In case anyone's interested, I returned to the question of "longitudinal" invariance with respect to three-force (via rapidity).
First: under a Lorentz boost, changes (or differences) in the component of rapidity that's parallel to the boost are invariant. For example, if the boost is along the ##x##-axis, then ##\Delta \phi_x = \Delta (\tanh^{-1} v_x)## is invariant (##c=1##).
Now, as for three-force, we have the following vector equation (##c=1##, ##\dot{m} = 0##):
## \vec f = \dfrac{m}{\cosh{\phi}} \, \dfrac{d}{d \tau} \left( \sinh{\phi} \, \hat{v} \right) = m \left( \dfrac{d \phi}{d \tau} \hat{v} + \tanh{\phi} \dfrac{d \hat{v}}{d \tau}
\right).##
If I'm not mistaken, the component vector of ##\vec f## that's parallel to the velocity ##\vec v## would then be:
## \vec f_{\parallel \vec v} = m \left[ \dfrac{d \phi_{\parallel \vec v}}{d \tau} \hat{v} + \tanh{\phi_{\parallel \vec v}} \left( \dfrac{d \hat{v}}{d \tau} \right)_{\parallel \vec v} \right].##
(Did I do that right? I'm unsure about the ##\phi_{\parallel \vec v}##'s.)
But the component vector ##( d \hat{v} / d \tau )_{\parallel \vec v}## is necessarily a zero vector, since the derivative of a vector of fixed magnitude (like ##\hat{v}##) must be perpendicular to the differentiated vector. Thus:
## \vec f_{\parallel \vec v} = m \dfrac{d \phi_{\parallel \vec v}}{d \tau} \hat{v} ##
##f_{\parallel \vec v} = m \dfrac{d \phi_{\parallel \vec v}}{d \tau} ##.
The ##m## and ##d \tau## are "fully" invariant, and the ##d \phi_{\parallel \vec v}## is "longitudinally" invariant (i.e., invariant under a boost in the ##\pm \hat{v}## direction, as discussed above). Therefore the component of three-force that's parallel to the three-velocity is likewise "longitudinally" invariant. Of course, in the special case that this is the only non-zero component of three-force, the three-force magnitude itself is invariant under a longitudinal boost:
##f = m \dfrac{d \phi}{d \tau} ##.
