Lorentz invariance of the Lorentz force law I'm self-studying Friedman and Susskind's book Special Relativity and Classical Field Theory. The following question popped up while reading section 6.3.4 Lorentz Invariant Equations.
In this Lecture, they derive the Lorentz force law from the Lagrangian given by
$$\mathcal L(t, X^i, \dot X^i) = -m\sqrt{1-(\dot X^i)^2} + e\dot X^\mu A_\mu(t, X^i),\tag{6.13}$$
where $A$ is a 4-vector field. Now solving the Euler-Lagrange equations, they get
$$m {dU_\mu\over d\tau} = e F_{\mu\nu}U^\nu,\tag{6.34}$$
for each $\mu$. Here, $U$ is the 4-velocity and $F_{\mu\nu}:=\partial_\mu A_\nu-\partial_\nu A_\mu$ for all $\mu, \nu$. Also, Einstein's sum notation's been used.
Now, they claim that this equation is "manifestly invariant" under Lorentz transformations and they give the reason that "all the objects in the equations are 4-vectors and all the repeated indices are properly contracted." I don't understand a thing by this.
Question: I think that to show the Lorentz invariance of any equation, one way is to just ensure that all the quantities appearing in the equation are either scalars or 4-vectors. Hence for the above equation, we need to show that the complex of four numbers $(F_{\mu\nu} U^\nu)_{\mu = 0}^3$ is in fact a 4-vector (i.e., it indeed transforms like a 4-vector upon a Lorentz boost in $x$-direction, and under spatial rotations).
But I'm stuck in proving this. Any help is appreciated.
 A: You should split the worded argument into two parts.
Part 1: "All objects appearing in the equation are 4-vectors or Lorentz tensors"
Part 2: "If part 1 holds and indices are contracted, Lorentz covariant quantities are obtained"
I will try rephrasing what is to be shown. An equation is said to be (Lorentz) covariant if under an arbitrary Lorentz transformation the functional form of the equation is the same. So let us first attack part 2 of the statement which is simpler. Let us say we start with quantities $U_\mu, F_{\mu\nu}$, which we assume are a 4-vector and a Lorentz-tensor. Let us first recall the defining property of a Lorentz transformation, $\Lambda^\mu_{\;\;\nu}$:
$$\eta_{\mu\nu} \Lambda^{\mu}_{\;\;\alpha}  \Lambda^\nu_{\;\;\beta} = \eta_{\alpha\beta}$$
where $\eta$ is the Minkowski metric. Then we have that 4-vectors and (Lorentz)-tensors are transformed like this:
$$U'^\mu = \Lambda^\mu_{\;\;\nu}U^\nu$$
and
$$F'_{\mu\nu} = \Lambda_\mu^{\;\;\alpha} \Lambda_\nu^{\;\;\beta}F_{\alpha\beta}= \Lambda_\mu^{\;\;\alpha} F_{\alpha\beta} (\Lambda^{-1})^{\beta}_{\;\;\nu}$$
where we have used the conventional notation $\Lambda_\nu^{\;\;\mu} = (\Lambda^{-1})^\mu_{\;\;\nu}$.
Let us then take your equation and apply $\Lambda_\sigma^{\;\;\mu}$ on both sides (recall this Lorentz transformation does not depend on $\tau$), and try rewriting everything in terms of prime quantities:
\begin{align}
m \Lambda_\sigma^{\;\;\mu}\frac{{\rm d}U_\mu}{{\rm d}\tau} &= e \Lambda_\sigma^{\;\;\mu} F_{\mu\nu} U^\nu = e \Lambda_\nu^{\;\;\mu} F_{\mu\nu}\eta^{\nu\alpha} U_\alpha\\[7pt]
%%%
m\frac{{\rm d}U'_\sigma}{{\rm d}\tau} &= e \Lambda_\sigma^{\;\;\mu} F_{\mu\alpha}\eta^{\nu\alpha} ((\Lambda^{-1})^{\beta}_{\;\;\nu} \Lambda^\alpha_{\;\;\beta}) U_\alpha\\[7pt]
%%%
m\frac{{\rm d}U'_\sigma}{{\rm d}\tau} &= e \Big(\Lambda_\sigma^{\;\;\mu} F_{\mu\alpha}\eta^{\nu\alpha}(\Lambda^{-1})^{\beta}_{\;\;\nu}\Big) \Big(\Lambda^\alpha_{\;\;\beta} U_\alpha\Big)\\[7pt]
%%%%
m\frac{{\rm d}U'_\sigma}{{\rm d}\tau} &= e \Big(\Lambda_\sigma^{\;\;\mu} F_{\mu\alpha}\eta^{\nu\alpha}\Lambda^{\;\;\beta}_\nu\Big) U'_\beta\\[7pt]
%%%
m\frac{{\rm d}U'_\sigma}{{\rm d}\tau} &= e \Big(\Lambda_\sigma^{\;\;\mu} F_{\mu\alpha}\eta^{\nu\beta}\Lambda^{\;\;\alpha}_\nu\Big) U'_\beta\\
%%%
m\frac{{\rm d}U'_\sigma}{{\rm d}\tau} &= e F'_{\sigma\nu}\eta^{\nu\beta} U'_\beta = e F'_{\sigma\nu} U'^\nu
\end{align}
Where I have inserted an identity in the second line and at the previous to last, I used the defining equation of a Lorentz transformation.
So as you can see the equation looks functionally exactly the same as before in the transformed (boosted or rotated) quantities.
This "game" can always be done with contracted indices, that is why contractions represent covariant quantities (full contractions, meaning no free indices, represent scalars and therefore invariants).
Now for part 1, it is trickier to show that these are well defined Lorentz-vectors and tensors. The four-velocity is perhaps simpler to understand, if you realize that a wordline is a geometric object, therefore independent of coordinates and gives a well defined tangent vector with respect to its proper time, so it must transform as a Lorentz vector by construction.
Something similar can be said about the field strength tensor. It is by construction an antisymmetric two form formally. So it transforms as a tensor in general under any coordinate transformation. Under this view again Lorentz transformations are just a specific coordinate change that respect the metric.
More formal statements about them can be said digging more into the geometry of the theory, but I believe with this information you can understand what is asked, namely what I wrote above proves at the same time why $F_{\mu\nu}U^\nu$ behaves as a 4-vector.
A: You just need to first contact $F_{\mu\nu}$ and $U^\mu$ with the index $\nu$. Then, the derivative can be contracted with the 4-vector from above, using index $\mu$. That gives you a scaler, which is invariant under the Lorentz transformation.
