Pre-Quantum Electrodynamics

Newton's second law remains valid provided we use the relativistic momentum:

Example: motion under a constant force

A particle of mass \(m\) is subjected to a constant force \(F\). If it starts at the origin at \(t=0\), what is it's position as a function of time?

Solution: \[ \frac{dp}{dt} = F, ~~p(0) = 0 ~~\longrightarrow~~ p = Ft \] and thus \[ p(t) = \frac{m u(t)}{\sqrt{1 - u(t)^2/c^2}} = Ft ~~\longrightarrow~~ u(t) = \frac{(F/m)t}{\sqrt{1 + (Ft/mc)^2}}. \] Integrating again to get the displacement, \[ x(t) = \frac{F}{m}\int_0^t dt' \frac{t'}{\sqrt{1 + (Ft'/mc)^2}} = \frac{mc^2}{F} \left[ \sqrt{1 + (Ft/mc)^2} - 1 \right]. \] The particle's world line thus shows hyperbolic motion.

Work and energy

In the context of relativity, work is still the line integral of the force: \[ W \equiv \int {\boldsymbol F} \cdot d{\boldsymbol l} \label{eq:RelativisticWork} \] The relationship between work done and increased energy also still holds: \[ W = \int d{\boldsymbol l} \cdot \frac{d{\boldsymbol p}}{dt} = \int dt ~\frac{d {\boldsymbol l}}{dt} \cdot \frac{d{\boldsymbol p}}{dt} = \int dt ~{\boldsymbol u} \cdot \frac{d{\boldsymbol p}}{dt} \] but \[ \frac{d{\boldsymbol p}}{dt} \cdot {\boldsymbol u} = \frac{d}{dt} \left( \frac{m {\boldsymbol u}}{\sqrt{1 - u^2/c^2}} \right) \cdot {\boldsymbol u} = \frac{d}{dt} \left( \frac{m c^2}{\sqrt{1 - u^2/c^2}} \right) = \frac{dE}{dt} \] and we thus get \[ W = \int dt \frac{dE}{dt} = \Delta E \]

Force

Since \({\boldsymbol F}\) involves a derivative with respect to ordinary time, it does not transform well under Lorentz transformations. For the example of motion along \(\hat{\boldsymbol x}\), the transverse components transform similarly: \[ \bar{F}_y = \frac{d\bar{p}_y}{d\bar{t}} = \frac{dp_y}{\gamma dt - \frac{\gamma \beta}{c} dx} = \frac{dp_y/dt}{\gamma \left( 1 - \frac{\beta}{c} \frac{dx}{dt}\right)} = \frac{F_y}{\gamma (1 - \beta u_x/c)}, \hspace{6mm} \bar{F}_z = \frac{F_z}{\gamma (1 - \beta u_x/c^2)} \] whereas the longitudinal component transforms in a complicated way: \[ \bar{F}_x = \frac{d\bar{p}_x}{d\bar{t}} = \frac{\gamma dp_x - \gamma \beta dp^0}{\gamma dt - \frac{\gamma \beta}{c}dx} = \frac{F_x - \frac{\beta}{c} \frac{dE}{dt}}{1 - \beta u_x/c} = \frac{F_x - \beta({\boldsymbol u} \cdot {\boldsymbol F})/c}{1 - \beta u_x/c}. \] For the specific case where the particle is instantaneously at rest in the original frame, then

The way to avoid complicated transformation rules is to define a four-vector as the derivative of momentum with respect to proper time, which leads to the definition of the

whose zeroth component is defined as \(1/c\) times the rate of increase of energy, \[ K^0 = \frac{dp^0}{d\tau} = \frac{1}{c} \frac{dE}{d\tau}. \]