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Recently I found some publications on Cosmologies with variable cosmological constant. The Bianchi Identity then implies that the divergence of the modified Energy Stress: Tensor $$\hat{T}_{ab}=T_{ab}+\Lambda g_{ab}$$ has to vanish. I asked my professor about this and he said that variable cosmological constants don't really make sense, because one could always absorb the lambda term into the standard energy stress tensor and not be able to distinguish between the two. I thought about a counter argument. The only thing I could come up with was that if you consider an ideal fluid $$T_{ab}=(\rho+p)u_au_b-pg_{ab}$$ Then the modified energy stress tensor will be: $$T_{ab}=(\rho+p)u_au_b-(p-\Lambda)g_{ab}$$ At this point I could rewrite $p'=p-\Lambda$ At a first glance then the Lambda term would reappear in the first term: $$T'_{ab}=(\rho+p'+\Lambda)u_au_b-(p')g_{ab}$$ One might be tempted to think that this is a counter argument however I could make another change and label $\rho'=\rho+\Lambda$ such that I recover the standard form of the Energy stress tensor of an ideal fluid. Is my reasoning correct or is there a valid counter argument to what my professor said?

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Vacuum energy is an energy density characteristic of empty space. As we assume it does not pick up a preferred direction, the associated energy-momentum tensor should be Lorentz invariant in locally inertial coordinates, that is proportional to the metric $T_{\mu \nu}^{vac} = -\rho_{vac} \eta_{\mu \nu}$, as $\eta_{\mu \nu}$ is the only Lorentz invariant tensor. Generalizing to arbitrary coordinates $T_{\mu \nu}^{vac} = -\rho_{vac} g_{\mu \nu}$.

Comparing to the energy-momentum tensor of a perfect fluid $T_{\mu \nu} = (\rho + p) U_\mu U_\nu + p g_{\mu \nu}$, we have $p_{vac} = -\rho_{vac}$. The energy density should be constant throughout spacetime, since a gradient would not be Lorentz invariant.

If we add a cosmological constant $\Lambda$ to the L.H.S. of the Einstein field equations we have $R_{\mu \nu} -\frac12 R g_{\mu \nu} + \Lambda g_{\mu \nu} = 8 \pi T_{\mu \nu}$. By comparing we get $\Lambda = 8 \pi \rho_{vac}$.

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As per how the vacuum energy was conceived, the cosmological constant $\Lambda$ can not be variable.
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You can not embed the cosmological constant $\Lambda$ in a perfect fluid $f$ with equation of state $p = w \rho$ and rename it as $f'$ with equation of state $p' = w' \rho'$, because if $w'$ is not constant it is not legitimate to talk of equation of state. The perfect fluid models used in general relativity are matter: $w = 0$, radiation: $w = 1/3$ and vacuum: $w = -1$.
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Nevertheless you can keep the $\Lambda$ on the R.H.S. as a separate term compounding the total energy-momentum tensor which shapes the geometry of spacetime. This is an essential constituent of the current $\Lambda CDM$ cosmological model.

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When people say that "you can absorb the cosmological constant into the definition of the stress-energy tensor," they don't mean that its effect is not physically measurable, they just been that you're free to decide whether or not you consider vacuum energy to be "stress-energy". Either way, its effect is physically measurable.

Also, $p$ and $\rho$ are set by the physics of the fluid that you're considering - you can't just redefine them at will. It's conceivable that a perfect fluid plus dark energy might correspond to another perfect fluid, but that doesn't mean that the dark energy is physically unmeasurable, because they'd be two different fluids that could be distinguished experimentally.

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The energy tensor (and thus the Friedmann equations) shall be supplemented with equation of state: $$ p = w \rho $$ where $w = -1, 0, \frac{1}{3}$, for CC, baryonic matter, radiation, respectively.

When you cavalierly make the shifting $$ p′ =p−Λ $$ or $$ ρ′ =ρ+Λ $$ you got a mixed bag of thingamajigs.

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  • $\begingroup$ well then you could just say: $p'=w\rho'$ $\endgroup$
    – eeqesri
    Nov 28, 2018 at 18:40
  • $\begingroup$ No you can't. If you can, thousands of folks in $\Lambda CDM$ business would thank you for it. $\endgroup$
    – MadMax
    Nov 28, 2018 at 18:45
  • $\begingroup$ Could you please elaborate your statement. I didn't understand. $\endgroup$
    – eeqesri
    Nov 28, 2018 at 19:33

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