Consider the de Sitter metric:

$$ds^2 = (1-\frac{r^2}{a^2})dt^2 -(1-\frac{r^2}{a^2})^{-1}dr^2-r^2d\Omega^2$$

I know that we rewrite the metric as $(u,r,\theta \phi)$ using the substitution $$u = t-atanh^{-1}(r/a)$$ as follows: $$ds^2 = (1-\frac{r^2}{a^2})du^2 +2drdu{}-r^2d\Omega^2$$

My question is why is the contravariant derivative satisfy the relation (this is something stated in my notes and am unsure why it is the case)

$$\nabla^au = (1-r^2/a^2)^{-1}\partial_t-\partial_r$$


$$\nabla^{a}u=g^{a\alpha }\nabla_{\alpha }u\overset{u\space is\space scale\space,\Gamma=0}{=}g^{a\alpha }\partial_{\alpha }u= g^{at }\partial_{t }u+g^{ar }\partial_{r }u=\frac{a^{2}}{a^2-r^2}(\frac{\partial}{\partial t})^{a}+(\frac{\partial}{\partial r})^{a}$$

  • $\begingroup$ can you just explain what that a around the derivative is exactly? $\endgroup$ – DJA Apr 8 at 1:38
  • $\begingroup$ Hi, $\nabla$is covariant derivative, but here it has upper index ,which could be risen by construction of metric $\endgroup$ – explorer Apr 8 at 4:47

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