I'm studying Friedman and Susskind's Special Relativity and Classical Field Theory.
In Lecture 6, they derive $$m{dU_\mu\over d\tau}=eF_{\mu\nu}U^\nu,\tag{6.33}$$ but only when $\mu=1, 2, 3$. ($F_{\mu\nu}$ stands for $\partial_\mu A_\nu-\partial_\nu A_\mu$.) To prove that this equation also holds for $\mu=0$, they give the following argument.
If the equations are Lorentz invariant, and the three space comonents of a certain 4-vector are equal to the three space components of some other 4-vector, then we know automatically that their zeroth components (the time components) must also match.
I can see that both, the LHS and the RHS of the above equation are 4-vectors. But I don't see how the equality of the LHS and the RHS for $\mu=1,2,3$ implies their equality for $\mu=0$. Please help me prove it.
I found that this question has been asked before on SE. But I couldn't understand the posted answer (by Prahar) there—it used gauge invariance concepts and I'm not yet familiar with that. But I'll appreciate if someone can explain me enough to understand that answer.