Four-acceleration

Remarks

Since $\eta(U, U) = -c^2$ is constant, differentiating gives $\eta(A, U) = 0$; so $A$ is orthogonal to a timelike vector, hence spacelike or zero, and $\alpha$ is real. In an inertial frame, with velocity $\mathbf{u}$, ordinary acceleration $\mathbf{a} = d\mathbf{u}/dt$ and Lorentz factor $\gamma = \gamma(\abs{\mathbf{u}})$, $$A = \left( \frac{\gamma^4 \, \mathbf{u} \cdot \mathbf{a}}{c}, \ \gamma^2 \mathbf{a} + \frac{\gamma^4 (\mathbf{u} \cdot \mathbf{a})}{c^2} \mathbf{u} \right).$$ Example: in the instantaneous rest frame ($\mathbf{u} = 0$, $\gamma = 1$) one gets $A = (0, \mathbf{a})$ and $\alpha = \abs{\mathbf{a}}$. Non-example: $d^2 x / dt^2 = (0, \mathbf{a})$, the second derivative in coordinate time, is not the four-acceleration. It is not a four-vector: in a frame where the particle moves, it is not the Lorentz transform of its rest-frame value. For a particle of mass $m$ under a four-force $F$, the equation of motion is $m A = F$.