Relativistic charged-particle orbits

Continue with signature $(+,-,-,-)$, $x^0=ct$, and the force convention $dp^\mu/d\tau=qF^{\mu\nu}u_\nu$. Let E vanish and let $\mathbf B=B\hat{\mathbf z}$ be uniform and time independent. Since magnetic force does no work, gamma is constant. The laboratory equations for transverse momentum become $\dot p_x=qBp_y/(\gamma m)$ and $\dot p_y=-qBp_x/(\gamma m)$. They rotate the momentum without changing its length. A positive charge initially moving along positive x starts turning toward negative y.

The radius is transverse momentum divided by the magnitude of qB. The laboratory angular frequency is reduced by gamma compared with the nonrelativistic cyclotron frequency. Motion parallel to B has constant momentum and velocity, giving a helix rather than a circle. The same gamma includes both parallel and perpendicular velocity components. At fixed transverse momentum the radius is unchanged by how that momentum was acquired; at fixed transverse speed relativistic momentum increases the radius. Stating what is held fixed prevents misleading comparisons.

Now let B vanish and take a uniform constant electric field along positive x. Assume a positive charge initially at rest at the origin. The momentum equation integrates immediately to $p_x=qEt$. Define the dimensionless impulse $\xi=qEt/(mc)$. The mass-shell relation then gives $\gamma=\sqrt{1+\xi^2}$ and $v_x/c=\xi/\sqrt{1+\xi^2}$. Momentum continues to grow linearly, but speed remains below c at every finite time. The Newtonian formula for velocity is recovered only when the impulse is small compared with mc.

Integrating the velocity with the stated initial position gives the position formula below. Its small-time expansion yields $x\simeq qEt^2/(2m)$, providing a useful limit check. At late times the laboratory acceleration decreases even though force stays constant. The electric field supplies energy at the rate qEv, and the gained energy equals qEx for this static uniform potential choice. A constant force is therefore not a constant coordinate acceleration in relativity. No special velocity cap needs to be inserted by hand.

For the electric orbit, set $a_0=qE/m$. The proper-time description is $ct=(c^2/a_0)\sinh(a_0\tau/c)$ and $x=(c^2/a_0)[\cosh(a_0\tau/c)-1]$. Its four-velocity stays normalized and the proper acceleration magnitude is constant. The rapidity grows linearly with proper time. This hyperbolic worldline belongs to the electric problem; the laboratory magnetic orbit is circular at constant gamma. Confusing their parameterizations can produce a false claim that both have the same laboratory frequency or acceleration behavior.

These are exact solutions of an ideal test-particle equation, not universal descriptions of real accelerators. The fields are prescribed, uniform over the orbit, and unaffected by the particle. Radiation reaction, collisions, boundaries, quantum effects, and collective self-fields are omitted. Synchrotron losses can make the constant-energy magnetic model fail for real high-energy charges. Spatially varying fields can destroy simple circular or hyperbolic motion. Before applying a closed formula, check its field geometry, initial conditions, time parameter, and whether its neglected energy losses remain small.