Talk:Metric tensor (general relativity)

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edit·history·watch·refresh Stock post message.svg 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[edit]

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}.

See also volume form.

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

Amusing Veblen/Einstein anecdote[edit]

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


Question on Metric Equation[edit]

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))