- #1

Markus Kahn

- 112

- 14

- TL;DR Summary
- I'm having a hard time relating two things that should be the same, but don't appear to be the same...

I've stumbled over this article and while reading it I saw the following statement (##\xi## a vectorfield and ##d/d\tau## presumably a covariant derivative***):

$$\begin{align*}\frac{d \xi}{d \tau}&=\frac{d}{d \tau}\left(\xi^{\alpha} \mathbf{e}_{\alpha}\right)=\frac{d \xi^{\alpha}}{d \tau} \mathbf{e}_{\alpha}+\xi^{\alpha} \frac{d \mathbf{e}^{\alpha}}{d \tau} \\& =\frac{d \xi^{\alpha}}{d \tau} \mathbf{e}_{\alpha}+\xi^{\alpha} \frac{d x^{\mu}}{d \tau} \frac{\partial \mathbf{e}^{\alpha}}{\partial x^{\mu}} \\& =\frac{d \xi^{\alpha}}{d \tau} \mathbf{e}_{\alpha}+ \xi^{\sigma} u^{\mu} \Gamma_{\mu \sigma}^{\alpha} \mathbf{e}_{\alpha}.\end{align*}$$

This gives me some headaches... especially the

$$\frac{d \mathbf{e}^{\alpha}}{d \tau}$$

part. First of all, why does ##\alpha## gets shifted up? I don't see any metric or the like to compensate for that... Second, how exactly does one come to

$$ \frac{d \mathbf{e}^{\alpha}}{d \tau} =\frac{d x^{\mu}}{d \tau} \frac{\partial \mathbf{e}^{\alpha}}{\partial x^{\mu}}= u^{\mu} \Gamma_{\mu \sigma}^{\alpha} \mathbf{e}_{\alpha}.$$

(just to make it clear, I don't mean using the chain rule, this much would be obvious, see below for a clearer explanation...)

I know that for any function ##f## and vector field ##X## we have ##\nabla_{X}f = Xf## so in particular ##\nabla_{\frac{\partial}{\partial x^\mu}}f = \frac{\partial}{\partial x^\mu}f##. I therefore just assumed that ##\frac{d}{d\tau} \mathbf{e}_\alpha = \nabla_{\frac{d}{d\tau}} \mathbf{e}_\alpha##. Now if ##\mathbf{e}_\alpha## is a function of ##x##, which is parametrized over ##\tau## , then how can you show that

$$\nabla_{\frac{d}{d\tau}} \mathbf{e}_\alpha(x(\tau)) = u^{\mu} \Gamma_{\mu \sigma}^{\alpha} \mathbf{e}_{\alpha}$$

---------------------

*** I'm pretty sure it is b.c. we are trying to prove that

$$\frac{D^{2}}{D \tau^{2}} \delta x^{\mu}=R_{\nu \sigma \rho}^{\mu} \frac{d x^{\nu}}{d \tau} \frac{d x^{\sigma}}{d \tau} \delta x^{\rho}$$

and apparently

$$\frac{\partial \mathbf{e}^{\alpha}}{\partial x^{\mu}}= \Gamma_{\mu \sigma}^{\alpha} \mathbf{e}_{\alpha}.$$

$$\begin{align*}\frac{d \xi}{d \tau}&=\frac{d}{d \tau}\left(\xi^{\alpha} \mathbf{e}_{\alpha}\right)=\frac{d \xi^{\alpha}}{d \tau} \mathbf{e}_{\alpha}+\xi^{\alpha} \frac{d \mathbf{e}^{\alpha}}{d \tau} \\& =\frac{d \xi^{\alpha}}{d \tau} \mathbf{e}_{\alpha}+\xi^{\alpha} \frac{d x^{\mu}}{d \tau} \frac{\partial \mathbf{e}^{\alpha}}{\partial x^{\mu}} \\& =\frac{d \xi^{\alpha}}{d \tau} \mathbf{e}_{\alpha}+ \xi^{\sigma} u^{\mu} \Gamma_{\mu \sigma}^{\alpha} \mathbf{e}_{\alpha}.\end{align*}$$

**Question:**This gives me some headaches... especially the

$$\frac{d \mathbf{e}^{\alpha}}{d \tau}$$

part. First of all, why does ##\alpha## gets shifted up? I don't see any metric or the like to compensate for that... Second, how exactly does one come to

$$ \frac{d \mathbf{e}^{\alpha}}{d \tau} =\frac{d x^{\mu}}{d \tau} \frac{\partial \mathbf{e}^{\alpha}}{\partial x^{\mu}}= u^{\mu} \Gamma_{\mu \sigma}^{\alpha} \mathbf{e}_{\alpha}.$$

(just to make it clear, I don't mean using the chain rule, this much would be obvious, see below for a clearer explanation...)

I know that for any function ##f## and vector field ##X## we have ##\nabla_{X}f = Xf## so in particular ##\nabla_{\frac{\partial}{\partial x^\mu}}f = \frac{\partial}{\partial x^\mu}f##. I therefore just assumed that ##\frac{d}{d\tau} \mathbf{e}_\alpha = \nabla_{\frac{d}{d\tau}} \mathbf{e}_\alpha##. Now if ##\mathbf{e}_\alpha## is a function of ##x##, which is parametrized over ##\tau## , then how can you show that

$$\nabla_{\frac{d}{d\tau}} \mathbf{e}_\alpha(x(\tau)) = u^{\mu} \Gamma_{\mu \sigma}^{\alpha} \mathbf{e}_{\alpha}$$

---------------------

*** I'm pretty sure it is b.c. we are trying to prove that

$$\frac{D^{2}}{D \tau^{2}} \delta x^{\mu}=R_{\nu \sigma \rho}^{\mu} \frac{d x^{\nu}}{d \tau} \frac{d x^{\sigma}}{d \tau} \delta x^{\rho}$$

and apparently

$$\frac{\partial \mathbf{e}^{\alpha}}{\partial x^{\mu}}= \Gamma_{\mu \sigma}^{\alpha} \mathbf{e}_{\alpha}.$$