# Geodesic eqn for non-affine parameterization

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

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
$$\eta(\xi)~=~\frac{d}{d\xi}\ln\left|\frac{d\lambda}{d\xi}\right| \qquad\Leftrightarrow\qquad \lambda ~=~\pm \int e^{\int \eta}.$$
• does it also mean from an arbitarary parameterized geodesic, we should get an affine parameter where the geodesic will look like $u^a\nabla_au^b=0$? Apr 18, 2023 at 11:12
• 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 Apr 18, 2023 at 11:21