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In general, I prefer the other form of this transformation rule as I find it easier to remember

$$

\newcommand{\tc}[2]{\frac{\partial x^{#1}}{\partial x^{#2}}}

\Gamma^{\nu'}_{\mu'\lambda'} = \tc{\mu}{\mu'}\tc{\lambda}{\lambda'} \tc{\nu'}{\nu} \Gamma^\nu_{\mu\lambda} + \tc{\nu'}{\nu}\frac{\partial^2 x^\nu}{\partial x^{\mu'}\partial x^{\lambda'}}.

$$

The equivalence of the two is rather straightforward to show.

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- #4

strangerep

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In my copy of Carroll's book (Pearson2014 edition), it[...] but in the book, he has a ##+## between the two terms on the RHS. Is this simply a typo, or am I missing something?

Maybe it was a typo in the original edition?

Btw, there is an ongoing errata list here.

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Dang... yeah, immediately after posting this thread, I thought, “Shoot, I should have searched the forum first, but oh well, too late now.” Sorry guys!

Good to know, thanks!Btw, there is an ongoing errata list here.

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In general, I prefer the other form of this transformation rule as I find it easier to remember

$$

\newcommand{\tc}[2]{\frac{\partial x^{#1}}{\partial x^{#2}}}

\Gamma^{\nu'}_{\mu'\lambda'} = \tc{\mu}{\mu'}\tc{\lambda}{\lambda'} \tc{\nu'}{\nu} \Gamma^\nu_{\mu\lambda} + \tc{\nu'}{\nu}\frac{\partial^2 x^\nu}{\partial x^{\mu'}\partial x^{\lambda'}}.

$$

The equivalence of the two is rather straightforward to show.

There definitely is a typo in the new book in equation (3.10). There's a plus sign instead of a minus sign.

Also, Carroll invites the reader to check that the object defined by equation (3.27):

$$\Gamma^{\sigma}_{\mu \nu} = \frac 1 2 g^{\sigma \rho}(\partial_{\mu}g_{\nu \rho} + \partial_{\nu}g_{\rho \mu} - \partial_{\rho}g_{\mu \nu})$$

transforms as a connection.

There's no single way to do this, but when I tried it the form given by @Orodruin above (eventually) came out. I wasted a bit of time before I figured out that this is equivalent to the corrected form of (3.10).

Perhaps Carroll intended adding this alternative form in addition to (3.10) and that's why the rogue plus sign appeared. In any case, it's a bit of a trap for the unwary reader.

- #7

strangerep

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Er,... which "new" book are you looking at? Carroll's 2014 edition (Pearson) does have the (correct, afaict) minus sign.There definitely is a typo in the new book in equation (3.10). There's a plus sign instead of a minus sign.

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The latest 2019 edition from CUP.Er,... which "new" book are you looking at? Carroll's 2014 edition (Pearson) does have the (correct, afaict) minus sign.

- #9

strangerep

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Hmm, that's interesting. CUP seems to have taken an older (pre-2014) version as the basis for their 2019 edition (sigh), even though Sean says (on his website) that it's the "same book, just with a different cover".[...] The latest 2019 edition from CUP.

[Edit: I've sent him (Sean Carroll) an email pointing this out.]

Last edited:

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Yes. This one has a minus.But @Orodruin 's transformation formula in #2 with the + sign is correct, isn't it?

$$\Gamma^{\nu'}_{\mu' \lambda'} = \frac {\partial x^\mu} {\partial x^{\mu'}} \frac {\partial x^\lambda} {\partial x^{\lambda'}} \frac {\partial x^{\nu'}} {\partial x^\nu} \Gamma^\nu_{\mu \lambda} - \frac {\partial x^\mu} {\partial x^{\mu'}} \frac {\partial x^\lambda} {\partial x^{\lambda'}} \frac {\partial^2 x^{\nu'}} {\partial x^\mu \partial x^\lambda}$$but in the book, he has a ##+## between the two terms on the RHS. Is this simply a typo, or am I missing something?

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https://en.wikipedia.org/wiki/Christoffel_symbols#Transformation_law_under_change_of_variable

or any textbook on vector calculus or GR.

- #13

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There are two different but equivalent formulas, related by:

https://en.wikipedia.org/wiki/Christoffel_symbols#Transformation_law_under_change_of_variable

or any textbook on vector calculus or GR.

$$

\tc{\nu'}{\nu}\frac{\partial^2 x^\nu}{\partial x^{\mu'}\partial x^{\lambda'}} =

- \frac {\partial x^\mu} {\partial x^{\mu'}} \frac {\partial x^\lambda} {\partial x^{\lambda'}} \frac {\partial^2 x^{\nu'}} {\partial x^\mu \partial x^\lambda}$$

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Argh! Yes, careful reading helps!

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