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In Newtonian physics, the force acting on a particle is defined as the rate of change of momentum $$F=\frac{dp}{dt}.$$

Also, the force can be defined as the derivative of the potential $$F=-\frac{dV}{dr}.$$

However, I am curious whether in General Relativity, we can define the four-force as the rate of change of the four-momentum as follows: $$F^r=\frac{dp^r}{d\tau},$$ where $p^r$ is the radial component of the four-momentum and $\tau$ is the proper time. If this definition is correct, it would be helpful if someone provide links to relevant references.

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The equations of Special and General Relativity are often not linear like Newton's Second Law is. To fit Special Relativity, you have to modify it to the form: \begin{equation} \mathbf{F} =\gamma(\mathbf{v})^3m\mathbf{a}_\parallel +\gamma(\mathbf{v})m\mathbf{a}_\perp, \end{equation} where $\mathbf{a}_\parallel$ and $\mathbf{a}_\perp$ is a decomposition of the acceleration $\mathbf{a}$ into a parallel and perpendicular part to the velocity $\mathbf{v}$. You can find this derivation here.

In General Relativity, the geodesic equation is also not linear, so you can't define such a four-force vector for gravity, but you can do so for all the other forces (for example electromagnetic ones) using said geodesic equation: \begin{equation} F^\sigma =m\left(\ddot{x}^\sigma+\Gamma_{\mu\nu}^\sigma\dot{x}^\mu\dot{x}^\nu\right). \end{equation}

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