# Laplace-Einstein metrics

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If $(M,g)$ is a riemannian manifold, I define the Laplace-Einstein equations:

$$\mu ric(g)+\mu' \Delta (ric(g))=\lambda g + \lambda' \Delta (g)$$

where $ric$ is the Ricci curvature and $\Delta$ is the Laplacian.

Have we black holes solutions of the Laplace-Einstein equations?

asked Dec 4, 2022
edited Dec 4, 2022

Is there any physical motivation for this choice of equations? The equations contain 4th-order derivatives of the metric; one motivation for Einstein's equations was to have equations with at most second order derivatives of the metric.

As for black hole solutions: Why don't you start out like Karl Schwarzschild and try to derive a spherically symmetric solution?

It may be interesting to note that with the Laplace-Einstein equations, the Einstein tensor $ric(g)-\lambda g$ can be a proper vector of the Laplacian operator.

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