User:JRSpriggs/Force in general relativity

In general relativity, force is a non-tensor, the time derivative of (kinetic) linear momentum,
 * $$F_{\alpha} = \frac{d p_{\alpha}}{d t} \,$$

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
 * $$p_{\alpha} = m \, g_{\alpha \beta} \, \frac{d x^{\beta}}{d \tau} = m \, g_{\alpha \beta} \, v^{\beta} \frac{ d t }{d \tau} \,$$

where: m is the mass of the particle; g&alpha;&beta; is the metric tensor which is also the gravitational potential field; x&beta; is the position vector in 4D space-time; and &tau; is the proper time measured along the trajectory of the particle
 * $$\frac{d \tau}{d t} = \sqrt{ \frac{- 1}{c^2} g_{\mu \nu} \frac{d x^{\mu}}{d t} \frac{d x^{\nu}}{d t} } \,.$$

For particles without mass, all we can say about momentum is that it is parallel to velocity
 * $$p_{\alpha} \parallel g_{\alpha \beta} \, \frac{d x^{\beta}}{d t} \,.$$

If t = x0, then
 * $$ v^{\alpha} = \frac{ d x^{\alpha} }{ d t } = \frac{ v^{\alpha} }{ v^0 } = \frac { g^{\alpha \beta} p_{\beta} }{ g^{0 \gamma} p_{\gamma} } \,. $$

Force in special relativity
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. Unfortunately, the resulting object is not a tensor even with respect to Lorentz transformations.

In a continuous medium, the 3D density of force combines with the density of power to form a covariant 4-vector density. The spatial part is the result of dividing the force on a small cell (in 3-space) by the volume of that cell. The time component is negative of the power transferred to that cell divided by the volume of the cell.

Weight and fictitious force
A point particle in free fall obeys the equation
 * $$\frac{d p_\mu}{d t} = \Gamma^{\lambda}_{\mu \nu} p_{\lambda} \frac{d x^\nu}{dt} \,$$

where p&mu; is the momentum 4-vector and &Gamma;&lambda;&mu;&nu; is the Christoffel symbol (which is the gravitational force field). Any separation of the inertial force on the right hand side into a part which is weight and a part which is fictitious force must be based on an arbitrary choice of a preferred reference frame where there are no fictitious forces by definition.
 * $$\Gamma^{\lambda}_{\mu \nu} p_{\lambda} \frac{d x^\nu}{dt} = \big( \Gamma^{\lambda}_{\mu \nu} - \widehat{\Gamma}^{\lambda}_{\mu \nu} \big) p_{\lambda} \frac{d x^\nu}{dt} + \widehat{\Gamma}^{\lambda}_{\mu \nu} p_{\lambda} \frac{d x^\nu}{dt} \,$$

where the first term on the right side is the weight (gravitational force, a tensor (except for the dt)) and the second term on the right side is the fictitious force (a non-tensor which is zero in the preferred reference frame). The hatted Christoffel symbol is defined by using the Minkowski metric in the preferred reference frame rather than the physical metric used by the regular Christoffel symbol.

Classical force equation
The non-relativistic version of the force equation is
 * $$\frac{d \vec{p}}{d t} = m \vec{g} + q \vec{E} + \text{ other forces } \,$$

where
 * $$\vec{p} = m \vec{v} \,$$

is the linear momentum,
 * $$\vec{g} = - \vec{\nabla} \zeta \,$$

is the negative of the gradient of the gravitational potential field, and
 * $$\vec{E} = - \vec{\nabla} \phi \,$$

is the negative of the gradient of the electrostatic potential field.

Force equation
A test particle obeys the equation
 * $$\frac{d p_\mu}{d t} = \Gamma^{\lambda}_{\mu \nu} p_{\lambda} \frac{d x^\nu}{dt} + F_{\mu \nu} q \frac{d x^\nu}{dt} + \text{ other forces } \,.$$

The linear momentum of a particle is a covariant tensor. For particles which have mass, linear momentum is
 * $$p_{\alpha} = m \, g_{\alpha \beta} \, \frac{d x^{\beta}}{d \tau} \,$$

where: m is the mass of the particle; g&alpha;&beta; is the metric tensor which is also the gravitational potential field; x&beta; is the position vector in 4D space-time; and &tau; is the proper time measured along the trajectory of the particle
 * $$\frac{d \tau}{d t} = \sqrt{ \frac{- 1}{c^2} g_{\mu \nu} \frac{d x^{\mu}}{d t} \frac{d x^{\nu}}{d t} } \,.$$

&Gamma;&lambda;&mu;&nu; is the Christoffel symbol (which is the gravitational force field)
 * $$\Gamma^{\lambda}_{\mu \nu} = \frac{1}{2} g^{\lambda \sigma} \left( g_{\sigma \mu, \nu} + g_{\nu \sigma, \mu} - g_{\mu \nu, \sigma} \right) \,.$$

q is the electric charge of the particle; and F&mu;&nu; is the electromagnetic field
 * $$F_{\alpha \beta} \, = \, A_{\beta, \alpha} \, - \, A_{\alpha, \beta} \,$$

where A&beta; is the electromagnetic potential 4-vector.

Derivation from Lagrangian
The Lagrangian of a general relativistic test particle in an electromagnetic field is
 * $$ L[t] = - m c^2 \frac{d \tau [t]}{d t} + q \frac{d x^{\gamma}[t]}{d t} A_{\gamma}[x[t]] \,$$

where
 * $$ \frac{d \tau [t]}{d t} = \sqrt {\frac{-1}{c^2} \, g_{\alpha\beta}[x[t]] \frac{d x^{\alpha}[t]}{d t} \frac{d x^{\beta}[t]}{d t}} \,.$$

Taking the variation of the Lagrangian with respect to variations, &delta;x&sigma;, in the trajectory of the particle, we get
 * $$\delta L = - m c^2 \frac{d t}{2 d \tau} \delta \left( \frac{-1}{c^2} \, g_{\alpha\beta} \frac{d x^{\alpha}}{d t} \frac{d x^{\beta}}{d t} \right) + q \frac{d \delta x^{\gamma}}{d t} A_{\gamma} + q \frac{d x^{\gamma}}{d t} \delta A_{\gamma} \,$$


 * $$\delta L = \frac{m d t}{2 d \tau} \left( \delta g_{\alpha\beta} \frac{d x^{\alpha}}{d t} \frac{d x^{\beta}}{d t} + g_{\alpha\beta} \frac{d \delta x^{\alpha}}{d t} \frac{d x^{\beta}}{d t} + g_{\alpha\beta} \frac{d x^{\alpha}}{d t} \frac{d \delta x^{\beta}}{d t} \right) + q \frac{d \delta x^{\gamma}}{d t} A_{\gamma} + q \frac{d x^{\gamma}}{d t} \delta A_{\gamma} \,$$


 * $$\delta L = \frac{m d t}{2 d \tau} \left( g_{\alpha\beta,\sigma} \delta x^{\sigma} \frac{d x^{\alpha}}{d t} \frac{d x^{\beta}}{d t} + 2 g_{\sigma\beta} \frac{d \delta x^{\sigma}}{d t} \frac{d x^{\beta}}{d t} \right) + q \frac{d \delta x^{\sigma}}{d t} A_{\sigma} + q \frac{d x^{\gamma}}{d t} A_{\gamma,\sigma} \delta x^{\sigma} \,$$


 * $$\delta L = \frac{m}{2} g_{\alpha\beta,\sigma} \frac{d t}{d \tau} \delta x^{\sigma} \frac{d x^{\alpha}}{d t} \frac{d x^{\beta}}{d t} - \frac{d}{d t} \left( m g_{\sigma\beta} \frac{d x^{\beta}}{d \tau} \right) \delta x^{\sigma} - q \frac{d A_{\sigma}}{d t} \delta x^{\sigma} + q \frac{d x^{\gamma}}{d t} A_{\gamma,\sigma} \delta x^{\sigma} + \frac{d (\ldots)}{d t} \,$$


 * $$\delta L = m g_{\rho\beta} \Gamma^{\rho}_{\alpha\sigma} \frac{d t}{d \tau} \delta x^{\sigma} \frac{d x^{\alpha}}{d t} \frac{d x^{\beta}}{d t} - \frac{d}{d t} \left( m g_{\sigma\beta} \frac{d x^{\beta}}{d \tau} \right) \delta x^{\sigma} - q A_{\sigma,\gamma} \frac{d x^{\gamma}}{d t} \delta x^{\sigma} + q \frac{d x^{\gamma}}{d t} A_{\gamma,\sigma} \delta x^{\sigma} + \frac{d (\ldots)}{d t} \,.$$

Thus the equation of motion is


 * $$ 0 = m g_{\rho\beta} \Gamma^{\rho}_{\alpha\sigma} \frac{d t}{d \tau} \frac{d x^{\alpha}}{d t} \frac{d x^{\beta}}{d t} - \frac{d}{d t} \left( m g_{\sigma\beta} \frac{d x^{\beta}}{d \tau} \right) - q A_{\sigma,\gamma} \frac{d x^{\gamma}}{d t} + q \frac{d x^{\gamma}}{d t} A_{\gamma,\sigma} \,$$


 * $$ \frac{d p_{\sigma}}{d t} = \Gamma^{\rho}_{\alpha\sigma} p_{\rho} \frac{d x^{\alpha}}{d t} + q F_{\sigma\gamma} \frac{d x^{\gamma}}{d t} \,$$

Acceleration
Changing notation and letting f&sigma; be the total of all non-gravitational forces, we get
 * $$ \frac{d}{d t} \left( m g_{\sigma\beta} \gamma v^{\beta} \right) = m g_{\rho\beta} \Gamma^{\rho}_{\alpha\sigma} \gamma v^{\alpha} v^{\beta} + f_{\sigma} \,$$
 * $$ m g_{\sigma\beta,\alpha} v^{\alpha} \gamma v^{\beta} + m g_{\sigma\beta} \frac{d \gamma}{d t} v^{\beta} + m g_{\sigma\beta} \gamma a^{\beta} = m g_{\rho\beta} \Gamma^{\rho}_{\alpha\sigma} \gamma v^{\alpha} v^{\beta} + f_{\sigma} \,$$
 * $$ m g_{\sigma\beta} \gamma a^{\beta} = - m g_{\sigma\mu,\alpha} v^{\alpha} \gamma v^{\mu} - m g_{\sigma\beta} \frac{d \gamma}{d t} v^{\beta} + m g_{\rho\mu} \Gamma^{\rho}_{\alpha\sigma} \gamma v^{\alpha} v^{\mu} + f_{\sigma} \,$$
 * $$ a^{\beta} = - g^{\beta\sigma} g_{\sigma\mu,\alpha} v^{\alpha} v^{\mu} - \frac{d \gamma}{\gamma d t} v^{\beta} + g^{\beta\sigma} g_{\rho\mu} \Gamma^{\rho}_{\alpha\sigma} v^{\alpha} v^{\mu} + \frac{1}{m \gamma} g^{\beta\sigma} f_{\sigma} \,$$

using
 * $$ g_{\sigma\mu,\alpha} = g_{\rho\mu} \Gamma^{\rho}_{\sigma\alpha} + g_{\sigma\rho} \Gamma^{\rho}_{\mu\alpha} \,$$

gives
 * $$ a^{\beta} = - \Gamma^{\beta}_{\alpha\mu} v^{\alpha} v^{\mu} - \frac{d \gamma}{\gamma d t} v^{\beta} + \frac{1}{m \gamma} g^{\beta\sigma} f_{\sigma} \,$$

substituting &beta;=0 and v0=1 and a0=0 gives
 * $$ 0 = - \Gamma^{0}_{\alpha\mu} v^{\alpha} v^{\mu} - \frac{d \gamma}{\gamma d t} + \frac{1}{m \gamma} g^{0\sigma} f_{\sigma} \,$$

subtracting v&beta; times this from the previous equation gives
 * $$ a^{\beta} = - \Gamma^{\beta}_{\alpha\mu} v^{\alpha} v^{\mu} + \Gamma^{0}_{\alpha\mu} v^{\alpha} v^{\mu} v^{\beta} + \frac{1}{m \gamma} g^{\beta\sigma} f_{\sigma} - \frac{1}{m \gamma} g^{0\sigma} f_{\sigma} v^{\beta} \,$$

which is the formula for the acceleration of a particle in general relativity.

An alternative derivation
Let the Lagrangian be:
 * $$ L = p_\mu [t] \, v^{\mu} [t] + \Lambda [t] \, ( p_{\alpha} [t] \, g^{\alpha \beta} [x^{.} [t]] \, p_{\beta} [t] + m^2 c^2 ) + L_{\text{int}} [x^{.} [t], v^{.} [t]] $$

where
 * $$ v^{\mu} [t] = \frac{d x^{\mu} [t]}{d t} $$.

Variation of the integral of L with respect to p gives:
 * $$ 0 = v^{\mu} + 2 \, \Lambda \, g^{\mu \beta} \, p_{\beta} $$
 * $$ \Lambda \, p_{\beta} = \frac{-1}{2} g_{\beta \mu} \, v^{\mu} $$.

Variation of the integral of L with respect to &Lambda; gives the constraint equation:
 * $$ 0 = p_{\alpha} \, g^{\alpha \beta} \, p_{\beta} + m^2 c^2 $$
 * $$ - m^2 c^2 \, \Lambda^2 = \frac{1}{4} v^{\alpha} \, g_{\alpha \beta} \, v^{\beta} = \frac{- c^2}{4} \left( \frac{d \tau}{d t} \right)^2 $$
 * $$ m^2 \, \Lambda^2 = \frac{1}{4} \left( \frac{d \tau}{d t} \right)^2 $$
 * $$ m \, \Lambda = - \frac{1}{2} \frac{d \tau}{d t} $$
 * $$ p_{\beta} = m \, g_{\beta \mu} \, v^{\mu} \, \frac{d t}{d \tau} $$.

Variation of the integral with respect to x gives:
 * $$ 0 = - \frac{d p_{\mu}}{d t} + \Lambda \, p_{\alpha} \, g^{\alpha \beta}{}_{, \mu} \, p_{\beta} + \frac{\partial L_{\text{int}}}{\partial x^{\mu}} - \frac{d}{d t} \frac{\partial L_{\text{int}}}{\partial v^{\mu}} $$
 * $$ \frac{d p_{\mu}}{d t} = \Lambda \, p_{\alpha} \, \left( - \Gamma^{\alpha}_{\rho \mu} g^{\rho \beta} - \Gamma^{\beta}_{\sigma \mu} g^{\alpha \sigma} \right) \, p_{\beta} + f_{\mu} $$

where the non-gravitational forces on the particle are:
 * $$ f_{\mu} = \frac{\partial L_{\text{int}}}{\partial x^{\mu}} - \frac{d}{d t} \frac{\partial L_{\text{int}}}{\partial v^{\mu}} $$
 * $$ \frac{d p_{\mu}}{d t} = \Gamma^{\alpha}_{\rho \mu} \, p_{\alpha} \, v^{\rho} + f_{\mu} $$

Is the equation of motion consistent with the constraint?
 * $$ 0 = \frac{d (- m^2 c^2)}{d t} = \frac{d (p_{\alpha} \, g^{\alpha \beta} \, p_{\beta})}{d t} = (\Gamma^{\sigma}_{\rho \alpha} \, p_{\sigma} \, v^{\rho} + f_{\alpha}) \, g^{\alpha \beta} \, p_{\beta} + p_{\alpha} \, g^{\alpha \beta}{}_{, \gamma} v^{\gamma} \, p_{\beta} + p_{\alpha} \, g^{\alpha \beta} \, (\Gamma^{\sigma}_{\rho \beta} \, p_{\sigma} \, v^{\rho} + f_{\beta}) $$
 * $$ = 2 f_{\alpha} \, g^{\alpha \beta} \, p_{\beta} + \Gamma^{\sigma}_{\rho \alpha} \, p_{\sigma} \, v^{\rho} \, g^{\alpha \beta} \, p_{\beta} + p_{\alpha} \, \left( - \Gamma^{\alpha}_{\rho \gamma} g^{\rho \beta} - \Gamma^{\beta}_{\sigma \gamma} g^{\alpha \sigma} \right) v^{\gamma} \, p_{\beta} + p_{\alpha} \, g^{\alpha \beta} \, \Gamma^{\sigma}_{\rho \beta} \, p_{\sigma} \, v^{\rho} = 2 f_{\alpha} \, g^{\alpha \beta} \, p_{\beta} $$
 * $$ 0 = f_{\alpha} v^{\alpha} $$

Only if the non-gravitational force is orthogonal to the velocity 4-vector! That is, in an inertial frame where the particle is instantaneously at rest, the non-gravitational force has no time component. The only interaction Lagrangian consistent with this restriction on the resulting non-gravitational forces is a sum of terms of the form
 * $$ q A_{\alpha} [x^{.} [t]] v^{\alpha} [t] $$

where q is a constant charge or coupling constant, and A is a 4-vector potential field. In this case, the resulting forces will have the same form as the Lorentz force.

Null geodesics from what?
Assume that a future-oriented path (hopefully a geodesic) is smoothly parameterized by strictly increasing &lambda;, t=x0, and &theta;. Consider the integral
 * $$ s^{*} = \int_{\lambda_0}^{\lambda_1} g_{\alpha \beta} [x^{.} [\lambda]] \frac{d x^{\alpha} [\lambda]}{d \lambda} \frac{d x^{\beta} [\lambda]}{d \lambda} \frac{d \lambda}{d \theta [\lambda]} \, d \lambda $$.

Suppose its variations with respect to x&rho; and &theta; are zero (while holding the end-points fixed):
 * $$ 0 = g_{\alpha \beta, \rho} \frac{d x^{\alpha}}{d \lambda} \frac{d x^{\beta}}{d \lambda} \frac{d \lambda}{d \theta} - 2 \frac{d}{d \lambda} \left( g_{\rho \beta} \frac{d x^{\beta}}{d \lambda} \frac{d \lambda}{d \theta} \right) $$
 * $$ 0 = \frac{d}{d \lambda} \left( g_{\alpha \beta} \frac{d x^{\alpha}}{d \lambda} \frac{d x^{\beta}}{d \lambda} \left( \frac{d \lambda}{d \theta} \right)^2 \right) $$.

The second equation is satisfied, if there is a constant K such that:
 * $$ g_{\alpha \beta} \frac{d x^{\alpha}}{d \theta} \frac{d x^{\beta}}{d \theta} = K $$

that is,
 * $$ g_{\alpha \beta} \frac{d x^{\alpha}}{d t} \frac{d x^{\beta}}{d t} = K \frac{d \theta}{d t} \frac{d \theta}{d t} $$.

The first equation gives
 * $$ 0 = \tfrac12 g_{\alpha \beta, \rho} \frac{d x^{\alpha}}{d \lambda} \frac{d x^{\beta}}{d \lambda} \frac{d \lambda}{d \theta} - g_{\rho \beta, \alpha} \frac{d x^{\alpha}}{d \lambda} \frac{d x^{\beta}}{d \lambda} \frac{d \lambda}{d \theta} - g_{\rho \beta} \frac{d^2 x^{\beta}}{{d \lambda}^2} \frac{d \lambda}{d \theta} + g_{\rho \beta} \frac{d x^{\beta}}{d \lambda} \frac{d \lambda}{d \theta} \frac{d \lambda}{d \theta} \frac{d^2 \theta}{{d \lambda}^2} $$
 * $$ 0 = \tfrac12 g_{\alpha \beta, \rho} \frac{d x^{\alpha}}{d \lambda} \frac{d x^{\beta}}{d \lambda} - g_{\rho \beta, \alpha} \frac{d x^{\alpha}}{d \lambda} \frac{d x^{\beta}}{d \lambda} - g_{\rho \beta} \frac{d^2 x^{\beta}}{{d \lambda}^2} + g_{\rho \beta} \frac{d x^{\beta}}{d \lambda} \frac{d \lambda}{d \theta} \frac{d^2 \theta}{{d \lambda}^2} $$.

Use the Christoffel symbols:
 * $$ 0 = - g_{\rho \sigma} \Gamma^{\sigma}_{\alpha \beta} \frac{d x^{\alpha}}{d \lambda} \frac{d x^{\beta}}{d \lambda} - g_{\rho \beta} \frac{d^2 x^{\beta}}{{d \lambda}^2} + g_{\rho \beta} \frac{d x^{\beta}}{d \theta} \frac{d^2 \theta}{{d \lambda}^2} $$.

Use the chain rule to put this in terms of t:
 * $$ 0 = - g_{\rho \sigma} \Gamma^{\sigma}_{\alpha \beta} \frac{d x^{\alpha}}{d t} \frac{d x^{\beta}}{d t} - g_{\rho \beta} \frac{d^2 x^{\beta}}{{d t}^2} - g_{\rho \beta} \frac{d x^{\beta}}{d t} \frac{d^2 t}{{d \lambda}^2} \frac{d \lambda}{d t} \frac{d \lambda}{d t} + g_{\rho \beta} \frac{d x^{\beta}}{d \theta} \frac{d^2 \theta}{{d t}^2} + g_{\rho \beta} \frac{d x^{\beta}}{d \theta} \frac{d \theta}{d t} \frac{d^2 t}{{d \lambda}^2} \frac{d \lambda}{d t} \frac{d \lambda}{d t} $$
 * $$ 0 = - g_{\rho \sigma} \Gamma^{\sigma}_{\alpha \beta} \frac{d x^{\alpha}}{d t} \frac{d x^{\beta}}{d t} - g_{\rho \beta} \frac{d^2 x^{\beta}}{{d t}^2} + g_{\rho \beta} \frac{d x^{\beta}}{d \theta} \frac{d^2 \theta}{{d t}^2} $$.
 * $$ 0 = - \Gamma^{\sigma}_{\alpha \beta} \frac{d x^{\alpha}}{d t} \frac{d x^{\beta}}{d t} - \frac{d^2 x^{\sigma}}{{d t}^2} + \frac{d x^{\sigma}}{d \theta} \frac{d^2 \theta}{{d t}^2} $$.
 * $$ \frac{d^2 x^{\sigma}}{{d t}^2} = - \Gamma^{\sigma}_{\alpha \beta} \frac{d x^{\alpha}}{d t} \frac{d x^{\beta}}{d t} + \frac{d x^{\sigma}}{d \theta} \frac{d^2 \theta}{{d t}^2} $$.

Replacing &sigma; by 0, x0 by t, dt/dt by 1, d2t/dt2 by 0 gives:
 * $$ 0 = - \Gamma^{0}_{\alpha \beta} \frac{d x^{\alpha}}{d t} \frac{d x^{\beta}}{d t} + \frac{d t}{d \theta} \frac{d^2 \theta}{{d t}^2} $$.
 * $$ 0 = - \Gamma^{0}_{\alpha \beta} \frac{d x^{\alpha}}{d t} \frac{d x^{\beta}}{d t} \frac{d x^{\sigma}}{d t} + \frac{d x^{\sigma}}{d \theta} \frac{d^2 \theta}{{d t}^2} $$.

Subtract this from the second previous equation to get
 * $$ \frac{d^2 x^{\sigma}}{{d t}^2} = - \Gamma^{\sigma}_{\alpha \beta} \frac{d x^{\alpha}}{d t} \frac{d x^{\beta}}{d t} + \Gamma^{0}_{\alpha \beta} \frac{d x^{\alpha}}{d t} \frac{d x^{\beta}}{d t} \frac{d x^{\sigma}}{d t} $$

which is the geodesic equation. We also get a formula for &theta;:
 * $$ \frac{d}{d t} \ln \frac{d \theta}{d t} = \Gamma^{0}_{\alpha \beta} \frac{d x^{\alpha}}{d t} \frac{d x^{\beta}}{d t} $$.
 * $$ \theta = \int \exp \left( \int \Gamma^{0}_{\alpha \beta} \frac{d x^{\alpha}}{d t} \frac{d x^{\beta}}{d t} d t \right) d t $$.