LASER-ACCELERATED CHARGE: MOTION AND TRAJECTORIES

Consider a particle of mass $m$ and charge $q$ in an electromagnetic field $F_{\alpha\beta}$. The motion of such a particle is governed by the relativistic version of Newton's second law,

$$m\frac{d^2x^\alpha}{d\tau^2}=qF^\alpha_{~\beta}\frac{dx^\beta}{d\tau}~ \alpha=0,1,2,3~.$$ (1)

Here

$$\left\{\frac{dx^\beta}{d\tau}\right\}=\left\{ \frac{dt}{d\tau}, \frac{dx}{d\tau},\frac{dy}{d\tau},\frac{dz}{d\tau} \right\}.$$

are the components of the particle's four-velocity. The particle world line is parametrized by its proper time $\tau$, a fact which is expressed by the equation

$$\left( \frac{dt}{d\tau} \right)^2 -\left( \frac{dx}{d\tau} \... ...rac{dy}{d\tau} \right)^2 -\left( \frac{dz}{d\tau} \right)^2=1.$$ (2)

Suppose the e.m. field is that of a laser beam along the $z$-direction with field polarization directions in the $x$-$y$ plane. The field components

$$\begin{eqnarray*} \vec E:&&(F_{10},F_{20},F_{30})\equiv (E_x(t,z),0,0)\\ \vec B:&&(F_{23},F_{31},F_{12})\equiv (0,B_y(t,z),0) \end{eqnarray*}$$

are derived from a single function $A_x(t,z)$ which satisfies the wave equation,

$$\frac{\partial^2A_x}{\partial t^2}-\frac{\partial^2A_x}{\partial z^2}=0,$$

namely

$$E_x=-\frac{\partial A_x(t,z)}{\partial t} \textrm{ and } B_y=+\frac{\partial A_x(t,z)}{\partial z}.$$ (3)

a)

VERIFY that the equations of motion, Eqs.(1), are consistent with the constraint Eq.(2); i.e. SHOW that Eq.(2) is an ``integral of motion''.

b)

EXHIBIT the four equations of motion and OBTAIN an integral of motion associated with the $x$-translation invariance of the system.

 

Answer:

 

The equations of motion are

$$\frac{d^2t}{d\tau^2} = \frac{q}{m}E_x\frac{dx}{d\tau}$$ (4)

$$\frac{d^2z}{d\tau^2} = \frac{q}{m}B_y\frac{dx}{d\tau}$$ (5)

$$\frac{d^2y}{d\tau^2} = 0$$

$$\frac{d^2x}{d\tau^2} = \frac{q}{m}E_x\frac{dt}{d\tau}-\frac{q}{m}B_y\frac{dz}{d\tau} =-\frac{q}{m}\frac{dA_x(t,z)}{d\tau}.$$

 

The integral of motion associated with $x$-translation invariance is

$$\frac{dx}{d\tau}+\frac{q}{m}A_x(t,z)\equiv u_x=const.$$

The other (trivial) integral of motion is

$$\frac{dy}{d\tau}\equiv u_y=const.$$

Nota bene: The integral $u_x$ is the ``canonical'' momentum (divided by $m$) along the $x$-direction.

c)

EXHIBIT the equations of motion and the integral of motion, Eq.(2), in terms of the potential function $A_x(t,z)$ only, and show that any two of Eqs.(2), (4), and (5) imply the third.

 

Answer: The equations of motion are

 

$$\frac{d^2t}{d\tau^2} = +\frac{1}{2}\frac{\partial}{\partial t} [u_x-\frac{q}{m}A_x(t,z)]^2$$ (6)

$$\frac{d^2z}{d\tau^2} = -\frac{1}{2}\frac{\partial}{\partial z} [u_x-\frac{q}{m}A_x(t,z)]^2$$ (7)

$$\frac{d^2y}{d\tau^2} = 0$$

$$\frac{dx}{d\tau} = [u_x-\frac{q}{m}A_x(t,z)],$$

while the integral of motion, Eq.(2), assumes the form

$$\left( \frac{dt}{d\tau}\right)^2 -\left( \frac{dz}{d\tau}\right)^2 =1+[u_x-\frac{q}{m}A_x(t,z)]^2~.$$ (8)

Multiplying Eq.(6) by $\frac{dt}{d\tau}$ and Eq.(7) by $\frac{dz}{d\tau}$, one finds that their difference is the $\tau$-derivative of Eq.(8). Conversely, differentiating Eq.(8), one finds with the help of one of the two Eqs.(6)-(7) that the other is satisfied.

In contrast to the particle's proper time $\tau$, the lab time $t$ of a laboratory clock is considerably more accessible as a standard of measurement. For this reason it is appropriate and better to recast the equations with $t$ as the independent parameter. Thus

d)

EXHIBIT the differential equations of motion with lab time $t$ as the independent variable. In other words, ignoring motion into the $y$-direction and using Eq.(2), show that

$$\begin{eqnarray*} \frac{dx}{dt}&=& f(t,z,\frac{dz}{dt},u_x)\\ \frac{d^2z}{dt^2}&=& g(t,z,\frac{dz}{dt},u_x), \end{eqnarray*}$$

where $u_x$ is the integration constant obtained in part b). Exhibit $f$ and $g$ in terms of the potential $A_x(t,z)$.

 

Answer:

 

$$\begin{eqnarray*} \frac{dx}{dt}&=&(u_x-\frac{q}{m}A_x)\sqrt{\frac{1-\left(\frac{... ...\frac{q}{m}A_x)^2} \left( 1-\left(\frac{dz}{dt}\right)^2 \right) \end{eqnarray*}$$

e)

Consider the circumstance where the laser beam is a standing wave with potential function

$$A_x=-\frac{E}{\omega}\sin \omega t~\cos\omega z$$ (9)

of frequency $\omega$ and wavelength $k_z=\omega$. Introduce the dimensionless relativistic factor

$$\frac{qE}{m\omega}\equiv\beta.$$ (10)

This number (``acceleration $\times$ time'') is the velocity (in units of the speed of light) a charge $q$ would acquire if a steady electric field $E$ were to act on it for the duration of one ``rationalized'' oscillation period ( $\frac{T}{2\pi}=\omega^{-1}$). In an oscillating field this number is an order of magnitude estimate of the maximum velocity a particle can acquire.

With this dimensionless factor in place, VERIFY that the the equations of motion for a charge in the standing wave field of a laser become

$$\frac{dx}{dt} = \left[ \frac{1-\left(\frac{dz}{dt}\right)^2}{1+(\beta \sin \omega t~\cos \omega z+u_x)^2} \right]^{1/2} (\beta \sin \omega t~\cos \omega z+u_x)$$ (11)

$$\frac{d^2z}{dt^2} = \left[ \frac{1-\left(\frac{dz}{dt}\right)^2}{1+(\beta \sin \omega t~\cos \omega z+u_x)^2} \right] (\beta \sin \omega t~\cos \omega z+u_x)$$

$$~ \times ~\omega \left(\beta \sin \omega t~\sin \omega z -\beta \cos \omega t~\cos \omega z~\frac{dz}{dt}\right).$$ (12)

POINT OUT why the shape of the particle's trajectories in spacetime, as determined by these nonlinear equations of motion in the lab, does not depend on the laser frequency $\omega$, but instead depends only on $\beta$ and on $u_x$, the integral of motion (initial $x$-momentum $\times~ m^{-1}$) found in b).

 

Validation:

In terms of the rescaled time and space variables

$$T=\omega t,~Z=\omega z,~X=\omega x$$

the equations of motion have the $\omega$-independent form

$$\begin{eqnarray*} \frac{dX}{dT}&=&\left[ \frac{1-\left(\frac{dZ}{dT}\right)^2}{1... ...(\beta \sin T~\sin Z -\beta \cos T~\cos Z~\frac{dZ}{dT}\right). \end{eqnarray*}$$

Thus one has to solve a two-parameter family of problems which depends on two parameters only. The key to understanding is Eq.(12). Once the solution to this second order nonlinear equation with periodic coefficients has been obtained, the solution to Eq.(11) is immediate.

The method of choice for identifying the solutions to Eq.(12) is to characterize them as trajetories in the phase space spanned by $z$ and $dz/dt$. Such a characterization yields a vector field of tangents and it reveals the key aspects of the particle motion, including the existence of periodic solutions, bounded solutions, solutions which are stable or unstable, etc. The family of solution curves which fill up the whole phase space make up the phase portrait of the differential equation of motion.

The fact that one has a $(\beta,u_x)$-parametrized family of differential Eqs.(12) implies that one has a $(\beta,u_x)$-parametrized family of phase portraits. The important question is this: What happens to a phase portrait as one varies the parameters? Does there occur a qualitative change in its appearance as one goes from weak laser fields ( $0<\vert\beta\vert \ll 1$), which accomodate nonrelativistic motion, to extremely strong laser fields ( $1\ll\vert\beta\vert$), which give rise to relativistic motion? The answer to such questions requires that one use diverse and powerful mathematical methods, which are readily available.

f)

With some exceptions, most lasers used nowadays are characterized by

$$0< \vert\beta\vert \ll 1.$$ (13)

Such laser fields accomodate nonrelativistic particle motion, which is characterized by

$$\left( \frac{dx}{d\tau} \right)^2+\left( \frac{dy}{d\tau} \right)^2 +\left( \frac{dz}{d\tau} \right)^2 \ll 1,$$ (14)

and whose description lends itself to a simple mathematical analysis.

SHOW that inequalities (13) and (14) imply

$$\left( \frac{dz}{dt} \right)^2\ll 1\textrm{~~and~~} \vert\beta u_x\vert \ll 1.$$ (15)

POINT OUT (i) how these inequalities lead from Eq.(12) to

$$\frac{d^2z}{dt^2} = \omega \beta^2 \left( \frac{1-\cos 2\omega t}{2}\cos \omega z ~\sin \omega z-\frac{\sin 2\omega t}{2}\cos^2 \omega z ~\frac{dz}{dt}\right),$$ (16)

(ii) why $\cos \omega z(t)$ and $\sin \omega z(t)$ are slowly varying functions of $t$, while $\cos \omega t$ and $\sin \omega t$ are rapidly varying, and (iii) why, as a consequence, the rapidly wiggling terms in Eq.(16) can be dropped, with the result that

$$\frac{d^2\overline z}{dt^2}+\frac{\omega\beta^2}{4} \sin 2\o... ...e z=0, \textrm{~~ where ~~ }z=\frac{\pi}{2\omega}+\overline z.$$ (17)

g)

By comparing this equation with the equation for a spherical pendulum of length $\ell$,

$$\frac{d^2\theta}{dt^2}+\frac{g}{\ell}\sin~\theta=0$$

POINT OUT why

$$\overline z=0,\pm\frac{\pi}{\omega},\pm 2\frac{\pi}{\omega},\cdots$$

are stable equilibrium points while

$$\overline z=\pm\frac{\pi}{2\omega},\pm 3\frac{\pi}{2\omega},\cdots$$

are unstable equilibrium points of oscillation for the $z$-motion of the charged particle. What is the electric field $E_x$ and the magnetic field $B_y$ experienced by the particle at a stable equilibrium point? At an unstable equilibrium point?

Answer:

The electromagnetic field for the standing wave mode is

$$\begin{eqnarray*} E_x&=&-\frac{\partial A_x(t,z)}{\partial t}=E\cos\omega t\,\co... ...frac{\partial A_x(t,z)}{\partial z}=E\sin\omega t\,\sin\omega z. \end{eqnarray*}$$

Figure: Standing wave pattern of a linearly polarized laser beam along the z-direction. The x-direction is upward, and the electric field has maxima at $z=0,2\pi,4\pi,\cdots$ . The magnetic field vanishes at these points.

At a stable point, namely, where

$$\omega z=\frac{\pi}{2},~\frac{\pi}{2} \pm\pi,~\frac{\pi}{2} \pm 2\pi,~\cdots$$

one has

$$\begin{eqnarray*} E_x&=&0\\ B_y&=&E \sin \omega t \times\left\{ \begin{array}{l... ...extrm{if }\omega z=\frac{3\pi}{2}+2\pi n \end{array} \right. ~. \end{eqnarray*}$$

At a unstable point, namely, where

$$\omega z=0,~\pm\pi,~\pm 2\pi,~\cdots$$

one has

$$\begin{eqnarray*} E_x&=&E \cos \omega t \times\left\{ \begin{array}{ll} +1&\tex... ...xtrm{if }\omega z=\pi +2\pi n \end{array} \right. ~.\\ B_y&=&0 \end{eqnarray*}$$

h)

POINT OUT why for small oscillations ( $\vert\omega \overline z\vert \ll 1$) the ratio of the laser frequency $\omega$ to the frequency $\Omega$ of the $z$- motion of the charged particle is

$$\frac{\omega}{\Omega}= \sqrt2~\beta^{-1}~\left(=\sqrt2~\left(\frac{qE}{m\omega}\right)^{-1}\right).$$

Answer:

For $\vert\omega \overline z\vert \ll 1$, Eq.(17), which governs the particle's z-motion, becomes

$$\frac{d^2\overline z}{dt^2}+\frac{\omega^2\beta^2}{2}\overline z=0.$$ (18)

Its oscillation frequency is evidently

$$\Omega=\frac{\omega\beta}{\sqrt{2}}$$

in terms of the laser frequency $\omega$ and laser's relativistic factor $\beta$. The ratio of the corresponding time periods is

$$\frac{z\textrm{-motion period}}{\textrm{laser period}}=\frac{\sqrt{2}}{\beta}$$

i)

Applying the vector potential, Eq.(9), to the equations of motion found in part c), setting the integral of motion associated with the $x$-motion (the ``x-momentum') equal to zero, POINT OUT why the resulting equations of motion have the mathematical form of two parametrically coupled spherical pendulums. POINT OUT the physical nature of their coupling. POINT OUT why one of the pendulums always executes rotational motion, and why the other executes librations or rotation depending on whether the charged particle moves nonrelativistically or relativistically along the $z$-direction.

 

Answer:

 

Setting $u_x=0$, one finds that the equations of motion, Eqs.(4)-(5), in part c) are

$$\frac{d^2t}{d\tau^2} = \frac{\omega\beta^2}{4}(1+\cos 2\omega z) \sin 2\omega t$$ (19)

$$\frac{d^2z}{d\tau^2} = \frac{\omega\beta^2}{4}(1-\cos 2\omega t) \sin 2\omega z~.$$ (20)

Their integral of motion, Eq.(2), is

$$\left( \frac{dt}{d\tau}\right)^2 -\left( \frac{dz}{d\tau}\right)^2 =1+\frac{\beta^2}{4}(1-\cos2\omega t)(1+\cos2\omega z)~.$$

Introducing the two deflection angles

$$\begin{eqnarray*} \theta_t&=& 2\omega t -\pi\\ \theta_z&=& 2\omega z -\pi\\ \end{eqnarray*}$$

for the ``$t$-pendulum'' and the ``$z$-pendulum'', one has

$$\frac{d^2\theta_z}{d\tau^2}+\frac{\omega^2\beta^2}{2}(1+\cos... ...au^2}+\frac{\omega^2\beta^2}{2}(1-\cos \theta_z)\sin\theta_t=0$$ (21)

The integral of motion, Eq.(2), assumes the form

$$\frac{1}{2}\left[ \left( \frac{d\theta_t}{d\tau}\right)^2 -... ...2+\frac{\omega^2\beta^2}{2} (1+\cos\theta_t)(1-\cos\theta_z)~.$$ (22)

The three equations, Eqs.(21)-(22), are not independent. As pointed out in part c), any two imply the third. The first two equations are those of two identical parametrically coupled pendulums in respective force fields which are under the mutual control of their vertical amplitudes. As a consequence, the third equation expresses that the DIFFERENCE in their kinetic energies is constrained by the product of their vertical amplitudes. However, there is no bound on the SUM of their kinetic energies.

Figure2: Dynamics of two parametrically coupled pendulums as the dynamics of a charge in the field of a standing e.m. wave. The rotational motion of the right hand pendulum establishes the relation between the charge's comoving (proper, $\tau$) time and the labtime ( $\propto(\theta_t+\pi)$, in units of (half) the laser period). Each pendulum moves in a force field which is proportional to the squared laser amplitude $\beta^2$, which gets modulated by the vertical amplitude ( $1\pm\cos\theta_{\!\!~^t_z}$) of the other pendulum.

• Furthermore, the ``time'' pendulum always rotates. It never librates. The rotational velocity always equals or exceeds $2\omega$ (twice the laser frequency). The concomitant vertical amplitude oscillations ( $1+\cos\theta_t$) produce a correspondingly variable force field experienced by the other (``$z$'') pendulum. The magnitude of this force field is proportional to the laser intensity expressed by $\beta^2=(qE/m\omega)^2$. This force field is responsible for the parametric excitation of the $z$-pendulum, and hence for the parametric control over the particle's motion into the $z$-direction.

  If the force field and its variations are small ($\beta^2\ll1$) and the $z$-pendulum starts out from its equilibrium ($\theta_z=0$) with small velocity, then the pendulum will librate very slowly even though the force field oscillates rapidly by comparison. In this circumstance the force field
  

  $$\frac{\omega^2\beta^2}{2}(1+\cos\theta_t) ~~~~~~~~~(\textrm{\lq\lq oscillating force field''})$$
  
  
  oscillates many times during the slow motion of the $z$-pendulum. Consequently, the force field averages to its mean value
  
  $$\frac{\omega^2\beta^2}{2} ~~~~~~~~~~~~~~~~~~~~ (\textrm{\lq\lq mean force field''}).$$
  
  
  The corresponding period of the $z$-pendulum is therefore well-defined and has a value which is much longer than the (half) period of the laser,
  
  $$\left( \begin{array}{c} \textrm{period of } z\textrm{-pendu... ...{c} \textrm{ period}\\ \textrm{of laser} \end{array}\right) .$$
  
  
  The fact that $z$-pendulum's motion is slow is expressed by the fact that
  
  $$\frac{d\theta_z}{d\tau}\equiv \left\vert \frac{d\, 2\omega z}{d\tau}\right\vert \ll 2\omega$$
  
  
  Consequently, $\vert dz/d\tau\vert\ll 1$, ie. the particles motion is nonrelativistic. However it is not clear whether parametric resonance is capable of depositing enough energy into the $z$-pendulum to have it go from a state of libration to a state of rotation, and hence express the particle's $z$-motion become relativistic.