# Lorentz Force Equation

We know that from relativistic Lagrangian for a charged particle is

$$L = - m_0 c^2 \sqrt{1 - \frac{u^2}{c^2}} + \frac{q}{c} (\vec u \cdot \vec A) - q \Phi$$

leads to the Lorentz force equation, but how can we know that the 4-Vector of Lorentz force is perpendicular to 4-vector velocity of particle ?

Dot products. If $$\vec{F}\cdot\vec{u} = 0$$, then $$\vec{F}$$ and $$\vec{u}$$ are perpendicular, where $$\vec{F}$$ is the Lorentz force and $$\vec{u}$$ is the particle 4-velocity.
Since $$\vec{F}\cdot\vec{u}$$ is a scalar, it's the same in all reference frames. That means you can calculate it in any frame that is convenient. The rest frame of the particle where $$\vec{u} \rightarrow (c, 0, 0, 0)$$ is usually a good place to start.