- #71
PAllen
Science Advisor
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PeterDonis said:Just to be clear, the complete action in question is:
[tex]S = \int \left[ \frac{R}{16 \pi} + L_{M} \right] \sqrt{-g} d^4 x[/tex]
where [itex]R[/itex] is the Ricci scalar, [itex]L_{M}[/itex] is the Lagrangian due to matter fields, and [itex]g[/itex] is the determinant of the metric tensor. The variation of the first term with respect to the metric gives the Einstein tensor, and the variation of the second term gives (minus) the SET. The total variation must be zero, which yields the Einstein Field Equation.
It's true that the variation of the second term with respect to the metric will include the metric. But it includes no *derivatives* of the metric, so it contains no information about the curvature (or even about the connection, which is first derivatives of the metric--curvature is second derivatives).
Is it so simple as that?
y'' = y * g(x)
has very different solutions than:
y'' = g(x)
The metric, which describes aspects of geometry directly, is buried in the source term.