Christoffel |

I followed equations (3.5)-(3.10) carefully because I fell into the same tramp as I had before on one step and found an error of a sign in Carroll's (3.10) which is the important equation for the transformation of the connection. This is fairly obvious because it comes from (3.9) and a + term has gone to the other side of the equation without changing sign. It was also confirmed by notes I found at Physics 171 and the proof about Carroll's (3.26), see below. I struggled with that proof (and part 2 of exercise 1) for too long. Eventually I found it after I found another erroneous proof on another website which nevertheless gave me a great new indexing trick (note d). I have corrected the error here.

I still do not understand Carroll's third rule for covariant derivatives that they commute with contractions but he never seems to use it. Its meaning provoked a discussion on physics forums which did not help me. In another discussion my false assumption about commuting partial derivatives was exposed.

The three most important equations here are$$

{\mathrm{\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 }}{\mathrm{\Gamma }}^{\nu }_{\mu \lambda }-\frac{\partial x^{\mu }}{\partial x^{\mu '}}\frac{\partial x^{\lambda }}{\partial x^{\lambda '}}\frac{{\partial }^2x^{\nu '}}{\partial x^{\mu }\partial x^{\lambda }}

$$ and $$

{\mathrm{\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 }}{\mathrm{\Gamma }}^{\nu }_{\mu \lambda }+\frac{\partial x^{\nu '}}{\partial x^{\lambda }}\frac{{\partial }^2x^{\lambda }}{\partial x^{\mu '}\partial x^{\lambda '}}

$$ which are alternatives for the transformation of a connection. The first one was Carroll's (3.10) (corrected.)

The third is Carroll's (3.27). He writes it is "one of the most important equations in this subject; commit it to memory." It is for a torsion-free (##{\mathrm{\Gamma }}^{\lambda }_{\mu \nu }={\mathrm{\Gamma }}^{\lambda }_{\nu \mu }##) metric-compatible (##{\mathrm{\nabla }}_{\rho }g_{\mu \nu }=0##) connection and is$$

{\mathrm{\Gamma }}^{\sigma }_{\mu \nu }=\frac{1}{2}g^{\sigma \rho }\left({\partial }_{\mu }g_{\nu \rho }+{\partial }_{\nu }g_{\rho \mu }-{\partial }_{\rho }g_{\mu \nu }\right)

$$For some reason Carroll writes ##{\mathrm{\Gamma }}^{\lambda }_{\mu \nu }={\mathrm{\Gamma }}^{\lambda }_{\nu \mu }## as ##{\mathrm{\Gamma }}^{\lambda }_{\mu \nu }={\mathrm{\Gamma }}^{\lambda }_{(\mu \nu )}## which is the same but more complicated. The brackets are the symmetrisation operator.

Read all 12 pages at Commentary 3.2 Christoffel Symbol.pdf.

## No comments:

## Post a Comment