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2008 II 3

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This problem asks us to consider the motion of a relativistic electron in a static magnetic field generated by a helical "wiggler" magnet configuration. This field is given by

$\mathbf{B}^0_w(\mathbf{x}) = -B_w\left[\cos(k_w z)\mathbf{\hat{e}}_x + \sin(k_w z)\mathbf{\hat{e}}_y\right],$

where $\displaystyle B_w = \mbox{const.}$ and $\displaystyle \lambda_w = 2\pi/k_w = \mbox{const.}$ are the wiggler field amplitude and wavelength. The relativistic mass factor $\displaystyle\gamma$ is given by

$\displaystyle \gamma = \sqrt{1 + \frac{\mathbf{p}^2}{m^2 c^2}},$

where $\displaystyle\mathbf{p} = \gamma m \mathbf{v}$ is the kinematic momentum.

Part (a)

Solve for that particle velocities in the particle orbit (primed) frame using the relativistic Lorentz force equation, which in Gaussian units is given by

$\frac{d\mathbf{p}}{dt} = -e\left[\mathbf{E}(\mathbf{x},t) + \frac{\mathbf{p}\times\mathbf{B}(\mathbf{x},t)}{\gamma m c}\right].$

First, show that $\displaystyle \gamma'(t') = \gamma = \mbox{const.}$ by taking its time derivative and applying the Lorentz force equation:

$\frac{d\gamma'(t')}{dt'} = \frac{d}{dt'}\left(\sqrt{1 + \frac{(\mathbf{p}')^2}{m^2 c^2}}\right) = \frac{1}{2\gamma'}\,\mathbf{p'}\cdot\frac{d\mathbf{p'}}{dt'} = -\frac{e}{2\gamma'}\,\mathbf{p'}\cdot\left[\mathbf{E}(\mathbf{x}',t') + \frac{\mathbf{p'}\times\mathbf{B}(\mathbf{x}',t')}{\gamma' m c}\right]$

The notation in the above expression implicitly assumes that $\displaystyle\mathbf{p}' = \mathbf{p}'(t')$ and $\displaystyle\mathbf{x}' = \mathbf{x}'(t')$ . The first term vanishes because $\displaystyle\mathbf{E}(\mathbf{x}',t')=0$ in this problem, and the second term vanishes due to orthogonality. This gives the result that $\displaystyle d\gamma'/dt' = 0$ such that

$\displaystyle \gamma'(t') = \gamma = \mbox{const.}$

Now solve for the components of the particle orbit velocities in the primed frame. Use the method of characteristics to directly integrate both sides of the Lorentz force equation, which after inserting the wiggler magnetic field is given by

$\frac{d\mathbf{p}'}{dt'} = -e\left[\frac{\mathbf{p}\times\mathbf{B}^0_w(\mathbf{x}')}{\gamma' m c}\right] = -\frac{e}{c}\left[\left(\frac{\mathbf{p'}}{\gamma'm}\right)\times\mathbf{B}^0_w(\mathbf{x}')\right]$

Because $\displaystyle\gamma'$ is constant, it can be brought out through the time derivative, giving

$\frac{d\mathbf{p}'}{dt'} = \gamma m \frac{d\mathbf{v}'}{dt'} = -\frac{e}{c}\left[\left(\frac{\mathbf{p'}}{\gamma'm}\right)\times\mathbf{B}^0_w(\mathbf{x}')\right] = -\frac{e}{c}\bigg[\mathbf{v}'\times\mathbf{B}^0_w(\mathbf{x}')\bigg].$

Evaluating the cross product gives

$\mathbf{v}'\times\mathbf{B}^0_w(\mathbf{x}') = \mathbf{v}'\times\left[B^0_{w,x}(\mathbf{x}')\mathbf{\hat{e}}_x + B^0_{w,y}(\mathbf{x}')\mathbf{\hat{e}}_y\right] = -v'_z B^0_{w,y}\mathbf{\hat{e}}_x + v'_z B^0_{w,x}\mathbf{\hat{e}}_y + \left(v'_x B^0_{w,y} - v'_y B^0_{w,x} \right)\mathbf{\hat{e}}_z,$

which recombines into the Lorentz force equation giving

$\gamma m \frac{d\mathbf{v}'}{dt'} = -\frac{e}{c}\bigg[\mathbf{v}'\times\mathbf{B}^0_w(\mathbf{x}')\bigg] = \frac{e}{c}\bigg[v'_z B^0_{w,y}\mathbf{\hat{e}}_x - v'_z B^0_{w,x}\mathbf{\hat{e}}_y - \left(v'_x B^0_{w,y} - v'_y B^0_{w,x} \right)\mathbf{\hat{e}}_z\bigg]$

Separating the $\displaystyle\mathbf{\hat{e}}_x$ and $\displaystyle\mathbf{\hat{e}}_y$ components of the above equation gives

$\displaystyle \gamma m \frac{dv'_x}{dt'} = \;\;\,\displaystyle\frac{e}{c}\,v_z'(t')B^0_{w,y}(\mathbf{x}') = -\displaystyle\frac{e}{c}\,v_z'(t')B_w\sin[k_wz'(t')]$

$\displaystyle \gamma m \frac{dv'_y}{dt'} = -\displaystyle\frac{e}{c}\,v_z'(t')B^0_{w,x}(\mathbf{x}') = \;\;\,\displaystyle\frac{e}{c}\,v_z'(t')B_w\cos[k_wz'(t')]$

and the $\displaystyle\mathbf{\hat{e}}_z$ component gives

$\gamma m \frac{dv'_z}{dt'} = \frac{e}{c}\bigg[v'_x(t') B^0_{w,y}(\mathbf{x}') - v'_y(t') B^0_{w,x}(\mathbf{x}')\bigg] = -\frac{e}{c}\,B_w\bigg[v'_x(t')\sin[k_wz'(t')] - v'_y(t')\cos[k_wz'(t')]\bigg].$

The goal is to express the right hand sides of the $\displaystyle\mathbf{\hat{e}}_x$ and $\displaystyle\mathbf{\hat{e}}_y$ equations in terms of $\displaystyle d/dt'$ so that these equations can be directly integrated in $\displaystyle t'$ along the characteristics of the system. Noting that

$\displaystyle \frac{d}{dt'}\cos[k_wz'(t')] = \displaystyle -k_w\left(\frac{dz'(t')}{dt'}\right)\sin[k_wz'(t')] = -k_w v'_z(t')\sin[k_wz'(t')]$

$\displaystyle \frac{d}{dt'}\sin[k_wz'(t')] = \;\;\,\displaystyle k_w\left(\frac{dz'(t')}{dt'}\right)\cos[k_wz'(t')] = \;\;\,k_w v'_z(t')\cos[k_wz'(t')]$

Substituting these expressions into the $\displaystyle\mathbf{\hat{e}}_x$ and $\displaystyle\mathbf{\hat{e}}_y$ equations yields

$\displaystyle \gamma m \frac{dv'_x}{dt'} = -\displaystyle\frac{e}{c}\,v_z'(t')B_w\sin[k_wz'(t')] = \displaystyle\left(\frac{eB_w}{c k_w}\right)\frac{d}{dt'}\cos[k_wz'(t')]$

$\displaystyle \gamma m \frac{dv'_y}{dt'} = \;\;\,\displaystyle\frac{e}{c}\,v_z'(t')B_w\cos[k_wz'(t')] = \displaystyle\left(\frac{eB_w}{c k_w}\right)\frac{d}{dt'}\sin[k_wz'(t')]$

Directly integrating both sides with respect to $\displaystyle t'$ gives the final expressions for $\displaystyle v'_x(t')$ and $\displaystyle v'_y(t')$ , which are

$v'_x(t') = \displaystyle\frac{ca_w}{\gamma}\cos[k_wz'(t')]$

$v'_y(t') = \displaystyle\frac{ca_w}{\gamma}\sin[k_wz'(t')]$

where $\displaystyle a_w \equiv e B_w/mc^2k_w$ .

Going back to the equation for $\displaystyle dv'_x(t')/dt'$ from the $\displaystyle\mathbf{\hat{e}}_z$ component of the Lorentz force equation, which is,

$\gamma m \frac{dv'_z}{dt'} = -\frac{e}{c}\,B_w\bigg[v'_x(t')\sin[k_wz'(t')] - v'_y(t')\cos[k_wz'(t')]\bigg],$

and substituting for $\displaystyle v'_x(t')$ and $\displaystyle v'_y(t')$ from above gives

$\gamma m \frac{dv'_z}{dt'} = -\frac{ea_w}{\gamma}\,B_w\bigg[\cos[k_wz'(t')]\sin[k_wz'(t')] - \sin[k_wz'(t')]\cos[k_wz'(t')]\bigg] = 0$

Thus, because $\displaystyle dv'_z/dt'=0$ , $\displaystyle v'_z(t')$ is simply

$\displaystyle v'_z(t') = v_z = \mbox{const.}$

The above condition directly implies that

$\displaystyle z'(t') = z + v_z(t'-t),$

which is a quantity that is present in the expressions for $\displaystyle v'_x(t')$ and $\displaystyle v'_y(t')$ . Note: This last equation is already given in the problem.

Part (b)

The spontaneous emission spectrum $\displaystyle \eta(\omega)$ is given in the problem to be of the form

$\eta(\omega) \equiv \frac{1}{T}\frac{d^2I}{d\omega d\Omega} = \frac{e^2\omega^2}{4\pi^2c^3T} \Bigg|\int_0^T\!d\tau\,\Big[v'_x(\tau)\mathbf{\hat{e}}_x + v'_y(\tau)\mathbf{\hat{e}}_y\Big]\exp\Big\{ik_z z'(\tau) - i\omega\tau\Big\}\Bigg|^2,$

where $\displaystyle\omega$ is the emission frequency, $\displaystyle k_z$ is the wavenumber, $\displaystyle T = L/v_z$ is the length of time that the electron is in the interaction region of length $\displaystyle L$ , and $\displaystyle d^2 I/d\omega d\Omega$ is the energy radiated per unit frequency interval per unit solid angle. The integration variable is $\displaystyle\tau\equiv t'-t$ such that $\displaystyle z'(\tau) = z + v_z\tau$ .

Defining $\displaystyle\mathcal{I}$ as the integral in the $\displaystyle \eta(\omega)$ expression such that

$\mathcal{I}\equiv\int_0^T\!d\tau\,\Big[v'_x(\tau)\mathbf{\hat{e}}_x + v'_y(\tau)\mathbf{\hat{e}}_y\Big]\exp\Big\{ik_z z'(\tau) - i\omega\tau\Big\} = \mathcal{I}_x\mathbf{\hat{e}}_x + \mathcal{I}_y\mathbf{\hat{e}}_y,$

where the separated integrals $\displaystyle\mathcal{I}_x$ and $\displaystyle\mathcal{I}_y$ are given by

$\mathcal{I}_x = \int_0^T\!d\tau\,v'_x(\tau)\exp\Big\{ik_z z'(\tau) - i\omega\tau\Big\} = \frac{ca_w}{\gamma}\int_0^T\!d\tau\,\cos[k_wz'(\tau)]\exp\Big\{ik_z z'(\tau) - i\omega\tau\Big\}$

$\mathcal{I}_y = \int_0^T\!d\tau\,v'_y(\tau)\exp\Big\{ik_z z'(\tau) - i\omega\tau\Big\} = \frac{ca_w}{\gamma}\int_0^T\!d\tau\,\sin[k_wz'(\tau)]\exp\Big\{ik_z z'(\tau) - i\omega\tau\Big\}.$

Exponentiating the $\displaystyle\cos$ and $\displaystyle\sin$ factors gives

$\mathcal{I}_x = \frac{ca_w}{2\gamma}\Big(\mathcal{I}_+ + \mathcal{I}_-\Big) ~~\mbox{and}~~ \mathcal{I}_y = \frac{ca_w}{2i\gamma}\Big(\mathcal{I}_+ - \mathcal{I}_-\Big),$

where

$\mathcal{I}_{\pm} \equiv \int_0^T\!d\tau\,\exp\Big\{\pm i k_wz'(\tau)\Big\}\exp\Big\{ik_z z'(\tau) - i\omega\tau\Big\}.$

Substituting for $\displaystyle z'(\tau) = z + v_z\tau$ and combining terms gives

$\mathcal{I}_{\pm} = e^{i(k_z\pm k_w)z}\int_0^T\!d\tau\,\exp\Big\{-i[\omega - (k_z\pm k_w)v_z]\tau\Big\},$

which can be directly integrated to give

$\mathcal{I}_{\pm} = - \frac{e^{i(k_z\pm k_w)z}}{i[\omega - (k_z\pm k_w)v_z]}\exp\Big\{-i[\omega - (k_z\pm k_w)v_z]\tau\Big\}\bigg|_0^T.$

This expression has resonant peaks near $\displaystyle \omega = (k_z \pm k_w)v_z$ . The problem asks us to focus on the upshifted resonant peak near $\displaystyle \omega = (k_z + k_w)v_z$ . Here, $\displaystyle\mathcal{I}_{+}\gg\mathcal{I}_{-}$ due to the resonant denominator, so only the $\displaystyle\mathcal{I}_{+}$ terms need to be considered when evaluating $\displaystyle\mathcal{I}$ to find $\displaystyle \eta(\omega)$ near the upshifted resonant peak. The reduced expressions for $\displaystyle\mathcal{I}_{x}$ and $\displaystyle\mathcal{I}_{y}$ are now

$\mathcal{I}_x \simeq \frac{ca_w}{2\gamma}\mathcal{I}_+ ~~\mbox{and}~~ \mathcal{I}_y \simeq \frac{ca_w}{2i\gamma}\mathcal{I}_+.$

Defining

$\displaystyle\omega_{eff}\equiv\frac{1}{2}\Big[\omega - (k_z + k_w)v_z \Big],$

the expression for $\displaystyle\mathcal{I}_{+}$ evaluates to

$\mathcal{I}_{+} = -\frac{e^{i(k_z+k_w)z}}{2i\omega_{eff}}e^{-2i\omega_{eff}\tau}\bigg|_0^T = -\frac{e^{i(k_z+k_w)z}}{2i\omega_{eff}}\bigg[e^{-2i\omega_{eff}T} - 1 \bigg].$

Factoring to find a $\displaystyle\sin$ function gives

$\mathcal{I}_{+} = \frac{1}{\omega_{eff}}e^{i(k_z+k_w)z}e^{-i\omega_{eff}T}\left\{\frac{1}{2i}\bigg[e^{i\omega_{eff}T} - e^{-i\omega_{eff}T} \bigg]\right\} = Te^{i(k_z+k_w)z}e^{-i\omega_{eff}T}\left[\frac{\sin(\omega_{eff}T)}{\omega_{eff}T}\right]$

Now going back to evaluate $\displaystyle \eta(\omega)$ ,

$\eta(\omega) = \frac{e^2\omega^2}{4\pi^2c^3T}\big|\mathcal{I}\big|^2 = \frac{e^2\omega^2}{4\pi^2c^3T}\bigg[\big|\mathcal{I}_x\big|^2 + \big|\mathcal{I}_y\big|^2\bigg] = \frac{e^2\omega^2}{4\pi^2c^3T}\bigg[\frac{c^2a_w^2}{2\gamma^2}\big|\mathcal{I}_+\big|^2\bigg] = \frac{e^2\omega^2a_w^2T}{8\pi^2c\gamma^2}\frac{1}{T^2}\big|\mathcal{I}_+\big|^2.$

Substituting for $\displaystyle\mathcal{I}_{+}$ gives the final expression for $\displaystyle \eta(\omega)$ to be

$\eta(\omega) \equiv \frac{1}{T}\frac{d^2I}{d\omega d\Omega} = \frac{e^2\omega^2a_w^2T}{8\pi^2c\gamma^2}\left[\frac{\sin^2(\omega_{eff}T)}{(\omega_{eff}T)^2}\right],$

where $\displaystyle\omega_{eff}\equiv[\omega - (k_z + k_w)v_z]/2$ .

Part (c)

In this last part, we are asked to show that if the emitted radiation corresponds to light wave in vacuum such that $\displaystyle\omega=ck_z$ , then the radiation spectrum $\displaystyle \eta(\omega)$ that we calculated above is maximized for a wavelength $\displaystyle \lambda_z = 2\pi/k_z$ given by

$\lambda_z = \frac{1+a_w^2}{\gamma^2\beta_z(1+\beta_z)}\lambda_w,$

where $\displaystyle \beta_z\equiv v_z/c$ , $\displaystyle \lambda_w = w\pi/k_w$ , and $\gamma = (1+\beta_z^2\gamma^2+a_w^2)^{1/2}$ .

Because the final expression for $\displaystyle\eta(\omega)$ has a $\displaystyle \sin x/x$ dependence, it will be maximized when $\displaystyle x = \omega_{eff}T = 0$ . Thus, examining

$\omega_{eff} = \frac{1}{2}\Big[\omega - (k_z + k_w)v_z \Big] = \frac{1}{2}\left[\frac{2\pi c}{\lambda_z} - \left(\frac{2\pi}{\lambda_z} + \frac{2\pi}{\lambda_w}\right)v_z \right] = \pi\left[\frac{1}{\lambda_z}\big(c-v_z\big) - \frac{1}{\lambda_w}v_z \right] = 0$

gives

$\lambda_z = \left(\frac{c-v_z}{v_z}\right)\lambda_w = \left(\frac{c}{v_z}-1\right)\lambda_w \left(\frac{1}{\beta_z}-1\right)\lambda_w = \left(\frac{1-\beta_z}{\beta_z}\right)\lambda_w.$