2
$\begingroup$

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?)

$\endgroup$

1 Answer 1

2
$\begingroup$

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 $\nabla_\mu A^\sigma:=\partial_\mu A^\sigma+\Gamma_{\mu\nu}^\sigma A^\nu$.

$\endgroup$
4
  • 1
    $\begingroup$ 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
    Commented 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! $\endgroup$ Commented 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
    Commented 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. $\endgroup$ Commented Apr 7, 2022 at 12:30

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.