# Talk:Metric tensor (general relativity)

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 To-do list for Metric tensor (general relativity): Discussion of metric as gravitational potential (+ link to weak-field approximation). Raising and lowering indices. Cut down the 'volume' section (and shift that derivation to another article - or just get rid of it).

## Cut material

I cut out the material on the volume form which properly belongs at volume form. Here it is for reference:

Let [g] be the matrix of elements $g_{\mu\nu}$. Matrix [g] is symmetric, so due to a corollary of the spectral theorem, there exists an orthogonal transformation matrix Λ which diagonalizes [g], e.g.

$D = \Lambda^\top [g] \Lambda$

where D is a diagonal matrix whose diagonal elements are eigenvalues of [g]: $D_{\alpha\alpha} = \lambda_\alpha$. (Note that Λ can be chosen so that the eigenvalues are in numerical order, D00 being the smallest.) Then there is a diagonal matrix V which "unitizes" D, i.e. which applies the mapping $\lambda_\alpha \mapsto \mbox{sgn} (\lambda_\alpha)$ to the diagonal elements of D. Such matrix V has diagonal elements

$V_{\alpha\alpha} = \left\{ \begin{matrix} {1 \over \sqrt{| \lambda_\alpha |}} & \quad \mbox{if} \, \lambda_\alpha \ne 0 \\ 0 & \quad \mbox{if} \, \lambda_\alpha = 0 \end{matrix} \right.$

Then

$[\eta] = V^\top \Lambda^\top [g] \Lambda V$

and for a given manifold, the trace of [η] will be the same for all points and is referred to as the signature of the metric. (A signature of +2 is synonymous with a signature of (− + + +). ) This matrix [η] has the components of the Minkowski metric, which means that the manifold is, at each one of its points, locally smooth.

The matrix $(V \Lambda)^\top$ is a Jacobian (a multivariate differential, or push forward) which transforms [η] to [g],

$[g] = V \Lambda [\eta] \Lambda^\top V^\top$

and taking determinants

$g := \mbox{det}([g]) = \mbox{det}\,(V \Lambda) \,\mbox{det}([\eta]) \,\mbox{det}(\Lambda^\top V^\top)$
$= \mbox{det}^2 (V \Lambda) \, \mbox{det}([\eta]), \$
$g = -\mbox{det}^2 (V \Lambda), \$
$\mbox{det}(V \Lambda) = \sqrt{-g},$

but due to a property of diffeomorphisms, a volume element $dx^0 dx^1 dx^2 dx^3$ whose factors are components of an orthonormal basis (locally), when transformed to components $dx^{\bar\mu}$, has the determinant of the Jacobian matrix J as conversion factor:

$G = dx^0 dx^1 dx^2 dx^3 = \mbox{det}(J) \, dx^{\bar 0} dx^{\bar 1} dx^{\bar 2} dx^{\bar 3}.$

-- Fropuff 18:02, 22 February 2006 (UTC)

## Amusing Veblen/Einstein anecdote

See Sign convention ---CH 01:54, 25 May 2006 (UTC)

## Question on Metric Equation

$g_{\bar \mu \bar \nu} = \frac{\partial x^\rho}{\partial x^{\bar \mu}}\frac{\partial x^\sigma}{\partial x^{\bar \nu}} g_{\rho\sigma} = \Lambda^\rho {}_{\bar \mu} \, \Lambda^\sigma {}_{\bar \nu} \, g_{\rho \sigma} .$

should perhaps be:

$g_{\bar \mu \bar \nu} = \frac{\partial x^\rho}{\partial x^{\bar \mu}}\frac{\partial x^\sigma}{\partial x^{\bar \nu}} g_{\rho\sigma}\overset ? = \Lambda_{\bar \mu} { }^ \rho \, \Lambda_{\bar \nu}{ }^\sigma \, g_{\rho \sigma}.$

as per discussion

http://www.physicsforums.com/showthread.php?p=3398120#post3398120 (especially post #36) JDoolin (talk) 14:51, 11 July 2011 (UTC)

I withdraw the question based on post #39 in the same thread. Thanks. (JDoolin (talk) 15:11, 12 July 2011 (UTC))