1
$\begingroup$

Let $(\mathcal{M},g)$ be a globally hyperbolic manifold equipped with a metric $g$ and a Levi-Civita connection $\nabla$. Suppose $f:\mathcal{M}\times\mathcal{M}\to\mathbb{R}$ is a smooth scalar function dependent on two points of the manifold. Suppose the diffeomorphism $\phi:\mathcal{M}\to\mathcal{M}$ is a symmetry transformation of the metric, i.e., an isometry. Let $p,q\in\mathcal{M}$ be distinct spacetime points. If $\mathbb{X}_p$ is a vector field belonging to the tangent space at $p$, then let $\phi_t$ be a diffeomorphic map which sends the point $p$ a parameter distance $t$ along the integral curve of $\mathbb{X}$ to $q$. Given this structure one can define a the Lie derivative as the difference between the pushforward of a tensor field and itself in the limit that the parameter $t\to0$, in math words;$$\mathcal{L}_\mathbb{X}\mathbb{T} = \lim_{t\to0}\frac{((\phi_{-t})_\star\mathbb{T})-\mathbb{T}}{t},$$ where ${}_\star$ represents the pushforward operation.

With this structure, I would like to know if the following are possible;

Question 1: If the scalar function is of two variables then is it possible to define $f(x,t)$ with boundary conditions $f(p)\equiv f(p,0)$ and $f(q)\equiv f(p,t_1)$ and transport $\mathbb{T}$ along the gradient of $\nabla f$? That is, is this a valid procedure;$$\mathbb{X}=\mathbb{X}^\mu\nabla_\mu=\frac{\text{d}x^\mu}{\text{d}t}\nabla_\mu\overset{?}{=}g^{\mu\nu}(\nabla f)_\nu \nabla_\mu=(\nabla f)^\mu\nabla_\mu,$$ where we have use the coordinate basis at $p$ as our starting point and will transport to $q$?

Question 2: If $\mathbb{X}$ is a Killing vector, then if the above holds, does this translate to $$(\mathcal{L}_\mathbb{X}(g))_{ab}=\nabla_a\mathbb{X}_b+\nabla_b\mathbb{X}_a\overset{?}{=}\nabla_a\nabla_b f+\nabla_b\nabla_a f?$$

$\endgroup$

migrated from physics.stackexchange.com Apr 2 '18 at 17:13

This question came from our site for active researchers, academics and students of physics.

  • 1
    $\begingroup$ Would Mathematics be a better home for this question? $\endgroup$ – Qmechanic Mar 31 '18 at 18:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.