User:JRSpriggs/Force in general relativity

In general relativity, force is a non-tensor, the time derivative of (kinetic) linear momentum,

${\displaystyle F_{\alpha }={\frac {dp_{\alpha }}{dt}}\,}$

where t is any 'time' coordinate which parameterizes the trajectory of the particle (it does not have to be the same as any of x⁰, x¹, x², or x³).

To use Newton's third law of motion, both forces must be defined as the rate of change of momentum with respect to the same time coordinate. Consequently, the differentiation cannot be with respect to proper time for both particles since it is usually different for two particles.

The linear momentum of a particle is a covariant tensor. For particles which have mass, linear momentum is

${\displaystyle p_{\alpha }=m\,g_{\alpha \beta }\,{\frac {dx^{\beta }}{d\tau }}=m\,g_{\alpha \beta }\,v^{\beta }{\frac {dt}{d\tau }}\,}$

where: m is the mass of the particle; g_αβ is the metric tensor which is also the gravitational potential field; x^β is the position vector in 4D space-time; and τ is the proper time measured along the trajectory of the particle

${\displaystyle {\frac {d\tau }{dt}}={\sqrt {{\frac {-1}{c^{2}}}g_{\mu \nu }{\frac {dx^{\mu }}{dt}}{\frac {dx^{\nu }}{dt}}}}\,.}$

For particles without mass, all we can say about momentum is that it is parallel to velocity

${\displaystyle p_{\alpha }\parallel g_{\alpha \beta }\,{\frac {dx^{\beta }}{dt}}\,.}$

If t = x⁰, then

${\displaystyle v^{\alpha }={\frac {dx^{\alpha }}{dt}}={\frac {v^{\alpha }}{v^{0}}}={\frac {g^{\alpha \beta }p_{\beta }}{g^{0\gamma }p_{\gamma }}}\,.}$