The covariant Lorentz force

Keep $x^0=ct$, signature $(+,-,-,-)$, $F^{0i}=-E_i/c$, and $F^{ij}=-\epsilon^{ijk}B_k$. For a point particle with fixed rest mass m and charge q, write $u^\mu=\gamma(c,\mathbf v)$ and $p^\mu=mu^\mu$. The proper-time derivative of momentum is the four-force. The velocity being contracted has a lower index, so its spatial components carry minus signs. This is essential: the same component table would give the wrong magnetic force if contracted with unlowered spatial velocity components.

The spatial part becomes $d\mathbf p/d\tau=\gamma q(\mathbf E+\mathbf v\times\mathbf B)$. Divide by $dt/d\tau=\gamma$ to recover the laboratory momentum equation. Relativity changes the momentum to $\gamma m\mathbf v$, not the three-force expression. In particular, replacing the left side by $m\,d\mathbf v/dt$ is generally incorrect. Acceleration parallel to the velocity and acceleration perpendicular to it involve different powers of gamma when expressed in terms of ordinary force.

The temporal momentum component is $p^0=\mathcal E/c$, where $\mathcal E=\gamma mc^2$ is particle energy. Evaluating $qF^{0i}u_i$ gives $\gamma q\mathbf E\cdot\mathbf v/c$. Converting from proper time to laboratory time produces the usual electrical power. A magnetic field makes no direct contribution because velocity is perpendicular to its own cross product with B. A magnetic field can therefore turn momentum while preserving its magnitude and the particle's energy, provided no electric field or additional force acts.

Zero instantaneous electric power does not mean zero electric force. An electric field perpendicular to the velocity can initially change the momentum direction without changing energy at that instant. As the velocity develops a component along the field, power can become nonzero. Conversely, work and energy are frame dependent: an observer moving relative to a purely magnetic laboratory setup may see electric work. The full tensor equation ensures the worldline descriptions agree, even though each observer partitions the same force into different electric and magnetic contributions.

Contract the force law with $u_\mu$. The product $u_\mu u_\nu$ is symmetric, while $F^{\mu\nu}$ is antisymmetric, so the contraction vanishes. This makes the derivative of $p^\mu p_\mu=m^2c^2$ vanish for constant m. The argument works in any inertial frame without choosing field directions. It also supplies a numerical diagnostic: a computed electromagnetic four-force that is not Minkowski-orthogonal to the four-velocity signals an inconsistent convention or integration error under this model.

Mass-shell preservation is not the same as preservation of kinetic energy. An electric field changes gamma and energy while leaving the invariant rest mass fixed. The model assumes a classical structureless test charge in a prescribed field, with negligible self-force and quantum effects. Accelerated real charges radiate; when radiation losses matter, particle energy and prescribed-field work require extra accounting. The simple Lorentz equation is then an approximation rather than a complete closed description of the particle plus its own electromagnetic field.