# Charged particle under a uniform electric field

Suppose a charge particle $q$ starts to move without initial velocity under the influence of a uniform electric field $E$ pointing in the positive $x$ direction. Express its position vector in terms of proper time $\tau$.

According to wiki:http://en.wikipedia.org/wiki/Lorentz_force#Relativistic_form_of_the_Lorentz_force, The Lorentz Force is given by $\frac {dp^{\alpha}}{d\tau}=qU_{\beta}F^{\alpha\beta}$. In this case $F^{\alpha\beta}$ reduces to $$\begin{bmatrix} 0 & \frac{E}{c} \\ -\frac{E}{c} & 0\end{bmatrix}.$$ Let $U_{\beta}=(u_0,u_1)$, then $p^{\alpha}=m_0(u_0,u_1)$, so we have $$\left(\begin{array}{cccc}m_0\dot u_0&\\m_0\dot u_1\end{array}\right)=q\left(\begin{array}{cccc}0&-\frac Ec&\\\frac Ec&0\end{array}\right)\times\left(\begin{array}{cccc} u_0&\\u_1\end{array}\right).$$ Since $$m_0\dot u_1=q\frac Ecu_0,$$ $$u_1=\frac{-m_0\dot u_0c}{qE},$$ we get $$m_0(\frac{-m_0\ddot u_0c}{qE})=\frac{qE}cu_0,$$ $$\ddot u_0+\frac{q^2E^2}{m_0^2c^2}u_0=0.$$ The characteristic polynomial is $$r^2+\frac{q^2E^2}{m_0^2c^2}=0.$$ Obviously the determinant is negative and $u_0$ is a trigonometric function of the proper time $\tau$. It can also be deduced that $u_1$ is the same kind of function. But this is clearly not the case. I hope someone can tell me where I did it wrong.

As you know the answer should be a hyper trigonometric function instead of a trigonometric one. Your mistake is with lowering/raising of vector components $$p^\alpha = m_0 \left( u^0, u^1 \right) = m_0 \left( \eta^{00}u_0, \eta^{11}u_1 \right) = \pm \left( - u_0, u_1\right)$$ Where the $\pm$ comes from your metric convention. This will lead to $$r^2 + \frac{q^2 E^2}{m_0^2 c^2} \longrightarrow r^2 - \frac{q^2 E^2}{m_0^2 c^2}$$ and you get the expected solution