Chapter 11 Charged particle in electromagnetic field

11.1 Electromagnetic fields, Maxwell’s equations and gauge potentials

Maxwell’s equations for the electric and magnetic fields \(\vec E(\vec x,t)\), \(\vec B(\vec x,t)\) are

\begin{eqnarray} \vec \nabla \cdot \vec E & =\frac {\rho }{\epsilon _0},\label {M1}\\ c^2 \vec \nabla \times \vec B-\partial _t \vec E & =\frac {\vec j}{\epsilon _0},\label {M2}\\ \vec \nabla \cdot \vec B & =0,\label {M3}\\ \vec \nabla \times \vec E+\partial _t \vec B &= 0,\label {M4} \end{eqnarray}

where \(\rho (\vec x,t)\) is the electric charge density and \(\vec j(\vec x,t)\) the electric current density.

In particular, taking the gradient of (11.2), i.e., applying \(\vec \nabla \cdot \) to this equation gives \(-\partial _t\vec \nabla \cdot \vec E=\vec \nabla \frac {\vec j}{\epsilon _0}\). When using (11.1), this gives the continuity equation \(\partial _t \rho +\vec \nabla \cdot \vec j=0\), which expresses local conservation of electric charge.

In a manifestly relativistic notation, \(\mu =0,1,2,3\), \(i=1,2,3\) \(x^0=ct\), \(\frac {\partial }{\partial t}=c\frac {\partial }{\partial x^0}\), \(\partial _\mu =\frac {\partial }{\partial x^\mu }\)

When introducing the electric four-vector current density \(J^\mu \), and the antisymmetric electromagnetic field \(F^{\mu \nu }=-F^{\nu \mu }\) tensor defined through

\begin{equation} J^{\mu }=\begin{pmatrix} \frac {\rho }{c\epsilon _0} \\ \frac {j^1}{c^2\epsilon _0} \\ \frac {j^2}{c^2\epsilon _0} \\ \frac {j^3}{c^2\epsilon _0} \end {pmatrix},\quad F^{\mu \nu }=\begin{pmatrix} 0 & \frac {E^1}{c} & \frac {E^2}{c} & \frac {E^3}{c} \\ -\frac {E^1}{c} & 0 & B^3 & -B^2 \\ -\frac {E^2}{c} & -B^3 & 0 & B^1 \\ -\frac {E^3}{c} & B^2 & -B^1 & 0 \end {pmatrix}, \end{equation}

Maxwell’s equations and the continuity equation can be written in the equivalent (manifestly Lorentz covariant) form as

\begin{equation} \boxed {\partial _\nu F^{\mu \nu } = J^\mu ,\quad \epsilon ^{\mu \nu \lambda \rho }\partial _\nu F_{\lambda \rho } =0,\quad \partial _\mu J^\mu =0}, \end{equation}

where \(\epsilon ^{\mu \nu \lambda \rho }\) is the Levi-Civita pseudo-tensor defined to be completely antisymmetric in the exchange of \(\mu ,\nu ,\lambda ,\rho \) with \(\epsilon ^{0123}=1\), and one remembers that indices are lowered and raised with \(\eta _{\mu \nu }\) and its inverse \(\eta ^{\nu \lambda }\) (so that \(\eta _{\mu \nu }\eta ^{\nu \lambda }=\delta _\mu ^\lambda \)), where

\begin{equation} \eta _{\mu \nu }=\begin{pmatrix} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end {pmatrix}=\eta ^{\mu \nu }. \end{equation}

First, we note that

\begin{equation*} \epsilon ^{0ijk}=\epsilon ^{ijk},\quad \epsilon _{ijk}\epsilon ^{lmk}=\delta _i^l\delta _j^m-\delta _i^m\delta _j^l,\quad \epsilon _{ijk}\epsilon ^{ljk}=2\delta _i^l. \end{equation*}

In components, the electromagnetic field tensor is determined by \(F^{0i}=E^i\), \(F^{ij}=\epsilon ^{ijk}B_k\), and \(\partial _i F^{0i}=J^0\) is (11.1), while \(\partial _0 F^{i0}+\partial _j F^{ij}=J^i\) is equivalent to \(-\partial _0 E^i+\epsilon ^{ijk}\partial _jB_k=J^i\) which is equivalent to (11.2). For the second set of equations, one starts with \(\mu =0\). In this case, they reduce to \(0=\epsilon ^{ijk}\partial _i F_{jk}=\epsilon ^{ijk}\partial _i(\epsilon _{jkl} B^l)=2\partial _i B^i\), and thus to (11.3). If \(\mu =i\), we get \(0=\epsilon ^{i0jk}\partial _0 F_{jk}+\epsilon ^{ij0k}\partial _j F_{0k}+\epsilon ^{ijk0}\partial _j F_{k0}=\epsilon ^{0ijk}(\partial _0 F_{kj}+\partial _j F_{0k}+\partial _k F_{j0})=2(-\partial _0 B^i-\epsilon ^{ijk}\partial _j E_k)\) which is equivalent to (11.4).

On \(\mathbb R^3\) with suitable fall-off conditions, every vector field \(\vec v\) admits a unique decomposition into a longitudinal and a transverse part,

\begin{equation} \label {eq:534} \vec v=\vec \nabla \psi +\vec \nabla \times \vec w. \end{equation}

For a divergence-free vector field, the longitudinal part vanishes, while for a curl-free vector field, it is the transverse part that vanishes.

When the Laplacian \(\Delta \) is invertible (which we assume to be the case and which holds if one considers fields on \(\mathbb R\times \mathbb R^3\) that decrease at least as fact as \(r^{-1}\) when \(r\to \infty \)) the decomposition of a \(\vec v(x)\) into its longitudinal and transversal components follows from \(\vec \nabla \times (\vec \nabla \times \vec v)=\vec \nabla (\vec \nabla \cdot \vec v)-\Delta \vec v\):

\[\vec v=\vec \nabla \psi +\vec \nabla \times \vec w,\quad \psi =\Delta ^{-1}(\vec \nabla \cdot \vec v),\quad \vec w=-\Delta ^{-1}(\vec \nabla \times \vec v). \]

(if \(\Delta ^{-1}\) commutes with the gradient \(\vec \nabla \) and the curl \(\vec \nabla \times \)). NB: These properties can be explicitly checked when using that \(\Delta \phi (\vec x)=-\delta ^{(3)}(\vec x-\vec y)\) iff \(\phi (\vec x)=\frac {1}{4\pi } \frac {1}{|\vec x -\vec y|}\) so that \(\Delta \phi (\vec x)=j(\vec x)\iff \phi (\vec x)= -\frac {1}{4\pi }\int d^3y \frac {j(\vec y)}{|\vec x -\vec y|}\), which provides an explicit expression for \(\Delta ^{-1}\).

Equation (11.3) then implies the existence of a vector potential \(\vec A\) such that \(\vec B=\vec \nabla \times \vec A\). When injecting into (11.4), this implies \(\vec \nabla \times (\vec E+\partial _t\vec A)=0\), and in turn that \(\vec E=-\partial _t \vec A-\vec \nabla \phi \) for some scalar potential \(\phi \).

NB: the vector and scalar potentials that give rise to a given electric and magnetic field are not uniquely defined. If \(\vec B=\vec \nabla \times \vec A'\), \(\vec E=-\partial _t\vec A' -\vec \nabla \phi '\), it follows by substraction that \(\vec \nabla \times (\vec A'-\vec A)=0\), so that \(\vec A'=\vec A+\vec \nabla \chi \) for some \(\chi \), and then, when subtracting the two expressions for \(\vec E\), that \(\vec \nabla (\partial _t\chi +\phi '-\phi )=0\), so that \(\phi '=\phi -\partial _t\chi \) up to a function of time alone, which needs to vanish on account of the boundary conditions imposed on the fields.

Defining

\begin{equation} \label {eq:67} A_\mu =(-\frac {\phi }{c}, A_1, A_2, A_3), \end{equation}

it follows that

\begin{equation} \label {eq:88a} \boxed {F_{\mu \nu }=\partial _\mu A_\nu -\partial _\nu A_\mu }, \end{equation}

while gauge transformations of the potentials giving rise to identical electric and magnetic fields can be written as

\begin{equation} \label {eq:214a} \boxed {A'_\mu =A_\mu +\partial _\mu \chi }. \end{equation}

Indeed, \(F_{0i}=-E_i=\partial _0 A_i+\partial _i\phi \), as it should. For the other components, we have \(F_{ij}=\epsilon _{ijk}B^k\) by definition, and \(F_{ij}=\partial _i A_j-\partial _j A_i\) according to (11.10). Contracting both with \(\epsilon ^{ljk}\) on the other gives \(2B^l=2\epsilon ^{ljk}\partial _j A_k\) as it should.

In these terms, equations (11.1) and(11.2), which have been shown to be equivalent to \(\partial _\nu F^{\mu \nu }=J^\mu \) read explicitly \(\partial ^\mu \partial _\nu A^\nu -\Box A^\mu =j^\mu \), or \(\partial ^0(\cancel {\partial _0 A^0}+\partial _i A^i)-\cancel {\partial _0\partial ^0 A^0}-\Delta A^0=\rho \) and \(\partial _0 (\partial ^i A^0-\partial ^0 A^i)+\partial _j(\partial ^i A^j-\partial ^j A^i)=j^j\). When taking into account that \(\partial _j(\partial ^i A^j-\partial ^j A^i)=\partial _j (\epsilon ^{ijk}\epsilon _{klm} \partial ^l A^m)\) these can be written as

\begin{equation} -(\partial _0\vec \nabla \cdot \vec A+\Delta \phi )=J^0,\quad \label {eq:147a} \partial _0(\vec \nabla \phi +\partial _0\vec A)+\vec \nabla \times (\vec \nabla \times \vec A)=\vec J. \end{equation}

The important point is that while equations (11.3) and (11.4) have been solved by the introduction of the potentials, the remaining equations (11.1) and(11.2) derive from the action principle,

\begin{equation} \label {eq:2} \boxed {S_M[A_\mu ;j^\mu ]=\int d^4x\, \mathcal L_M=\int d^4x \, (-\frac {1}{4}F^{\mu \nu }F_{\mu \nu }+j^\mu A_\mu )}. \end{equation}

Indeed, varying the gauge potentials and neglecting boundary terms in the action gives

\begin{equation*} \int d^4x\ [-\frac {1}{2}F^{\mu \nu }(\partial _\mu \delta A_\nu -\partial _\nu \delta A_\mu )+j^\mu \delta A_\mu ]=\int d^4x\ \delta A_\mu (x)[-\partial _\nu F^{\mu \nu }(x)+j^\mu (x)]. \end{equation*}

The action principle then requires that

\begin{equation} 0=\frac {\delta S_M}{\delta A_\mu (x)}=\frac {\delta \mathcal L^M}{\delta A_\mu }(x)=[\frac {\partial \mathcal L_M}{\partial A_\mu }-\partial _\lambda \frac {\partial \mathcal L_M}{\partial \partial _\lambda A_\mu }](x)=[-\partial _\nu F^{\mu \nu }+j^\mu ](x). \end{equation}

11.2 Newtonian formulation

Empirical considerations have shown that the force that acts on a charged point particle under the influence of an electro-magnetic field is given by the Lorentz force,

\begin{equation} \label {eq:405} \vec F= q\Big (\vec E+\dot {\vec x}\times \vec B\Big ). \end{equation}

For a charged particle that moves at a non relativistic velocity \(v\ll c\), Newton’s second law thus implies

\begin{equation} \label {eq:406} m\ddot {\vec x}=q\Big (\vec E+\dot {\vec x}\times \vec B\Big ). \end{equation}

11.3 Lagrangian formulation

In order to derive the equations of motion (11.16) from a variational principle, one needs to describe the electromagnetic fields in terms of potentials, \(\phi (x^i,t), \vec A(x^i,t)\).

\begin{equation} \label {eq:407} \vec B=\vec \nabla \times \vec A,\quad \vec E=-\partial _t \vec A-\vec \nabla \phi , \end{equation}

the appropriate Lagrangian is

\begin{equation} \label {eq:409} \boxed {L=\frac {1}{2}m \dot {\vec x}^2-q\phi +q\dot {\vec x}\cdot \vec A}. \end{equation}

Indeed, the associated Euler-Lagrange derivatives are

\begin{equation} \begin{split} \label {eq:410} -\frac {\delta L}{\delta x^i}&=\frac {d}{dt}\frac {\partial L}{\partial \dot x^i}-\frac {\partial L}{\partial x^i}=\frac {d}{dt}(m\dot x_i+q A_i)+q\big (\frac {\partial \phi }{\partial x^i}-\dot x^j\frac {\partial A_j}{\partial x^i}\big )\\ &=m\ddot x_i+q\big (\frac {\partial A_i}{\partial x^j}\dot x^j+\frac {\partial A_i}{\partial t}\big )+q\big (\frac {\partial \phi }{\partial x^i}-\dot x^j\frac {\partial A_j}{\partial x^i}\big ). \end {split} \end{equation}

Re-organizing, the associated equations of motion are

\begin{equation} \label {eq:411} m\ddot x_i=q\big (\partial _i A_j-\partial _jA_i\big )\dot x^j-q\big (\partial _i\phi +\partial _tA_i\big )=q\epsilon _{ijk}\epsilon ^{klm}\partial _l A_m\dot x^j-q(\partial _i\phi +\partial _t A_i), \end{equation}

which coincide with (11.16).

11.4 Hamiltonian formulation

The canonical momenta are given by

\begin{equation} \label {eq:412} p_i=\frac {\partial L}{\partial \dot x^i}=m\dot x_i+qA_i \iff \dot x^i=\frac {1}{m}(p^i-q A^i), \end{equation}

while the Hamiltonian is

\begin{equation} \label {eq:413} H=\big [p_i\dot x^i-L\big ]\big |_{\dot x=\dot x(x,p)}=\frac {1}{m}(p^i-q A^i)p_i-\frac {1}{2m}(p^i-q A^i)(p_i-q A_i)+q\phi -q\frac {1}{m}(p^i-qA^i)A_i. \end{equation}

After regrouping terms,

\begin{equation} \label {eq:414} \boxed {H=\frac {1}{2m}(p^i-q A^i)(p_i-q A_i)+q\phi .} \end{equation}

In other words, as compared to the Hamiltonian of a free particle, \(H=\frac {1}{2m}p^i p_i\), the Hamiltonian of a charged particle in an electromagnetic field is obtained by adding the potential \(V=q\phi \) and the so-called rule of “minimal substitution”, \(p^i\to p^i-q A^i\).

Remarks:

In the case of \(N\) particles with coordinates \(x^i_a\), mass \(m_a\) and charge \(q_a\), the Lagrangian becomes

\begin{equation} \label {eq:415} L=\sum _{a=1}^{N}\Big [\frac 12 m_{(a)}\dot x^i_a \dot x_{a i}-q_{(a)} \phi (\vec x_a,t)+q_{(a)}\dot x^i_a A_i(\vec x_a,t)\Big ], \end{equation}

while the Hamiltonian is given by

\begin{equation} \label {eq:416} H=\sum _{a=1}^{N}\Big [\frac {1}{2m_{(a)}}\big [p^i_a -q_{(a)}A^i(\vec x_a,t)\big ]\big [p_{a i} -q_{(a)}A_i(\vec x_a,t)\big ]+q_{(a)}\phi (\vec x_a,t)\Big ]. \end{equation}

One may rewrite the action associated to the Lagrangian (11.24) as \(S=S_P+S_I\), where

\begin{equation} \label {eq:417} S_P=\int dt \sum _{a=1}^N \frac 12 m_{(a)}\dot x^i_a\dot x_{ai}, \end{equation}

is the action for \(N\) free particles, while

\begin{equation} \label {eq:418} S_I=\int d^4x A_\mu (x)J^\mu (x),\quad J^0=\sum _{a=1}^N q_{(a)}\delta ^{(3)}(\vec x-\vec x_a(t)),\quad J^i=\sum _{a=1}^N\frac {q_{(a)}}{c}\frac {dx^i}{dt}\delta ^{(3)}(\vec x-\vec x_a(t)), \end{equation}

is the action that describes the interaction of the particles with the electromagnetic field.

11.5 Relativistic particles

When one considers relativistic particles rather than non-relativistic ones, the action (11.26) for the free particles must be taken as

\begin{equation} \label {eq:419} S_P=-\sum _{a=1}^N m_{(a)}c\int d\tau \sqrt {-\frac {d x^\mu _a}{d \tau }\frac {dx_{a\mu }}{d\tau }}. \end{equation}

This action reduces to the non-relativistic one in the limit of small velocities as compared to the speed of light. Indeed, if we choose \(\tau =t=x^0/c\),

\begin{equation} \begin{split} \label {eq:420} &-m_{(a)}c\int d\tau \sqrt {-\frac {d x^\mu _a}{d \tau }\frac {dx_{a\mu }}{d\tau }}=-m_{(a)}c\int dt \sqrt {\frac {dx^0}{dt}\frac {dx^0}{dt}- \frac {d\vec x_a}{dt}\cdot \frac {d \vec x^a}{dt}}\\&= -m_{(a)}c\int dt c\sqrt {1-\frac {1}{c^2}\frac {d\vec x_a}{dt}\cdot \frac {d \vec x^a}{dt}} \approx -m_{(a)}c^2\int dt\big (1-\frac 12\frac {1}{c^2} \frac {d\vec x_a}{dt}\cdot \frac {d\vec x^a}{dt} \big ). \end {split} \end{equation}

The first term is a constant which does not contribute to the Euler-Lagrange equations and thus can be dropped, while the second terms reduces to the action for a collection of non-relativistic particles, (11.26).

Let us now set \(c=1\) and study in more details a single relativistic particle, whose worldline is described by \(x^\mu (\tau )\) in Minkowski spacetime. The associated action is

\begin{equation} \label {eq:408} S_P[x^\mu ]=\int d\tau \, L_P,\quad L_P=-m \sqrt {-\dot x^\nu \dot x_\nu }, \end{equation}

with \(\dot x^\nu =\frac {dx^\nu }{d\tau }\). For variations that vanish at the boundary of the \(\tau \) interval, the variation of the Lagrangian is

\begin{equation} \label {eq:425} \delta S_P=-m\int d\tau \, \frac {-\delta \dot x^\mu \dot x_\mu }{\sqrt {-\dot x^\nu \dot x_\nu }}= -m\int d\tau \, \delta x^\mu \frac {d}{d\tau }\Big (\frac {\dot x_\mu }{\sqrt {-\dot x^\nu \dot x_\nu }}\Big ), \end{equation}

and the Euler-Lagrange equations are given by

\begin{equation} \label {eq:426} \frac {\delta L_P}{\delta x^\mu }=-m\frac {d}{d\tau }\Big (\frac {\dot x_\mu }{\sqrt {-\dot x^\nu \dot x_\nu }}\Big )=0. \end{equation}

It is not possible to pass to the Hamiltonian formulation through a Legendre transformation: the Hessian matrix

\begin{equation} \label {eq:427} \frac {\partial ^2 L_P}{\partial \dot x^\mu \partial \dot x^\lambda } \end{equation}

is not invertibe. Indeed,

\begin{equation} \label {eq:428} \frac {\partial ^2 L_P}{\partial \dot x^\mu \partial \dot x^\lambda }= -m\frac {-\dot x^\nu \dot x_\nu \eta _{\mu \lambda }+\dot x_\mu \dot x_\lambda }{(\sqrt {-\dot x^\nu \dot x_\nu })^3},\quad \frac {\partial ^2 L_P}{\partial \dot x^\mu \partial \dot x^\lambda }\dot x^\lambda =0, \end{equation}

so that \(\dot x^\lambda \) is an eigenvector of eigenvalue \(0\).

The correct first order action principle whose equation of motion are equivalent to the Lagrangian ones in (11.33) is given by

\begin{equation} \label {eq:429} S[x^\mu ,p_\mu ,\lambda ]=\int d\tau \,\big [\dot x^\mu p_\mu -\lambda (p^\mu p_\mu +m^2)\big ]. \end{equation}

The associated Euler-Lagrange equations with respect to \(x^\mu ,p_\mu ,\lambda \) are

\begin{equation} \label {eq:430} \dot p_\mu =0,\quad \dot x^\mu -2\lambda p^\mu =0,\quad p^\mu p_\mu +m^2=0. \end{equation}

It is a “constrained Hamiltonian system” in the sense that \(\lambda (\tau )\) is a Lagrange multiplier for the constraint \(p^\mu p_\mu +m^2\) that does not involve time derivatives of the phase space variables \(x^\mu ,p_\mu \).

The general solution to the equations of motion is

\begin{equation} \label {eq:431} p_\mu (\tau )=p_\mu (0)\equiv p_\mu ,\quad x^\mu (\tau )=x^\mu (0)+2 p^\mu (0)\int _0^\tau d\tau ' \lambda (\tau '),\quad p^0=\pm \sqrt {p_ip^i+m^2}. \end{equation}

where \(p_\mu \) are \(\tau \)-independent constants. Alternatively, if \(\lambda > 0\), the equations for \(p_\mu \) may be solved as

\begin{equation} \label {eq:432} p_\mu =-\frac {1}{2\lambda }\dot x_\mu . \end{equation}

When substituting in the equation for \(\lambda \), one finds

\begin{equation} \label {eq:433} \frac {\dot x^\mu \dot x_\mu }{4\lambda ^2}+m^2=0\iff \lambda = \frac {\sqrt {-\dot x^\nu \dot x_\nu }}{2m}. \end{equation}

When substituting both of these equations into the equation for \(x^\mu \), i.e., the first equation in (11.36), one recovers the Lagrangian equations (11.32). It is in this sense that the Lagrangian equations (11.32) and the first order equations (11.36) are equivalent. Finally, note also that the first order action (11.35) reduces to the Lagrangian action (11.30) when substituting both (11.38) and (11.39).

The \(x^\mu \) can be considered as coordinates on Minkowski spacetime with metric \(\eta _{\mu \nu }={\rm diag}(-1,1,1,1)\) that is used to lower indices. Its inverse is denoted by \(\eta ^{\mu \nu }={\rm diag}(-1,1,1,1)\), \(\eta _{\mu \nu }\eta ^{\nu \rho }=\delta _\mu ^\rho \) and used to raise indices.

Poincaré transformations are defined by

\begin{equation} \label {eq:436} x'^\mu ={\Lambda ^\mu }_\nu x^\nu +a^\mu , \end{equation}

where the matrix \({\Lambda ^\mu }_\nu \) satisfies

\begin{equation} \label {eq:437} {\Lambda ^\mu }_\nu \eta _{\mu \sigma }{\Lambda ^\sigma }_\rho =\eta _{\nu \rho } \iff \Lambda ^T\eta \Lambda =\eta \iff {\Lambda _\sigma }^\nu {\Lambda ^\sigma }_\rho =\delta ^\rho _\nu \iff \big ({\Lambda ^\sigma }_\rho \big )^{-1}={\Lambda _\sigma }^\rho , \end{equation}

The transformations with \({\Lambda ^\mu }_\nu =0\), \(x'^\mu =x^\mu +a^\mu \) are called translations, those with \(a^\mu =0\), \(x'^\mu ={\Lambda ^\mu }_\nu x^\nu \) Lorentz transformations.

For Poincaré transformations (that do not depend on \(\tau \)), the Lagrangian \(L_P\) and thus the action \(S_P\) is invariant,

\begin{equation} \label {eq:438} S_P[x'^\mu ]=S[x^\mu ]. \end{equation}

As in the discussion of Galilean invariance, infinitesimal Lorentz transformations are of the form

\begin{equation} \label {eq:439} {\Lambda ^\mu }_\nu =\delta ^\mu _\nu +{\omega ^\mu }_\nu +O(\omega ^2),\quad \omega _{\mu \nu }=-\omega _{\nu \mu }, \end{equation}

and contain \(6\) parameters, while infinitesimal translations are characterized by \(a^\mu =0+\alpha ^\mu \) and contain \(4\) parameters. The associated infinitesimal Poincaré transformations on Minkowski spacetime are given by

\begin{equation} \label {eq:440} \delta _{\omega ,\alpha } x^\mu ={\omega ^\mu }_\nu x^\nu +\alpha ^\mu . \end{equation}

According to Noether’s theorem, there are \(10\) conserved quantitites given by

\begin{equation} \label {eq:441} K_{\omega ,\alpha }=\frac {\partial L_P}{\partial \dot x^\mu }\delta _{{\omega ,\alpha }} x^\mu =m\frac {\dot x_\mu \delta _{\omega ,\alpha }x^\mu }{\sqrt {-\dot x^\nu \dot x_\nu }}. \end{equation}

The individual conserved quantities are denoted by \(P^\mu ,L^{\mu \nu }\),

\begin{equation} \label {eq:442} K_{\omega ,\alpha }=\frac 12 \omega _{\mu \nu }L^{\mu \nu }+\alpha _\mu P^\mu , \end{equation}

and are explicitly given by

\begin{equation} \label {eq:443} P^\mu =m\frac {\dot x^\mu }{\sqrt {-\dot x^\sigma \dot x_\sigma }},\quad L^{\mu \nu }=m\frac {\dot x^\mu x^\nu -\dot x^\nu x^\mu }{\sqrt {-\dot x^\sigma \dot x_\sigma }}. \end{equation}

For several particles, one finds instead,

\begin{equation} \label {eq:444} P^\mu =\sum _{a=1}^Nm_{(a)}\frac {\dot x^\mu _a}{\sqrt {-\dot x^\sigma _a\dot x_{\sigma ,a}}},\quad L^{\mu \nu }=\sum _{a=1}^Nm_{(a)} \frac {\dot x^\mu _a x^\nu _a-\dot x^\nu _a x^\mu _a}{\sqrt {-\dot x^\sigma _a\dot x_{\sigma ,a}}}. \end{equation}

relativistic invariance, to be continued

Furthermore, in relativistic notation, the current may be written as

\begin{equation} \label {eq:421} J^\mu =\sum _{a=1}^N \int d\tau \frac {q_{(a)}}{c}\frac {dx^\mu _a}{d\tau }\delta ^{(4)}(\vec x-\vec x_a(\tau )), \end{equation}

That this expression reduces to that in (11.27) can be seen as follows:

\begin{multline} \label {eq:423} \int d\tau \frac {q_{(a)}}{c}\frac {dx^\mu _a}{d\tau }\delta ^{(4)}(\vec x-\vec x_a(\tau ))=\dots =\frac {q_{(a)}}{c}\frac {dx^\mu _a}{d\tau }\delta ^{(3)}(\vec x-\vec x_a) \end{multline} where, in order to show the last equality, one uses the following property of the \(\delta \) function,

\begin{equation} \label {eq:422} \int dx f(x)\delta (g(x))=\frac {f(x_0)}{|g'(x_0)|}, \end{equation}

where \(g(x)\) is a monotonic function that vanishes in \(x_0\), \(g(x_0)=0\).

complete by using Felsager exercise 1.4.1 pages 20,28 and equation 1.34 in reverse.