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When your covariant derivative of your 4-velocity is zero, you are in geodesic. Now, what if I measure your 4-velocity from a non-inertial reference frame, and see that it is constant with respect to me? You are clearly in a non-inertial reference frame, yet moving in a 'geodesic' as measured by me. A good example of it can be passengers with respect to an accelerating car. Isn’t that paradoxical?

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  • $\begingroup$ Geodesics satisfy $u^{\mu} \nabla_{\mu} u^{\nu} = 0$. This is tensorial and independent of coordinates/frame of reference, therefore geodesics are not observer dependent. See for instance physics.stackexchange.com/q/604371 $\endgroup$
    – Eletie
    Oct 9, 2023 at 17:55
  • $\begingroup$ With respect to an accelerating car its passengers are at rest, so the covariant derivative of their 4-velocity is zero, and they will appear to follow geodesic equation - right? $\endgroup$
    – Nayeem1
    Oct 9, 2023 at 18:07
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    $\begingroup$ Wrong. A Rindler observer has non zero proper acceleration and therefore the covariant derivative of her own velocity is not zero. Regardless in what reference frame you express this. You seem to confuse coordinate acceleration with proper accelereation. $\endgroup$
    – Kurt G.
    Oct 9, 2023 at 18:43
  • $\begingroup$ so, a person in geodesic is in geodesic regardless of the observer, because to satisfy du/dTau = 0 one must not change direction with respect to his previous velocity? $\endgroup$
    – Nayeem1
    Oct 9, 2023 at 18:53
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    $\begingroup$ “With respect to an accelerating car its passengers are at rest”—the “rest” is unimportant. Inertial observers are only those which are freefalling, whose pocket accelerometer registers no acceleration. And free fall follows a geodesic. $\endgroup$ Oct 9, 2023 at 21:26

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In differential geometry, moving from an inertial to a non-inertial observer is just a matter of changing coordinates. It changes the metric and the Christoffel symbols that are used in the geodesic equation.

In the example of the accelerating car, being at rest inside it will not be a geodesic path.

A simpler example from differential geometry is to change from Cartesian to polar coordinates in the plane. The geodesic is a straight line for both cases, but it is simpler in the first one, with vanishing connections: $$\frac{\partial^2 X^{j}}{\partial \lambda^2} = 0$$

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