Timeline for Geometrical point of view of the harmonic constraints ($\Delta g_{ij}=0$) in General Relativity
Current License: CC BY-SA 3.0
11 events
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| Jul 4, 2016 at 17:29 | comment | added | Robert Bryant | @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. | |
| Jul 4, 2016 at 16:03 | comment | added | MathDG | Another little question...in this particular kind of metric (for example in dimension 2) where $\Delta g=0$ the curvature isn't necessary zero, is correct? | |
| Jul 4, 2016 at 16:00 | comment | added | user21349 | You say "in General Relativity," but there is no such constraint that is applied generically in general relativity. In the very early days of GR, it was thought that the metric's determinant needed to be constrained to be -1 everywhere. Today, it is often convenient to make a particular choice of gauge, such as harmonic coordinates en.wikipedia.org/wiki/Harmonic_coordinate_condition . But the only constraint on the metric that is absolutely mandatory in the standard modern formulation of GR is that it not be degenerate. | |
| Jul 4, 2016 at 14:17 | comment | added | MathDG | Thank prof.! my question arises because I was looking for a possible interest in the study of totally umbilical immersions with this particular type of metric and thought to find was something in GR. | |
| Jul 4, 2016 at 14:02 | comment | added | Robert Bryant | @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. | |
| Jul 4, 2016 at 13:44 | comment | added | MathDG | thanks prof. Bryant, then from what I understand, there is no interest in a metric of this type, am I right? ...I mean that there isn't interest in $g=h(x,y)(dx^2+dy^2)$ where $h>0$ and satisfies $h_{xx}+h_{yy}=0$, or wrong? | |
| Jul 4, 2016 at 13:19 | comment | added | Robert Bryant | @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.) | |
| Jul 4, 2016 at 13:08 | comment | added | MathDG | Thanks for your answers! and if we had harmonic coordinates where $g_{ij}=\lambda * \delta_{ij}$? | |
| Jul 4, 2016 at 12:43 | comment | added | Robert Bryant | 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. | |
| Jul 4, 2016 at 12:13 | comment | added | Igor Khavkine | 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. | |
| Jul 4, 2016 at 10:55 | history | asked | MathDG | CC BY-SA 3.0 |