Connection coefficient transformation law

  • #1
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Hello PF, in Carroll’s “Spacetime and Geometry”, he works out the transformation law for connection coefficients in his introduction to covariant derivatives, and I’m wondering if there is a typo in the final equation. He starts with$$\nabla_{\mu} V^{\nu} = \partial_{\mu} V^{\nu} + \Gamma^{\nu}_{\mu \lambda} V^{\lambda}$$and what the transformation law must be if we want the covariant derivative of a vector to be tensorial:$$\nabla_{\mu’} V^{\nu’} = \frac {\partial x^\mu} {\partial x^{\mu’}} \frac {\partial x^{\nu’}} {\partial x^\nu} \nabla_\mu V^\nu$$Starting with these, he eventually gets this:$$\Gamma^{\nu’}_{\mu’ \lambda’} \frac {\partial x^{\lambda'}} {\partial x^\lambda} V^\lambda + \frac {\partial x^\mu} {\partial x^{\mu'}} V^\lambda \frac {\partial^2 x^{\nu'}} {\partial x^\mu \partial x^\lambda} = \frac {\partial x^\mu} {\partial x^{\mu'}} \frac {\partial x^{\nu'}} {\partial x^\nu} \Gamma^\nu_{\mu \lambda} V^\lambda$$I followed him here no problem. The next thing he does is eliminate ##V^\lambda## from both sides, multiplies everything by ##\frac {\partial x^\lambda} {\partial x^{\sigma'}}##, then changes all the ##\sigma'##'s to ##\lambda'##'s (for aesthetics I guess). When I do this I end up with $$\Gamma^{\nu'}_{\mu' \lambda'} = \frac {\partial x^\mu} {\partial x^{\mu'}} \frac {\partial x^\lambda} {\partial x^{\lambda'}} \frac {\partial x^{\nu'}} {\partial x^\nu} \Gamma^\nu_{\mu \lambda} - \frac {\partial x^\mu} {\partial x^{\mu'}} \frac {\partial x^\lambda} {\partial x^{\lambda'}} \frac {\partial^2 x^{\nu'}} {\partial x^\mu \partial x^\lambda}$$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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  • #2
Orodruin
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I don't know what he writes in the book, but in his online lecture notes he has a minus sign (equation (3.6)).

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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[...] but in the book, he has a ##+## between the two terms on the RHS. Is this simply a typo, or am I missing something?
In my copy of Carroll's book (Pearson2014 edition), it is a minus sign -- assuming you're referring to eq(3.10).

Maybe it was a typo in the original edition?

Btw, there is an ongoing errata list here.
 
  • #6
PeroK
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I don't know what he writes in the book, but in his online lecture notes he has a minus sign (equation (3.6)).

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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There definitely is a typo in the new book in equation (3.10). There's a plus sign instead of a minus sign.
Er,... which "new" book are you looking at? Carroll's 2014 edition (Pearson) does have the (correct, afaict) minus sign.
 
  • #8
PeroK
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Er,... which "new" book are you looking at? Carroll's 2014 edition (Pearson) does have the (correct, afaict) minus sign.
The latest 2019 edition from CUP.
 
  • #9
strangerep
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[...] The latest 2019 edition from CUP.
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".

[Edit: I've sent him (Sean Carroll) an email pointing this out.]
 
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  • #11
PeroK
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But @Orodruin 's transformation formula in #2 with the + sign is correct, isn't it?
Yes. This one has a minus.
$$\Gamma^{\nu'}_{\mu' \lambda'} = \frac {\partial x^\mu} {\partial x^{\mu'}} \frac {\partial x^\lambda} {\partial x^{\lambda'}} \frac {\partial x^{\nu'}} {\partial x^\nu} \Gamma^\nu_{\mu \lambda} - \frac {\partial x^\mu} {\partial x^{\mu'}} \frac {\partial x^\lambda} {\partial x^{\lambda'}} \frac {\partial^2 x^{\nu'}} {\partial x^\mu \partial x^\lambda}$$but in the book, he has a ##+## between the two terms on the RHS. Is this simply a typo, or am I missing something?
 
  • #13
PeroK
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But @Orodruin has a plus, and in my opinion that's the correct one. See

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

or any textbook on vector calculus or GR.
There are two different but equivalent formulas, related by:
$$
\tc{\nu'}{\nu}\frac{\partial^2 x^\nu}{\partial x^{\mu'}\partial x^{\lambda'}} =
- \frac {\partial x^\mu} {\partial x^{\mu'}} \frac {\partial x^\lambda} {\partial x^{\lambda'}} \frac {\partial^2 x^{\nu'}} {\partial x^\mu \partial x^\lambda}$$
 
  • #14
vanhees71
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Argh! Yes, careful reading helps!
 

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