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.

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