There is a parameterized curve $\gamma(\tau)$ on a $4$-dim manifold. The self-parallel vector $X^{\alpha}(\tau)$ to the curve is to be found. By definition of auto parallel vectors, the covariant derivative of a vector along the curve must be zero.

In a textbook, it is given as follow:

$$\frac{\partial X^{\alpha}}{\partial\tau}+\Gamma^{\alpha}_{\beta\sigma}X^{\beta}\frac{d\gamma^{\sigma}}{d\tau}=0$$

I am confused about why the term $\frac{d\gamma^{\sigma}}{d\tau}$ is added? There is nothing similar to that in the definition of the covariant derivative.

  • $\begingroup$ Shouldn't $X^{\alpha}(\tau)$ be $\frac{d\gamma^{\alpha}}{d\tau}$ becuase $X^{\alpha}(\tau)$ is a vector to $\gamma(\tau)$? $\endgroup$
    – aitfel
    Jan 10, 2020 at 18:28
  • $\begingroup$ $X^{\alpha}=X^{\alpha}(\tau)$ is written for shorthand. $\endgroup$
    – Constantin
    Jan 10, 2020 at 18:31
  • $\begingroup$ Covariant derivative of $V^{\beta}$ along vector $U^{\alpha}$ is $\nabla_{U}V=U^{\alpha}V^{\beta}_{; \alpha}$. $\endgroup$
    – aitfel
    Jan 10, 2020 at 18:36

1 Answer 1


As you note, a vector is parallel-transported when its covariant derivative along a curve is zero. To put it another way, if $v^\alpha = d \gamma^\alpha/d\tau$ is the tangent to the curve, then the parallel transport equation is $$ v^\sigma \nabla_\sigma X^\alpha = v^\sigma \partial_\sigma X^\alpha + v^\sigma \Gamma^\alpha {}_{\beta \sigma} X^\beta = 0. $$ (In terms of conventional vector calculus, this is like saying that $(\vec{v} \cdot \vec{\nabla}) \vec{X} = 0$.) But $$ v^\sigma \partial_\sigma = \frac{d \gamma^\sigma}{d \tau} \frac{\partial}{\partial \gamma^\sigma} = \frac{d}{d\tau} $$ and so the above equation becomes $$ v^\sigma \nabla_\sigma X^\alpha = \frac{d X^\alpha}{d \tau} + \frac{d \gamma^\sigma}{d \tau}\Gamma^\alpha {}_{\beta \sigma} X^\beta = 0, $$ as desired.

  • $\begingroup$ Thank you! The key point here was to understand that $\gamma(\tau)=(\gamma^0(\tau),\gamma^1(\tau),\gamma^2(\tau),\gamma^3(\tau))$. Where $\gamma^{\alpha}$ are spacetime coordinates. $\endgroup$
    – Constantin
    Jan 10, 2020 at 18:53

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