# Proper force and relativistic force

Relativistic force of an object is $$F$$ = $$dp/dt$$ where the object is moving with velocity $$v$$ with respect to our frame of reference. $$p$$ is the relativistic momentum given by:

$$p = \frac{m_0v}{\sqrt{1-v^2}},$$

when observed from a resting frame of reference. If observed from another frame of reference moving with velocity $$u$$ relative to the original frame of reference, Newton's laws transform under Lorentz transformations as: $$F'= \frac{m_0v'}{\sqrt{1-v'^2}},$$

where $$v'= \frac{v-u}{1+uv}$$ is the relativistic velocity of the object with respect to the new frame of reference.

Then as forces should be same i.e. laws should be invariant in all inertial frames, will $$F'=F$$ by this transformation and by this definition of relativistic force?.

Also there's a definition of proper force $$K = m(du^\mu/d\tau$$) where $$K$$ and $$u^\mu$$ are 4-vector force and velocity respectively. Then under a similar transformation as mentioned above, will $$K'=K$$? Basically I want to know which definition of force keeps Newton's laws invariant!

Force (like all other three-vectors) is not a Lorentz invariant.

The laws of physics are covariant, not invariant. That means that you can write: $${\bf F} = \frac{d{\bf p}}{dt}$$ and $${\bf F}' = \frac{d{\bf p}'}{dt'}$$ in another frame of reference.

The transformation relation between the forces in different reference frames is $${\bf F}' = {\bf F}_{\parallel} + \gamma^{-1}{\bf F}_{\perp}\ ,$$ where the subscripts refer to the components that are parallel or perpendicular to the velocity difference between the frames.

How would you apply Newton's laws? Just interested since I have not seen many practical applications of Newton's laws in relativistic context. Could well be my ignorance, but I would start from the Lagrangian, and derive equations of motion from that, bypassing any talk about forces.

But let's see. Four-momentum is $$p^\mu=mu^\mu$$, where $$m$$ is the mass and $$u^\mu$$ is four-velocity. By formulating the Lagrangian for a free particle, you can show that four-momentum for such particle would be conserved (due to translational invariance of spacetime). That's Newton's first law.

Next the second law. Not even sure why call it law other than historical convention, it simply defines force as a rate of change of momentum. We can do the same for four-force

$$F^\mu=\frac{Dp^\mu}{D\tau}=\frac{d p^\mu}{d\tau}+\Gamma^\mu_{\kappa\rho}u^\kappa p^\rho$$

where $$\tau$$ is proper time, and $$\Gamma^\mu_{\kappa\rho}$$ is the connection. Such definition ensures that four-force is a vector in all cases (even in curvilinear coordinates)

Newton's third law is basically again about conservation of four-momentum, but now in case of a system with several interacting particles.

So that's it. Conservation of four-momentum and definition of the four-force is all that you need.

By the way you 'relativistic force' is not invariant. You need to work with four-vectors (or scalars) to formulate invariant laws.

First $$P'=\gamma'mv'$$ instead of $$F'=...$$

afterwards, we apply the Lorentz transformations for a 4-vector force $$\vec{\mathbf{F}}[\gamma\mathbf{F},\frac{\gamma}{c}(\mathbf{F}\mathbf{v})]$$:

$$F'_{x}=\frac{F_{x}-\frac{\beta}{c}(\mathbf{F}\mathbf{v})}{1-\frac{{V}}{c^{2}}v_{x}}\;\;\;\;,\;\;\mathbf{F'}\mathbf{v'}=\frac{(\mathbf{F}\mathbf{v})-V F_{x}}{1-\frac{{V}}{c^{2}}v_{x}}$$ $$F'_{y}=\frac{F_{y}\sqrt{1-\beta^{2}}}{1-\frac{{V}}{c^{2}}v_{x}}\;\;\;\;,\;\;F'_{z}=\frac{F_{z}\sqrt{1-\beta^{2}}}{1-\frac{{V}}{c^{2}}v_{x}}$$ with $$\vec{\mathbf{P}}[\gamma m_{0}\mathbf{v},\frac{E}{c}]$$

if $$\;\;\mathbf{v}=\mathbf{0}\;$$ the second relation gives us: $$\mathbf{F'}\mathbf{v'}=-VF_{x}$$

finally, if in $$\mathcal{R}$$ the particle is at rest, in $$\mathcal{R}'$$ it moves with a velocity $$-V$$, only the component of the force $$F_{x}$$ does the work.