Christoffel symbols transformation law

In summary, the transformation law for Christoffel symbols is derived from the requirement that the covariant derivative be tensorial. This is done by eliminating the vector V^{\lambda} and multiplying both sides by \partial x^{\lambda}/\partial x^{\sigma'} and relabeling \sigma' \to \lambda' to get the final equation with a minus sign. It is important to be careful with the indices when working with these equations.
  • #1
guitarphysics
241
7
In Carroll's GR book (pg. 96), the transformation law for Christoffel symbols is derived from the requirement that the covariant derivative be tensorial. I think I understand that, and the derivation Carroll carries out, up until this step (I have a very simple question here, I believe- something stupid I'm not seeing):
[tex] \Gamma^{\nu'}_{\mu'\lambda'} \frac{\partial x^{\lambda'}}{\partial x^{\lambda}}V^{\lambda} + \frac{\partial x^{\mu}}{\partial x^{\mu'}}V^{\lambda}\frac{\partial}{\partial x^{\mu}}\frac{\partial x^{\nu'}}{\partial x^{\lambda}} = \frac{\partial x^{\mu}}{\partial x^{\mu'}}\frac{\partial x^{\nu'}}{\partial x^{\nu}} \Gamma^{\nu}_{\mu\lambda}V^{\lambda} [/tex]

Since this must be true for any vector [itex] V^{\lambda} [/itex], that can be eliminated. We can then multiply by [itex] \partial x^{\lambda}/\partial x^{\sigma'} [/itex] on both sides, and relabel [itex] \sigma' \to \lambda' [/itex] to get:
[tex] \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}} [/tex]

Now, my question is just this: why the hell is there a plus sign in the last equation, instead of a minus sign? If I follow Carroll's steps directly from the first equation, I get a minus sign! :(
Sorry if it's obvious- any help is appreciated!
 
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  • #2
  • #3
guitarphysics said:
Now, my question is just this: why the hell is there a plus sign in the last equation, instead of a minus sign? If I follow Carroll's steps directly from the first equation, I get a minus sign!

My copy of Carroll's book has a minus sign (in (3.10) on page 96).
 
  • #5
As Mr-R noted, be careful with the indices!

I have to catch my bus now, but I will try to type in the details after I get home tonight.
 
  • #6
Here are the quantitative details:

$$\begin{align}
\frac{\partial x^{\mu }}{\partial x^{\mu ^{\prime }}}\frac{\partial x^{\lambda }}{\partial x^{\lambda ^{\prime }}}\frac{\partial ^{2}x^{\nu ^{\prime }}}{\partial x^{\mu }\partial x^{\lambda }} &= \frac{\partial x^{\mu }}{\partial x^{\mu ^{\prime }}}\left( \frac{\partial x^{\lambda }}{\partial x^{\lambda ^{\prime }}}\frac{\partial }{\partial x^{\lambda }}\right) \frac{\partial x^{\nu ^{\prime }}}{\partial x^\mu} \\
&= \frac{\partial x^{\mu }}{\partial x^{\mu ^{\prime }}}\left( \frac{\partial }{\partial x^{\lambda ^{\prime }}}\frac{\partial x^{\nu ^{\prime }}}{\partial x^{\mu }}\right) \\
&= \frac{\partial }{\partial x^{\lambda ^{\prime }}}\left( \frac{\partial x^{\mu }}{\partial x^{\mu ^{\prime }}}\frac{\partial x^{\nu ^{\prime }}}{\partial x^{\mu }}\right) -\frac{\partial x^{\nu ^{\prime }}}{\partial x^{\mu }}\left( \frac{\partial }{\partial x^{\lambda ^{\prime }}}\frac{\partial x^{\mu }}{\partial x^{\mu ^{\prime }}}\right) \\
&= \frac{\partial }{\partial x^{\lambda ^{\prime }}}\left( \frac{\partial x^{\nu ^{\prime }}}{\partial x^{\mu ^{\prime }}}\right) -\frac{\partial x^{\nu ^{\prime }}}{\partial x^{\mu }}\left( \frac{\partial }{\partial x^{\lambda ^{\prime }}}\frac{\partial x^{\mu }}{\partial x^{\mu ^{\prime }}}\right) \\
&= \frac{\partial }{\partial x^{\lambda ^{\prime }}}\left( \delta _{\mu ^{\prime }}^{\nu ^{\prime }}\right) -\frac{\partial x^{\nu ^{\prime }}}{\partial x^{\mu }}\left( \frac{\partial }{\partial x^{\lambda ^{\prime }}}\frac{\partial x^{\mu }}{\partial x^{\mu ^{\prime }}}\right) \\
&= -\frac{\partial x^{\nu ^{\prime }}}{\partial x^{\mu }}\frac{\partial ^{2}x^{\mu }}{\partial x^{\lambda ^{\prime }}\partial x^{\mu ^{\prime }}} .
\end{align}$$

Either the expression at the beginning or the term at the end can be used. Not the differing signs (and indices!).
 
  • #7
Sorry for the late response, I had been procrastinating going over the indices :s
I'm still not really seeing how that affects the original post; the last term you end on is not present, I believe, in Carroll's derivation at all. (It probably is and I'm just not seeing it- in which case I apologize!)
 

What are Christoffel symbols?

Christoffel symbols are a set of mathematical objects used in differential geometry to describe the curvature of a manifold. They are named after the German mathematician Elwin Bruno Christoffel.

What is the transformation law for Christoffel symbols?

The transformation law for Christoffel symbols is a rule that describes how these symbols change under a change of coordinates on a manifold. It is used to calculate the Christoffel symbols in a new coordinate system given the values in a previous coordinate system.

Why is the transformation law for Christoffel symbols important?

The transformation law for Christoffel symbols is important because it allows us to understand how the curvature of a manifold changes when we change the coordinate system. This is crucial in many areas of physics, such as general relativity, where the curvature of spacetime is described by the Christoffel symbols.

How is the transformation law for Christoffel symbols derived?

The transformation law for Christoffel symbols is derived using the concept of covariant differentiation, which is a way to differentiate vector fields on a manifold. By applying the covariant derivative to the metric tensor, we can obtain the transformation law for the Christoffel symbols.

Are there any special cases where the transformation law for Christoffel symbols is simplified?

Yes, there are special cases where the transformation law for Christoffel symbols is simplified. One example is in flat Euclidean space, where the Christoffel symbols are all equal to zero. Another example is in spherical coordinates, where the transformation law has a simpler form compared to Cartesian coordinates.

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