0
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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

2
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.