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So in flrw metric it's quite reasonable to take eigenvalue of the time-like component of the stress energy tensor and identify it with mass density.

Now, if someone argues the cosmological constant comes from the stress energy tensor. Wouldn't they also have to account for the change in mass density? In the Newtonian limit I'm pretty sure this is a change in mass density.

How do people who argue it doesn't matter where (stress energy or Einstein tensor) the cosmological constant comes from explain this discrepancy?

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There is a theoretical mass density associated with the cosmological constant, of roughly $10^{-26} \text{ kg}/\text{m}^3$. It is locally detectable in principle (by its gravitational effects) whichever side of the equation you put it on, so detecting it wouldn't settle the question of whether the quintessence is a cosmological constant or an ordinary field, unfortunately. It's also too small to detect, unfortunately.

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  • $\begingroup$ Wait .. I'm confused. When I modify the cosmological constant I change the value of the eigenvalue of the stress energy tensor. But if I do this from the Einstein tensor there's no eigenvalue change no? $\endgroup$ Oct 27, 2023 at 10:30
  • $\begingroup$ @MoreAnonymous Well, you can't measure the eigenvalue of the stress-energy tensor. By the same token, if you write $Λ=Λ_1+Λ_2$, you can't measure $Λ_1$ and $Λ_2$ separately. It's just algebraic manipulation and doesn't change what experiments are possible or their outcomes. $\endgroup$
    – benrg
    Oct 27, 2023 at 16:57
  • $\begingroup$ really? I could simply take the limit of special relativity (in the rest frame) and Taylor expand the force law. $F = mc^2 +1/2 mv^2 + 3/8 v^4/c^2 + 5/16 v^6/c^4 + O(10^{-32})$ Note because the speed of light is $10^8$ the error in the above equation is $8 \times -4$ and this equation only applies in the rest frame. $\endgroup$ Oct 27, 2023 at 20:09
  • $\begingroup$ Since in special relativity I can detect mass I don't see why in a limit of general relativity (the limit of the geodesic equation) a taylor expansion will not suffice $\endgroup$ Oct 27, 2023 at 20:11
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If you view the cosmological constant as dark energy related the field equations are $\rm G_{\mu \nu}= T_{\mu \nu} \ 8 \pi G/c^4$ and its associated constant density in the $\rm T^0_0$ is $\rm \rho_{\Lambda}=\Lambda c^2/(8 \pi G)$.

If you view it as a geometric term the field equations are $\rm G_{\mu \nu}\pm \Lambda_{\mu \nu}=T_{\mu \nu} \ 8 \pi G/c^4$ (the sign of the $\pm$ depending on the metric signature) and it doesn't contribute to the stress energy tensor since it is substracted on the left side of the equation.

Which one you choose is a rather philosophical decision where each of them has pros and contras, but as long as you stay consistent it doesn't really matter since the outcome doesn't depend on the interpretation.

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  • $\begingroup$ For more information see Eigenchris' video 110e on the subject. $\endgroup$
    – Yukterez
    Oct 28, 2023 at 21:09

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