4-velocities are normalized so that $u_\mu u^\mu = -1$, so
\[ 0 = \frac{d}{d \tau} \left( u_\mu u^\mu \right) = 2 w_\mu u^\mu\]This is one constraint on the 4-components of acceleration $w^\mu$. In the
momentarily comoving frame $u^\mu = (-1,\vec{0})$, and the constraint requires
$a^0 = 0$. A Newtonian accelerometer can be modeled as follows: Have the observer release a particle in the comoving frame and see how much velocity $d\vec{v}$ the observer picks up relative to it in a short time $d \tau$, then calculate $\vec{a}_\mathrm{Newton} = d\vec{v}/d \tau$. Of course the particle is really stationary in the momentarily comoving inertial frame, and we accelerate relative to it by an amount $d u^i = a^i d \tau$. Now
\[ u^i =\frac{v^i}{\sqrt{1-v^2}},\]so
\[ du^i = \frac{ d v^i}{\sqrt{1-v^2}} + v^i d \left( \frac{1}{\sqrt{1-v^2}} \right)\] . But $v = 0$ since the frames are momentarily comoving, and thus $du^i = dv^i$, so finally
\[a_\mathrm{Newton} = dv^i / d \tau = du^i / d \tau = a^i\]The 3 independent components of 4-acceleration in a comoving frame are just 3 Newtonian accelerations.