Connection coefficient transformation law

In summary, there is a typo in equation (3.10) of the new 2019 edition of Carroll's book, which incorrectly has a plus sign instead of a minus sign.
  • #1
Pencilvester
184
42
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?
 
Physics news on Phys.org
  • #2
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.
 
  • Like
Likes PeroK and Pencilvester
  • #4
Pencilvester said:
[...] 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
Orodruin said:
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
PeroK said:
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
strangerep said:
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
PeroK said:
[...] 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.]
 
Last edited:
  • #10
But @Orodruin 's transformation formula in #2 with the + sign is correct, isn't it?
 
  • #11
vanhees71 said:
But @Orodruin 's transformation formula in #2 with the + sign is correct, isn't it?
Yes. This one has a minus.
Pencilvester said:
$$\Gamma^{\nu'}_{\mu' \lambda'} = \frac {\partial x^\mu} {\partial x^{\mu'}} \frac {\partial x^\lambda} {\partial x^{\lambda'}} \frac {\partial x^{\nu'}} {\partial x^\nu} \Gamma^\nu_{\mu \lambda} - \frac {\partial x^\mu} {\partial x^{\mu'}} \frac {\partial x^\lambda} {\partial x^{\lambda'}} \frac {\partial^2 x^{\nu'}} {\partial x^\mu \partial x^\lambda}$$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
vanhees71 said:
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:
$$
\tc{\nu'}{\nu}\frac{\partial^2 x^\nu}{\partial x^{\mu'}\partial x^{\lambda'}} =
- \frac {\partial x^\mu} {\partial x^{\mu'}} \frac {\partial x^\lambda} {\partial x^{\lambda'}} \frac {\partial^2 x^{\nu'}} {\partial x^\mu \partial x^\lambda}$$
 
  • Like
Likes vanhees71
  • #14
Argh! Yes, careful reading helps!
 

1. What is the connection coefficient transformation law?

The connection coefficient transformation law is a mathematical formula that describes how the coefficients of a connection change when the basis of a vector space is transformed. It is used in differential geometry and tensor calculus to study the behavior of vector fields and tensors under coordinate transformations.

2. What is the purpose of the connection coefficient transformation law?

The purpose of the connection coefficient transformation law is to allow for the comparison of vectors and tensors in different coordinate systems. It helps to define a connection between different coordinate systems and allows for the calculation of quantities such as curvature and geodesics.

3. How is the connection coefficient transformation law derived?

The connection coefficient transformation law is derived using the concept of parallel transport, which is the idea of moving a vector or tensor along a path without changing its direction. By comparing the components of a vector or tensor in different coordinate systems, the transformation law can be derived.

4. What are the applications of the connection coefficient transformation law?

The connection coefficient transformation law has various applications in mathematics and physics, including in the study of general relativity, electromagnetism, and fluid dynamics. It is also used in engineering and computer graphics for modeling and simulating physical systems.

5. Are there any limitations to the connection coefficient transformation law?

Yes, the connection coefficient transformation law is limited to differentiable manifolds and cannot be applied to non-differentiable spaces. Additionally, it assumes that the coordinate systems being compared are continuously related, and does not account for discontinuous changes in the coordinate system.

Similar threads

  • Special and General Relativity
Replies
17
Views
1K
  • Special and General Relativity
Replies
2
Views
1K
  • Special and General Relativity
Replies
3
Views
2K
  • Special and General Relativity
Replies
7
Views
2K
  • Special and General Relativity
Replies
15
Views
1K
  • Special and General Relativity
Replies
7
Views
5K
  • Special and General Relativity
2
Replies
62
Views
3K
  • Special and General Relativity
Replies
14
Views
2K
  • Special and General Relativity
Replies
4
Views
3K
  • Special and General Relativity
Replies
34
Views
4K
Back
Top