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  Weyl transformation of geodesic distance

+ 4 like - 0 dislike

Consider a Riemannian manifold $M$ with a metric $g$. For two points $x,y \in M$ the geodesic distance $d(x,y)$ is defined in the usual way. 

I would like to know if there is a formula expressing how $d(x,y)$ transforms under a Weyl transformation
g(z) \mapsto e^{2 \sigma(z)} g(z).
Is there such a formula for arbitrary dimension of $M$, or at least for dimension 2?

asked Nov 20, 2015 in Theoretical Physics by no.palindromes (20 points) [ no revision ]

1 Answer

+ 4 like - 0 dislike

If you want to find the change in length of a single curve $\gamma$, the answer is of course trivial $$\int \sqrt{g(\gamma',\gamma')} \mathrm{d} \gamma \to  \int e^\sigma \sqrt{g(\gamma',\gamma')} \mathrm{d} \gamma $$ The problem is that a general conformal transformation changes the geodesic structure. This can be easily verified e.g. by taking a flat Euclidean metric, seeing that scalar curvature becomes non-zero under a conformal transform and the Levi-Civita connection is thus suddenly non-integrable (i.e. the geodesics are qualitatively different geodesics).

I am more of a dimension 4 person, but I am quite certain that a general $\sigma$ has no closed formula for a change in distances even in the all so special dimension 2.

There might be some luck for harmonic $\sigma: \nabla^i \partial_i \sigma=0$, because for those in dimension 2 the Ricci tensor is invariant and the scalar curvature only rescales by an $e^{-2 \sigma}$ factor. This means that e.g. in the case of a flat Euclidean metric the result is again flat because apart from Ricci tensor and it's trace, the Riemann curvature tensor has only the Weyl-tensor degree of freedom which is invariant with respect to conformal transformations.

answered Nov 21, 2015 by Void (1,645 points) [ no revision ]

Thank you for the reply! While I cannot assume that $\sigma$ is harmonic, I can assume that the Weyl transformation is infinitesimal. I tried to derive how a geodesic changes under the infinitesimal transformation but without success. Do you think this can be done?

Yes, it's possible to do this under some assumptions which I think would involve the requirement that the geodesic is stable (small perturbations of it stay small). This would suggest that this will not work in spaces with negative curvature because there all trajectories are unstable.

You simply take the formula for the rescaled distance, assume a perturbed trajectory $\gamma \to \gamma + \delta \gamma$, compute Lagrange equations for the $e^\sigma$ rescaled Lagrangian, assume that the original (unscaled, unperturbed) Lagrange equations are fulfilled, solve the resulting linearized equation for $\delta \gamma$, and compute the resulting change in geodesic distance.

This still involves 1) knowing the exact geodesic shape $\gamma(t)$, 2) solving a second order linear differential equation with non-constant coefficients and 3) computing two quadratures corresponding to the rescaling of length and the variation of length due to $\delta \gamma$. The reason why I am cautious to say what are the sufficient assumptions for this to work is that the equation to solve is in fact a boundary value problem with $\delta \gamma=0$ at the ends of the geodesic - the assumptions should simply secure the existence of a non-divergent solution to this problem.

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