Oscillation velocity of an electron in an electric field greater than c? In relativistic laser matter interaction it is usually defined a parameter $a_0$ as $$a_0 = \frac{e E}{m_e \omega c},$$ with $E$ the amplitude of the oscillating electric field of frequency $\omega$. The quiver (or oscillation) velocity of the electron in the electric field is $$v_{osc} = \frac{e E}{m_e \omega},$$ so that $$a_0 = v_{osc}/c.$$ The relativistic regime of interaction kicks in when $a_0 \geq 1$. This means $v_{osc} \geq c$: how is this possible? $v_{osc}$ shouldn't be always $< c$?
 A: Many formulas derived using Newtonian mechanics will fail when applied in a relativistic setting. The formula for Kinetic energy in terms of velocity is a good example: $\text{Ke}=\frac{1}{2}mv^2$. In a relativistic setting, it has to be modified to $$\text{Ke}=m c^2\left( \frac{1}{\sqrt{1-v^2/c^2}}-1\right)$$
Your quantity does have units of velocity though, and there's nothing wrong with quantities with units of velocity being greater in magnitude than $c$. You just have to keep in mind that it is not necessarily the actual velocity of something moving through spacetime.
Getting an actual relativistic velocity
If we wanted to find that quantity, then I would do the following. The most natural quantity to make with units of momentum is:
$$p_{c} = \frac{e E}{\omega}$$
where I write "c" to stand for "characteristic." Then, relativistic momentum is defined as $p=mv/\sqrt{1-v^2/c^2}$, so we can solve this for $v$ to get: $v=cp/\sqrt{(mc)^2+p^2}$. To make this look more like the Newtonian case, I can write this as $$v=\frac{p}{m}\cdot \frac{1}{\sqrt{1+p^2/(mc)^2}}$$ If I plug in the mass of the electron and the characteristic momentum on the right, I'll get what I could call the characteristic velocity on the left:
$$v_c=\frac{e E}{m_e \omega}\cdot \frac{1}{\sqrt{1+\left(\frac{e E}{\omega m_e c}\right)^2}}=v_{osc}\cdot \frac{1}{\sqrt{1+(v_{osc}/c)^2}}$$
You can convince yourself that this will always be slower than the speed of light $c$!
