Relativistic acceleration in sinusoidal electric field Consider a relativistic charge $q$ moving with an oscillating electric field $E_z$ with phase velocity $v_p=c$ in direction $\hat{z}$ (e.g. radially polarized laser coprogating with electron).  What is the energy gain of this charge as a function of time?
I set this up from the relativistic force 
$$F=\frac{dp}{dt}=\frac{d}{dt}(\gamma m \dot{z})= qE_0 \sin{((c-\dot{z})t k)}$$
where $t$ is time, and $k$ is the usual wave number $k=2\pi/\lambda$.
My confusion arises from the $\dot{z}$ on the RHS.  I don't have much experience with differential equations, and so I wonder if it is necessary to actually write it as $\dot{z}=\int_0^t dt' \ddot{z}$ or if the LHS' attribution of $\ddot{z}$ automatically leads to an appropriate $\dot{z}$?
Thank you for any help.
After some thinking the equation written above is actually wrong.  It should indeed be written like so: 
$$
F=\frac{dp}{dt}=\frac{d}{dt}(\gamma m \dot{z})= qE_0 \sin{(k(ct-\int_0^t\dot{z}dt' +z(t=0))}
$$ 
I would appreciate help solving this equation if someone has experience. TY!
 A: So actually this is not as easy as you assume, since for charged particle velocities being a significant fraction of c, you cannot omit the magnetic field in the Lorentz force anymore. Then the equation of motion has an additional term, leading to the so called figure of eight motion, which an electron in an intense laser field performs in the co-moving frame of reference. But however, you can solve the equation analytically, deriving a drift velocity of the particle, which is connected to its kinetic energy.
Have a look here, chapter 2.1.2:
PhD Thesis in the field of relativistic laser plasma interactions
A: The very first thing you should do is to Lorentz transform into the initial rest frame of the electron. This leads to a doppler shift in the electric field. If you really want to omit the magnetic field your equation of motions become 1 dimensional because the electron will only oscillate in the direction of the polarisation of the light field. The equations of motion become (I use a cosine field here):
$$\frac{d}{dt} p  = q E_0 \cos(\tilde{\omega} t)  $$
$$\Rightarrow p = \frac{E_0}{\omega} \sin(\tilde{\omega}t)$$
$$\frac{d}{dt} z = \frac{p}{m_e \sqrt{1+\frac{p^2}{m_e^2 c^2}}} $$
This can be solved after inserting the solution for $p$ by simple integration and yields
$$z(t) = \frac{c}{\tilde{\omega}}\left(\arctan{\alpha} - \arctan{\left( \frac{\alpha \cos(\tilde{\omega}t)}{\sqrt{1+\alpha^2 \sin^2(\tilde{\omega}t)}} \right)} \right)$$
with $\alpha = \frac{E_0}{m_e c \omega}$
Now in the last step we have to transform back into the lab frame. This can be done with the replacement 
$$t = \gamma_0 (t_\text{lab}+\frac{v_0 x_\text{lab}}{c^2}) $$
and trivially
$$z_\text{lab} = z$$
$$x_e = v_0 t_\text{lab} $$
finally
$$\tilde{\omega} = \sqrt{\frac{1-\frac{v_0}{c}}{1+\frac{v_0}{c}} }\omega $$
Below is a plot of the solution for different $\alpha$. You can clearly see that it goes from simple harmonic, which is the newtonian limit to a triangular function, which is the relativistic limit.
[1
A: So lets make this a little bit more detailed.
To make life easy, we consider the wave not in terms of its electrical field, but in terms of its vector potential.
$${\bf A}({\bf r},t)={\bf e}_y {\bf A}_0\sin(\varphi), \quad \varphi=kx-\omega t \qquad({\rm I})$$
To remember, in analogy to a potential in an conservative force field, you can define a vector potential for a force field, whose curl isn't vanishing by
$${\bf B}=\nabla\times{\bf A},\qquad({\rm II})$$
$${\bf E}=-\nabla\Phi-\frac{\partial}{\partial t}{\bf A},\qquad({\rm III})$$
Here we take $\Phi=0$ (Coulomb gauge). The first important thing to mention here, is the quantity
$$a_0=\frac{eA_0}{m_e c_0}=\frac{v_{class}}{c},\qquad({\rm IV})$$
which is called normalized vector potential. This is basically the classical quiver velocity of an electron in an electromagnetic field divided by the speed of light. This actually tells you, when you need to handle the motion of an electron in an em-field relativistically (i.e. for $a_0\ge 1$).
Okay, lets write down the equation of motion of the electron and the energy equation:
$$\frac{\mathrm{d}}{\mathrm{d}t}\left(\gamma m_e {\bf v}\right)=-e\left({\bf E} + {\bf v} \times {\bf B}\right),\qquad({\rm V})$$
$$\frac{\mathrm{d}}{\mathrm{d}t}\left(\gamma m_e c_0^2\right)=-e{\bf vE},\qquad({\rm VI})$$
If we now take the equation of motion (V), then we can rearrange it by using equations (II), (III) and the relations $\frac{\mathrm{d}{\bf A}}{\mathrm{d}t}=\frac{\partial {\bf A}}{\partial t}+(\frac{\mathrm{d}{\bf r}}{\mathrm{d}t} \cdot \nabla){\bf A}$,  ${\bf v}\times(\nabla\times{\bf A})=\nabla({\bf vA})-({\bf v}\nabla){\bf A}$ and $\frac{\partial {\bf A}}{\partial y}=\frac{\partial {\bf A}}{\partial z}=0$ to obtain
$$\frac{\mathrm{d}p_y}{\mathrm{d}t}=e\frac{\mathrm{d}A}{\mathrm{d}t}$$
Integration leads to $$p_y-eA=C_1=const.,\qquad({\rm VII)}$$ which is called the first invariant of the electron motion.
If you now take the energy equation (VI), you'll find with the help of ${\bf B_0}={\bf E_0}/c$ and ${\bf \tilde{p}}={\bf p}/(m_e c_0)$ the second invariant of electron motion
$$\frac{\mathrm{d}\tilde{p}_x}{\mathrm{d}t}=\frac{\mathrm{d}\gamma}{\mathrm{d}t} \quad  \Rightarrow \quad \gamma-\tilde{p}_x=C_2=const.,\qquad({\rm VIII)}$$
From both invariants (VII) and (VIII) you can derive the relation between the momentum in x and y direction, by taking into account that $\gamma^2=1+\tilde{p}^2$:
$$\tilde{p}_x=\frac{1-C^2_2+\tilde{p}_y^2}{2C_2}.\qquad({\rm IX)}$$
Actually everything is fine, up to here. If you want to go further, you need to determine the invariants, by setting the starting condition of your system.
For an electron in an intense laser field you simply can choose your starting conditions of your electron to be at rest at $t=0$ and at $x=0$. Then the invariants are $C_1=0$, $C_2=1$ and you'll get from (VII) and (IX)
$$\tilde{p}_y = \ \frac{eA}{m_e c_0}= a$$
$$\tilde{p}_x = \ \frac{\tilde{p}_y^2}{2}=\frac{a^2}{2}$$
$$\gamma = \ 1+\tilde{p}_x=1+\frac{a^2}{2}$$
And actually from this the kinetic energy of the particle computes as
$$\mathcal{E}_{kin}=(\gamma-1)m_e c_0^2=m_e c_0^2\frac{a^2}{2}$$
So be aware, that $a=a(x,t)$ is here a function of time and space!
If you want a solution for a relativistically particle in a weak em-field, then you need to set your starting conditions and see where you get from equation (IX) onwards.
A: Since I can't comment yet, here some thoughts to your initial question:
A changing electric field will always generate a magnetic field due to Maxwell:
$$\nabla \times {\bf H} = \varepsilon_0 \frac{\partial {\bf E}}{\partial t}$$
So you can't consider an electric field changing with time seperately! And you want to consider a particle moving with $\gamma>1$. The equation of motion for an electron in an em-field is 
$$\frac{{\rm d}}{{\rm d}t}(\gamma m_e c_0^2)=-e({\bf E + v\times B})$$.
The contribution of the magnetic field to the force is suppresed by a factor of $1/c_0$ against the electric field contribution, i.e. $B_0=\frac{E_0}{c_0}$. But this means, that the contribution to the force ${\bf v \times B}$ is not negligible, when $v\sim c$. If you want to neglect the magnetic field, you either have to consider a constant electric field, or a particle with $\gamma \ll 1$.
