Lorentz Force Equation

Dot products. If $vec{F}cdotvec{u} = 0$, then $vec{F}$ and $vec{u}$ are perpendicular, where $vec{F}$ is the Lorentz force and $vec{u}$ is the particle 4-velocity.

Since $vec{F}cdotvec{u}$ 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.

Lagrangian must have three properties.

1. it must be Lorentz scalar
2. it must be coordinate invariant
3. it must be gauge invariant

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 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 $u^mu = dx^mu/dtau$, where $x^mu(tau)$ is the particle's world-line and $tau$ its propertime. From the fact that

$$ds^2=eta_{munu}dx^mu dx^nu=-dtau^2$$

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

$$ frac{d}{dtau} (u^mu u_mu) = 2a^mu u_mu = 0, $$

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 $f_mu=partial_muphi$, you would immediately conclude that

$$ f_mu u^mu = frac{dx^mu}{dtau}partial_muphi = frac{dphi}{dtau} = 0,. $$

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.

The Lorentz force is $$f^mu=qF^{munu}u_nu ~.$$ Since $F$ is antisymmetric $$f^mu u_mu=qF^{munu}u_mu u_nu =0 ~.$$