# Parallel transport of a vector

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.

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

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.
• 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. Jan 10, 2020 at 18:53