S = \int \dd[4]{x} \mathcal{L} \thinspace , \end{equation}\] such that requiring the variation in action \(\delta S\), due to the variation in the dynamical variables \(A^{\nu}, \partial_{\mu} A^{\nu}\) of the system, to vanish: \[\begin{equation} \delta S = 0 \end{equation}\] leads to the so-called Euler-Lagrange equations, which are the equations of motion of the system: \[\begin{equation} \partial_{\mu} \qty( \pdv{ \mathcal{L} }{ (\partial_{\nu} A_{\mu}) } ) = \pdv{ \mathcal{L} }{A^\mu} \thinspace , \end{equation}\] By plugging in the Lagrangian density and doing some tensor calculus, we find the inhomogeneous Maxwell equations: \[\begin{equation} \partial_\mu F^{\mu \nu} = \mu_0 j^{\nu} \thinspace . \end{equation}\]
A charged particle subject to an electromagnetic field
If we would like to describe a relativistic charged particle subject to an electromagnetic field, we use the Lagrangian (with the dynamical variables of the system the particle’s position \(x^{\mu}\) and its velocity \(u^{\mu}\)): \[\begin{equation} \tilde{L}(x, u) = - m c \sqrt{u_{\mu} u^{\mu} } + \tilde{L}_{\text{int}}(x, u) \thinspace , \end{equation}\] in which \(\tilde{L}_{\text{int}}(x, u)\) is a yet unspecified Lagrangian for the interaction of the charged particle with the external electromagnetic field. Since charged particle moving on the trajectory \(\vb{r}(t)\) generates a \(4\)-current equal to (Reiher and Wolf 2015): \[\begin{equation} j^{\mu}(x') = q \int \dd{\tau} \delta(x' - x(\tau)) u^{\mu}(\tau) \thinspace , \end{equation}\] we can calculate the interaction action as: \[\begin{align} S_{\text{int}} & = \int \dd{x'}^4 \mathcal{L}_{\text{int}} \\ % & = - \int \dd{x'}^4 A_{\mu}(x') j^{\mu}(x') \\ % & = - q \int \dd{\tau} A_{\mu}(x) u^{\mu}(\tau) \thinspace , \end{align}\] which means that the interaction Lagrangian \(L_{\text{int}}(x, u)\) should become: \[\begin{equation} \tilde{L}_{\text{int}}(x, u) = - q A_{\mu}(x) u^{\mu} \thinspace . \end{equation}\]
Requiring the action to be stationary then leads to the Euler-Lagrange equations \[\begin{equation} \dv{\tau} \qty( \pdv{ \tilde{L}(x, u) }{ u^{\mu} } ) = \pdv{ \tilde{L}(x, u) }{ x^{\mu} } \thinspace , \end{equation}\] which are the equations of motion of the relativistic charged particle in an electromagnetic field: \[\begin{equation} \dv{p_{\mu}}{\tau} = q F_{\mu \nu} u^{\nu} \thinspace . \end{equation}\] The \(\mu = 0\) component leads to an equation that describes the change in energy over time (due to the work done by the electric field): \[\begin{equation} \dv{E}{t} = q \thinspace \vb{E}(\vb{r}, t) \vdot \vb{v}(t) \thinspace , \end{equation}\] while the other components recover the Lorentz force: \[\begin{equation} \dv{\vb{p}(t)}{t} = \vb{F}(\vb{r}, t) = q \qty( \vb{E}(\vb{r}, t) + \vb{v}(\vb{r}, t) \cross \vb{B}(\vb{r}, t) ) \thinspace , \end{equation}\] in which we should emphasize that the appearing velocities \(\vb{v}\) and momenta \(\vb{p}\) are the relativistic \(3\)-vectors.
The non-covariant form of the Lagrangian can be calculated as: \[\begin{equation} L(\vb{r}, \vb{v}, t) = - \frac{m c^2}{\gamma(\vb{v})} - q \qty( \phi(\vb{r}, t) - \vb{v} \vdot \vb{A}(\vb{r}, t) ) \thinspace . \end{equation}\] We can then calculate the conjugate momentum \(3\)-vector to be: \[\begin{equation} P_i = \pdv{L}{v_i} = p_i + q A_i \thinspace , \end{equation}\] which is the embodiment of the so-called minimal coupling substitution \[\begin{equation} \vb{p} \rightarrow \boldsymbol{\pi} = \vb{p} - q \vb{A} \thinspace , \end{equation}\] in which \(\boldsymbol{\pi}\) is then called the kinetic momentum.
The interaction of two moving charged particles
The non-covariant form of the Lagrangian for the interaction of a charged particle with an electromagnetic field can be calculated as: \[\begin{equation} L_{\text{int}}(\vb{r}, \vb{v}, t) = - q \qty( \phi(\vb{r}, t) - \vb{v}(t) \vdot \vb{A}(\vb{r}, t) ) \thinspace . \end{equation}\]
The benefit of using this Lagrangian is that we can make the connection with a Hamiltonian formalism: the interaction energy for a particle moving in an electromagnetic field is then \[\begin{equation} V_{\text{int}} = -L_{\text{int}} = q \qty( \phi(\vb{r}, t) - \vb{v}(t) \vdot \vb{A}(\vb{r}, t) ) \thinspace . \end{equation}\]
Since a second moving charged particle creates an external electromagnetic field for the first particle, we can use a reference frame IS’ in which the second particle is at rest and subsequently use the appropriate Lorentz transformation to obtain expressions for the scalar and vector potential, and consequently the interaction energy, in IS. For this system, the (inverse) Lorentz boost equations are \[\begin{align} & \phi_2(\vb{r}_1, t_1) = \gamma \phi_2'(\vb{r}_1', t_1') \\ % & \vb{A}_2(\vb{r}_1, t_1) = \gamma \frac{\dot{\vb{r}}_2}{c} \phi_2'(\vb{r}_1', t_1') \thinspace . \end{align}\] (Reiher and Wolf 2015) then finds the (Darwin) interaction energy \(V_{12}(t_1, \vb{r}_1, \vb{r}_2, \dot{\vb{r}_1}, \dot{\vb{r}_2})\) to be \[\begin{equation} V_{12} = q_1 q_2 \qty[ \frac{1}{ \norm{\vb{r}_{12}} } + \frac{1}{2c^2} \qty( \frac{ \norm{\dot{\vb{r}}_2}^2 - \vb{r}_{12} \vdot \ddot{\vb{r}}_2 - 2 \dot{\vb{r}}_1 \vdot \dot{\vb{r}}_2 }{ \norm{\vb{r}_{12}} } - \frac{ (\vb{r}_{12} \vdot \dot{\vb{r}}_2)^2 }{ \norm{\vb{r}_{12}}^3 }) ] + \mathcal{O}(c^{-3}) \thinspace , \end{equation}\] in which all quantities are to be evaluated at \(t_1\) and we have introduced the difference vector \[\begin{equation} \vb{r}_{12} = \vb{r}_1 - \vb{r}_2 \thinspace . \end{equation}\]
Upon inspection, however, we can see that this expression is not symmetric in the particle labels. We (Reiher and Wolf 2015) can add a total time derivative of a function to the Lagrangian (which doesn’t change the Euler-Lagrange equations), so in order to obtain the `gauge’-transformed potential energy \[\begin{equation} V_{12} = \frac{q_1 q_2}{r_{12}} - \frac{q_1 q_2}{2c^2} \qty( \frac{ \dot{\vb{r}}_1 \vdot \dot{\vb{r}}_2 }{ r_{12} } + \frac{ (\vb{r}_{12} \vdot \dot{\vb{r}}_1) (\vb{r}_{12} \vdot \dot{\vb{r}}_2) }{r_{12}^3} ) + \mathcal{O}(c^{-3}) \thinspace , \end{equation}\] in which we have used the short-hand notation \[\begin{equation} \norm{\vb{r}_{12}} = r_{12} \thinspace . \end{equation}\]
In terms of the momenta \(\vb{p} = m \dot{\vb{r}}\) of the particles, we can equivalently write for the interaction energy \(V_{12}(t_1, \vb{r}_1, \vb{r}_2, \vb{p}_1, \vb{p}_2)\): \[\begin{equation} V_{12} = \frac{q_1 q_2}{r_{12}} - \frac{q_1 q_2}{2 m_1 m_2 c^2} \qty( \frac{ \vb{p}_1 \vdot \vb{p}_2 }{ r_{12} } + \frac{ (\vb{r}_{12} \vdot \vb{p}_1) (\vb{r}_{12} \vdot \vb{p}_2) }{r_{12}^3} ) + \mathcal{O}(c^{-3}) \thinspace . \end{equation}\]