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Some of the mathematical aspects of the Liouville operator can be found in the second book by Reed and Simon, in section X.14 (it is not a comprehensive account, but it gives the basic ideas and proofs). In the notes at the end of chapter X, in the part dedicated to section X.14, there is also a quite extensive bibliography that may be useful.


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The answer posted by @KyleKanos and associated comments fully answered my question. Here, I am gathering that information together and hope to represent it in a way so that future readers with my limited physical/mathematical knowledge may more rapidly perceive the solution to my question. Note that I'll ignore the constants that I included in my question, ...


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As derivatives, the Lie and covariant derivatives involve comparing tensors at different points on the manifold. They differ in the prescription given for comparing the tensors at two different points. The key concept with a covariant derivative $\nabla_\xi = \xi^a\nabla_a$ is parallel transport. It is defined so that as you move along a geodesic in the ...


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The notion of derivative requires a notion of comparison. In a general manifold, tangent vectors at different points belong to totally different vector spaces (see footnote 1), so we must define a way of mapping one tangent vector to another tangent space that we shall take, by definition to be the the "invariant image" of the vector in the new tangent space ...


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There is an equivalent of the derivative for a vector field $\bf v$ - the gradient $\nabla {\bf v}$. This will work just fine on a gravitational field.


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Electric flux is given by: Here, S (vector area is same) while the strength of the electric field E is more in A (length of arrows). $\theta=0$ in both cases. Hence flux is not same. A > B


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You are making the problem too difficult for yourself. You should be looking for vector calculus identities and space-time orthogonality. Specifically, $$ \frac{\partial}{\partial t}\nabla\times\mathbf A=\nabla\times\frac{\partial\mathbf A}{\partial t}\\ \nabla^2\left(\nabla\times\mathbf A\right)=\nabla\times\left(\nabla^2\mathbf A\right) $$ You'll then ...


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After some more thought, I think I've realized what's going on. The short answer is that the above result is correct, and is just a specific case of a more general construction. Here's more explanation: the point is that the original tangent field $k^a$ is only defined on the geodesic $\gamma$, but the expression $\nabla_a k^b$ is only sensible if the ...



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