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Relation between 4-divergence and 3-divergence of vector on the hypersurface

Physics Asked by Perturbed on November 30, 2021

How the relation $nabla_{a} u^{a}=D_{a} u^{a}-epsilon v^{i} nabla_{i} v_{j} u^{j}$ where $D_{a}$ is covariant derivate on the hypersurface

between 4-divergence and 3-divergence of the vector on the hypersurface is obtained?

Which book should I look into for hypersurfaces?

One Answer

The ambient and the intrinsic connection satisfies the relation $$ h^kappa_nunabla_kappa X^mu=D_nu X^mu-epsilon K_{nukappa}X^kappa n^mu, $$ where $X$ is a tangential vector field along the hypersurface, $h$ is the induced metric/projector, $K$ is the extrinsic curvature, $n$ is the unit normal with length $g_{munu}n^mu n^nu=epsilon=pm 1$.

Using $delta^mu_nu=h^mu_nu+epsilon n^mu n_nu$, the divergence is $$ nabla_mu X^mu=delta^kappa_munabla_kappa X^mu=h^kappa_munabla_kappa X^mu+epsilon n^kappa n_munabla_kappa X^mu \ =D_mu X^mu-epsilon K_{mukappa}X^kappa n^mu+epsilon n^kappa n_munabla_kappa X^mu. $$

The second term vanishes because $K_{munu}n^mu=0$. In the third term, assuming that we are actually working with a local foliation instead of just one hypersurface, and both $n$ and $X$ has been extended off the hypersurface in a way to preserve their properties for all leaves (i.e. $n$ remains normal to all leaves of the foliation and $X$ remains tangent), we have $n_mu X^mu=0$ even off the hypersurface.

I leave the final step to be worked out by the OP.

Answered by Bence Racskó on November 30, 2021

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