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What does it mean, from the geometrical point of view, use (in General Relativity) of the constraints on the metric tensor's coefficients such that $\Delta g_{ij}=0$? (where $\Delta$ is the Beltrami-Laplace Operator, $g_{ij}$ the metric tensor).
With $\Delta g_{ij}=0$, I mean the Laplace-Beltrami operator, applied componentwise to the components of the metric tensor.

Thank you in advance!

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    $\begingroup$ Since you are applying $\Delta$ to individual coordinate components, this seems like a strongly coordinate dependent condition, which might be hard to interpret geometrically. If you take $\Delta_g$ to be defined by $g$ itself, via its Levi-Civita connection, $\Delta_g g = 0$ is an identity, since $g$ itself is covariantly constant. $\endgroup$
    – Igor Khavkine
    Commented Jul 4, 2016 at 12:13
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    $\begingroup$ Moreover, the equation $\Delta g_{ij}=0$ is an overdetermined system of equations for the coordinate system. For most metrics $g$, such coordinates don't exist, even locally. $\endgroup$
    – Robert Bryant
    Commented Jul 4, 2016 at 12:43
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    $\begingroup$ @AlexanderPigazzini: Well, there are almost none of those (i.e., harmonic, conformal coordinates). For example, in dimension $2$, this is equivalent to saying that the metric is what is called a Liouville metric, i.e., it admits a nontrivial quadratic first integral of its geodesic flow. Such metrics can be put in the local form $$g = h(x,y)(dx^2+dy^2)$$ where $h>0$ satisfies $h_{xx}+h_{yy}=0$ and, conversely, any such metric is a Liouville metric. (A nontrivial example is the metric on the general ellipsoid in $3$-space.) $\endgroup$
    – Robert Bryant
    Commented Jul 4, 2016 at 13:19
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    $\begingroup$ @AlexanderPigazzini: I wouldn't say that there is no interest, it's just of specialized interest. There was a time when Liouville metrics were of great interest indeed, but now they are mainly of interest to the integrable systems folks, not so much in general relativity. $\endgroup$
    – Robert Bryant
    Commented Jul 4, 2016 at 14:02
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    $\begingroup$ @AlexanderPigazzini: The condition $\Delta g_{ij} = 0$ does not constrain the curvature in any simple way. When $n=2$, there will be some very high order polynomial relation among $K$ and its first $m$ covariant derivatives that characterizes the existence of such a coordinate system, but I don't know what that is explicitly. $\endgroup$
    – Robert Bryant
    Commented Jul 4, 2016 at 17:29

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