# I'm trying to reproduce equation 4.6 from 4.5 in Carrolls Spacettime ;and Geometry

+1 vote

The following is my code. The last line has a partial with respect to nu and I want to use the chain rule to convert to d/dlamda.

I'm an newbie here so I may be doing something reallty dumb. I did convert the partial declaration to have the Derivative property so I don't end up with a double partial derivative of \nu and \lambda. Any suggestions would be appreciated.


{t,x,y,z,\lambda}::Coordinate.
{t,x,y,z,}::Depends(\lambda);
{\mu,\nu,\sigma,\rho,\alpha}::Indices(values={t,x,y,z},position=independent);
\nabla{#}::Derivative.
#\partial{#}::PartialDerivative;
\partial{#}::Derivative;

#deriv := \nabla_{\mu}{v?^{\nu}} -> \partial_{\mu}{v?^{\nu}} + \Gamma^{\nu}_{\sigma \mu} v?^{\sigma};

deriv := \nabla_{\nu}{\partial_{\lambda}{x^{\mu}}} -> \partial_{\nu}{\partial_{\lambda}{x^{\mu}}}
+ \Gamma^{\mu}_{\sigma \nu} \partial_{\lambda}{x^{\sigma}};

expr := \partial_{\lambda}{x^{\nu}} * \nabla_{\nu}{\partial_{\lambda}{x^{\mu}}};

substitute(expr,deriv);

distribute(expr);
sort_product(expr);


edited

+1 vote

Hi ChuckW.

I ran your code, compared the output with the result reported in the book, and it seems that the result is correct.

The last piece of the puzzle is the use of the chain rule. I complemented your code with the following:

chain_rule := \partial_{\lambda}{x^{\nu}} \partial_{\nu}{A??} -> \partial_{\lambda}{A??};

substitute(expr, chain_rule);