We have previously seen that the velocity does not transform from one reference frame to another using the standard Lorent transformation to 4-vectors, simply because the velocity is not a 4-vector. Similarly, the force is not one either (nor is it part of one, such as the 3-momentum). We therefore have to develop the transformation rule.

Let us develop the transformation of force from a system S where a particle is moving at a speed v to any system S' which is moving relative to system S at a speed V.

Calculating the transformation is very similar to the transformation of velocities. We begin with the momentum transformation (and not the coordinate transformation as we did for the velocities). We have:
$$P_x' = \gamma \left(P_x - {\beta \over c} E \right)~~~~P_y'=P_y~~~~P_z'=P_z$$
with β and γ defined using the coordinate system velocity V:
$$\gamma \equiv {1 \over \sqrt{ 1- {V^2 \over c^2} }}~~~~\beta \equiv {V \over c}.$$
Using the chain rule, we can write:
$$F_y' \equiv {d P_y' \over dt'} = {d P_y' \over dt}{dt \over dt'} = {dP_y / dt \over dt'/dt}.$$
However, we have from the Lorentz transformation for the time, that
$$t' = \gamma \left( t-{\beta x\over c}\right) ~~\Rightarrow ~~ {dt'\over dt} = \gamma - {\gamma\beta\over c} \underbrace{dx \over dt}_{v_x}$$
Thus,
$$F_y' = {F_y \over \gamma \left( 1- {V v_x \over c^2} \right) }.$$
For the x component, we have (again, using the Lorentz transformation) that:
$$F_x' \equiv {d P_x' \over dt'} = {dP_x' / dt \over dt'/dt} ={\gamma \left( {dP_x \over dt } - {V \over c^2} {d E \over dt } \right) \over \gamma \left( 1-{V v_x \over c^2} \right) }$$
We have seen, however, from the definition of force, that dE/dt=F⋅v, and thus
$$F_x' = \frac{F_x}{1-Vv_x/c^2}-\frac{V/c^2(F_xv_x+F_yv_y+F_zv_z)}{1-Vv_x/c^2}=F_x-\frac{V/c^2(F_yv_y+F_zv_z)}{1-Vv_x/c^2}$$
To summarize, the complete transformation is:
Transformation of Forces
$$\begin{eqnarray*} F_x' &=& F_x - {{V \over c^2} \left( F_y v_y + F_z v_z \right) \over 1-{V v_x \over c^2} } \\ F_y' &=& {F_y \over \gamma \left( 1- {V v_x \over c^2 } \right)} \\ F_z' &=& {F_z \over \gamma \left( 1- {V v_x \over c^2 } \right)}. \end{eqnarray*}$$
If the velocity at S vanishes at a given moment (namely, that S is the rest frame of the particle), then v=0, and we have $$F_x' = F_x ~~~~ F_y'=F_y/\gamma~~~~F_z'=F_z/\gamma$$ (where S' is the system moving relative to the particle and S is the rest-frame).

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