 Solution

Let $\mathbf{a}=d\mathbf{u}/dt$ denote coordinate acceleration and use $A^{\mu }$ for four-acceleration, to distinguish these two objects. Applying the chain rule to the Lorentz factor givesSince $dt/d\tau ={\gamma }_u$, differentiation of the four-velocity $U^{\mu }={\gamma }_u(c,\mathbf{u})$ with respect to proper time yieldsThis is also ${\gamma }_u^2[{\gamma }_u^2(\mathbf{u}\cdot \mathbf{a})/c,\mathbf{a}+{\gamma }_u^2(\mathbf{u}\cdot \mathbf{a})\mathbf{u}/c^2]$, as required.

The invariant Minkowski norm of the four-velocity is $U^{\mu }{U}_{\mu }=c^2$. Differentiating this identity gives $U^{\mu }{A}_{\mu }=0$. In the instantaneous rest frame, $U^{\mu }=(c,0)$, so $A^0=0$ and the spatial components equal the physically experienced proper acceleration $\mathbf{\alpha }$. Therefore ${\alpha }^2=-A^{\mu }{A}_{\mu }$ in the $+---$ metric signature.

For a direct calculation, put $q=\mathbf{u}\cdot \mathbf{a}$ and expand the temporal and spatial squares:The last step uses $1-u^2/c^2={\gamma }_u^{-2}$. This proper acceleration from coordinate velocity and acceleration also reads ${\alpha }^2={\gamma }_u^6{a}_{\parallel }^2+{\gamma }_u^4{a}_{\perp }^2$, providing the familiar one-dimensional limit $\alpha ={\gamma }_u^3|a_{\parallel }|$. Each expression has units ${\mathrm{m}}^2{\mathrm{s}}^{-4}$.