# How to write the force on a test particle as the negative gradient of potential energy in relativity?

In newtonian mechanics, the gravitational force on a test particle can be expressed as the gradient of the potential energy ($$\phi$$) as $$\vec{F}=-\nabla\phi.$$

Now, if we consider a relativistic system and we define the laws of motion using special relativity, how should the above expression be modified if $$\phi$$ is the relativistic gravitational potential? I expect that the force would be some function of the Lorentz factor $$\gamma=\frac{1}{\sqrt{1-v^2/c^2}},$$ where $$v^2=\sum_{i=1}^3v_iv^i.$$

• I believe this would need to be described with general relativity, not special relativity. Commented Aug 10, 2023 at 2:10
• There is no accepted relativistic theory of gravity where gravity is described by gravitational scalar potential $\phi$. Such theories were attempted by Nordström, but they had problems and were surpassed by GR. Commented Aug 10, 2023 at 3:37
• While relativistic gravity does have the problems mentioned by @JánLalinský, it's also worth noting that it's fairly straightforward to write down the electromagnetic Lorentz force on a particle in fully relativistic language. Commented Aug 10, 2023 at 12:31
• @MichaelSeifert note however that EM Lorentz force does not have the form of gradient of scalar potential - instead, it is antisymmetric derivative of a four-potential. Commented Aug 10, 2023 at 13:14

The force observed in a gravitational field arises in spherical coordinates from a central mass $$m$$ with Schwarzschild radius $$r_s = \frac{2 G m}{c^2}$$

by a metric

$$ds^2 = \frac{dct^2} {1- \frac{2 r_s }{ r}} -dr^ 2 \ \left(1- \frac{2 r_s}{r}\right) - \frac{1}{r^2}\text{angular terms}$$

With the numerical SI magnitudes at the earth's surface

$$dt = 1 s , dct = 300.000 km , dr = 1km , \frac{ G m}{r} = \frac{1 cm}{7000 km}$$

the nonconstant metrics with respect to $$r$$ reduces to an r-dependent inertial time scaling for clocks nearly at rest at $$(r, \theta, \phi)$$

$$d\tau = dt \left( 1- \frac{G m}{r} \right)$$

The euclidean geometry is practically flat for motions with variations of r in the kilometer range. This immediately changes the equation of motions to a Hamiltonian

$$H + \frac{G m}{r} == \frac{1}{2m} p^2$$

with the distribution of the Euclidean space metrics negligible.

simply by $$\dfrac{d}{dt} = \left(1+\frac{2 G m}{r}\right) \dfrac{d}{d\tau}$$

So, the gravitational potential is a distribution of the time derivative in the same way as the electromagnetic potential $$e A_0$$ is a distribution of the time component of the 4-momentum in the non-relativistic approximation of weak fields.