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Let I have a contravariant tensor $A^\alpha$, I want to find covariant derivative of the contravariant tensor, From the transformation of the contravariant tensor ($A^\alpha=\partial_\gamma x^\alpha A^\gamma$): $$\partial_\mu A^\alpha=\partial_\mu(\partial_\gamma x^\alpha A^\gamma)$$ $$=A^\gamma \partial_\mu\partial_\gamma x^\alpha +\partial_\gamma x^\alpha \partial_\mu A^\gamma$$

Taking, $\partial_\mu\partial_\gamma x^\alpha=\Gamma_{\mu\gamma}^\alpha$ $$\partial_\mu A^\alpha=A^\gamma \Gamma_{\mu\gamma}^\alpha +\partial_\gamma x^\alpha \partial_\mu A^\gamma$$ Looking at internet sources, I can say that $$D_\mu A^\alpha=\partial_\mu A^\alpha+ \Gamma^{\alpha}_{\alpha\mu}A^\alpha$$ But how it's true? I can't find it anyhow. In YT, most of people were writing it from their head (or rather conceptually?)

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First of all, the equation $\partial_\mu A^\alpha=\partial_\mu(\partial_\beta x^\gamma A_\gamma)$ cannot be true as the contravariant index $\alpha$ disappears and the covariant index $\beta$ appears. Furthermore the definition of the Christoffel symbols isn't $\Gamma_{\mu\beta}^\gamma=\partial_\mu\partial_\beta x^\gamma$, but: \begin{equation} \Gamma_{\mu\beta}^\gamma :=\frac{\partial x^\gamma}{\partial\xi^\nu}\frac{\partial^2\xi^\nu}{\partial x^\mu\partial x^\beta}, \end{equation} where $\xi$ is a free-falling coordinate system without gravity according to the equivalence principle.

The construction of the covariant derivative arises from the problem that $\partial_\mu A^\sigma$ does not transform like a tensor. Let $\alpha$ and $\overline{\alpha}$ denote the transformation matrices, then: \begin{align*} \partial_\kappa'{A'}^\rho &=\frac{\partial x^\mu}{\partial{x'}^\kappa}\frac{\partial}{\partial x^\mu} \left(\frac{\partial{x'}^\rho}{\partial x^\sigma}A^\sigma\right) =\frac{\partial x^\mu}{\partial{x'}^\kappa}\frac{\partial{x'}^\rho}{\partial x^\sigma}\frac{\partial A^\sigma}{\partial x^\mu} +\frac{\partial x^\mu}{\partial{x'}^\kappa}\frac{\partial^2{x'}^\rho}{\partial x^\mu\partial x^\sigma}A^\sigma \\ &=\overline{\alpha}_\kappa^\mu\alpha_\sigma^\rho\partial_\mu A^\sigma -\frac{\partial^2x^\xi}{\partial{x'}^\kappa\partial{x'}^\lambda}\alpha_\sigma^\lambda\alpha_\xi^\rho A^\sigma. \end{align*} The first term is that of a tensor transformation, but there is a additional second term. It can be eliminated using the Christoffel symbols, that also do not transform like a tensor: \begin{align*} {\Gamma'}_{\kappa\lambda}^\rho &=\frac{\partial{x'}^\rho}{\partial \xi^\alpha}\frac{\partial^2 \xi^\alpha}{\partial {x'}^\kappa{x'}^\lambda} =\frac{\partial{x'}^\rho}{\partial x^\sigma}\frac{\partial x^\sigma}{\partial \xi^\alpha}\frac{\partial}{\partial {x'}^\kappa}\left(\frac{\partial\xi^\alpha}{\partial x^\nu}\frac{\partial x^\nu}{\partial {x'}^\lambda}\right) \\ &=\frac{\partial{x'}^\rho}{\partial x^\sigma}\frac{\partial x^\sigma}{\partial \xi^\alpha}\left(\frac{\partial\xi^\alpha}{\partial x^\mu\partial x^\nu}\frac{\partial x^\mu}{\partial {x'}^\kappa}\frac{\partial x^\nu}{\partial {x'}^\lambda}+\frac{\partial\xi^\alpha}{\partial x^\nu}\frac{\partial^2 x^\nu}{\partial {x'}^\kappa\partial {x'}^\lambda}\right) \\ &=\frac{\partial {x'}^\rho}{\partial x^\sigma}\frac{\partial x^\mu}{\partial {x'}^\kappa}\frac{\partial x^\nu}{\partial {x'}^\lambda}\Gamma_{\mu\nu}^\sigma +\frac{\partial {x'}^\rho}{\partial x^\tau}\frac{\partial^2 x^\tau}{\partial {x'}^\kappa\partial {x'}^\lambda} =\alpha_\sigma^\rho\overline{\alpha}_\kappa^\mu\overline{\alpha}_\lambda^\nu\Gamma_{\mu\nu}^\sigma +\alpha_\tau^\rho\frac{\partial^2 x^\tau}{\partial {x'}^\kappa\partial {x'}^\lambda}. \end{align*} A suitable combination of both equations now yields the transformation of a tensor: \begin{align*} \partial_\kappa'{A'}^\rho+{\Gamma'}_{\kappa\lambda}^\rho{A'}^\lambda &=\overline{\alpha}_\kappa^\mu\alpha_\sigma^\rho\partial_\mu A^\sigma -\frac{\partial^2x^\xi}{\partial{x'}^\kappa\partial{x'}^\lambda}\alpha_\sigma^\lambda\alpha_\xi^\rho A^\sigma +\left(\overline{\alpha}_\kappa^\mu\overline{\alpha}_\lambda^\nu\alpha_\sigma^\rho\Gamma_{\mu\nu}^\sigma +\alpha_\tau^\rho\frac{\partial^2 x^\tau}{\partial {x'}^\kappa\partial {x'}^\lambda}\right)\alpha_\nu^\lambda A^\nu \\ &=\overline{\alpha}_\kappa^\mu\alpha_\sigma^\rho\left(\partial_\mu A^\sigma+\Gamma_{\mu\nu}^\sigma A^\nu\right). \end{align*} This tensor is the covariant derivative $\nabla_\mu A^\sigma:=\partial_\mu A^\sigma+\Gamma_{\mu\nu}^\sigma A^\nu$.

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    $\begingroup$ 1. the transformation wrong indices in transformation was a typo... 2. I had seen Christoffel symbol is written as you wrote... But, Take a look here, Prof Susskind had written as I wrote (actually I was following him when I couldn't understand others. and his method was looking easier to me so I chose to use that) $\endgroup$
    – Anonymous
    Apr 7, 2022 at 9:18
  • $\begingroup$ 1. Well, typos happen. 2. It looks like Prof. Susskind contracted the $\xi$, but he uses different symbols ($x$ and $y$) for the coordinate systems and therefore writes out the partial derivitates. The expression $\partial_\mu x^\nu$ doesn't show that, so I read it as $\delta_\mu^\nu$. By the way, his lectures are a really good choice for learning general relativity! $\endgroup$ Apr 7, 2022 at 9:24
  • $\begingroup$ But we can choose $y=x'$ and $\partial_\mu x^\nu=\dfrac{\partial x^\nu}{\partial x'^\mu}$, perhaps I misunderstood cause he was differentiating scalar quantity. $\endgroup$
    – Anonymous
    Apr 7, 2022 at 9:35
  • $\begingroup$ This is fine as long as you are clarifying you don't also use $x$ to derive. Considering partial derivatives, you can ignore the indices. They are only important when looking at the covariant derivative. $\endgroup$ Apr 7, 2022 at 12:30

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