# Geometrical point of view of the harmonic constraints ($\Delta g_{ij}=0$) in General Relativity

+ 3 like - 0 dislike
119 views

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.

This post imported from StackExchange MathOverflow at 2016-07-18 16:06 (UTC), posted by SE-user Alexander Pigazzini
retagged Jul 18, 2016
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.

This post imported from StackExchange MathOverflow at 2016-07-18 16:06 (UTC), posted by SE-user Igor Khavkine
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.

This post imported from StackExchange MathOverflow at 2016-07-18 16:06 (UTC), posted by SE-user Robert Bryant
Thanks for your answers! and if we had harmonic coordinates where $g_{ij}=\lambda * \delta_{ij}$?

This post imported from StackExchange MathOverflow at 2016-07-18 16:06 (UTC), posted by SE-user Alexander Pigazzini
@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.)

This post imported from StackExchange MathOverflow at 2016-07-18 16:06 (UTC), posted by SE-user Robert Bryant
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?

This post imported from StackExchange MathOverflow at 2016-07-18 16:06 (UTC), posted by SE-user Alexander Pigazzini
@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.
 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsOver$\varnothing$lowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.