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## New answers tagged vector-fields

0

The claims aren't true in complete generality. For example, $R^{2n}$ is a Kähler manifold and any vector $k$ is a Killing vector. But the Lie derivatives of $J$ and $\omega$ don't vanish for a general $k$. However, this example was special because it was insufficiently curved. For generic and curved enough Kähler manifold, the objects $J,\omega$ may be ...

0

If the connection you are using is torsion-free (as the Levi-Civita one), you can systematically replace the coordinate derivative for the covariant one. So (please, pay attention to signs and positions of indexes, since I could use a convention different from yours) $${\cal L}_\xi \nabla_a K^b = \xi^c\nabla_c \nabla_a K^b - (\nabla_c\xi^b) \nabla_a K^c + ... 0 It might be useful to know the formula for Lie derivative of metric itself. It is \mathcal{L}_Xg_{ab}=\nabla_a X_b+\nabla_b X_a. Then you might notice that covariant derivative can be split into symmetric and atisymmetric part:$$\nabla _a K_b = \nabla _{(a} K_{b)}+\nabla _{[a} K_{b]}$$The first part is half of lie derivative of metric with respect to ... 0 A slightly different and, perhaps, more simple proof follows. \def\lie{\mathit{£}} For K^a a Killing vector, we have (Kostant formula),$$\nabla_a \nabla_b K^c = -R_{bca}{}^d K^d.$$Using this, we may prove that the covariant and the Lie derivatives commute:$$\lie_K \nabla_a X^b = \nabla_a \lie_K X^b,$$for arbitrary X^a. We have: ... 1 The reason to consider such metrics was not a particular interpretation, but that for this ansatz one could hope to find some exact solutions. It is known as the Kerr-Schild metric, and its role in the process of finding exact solutions has been described by Kerr in Wiltshire, Visser, Scott (eds.), The Kerr Spacetime. One can, of course, try to develop ... 2 The commutator of two vector fields n^a and X^b is [n,X]^a = n^b \nabla_b X^a - X^b \nabla_b n^a. Since this vanishes, it follows that n^b \nabla_b X^a = X^b \nabla_b n^a, and the second step follows from there. I don't have my copy of Wald in front of me, but I'm 99% sure that the commutator of two vector fields is defined in terms of the ... 4 You are confused due to the vectorial form of the equation, so you should write it by components (I will use cartesian coordinates), and I will use \partial_x = \dfrac{d}{dx} for comfort, that being said, your equation can be written as:$$ F=-\int{(dr)}{(\vec{\nabla} \cdot \vec{P}) \vec{E} }=-\int{(dr)}{(\partial_x P_x +\partial_y P_y + \partial_z P_z ) ...

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The conservation of $\vec{k}\cdot\vec{u}$ only holds in the test particle limit. That is, it considers the metric to be unaffected by the motion of the particle. In this limit, there are no gravitational waves, since the metric has no time-varying quadrupole. If you want to see gravitational waves, you need to allow the metric to evolve dynamically, ...

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