- #1

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This image was taken from Sean Carroll's notes available here: preposterousuniverse.com/wp-content/uploads/grnotes-three.pdf

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- Thread starter accdd
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- #1

- 93

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This image was taken from Sean Carroll's notes available here: preposterousuniverse.com/wp-content/uploads/grnotes-three.pdf

- #2

- 20,004

- 10,655

- #3

- 93

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Thanks for the answer, could you show me the steps to get to the result?

- #4

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- #5

- 93

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$$

\nabla_{\mu'}V^{\nu'}=\partial_{\mu'}(\frac{\partial x^{\nu'}}{\partial x^{\nu}}V^{\nu})+\Gamma^{\nu'}_{\mu'\lambda'}(\frac{\partial x^{\lambda'}}{\partial x^{\lambda}}V^{\lambda}) =\partial_{\mu'}(\frac{\partial x^{\nu'}}{\partial x^{\nu}})V^{\nu}+\frac{\partial x^{\nu'}}{\partial x^{\nu}} \partial_{\mu'}V^{\nu}+\Gamma^{\nu'}_{\mu'\lambda'}\frac{\partial x^{\lambda'}}{\partial x^{\lambda}}V^{\lambda}=

$$

- #6

- 20,004

- 10,655

- #7

- 93

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Is this the correct transformation?

$$

\partial_{\mu'}=\frac{\partial x^\mu}{\partial x^{\mu'}}\partial_\mu

$$

$$

\partial_{\mu'}(\frac{\partial x^{\nu'}}{\partial x^{\nu}})V^{\nu}+\frac{\partial x^{\nu'}}{\partial x^{\nu}} \partial_{\mu'}V^{\nu}+\Gamma^{\nu'}_{\mu'\lambda'}\frac{\partial x^{\lambda'}}{\partial x^{\lambda}}V^{\lambda}= \frac{\partial x^\mu}{\partial x^{\mu'}}\partial_\mu(\frac{\partial x^{\nu'}}{\partial x^{\nu}})V^{\nu}+\frac{\partial x^{\nu'}}{\partial x^{\nu}} \frac{\partial x^\mu}{\partial x^{\mu'}}\partial_\mu V^{\nu}+\Gamma^{\nu'}_{\mu'\lambda'}\frac{\partial x^{\lambda'}}{\partial x^{\lambda}}V^{\lambda}

$$

so the first line of 3.3 should be more clearly:

$$

\nabla_{\mu'}V^{\nu'}=(\frac{\partial x^\mu}{\partial x^{\mu'}}\partial_\mu)(\frac{\partial x^{\nu'}}{\partial x^{\nu}}V^{\nu})+\Gamma^{\nu'}_{\mu'\lambda'}(\frac{\partial x^{\lambda'}}{\partial x^{\lambda}}V^{\lambda})

$$

$$

\partial_{\mu'}=\frac{\partial x^\mu}{\partial x^{\mu'}}\partial_\mu

$$

$$

\partial_{\mu'}(\frac{\partial x^{\nu'}}{\partial x^{\nu}})V^{\nu}+\frac{\partial x^{\nu'}}{\partial x^{\nu}} \partial_{\mu'}V^{\nu}+\Gamma^{\nu'}_{\mu'\lambda'}\frac{\partial x^{\lambda'}}{\partial x^{\lambda}}V^{\lambda}= \frac{\partial x^\mu}{\partial x^{\mu'}}\partial_\mu(\frac{\partial x^{\nu'}}{\partial x^{\nu}})V^{\nu}+\frac{\partial x^{\nu'}}{\partial x^{\nu}} \frac{\partial x^\mu}{\partial x^{\mu'}}\partial_\mu V^{\nu}+\Gamma^{\nu'}_{\mu'\lambda'}\frac{\partial x^{\lambda'}}{\partial x^{\lambda}}V^{\lambda}

$$

so the first line of 3.3 should be more clearly:

$$

\nabla_{\mu'}V^{\nu'}=(\frac{\partial x^\mu}{\partial x^{\mu'}}\partial_\mu)(\frac{\partial x^{\nu'}}{\partial x^{\nu}}V^{\nu})+\Gamma^{\nu'}_{\mu'\lambda'}(\frac{\partial x^{\lambda'}}{\partial x^{\lambda}}V^{\lambda})

$$

Last edited:

- #8

- 20,004

- 10,655

Yes. That is the chain rule for partial derivatives.Is this the correct transformation?

$$

\partial_{\mu'}=\frac{\partial x^\mu}{\partial x^{\mu'}}\partial_\mu

$$

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