When the same force is applied to identical objects moving at different relativistic speeds, how are they affected? For the purposes of this question we can assume these objects are in identical electric fields and have the same charge, but let their momentum 4-vectors be $p_1$ and $p_2$. I've seen force defined as the coordinate time derivative of the 3-momentum. If $\frac{dp_1}{dt}$ and $\frac{dp_2}{dt}$ have the same space components then they would have to have different time components, which we can see by differentiating $E^2-P^2=m_0^2$. (A geometric way to see this is that any derivative of a 4-momentum must lie tangent to the hyperbola of constant rest mass.) That seems unlikely to me that identical forces would produce the same change in the space components of the momentum but not the time component. We run into the same problem if we try to define force as the proper time derivative of momentum: if we claim that $\frac{dp_1}{d\tau}$ and $\frac{dp_2}{d\tau}$ have the same space components, then their time components must be different.
So what happens in reality when we apply the same force to two identical objects with different momentums? The change in the momentums considered as a 4-vector cannot be the same for the two objects, so what is the same between the two situations? Is it one of the two cases I outlined above, or is it something else?
 A: If by same force, you mean same change in momentum, $\Delta\vec p$, then the change in four momentum:
$$ \Delta p^{\mu} = (\Delta E/c, \Delta\vec p)$$
is orthogonal to the 4-momentum:
$$ \Delta p^{\mu}p_{\mu} = (\Delta E/c, \Delta\vec p)(E/c, \vec p)=\frac{E\Delta E}{c^2}-(\Delta\vec p)\cdot \vec p = 0 $$
which means:
$$ \Delta E = \frac{(\Delta\vec p)\cdot \vec p}{||\vec p^2||+mc^2}$$
which is not much different from differentiating $E^2+(pc)^2=(mc^2)^2$.
The other way to look at it is via:
$$ p^{\mu}=(\gamma m c, \gamma m \vec v)=mu^{\mu}$$
where the 4-velocity is:
$$ mu^{\mu} = (\gamma c, \gamma \vec v) $$
The 4-force:
$$ F^{\mu} = (\frac{\gamma \vec f \cdot \vec v} c, \gamma \vec f)$$
where $\vec f\cdot \vec u = \frac{dE}{dt}$, and equal 3-forces will pick up different $\gamma$ factors in the $F_{i}$. All this keeps the energy and momentum on-shell.
The 4-acceleration given by:
$$ F^{\mu} =ma^{\mu}$$
with the four-acceleration being orthogonal
$$ a^{\mu}u_{\mu} = 0 $$
so that no matter what
$$u^{\mu}u_{\mu}=c^2$$
