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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.
The fact that the Lorentz force and the particle 4-velocity are orthogonal follows directly from the fact that massive particles travel at the speed of light in spacetime. This is most easily seen in tensor notation.
Consider the 4-velocity $u^\mu = dx^\mu/d\tau$, where $x^\mu(\tau)$ is the particle's world-line and $\tau$ its propertime. From the fact that
$$ds^2=\eta_{\mu\nu}dx^\mu dx^\nu=-d\tau^2$$
it follows that $u^\mu u_\mu=-1$, as I imagine you can verify for yourself. Now, take the derivative of $u^\mu u_\mu=-1$ with respect to $\tau$. We have
$$ \frac{d}{d\tau} (u^\mu u_\mu) = 2a^\mu u_\mu = 0, $$
where $a^\mu$ is the particle's 4-acceleration. But acceleration is $f^\mu / m$, and therefore you conclude that, in relativity, the 4-velocity of a particle is necessarily orthogonal to the force acting on it. Note in particular that it holds for any 4-force, and not only the electromagnetic force. An interesting corollary of this result is that there cannot be conservative forces in relativity -- that is, forces deriving for a 4-gradient of a scalar. For if this were true, that is, if there was a force $f_\mu=\partial_\mu\phi$, you would immediately conclude that
$$ f_\mu u^\mu = \frac{dx^\mu}{d\tau}\partial_\mu\phi = \frac{d\phi}{d\tau} = 0\,. $$
That is, the variation of $\phi$ along the particle's trajectory is zero, which is only possible if $\phi$ is constant, and therefore $f_\mu=0$. Another way of saying this is that 4-forces are always path-dependent, i.e., the work they do on the particle does depend on the path.
Lagrangian must have three properties.
The objects that you have are $A$ and $u$. So one constructs all the possible combinations of $A$ and $u$ which have the above three properties.
The Lorentz force is $$f^\mu=qF^{\mu\nu}u_\nu ~.$$ Since $F$ is antisymmetric $$f^\mu u_\mu=qF^{\mu\nu}u_\mu u_\nu =0 ~.$$
