 Quote by Naty1
Was not aware of such a distinction....but it makes no sense to me....'intrinsic' usually means we have no good ideas!!!! In this case, assigning such a 'constant' factor of integration and ascribing a specific physical meaning amid the mess of GR is byond my paygrade!!!
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Well, you could think of a cosmological constant as being a geometric feature of the particular space-time. For example, a space with a cosmological constant and no matter is a de Sitter space. A de Sitter space just comes with this constant force (the cosmological constant). Nothing present in the space causes this, it's 'built in' to it's equations.
Specifically, you get the cosmological constant by inserting it into the Einstein-Hilbert action, and then deriving the Einstein Field Equations as usual.
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the exact mathematical relationship between vacuum energy and the cosmological constant is interesting....have not seen it....
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It comes from the normal Einstein Field equations. Take a look: [tex]R_{\mu \nu} - {1 \over 2}g_{\mu \nu}\,R + g_{\mu \nu} \Lambda = {8 \pi G \over c^4} T_{\mu \nu} [/tex] Now, say no matter is present, so we can ignore the the Ricci tensor and scalar. (Those are the two 'R's, the tensor has a subscript). Now the equation becomes [tex]g_{\mu \nu} \Lambda = {8 \pi G \over c^4} T_{\mu \nu}[/tex] Using a little algebra, you can solve for the stress-energy tensor (the 'T'), and get this [tex]T_{\mu \nu} = - \frac {\Lambda c^{4}} {8 \pi G} g_{ \mu \nu}[/tex] From which you could derive the expression I posted above.
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Do you think the assumptions that go into the FLRW metric solution
to the EFE apply within gravitationally bound systems?
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They most definitely don't apply for non-homogenous distributions of matter, such as galaxies. So, as John Baez explained on that page, we say that no expansion is occurring in the galaxies (dark energy is a different matter).