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I am confused about a point regarding parallel transport and geodesics. The basic idea of a geodesic is the unaccelerated test particle moves in a straight line, or the tangent vector of a curve $x^b(\lambda)$ will be parallelly transported along the curve. In curved spacetime or in general coordinates, the eqn looks $$ u^b\nabla_b u^c=0\tag{1} $$ where $u^b=\frac{dx^b}{d\lambda}$ tangent to the curve. The covariant derivative vanishes, means no external force acting on the particle, hence velocity is constant. But if we rewrite the same eqn with a non-affine parameter $\xi$ then $$ \exists \eta: u^b\nabla_bu^c=\eta(\xi)u^c\tag{2} $$ which means there is some acceleration working on the particle and the velocity $u^b=\frac{dx^b}{d\xi}$ is not of constant magnitude. Now this is not anymore a parallel transport of the tangent vector. My question is now can we say this is a geodesic?

My comment is I am not sure whether the appropriate definition of the geodesic is the magnitude of the tangent vector can be changed but the direction should be unchanged. If a curve is actually geodesic, we can find a unique affine parameter which will make the RHS zero. This makes sense for particle in EM field where the RHS is $F^c_du^d$ and we can not get some parameter which will make the RHS zero, hence it is not a geodesic. But this idea is not working for null geodesics at $r=2m$ sphere of the blackhole, where the RHS is $\frac{1}{2m}u^c$. Here also we can not find an affine parameter which makes the RHS zero. So why we call this as geodesic

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Yes, OP is right: Eq. (2) is the equation for an arbitrarily parametrized geodesic, while eq. (1) is the equation for an affinely parametrized geodesic. An affine parameter $\lambda$ can locally be found as the following nested indefinite integral
$$ \eta(\xi)~=~\frac{d}{d\xi}\ln\left|\frac{d\lambda}{d\xi}\right| \qquad\Leftrightarrow\qquad \lambda ~=~\pm \int e^{\int \eta}.$$

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  • $\begingroup$ does it also mean from an arbitarary parameterized geodesic, we should get an affine parameter where the geodesic will look like $u^a\nabla_au^b=0$? $\endgroup$ Apr 18, 2023 at 11:12
  • $\begingroup$ because the geodesic is the straightest curve and parallel transport demands covariant derivative should vanish. But as I have written for $r=2m$ radius of a Schwarzschild black hole, the geodesic eqn looks $u^a\nabla_au^b=\frac{1}{2m}u^b$, which does not have an affine parameter. SO how we can say this is a geodesic $\endgroup$ Apr 18, 2023 at 11:21
  • $\begingroup$ It is maybe worth to add that the affine parameter in the first answer is identical with the proper time only for vacuum or "dust particles" spacetimes, see "Quasi-geodesics in relativistic gravity" (arxiv.org/abs/2011.05891) and/or "Relativistic gravity train" arxiv.org/abs/1704.04026. $\endgroup$
    – JanG
    Apr 18, 2023 at 18:29
  • $\begingroup$ Hi @JanG. Thanks for the feedback. $\endgroup$
    – Qmechanic
    Apr 18, 2023 at 18:41

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