A Hyperbolic Metric

Classical Mechanics Level 3

The Poincare half-plane model for hyperbolic space puts the following metric on the plane:

$ds^2 = \frac{1}{y^2} (dx^2 + dy^2).$

Compute the Ricci scalar $R$ for this metric in matrix form. Is this a vacuum solution to Einstein's equations?

Give your answer as an $(R,$ Yes/No $)$ pair.

-2 ; \text{ No}

\frac1{y^2}; \text{ Yes}

0 ; \text{ No}

-2 ; \text{ Yes}

1 solution

Matt DeCross
May 10, 2016

The Ricci tensor for this metric is (skipping the many steps involved in obtaining the Ricci tensor, which are tedious):

$R_{\mu \nu} = \begin{pmatrix} -\frac{1}{y^2} & 0 \\ 0 & -\frac{1}{y^2} \end{pmatrix}.$

The Ricci scalar is thus $R = g^{\mu \nu} R_{\mu \nu} = -2$ . The hyperbolic half-space is one of constant negative curvature.

Computing the Einstein tensor, one finds:

$G_{\mu \nu} = R_{\mu \nu} - \frac12 R g_{\mu \nu} = 0.$

Thus, this metric is a solution for the vacuum Einstein equations. This seems strange because this is a space of constant negative curvature, which would seem to require some sort of matter/energy. However, the reason why it is allowed is that this is just a two-dimensional spatial metric, which turns out to be a degenerate case for the Einstein equations.

How was i supposed to know that R= g^( mu nu) R_( mu nu)? This should be in the article. I've never studied covariance and contravariance. The definition of the ricci scalar in tension from the ricci tensor was unclear

Bryce Cannon - 1 year, 1 month ago

So basically it's a solution because you say so, no need to show your work...? The question was to find it, and you skipped that part

David Löfqvist - 11 months ago

A Hyperbolic Metric

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