Non-relativistic limit of a particle in the uniform electric field 
For this problem, the solution is:
$$y=\frac{m}{qE}\gamma\left[\cosh\left(\frac{qEx}{mv\gamma}\right)-1\right]$$
where $\gamma = 1/\sqrt{1-v^2}$. Here something seems to be wrong. I think that for the non-relativistic limit where $v$ is very small in magnitude, the trajectory of $y$ must be a parabola. However, this kind of behavior happens when $v$ is close to $1$, when $1/(v\gamma)$ becomes small, so the taylor expansion works for some range of $x$. How should I account for this phenomenon? In Newtonian mechanics, it seems clear that the trajectory of $y$ is a parabola. 
 A: First, the Newtonian result is that 
$$
y = \frac{1}{2} a t^2 = \frac{1}{2} \frac{qE}{m} t^2, \qquad x = vt;
$$  combining these two equations, we obtain 
$$
y = \frac{1}{2} \frac{qE}{mv^2} x^2.
$$
Note that the $x$-velocity $v$ does in fact appear in this result.
To obtain this as a limit of the relativistic result, you need to take is the limit $x \ll m v / q E$, with $v \ll 1$ (but not zero) so that $\gamma \approx 1$.  This is always possible so long as $v \neq 0$.  In this limit, you can use the fact that $\cosh x \approx 1 +x^2/2$ to find that
$$
y(x) \approx \frac{m \gamma}{qE} \left[ 1 + \frac{1}{2} \left( \frac{qEx}{mv} \right)^2 - 1 \right] = \frac{1}{2} \frac{qE}{m v^2} x^2,
$$
in agreement with the Newtonian result.  Nowhere in this derivation is it assumed that $v \approx 1$;  in fact, it is necessary to have $v \ll 1$ so that $\gamma \approx 1$.
Note that you can't combine the Newtonian equations in the case $v = 0$ either, because the particle's trajectory is a straight line in this case.  Rather, you have to have a non-zero initial velocity that is much less than the speed of light, and only track the particle over ranges of time such that both its $x$-velocity and $y$-velocity remain small compared to the speed of light.  In particular, you'll probably find in the course of doing this problem that the $y$-component of the four-momentum of the particle is
$$
p^y = qEt,
$$
and if $v \ll 1$, then the non-relativistic limit is the limit in which this quantity is much less than the rest mass:  $qEt \ll m$.  Since $x = vt$ in the non-relativistic limit, this then implies that we must also have $x \ll mv/qE$, which is precisely the limit I stated at the top.
A: Electric and Magnetic fields transform for moving frames of reference:
$$\begin{align}
\vec E' &= \gamma( \vec E + \vec v \times \vec B ) - (\gamma - 1)(\vec E \cdot \hat v) \hat v \\
\vec B' &= \gamma( \vec B - \frac{\vec v \times \vec E}{c^2} ) - (\gamma - 1)(\vec B \cdot \hat v) \hat v \\
\end{align}$$
Note that these transformations have non-zero terms even for $\gamma \approx 1$. In your assignment, $\vec v(0) = v \hat x$, $\vec B = 0$ and $\vec E = E\hat y$ in the rest frame. Then in the frame of the electron, $\vec B'$ and $\vec E'$ are:
$$\begin{align}
\vec E' &= \gamma E \hat y \\
\vec B' &= -\frac{\gamma vE (\hat v \times \hat y)}{c^2} \\
\end{align}$$
The total force on the electron is then given by the Lorentz force law:
$$\vec F = q((1+\gamma {v^2 \over c^2}) E\hat y - \gamma {v^2 \over c^2} E(\hat v \cdot \hat y)\hat v )$$
This only reduces to the parabolic case when $v \approx 0$, but reduces to a non-relativistic (but still non-parabolic) case much earlier, when $\gamma \approx 1$. 
