0
$\begingroup$

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 ?

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.