Here we meet the covariant divergence and prove a thing or two about it. (Carroll said it was 'easy'.) We also have the curved-space version of Stokes's theorem using the covariant derivative and finally the exterior derivative and commutator, where Carroll seems to have made a very peculiar typo. Sir George Gabriel Stokes, 1st Baronet was a very clever but he did not discover the theorem. He popularised it.

Here's the very impressive Stokes's theorem, which applies to the diagram

$$\int^{\ }_{\mathrm{\Sigma }}{{\mathrm{\nabla }}_{\mu }V^{\mu }\sqrt{\left|g\right|}}d^nx=\int^{\ }_{\mathrm{\partial }\mathrm{\Sigma }}{n_{\mu }V^{\mu }\sqrt{\left|\gamma \right|}}d^{n-1}x

$$

At Carroll's (3.36) he says "if ##\mathrm{\nabla }## is the Christoffel symbol, ##{\omega }_{\mu }## is a one-form, and ##X^{\mu }## and ##Y^{\mu }## are vector fields, we can write

$${\left(\mathrm{d}\omega \right)}_{\mu \nu }=2{\partial }_{[\mu }{\omega }_{\nu ]}=2{\mathrm{\nabla }}_{[\mu }{\omega }_{\nu ]}

$$The phrase "if ##\mathrm{\nabla }## is the Christoffel symbol" is bizarre and it is easy to prove the equation without it, assuming the Christoffel connection is torsion-free (##{\mathrm{\Gamma }}^{\lambda }_{\mu \nu }={\mathrm{\Gamma }}^{\lambda }_{\nu \mu }##). I think our author meant "if the connection is torsion-free".

Read more at Commentary 3.2 Properties of covariant derivative.pdf (7 pages)

Here's the very impressive Stokes's theorem, which applies to the diagram

$$\int^{\ }_{\mathrm{\Sigma }}{{\mathrm{\nabla }}_{\mu }V^{\mu }\sqrt{\left|g\right|}}d^nx=\int^{\ }_{\mathrm{\partial }\mathrm{\Sigma }}{n_{\mu }V^{\mu }\sqrt{\left|\gamma \right|}}d^{n-1}x

$$

At Carroll's (3.36) he says "if ##\mathrm{\nabla }## is the Christoffel symbol, ##{\omega }_{\mu }## is a one-form, and ##X^{\mu }## and ##Y^{\mu }## are vector fields, we can write

$${\left(\mathrm{d}\omega \right)}_{\mu \nu }=2{\partial }_{[\mu }{\omega }_{\nu ]}=2{\mathrm{\nabla }}_{[\mu }{\omega }_{\nu ]}

$$The phrase "if ##\mathrm{\nabla }## is the Christoffel symbol" is bizarre and it is easy to prove the equation without it, assuming the Christoffel connection is torsion-free (##{\mathrm{\Gamma }}^{\lambda }_{\mu \nu }={\mathrm{\Gamma }}^{\lambda }_{\nu \mu }##). I think our author meant "if the connection is torsion-free".

Read more at Commentary 3.2 Properties of covariant derivative.pdf (7 pages)

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