What is the physical interpretation of four-force in SR? Four-force in SR is the ratio of the relativistic momentum to the proper time $\tau$, however, the three-force has a different definition so that the coordinate time of $t$ replaces the proper one. Inasmuch as $dt=\gamma d\tau$, it seems that the three-force indicates something completely different from that the four-force does. I want to know if these definitions are compatible with each other and which one is preferred to solving physical problems.
For instance, we know that if a force vector is measured to be $F'$ in the rest frame of reference, this force is measured to be Lorentz contracted by the observer who moves at a constant $v$ perpendicular to $F'$ according to the three-force definition for the transverse directions:
$$F=\alpha F'\space,$$ where $\alpha=\sqrt{1-v^2/c^2}$. I do want to know what the four-force predicts for these transverse components. Does, in this case, four-force agree with the three-force and predict the same $F=\alpha F'$? If not, which one is correct and physically measurable/applicable?
 A: The four-force is the only kind of force that makes physical sense.
You can always obtain the three-force from the four-force by discarding the $t$ component (projecting onto the $xyz$ hyperplane) and dividing the rest by $dt/dτ$. Neither of those operations is physically sensible, because $t$ is a coordinate on the same footing as $x,y,z$, and they shouldn't be treated differently, unless you're taking a Newtonian limit.
If the object on which the force acts is at rest in the primed system, and its (Lorentz-)invariant mass is constant, then the four-force is purely spatial in the primed system.
Supposing the four-force is $f = f_{x'} \hat{x'} + f_{y'} \hat{y'}$, and the relative velocity of the frames is in the $x$ (or equivalently $x'$) direction, then $\hat{x'} = γ(\hat x + v \hat t)$ and $\hat{y'} = \hat y$, so
$$f = γv f_{x'} \hat t + γ f_{x'} \hat x + f_{y'} \hat y$$
To get the three-force from this, you drop the $\hat t$ or $\hat{t'}$ component and divide the rest by $dt/dτ = γ$ or $dt'/dτ = 1$. This gives you
$$\begin{eqnarray}
F' &=& f_{x'} \hat{x'} + f_{y'} \hat{y'} \\
F &=& f_{x'} \hat x + (f_{y'}/γ) \hat y
\end{eqnarray}$$
which shows that the three-force in the unprimed frame is unchanged in the parallel direction and smaller by a factor of $γ=1/α$ in the perpendicular direction.
