5
$\begingroup$

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)?

$\endgroup$
  • $\begingroup$ Torsion doesn't take part in this. If you write $\frac{d^2x^\mu}{d\tau^2}+\Gamma^\mu_{\alpha\beta}\frac{dx^\alpha}{d\tau}\frac{dx^\beta}{d\tau}=0$ you see that the expression contracting the two lower indices of $\Gamma$ are symmetric, so they automatically annihilate any skew-symmetric part of the connection. $\endgroup$ – Bence Racskó Mar 12 '17 at 10:28
  • $\begingroup$ @Uldreth Agreed that only the symmetric part of the connection takes part in the auto-parallel transport equation but when the torsion is non-vanishing, the contorsion tensor is also non-vanishing (in the sense that it will not generically vanish) and thus, the symmetric part of the affine connection is not equal to the Christoffel symbol but differs from it precisely by the symmetric part of the contosion tensor. $\endgroup$ – Dvij Mankad Mar 12 '17 at 10:45
  • $\begingroup$ Since you've specified that the connection is metric-compatible, you should probably check if that imposes any conditions on the symmetric part of the contorsion $\endgroup$ – Ben Niehoff Mar 12 '17 at 11:32
  • $\begingroup$ You're right. Wikipedia claims that they have the same geodesics but I just calculated everything and there is indeed a nontrivial symmetric part. With that said, I seem to recall something about the torsion only affecting fermions in Einstein-Cartan theory, I'll look into this. $\endgroup$ – Bence Racskó Mar 12 '17 at 11:32
  • 1

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.