- #36
Mark M
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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.Naty1 said: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!
Specifically, you get the cosmological constant by inserting it into the Einstein-Hilbert action, and then deriving the Einstein Field Equations as usual.
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.the exact mathematical relationship between vacuum energy and the cosmological constant is interesting...have not seen it...
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).Do you think the assumptions that go into the FLRW metric solution
to the EFE apply within gravitationally bound systems?
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