Auto-parallel Transport or Principle of Extremum Action?

Question

In an affinely connected spacetime with a metric compatible connection, the equation of the curve in which the tangent vector at each point is the result of the parallel transport of every tangent vector to the curve along the curve (to that point) is $$\dfrac{d^2x^\lambda}{d\tau^2} + \dfrac{dx^\mu}{d\tau}\dfrac{dx^\nu}{d\tau}\Gamma_{{\mu}{\nu}}^{{ }{ }{\lambda}}=0,$$ where $x^\lambda$ represents the coordinates, $\tau$ is the affine parameter, and $\Gamma$ is the affine connection. We call such a curve an auto-parallel curve.

Now, the equation of the curve which extremizes the action (where action is defined as $S=\int\sqrt{g_{{\mu}{\nu}}dx^{\mu}dx^{\nu}}$) is given by $$\dfrac{d^2x^\lambda}{d\tau^2} + \dfrac{dx^\mu}{d\tau}\dfrac{dx^\nu}{d\tau}\begin{Bmatrix} \lambda \\ \mu\nu \end{Bmatrix}=0,$$ where $g_{{\mu}{\nu}}$ is the metric (and the connection $\Gamma$ is compatible with it) and $\begin{Bmatrix} \lambda \\ \mu\nu \end{Bmatrix}$ represents the Christoffel symbols. We call such a curve a geodesic curve.

If we impose the condition that the torsion must vanish then the two of the above equations represent the same class of curves but otherwise, they represent different classes of curves. So, in a generic case where torsion might be non-vanishing, should the trajectory of a free particle be described by the auto-parallel curves (which I feel should be the answer according to the Principle of Equivalence) or by the geodesic curves (which I feel should be the answer if we respect the Principle of Extremum Action)?

Answer 1

Both equations are equivalent. Recall that torsion is the antisymmetric part of the connection coefficients, but in the auto-parallel curve equation only the symmetric part contributes. Namely, $$\frac{d x^\mu}{d\tau}\frac{d x^\nu}{d\tau}\Gamma^{\lambda}{}_{\mu\nu} = \frac{d x^\mu}{d\tau}\frac{d x^\nu}{d\tau}\Gamma^{\lambda}{}_{(\mu\nu)}.$$ Hence, torsion does not contribute to either of these equations, and they actually coincide.

Answer 2

Short answer: yes. Auto-parallel. Why? Because ultimately, this is both an expression of the continuity equation for a singular source concentrated on a world-line, and the limit of the continuity equation for a distribution concentrated around a world tube.

The two equations $$ \frac{d^2x^ρ}{ds^2} + Γ_{μν}^ρ \frac{dx^μ}{ds} \frac{dx^ν}{ds} = 0 \quad↔\quad \frac{d^2x^ρ}{ds^2} + \begin{Bmatrix}ρ\\μ ν\end{Bmatrix} \frac{dx^μ}{ds} \frac{dx^ν}{ds} = 0 $$ are not equivalent in a [pseudo-]Riemann-Cartan geometry, though they are in a [pseudo-]Riemannian geometry. The difference $$ C_{μν}^ρ = Γ_{μν}^ρ - \begin{Bmatrix}ρ\\μ ν\end{Bmatrix} $$ is the contorsion, which is a linear function of the torsion.

Here: Test Bodies: Auto-Parallel or Geodesic? I do the derivation directly from the continuity equation. ("Without prejudicing the issue on either side, we can follow the Gandalf / Toucan Sam Strategy and follow our nose.") The result: auto-parallel. Notice the effect that the contorsion has on the local clock.