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## Einstein tensor with the cosmological constant present.

 Quote by Messenger I need to read up more on the derivation of the LaGrangian that you showed.
I don't have a handy online reference, but MTW discusses it. See below for a quick summary.

 Quote by Messenger I would think there would be some correlation between R and $\Lambda$ for them to show up together in it, similar to potential and kinetic energy.
No, there isn't. They have nothing to do with each other, other than appearing in the same Lagrangian.

The idea behind the Lagrangian is to construct the most general Lorentz scalar that is composed of no higher than second derivatives of the metric. There are only two such scalars possible: the scalar curvature $R$, composed of second derivatives of the metric, and a constant $\Lambda$, which is a "zeroth derivative" of the metric. So the most general Lagrangian is just the sum of those two terms (the factor of 2 in front of $\Lambda$ is there so that the field equation will just contain $\Lambda$ instead of $\Lambda / 2$). But there's no other connection between them; they are completely independent of each other as far as GR is concerned.

The $\sqrt{-g}$ is there to make a Lorentz invariant integration measure (since we're going to integrate the Lagrangian over all spacetime and then vary it with respect to the metric to obtain the field equation). The factor of $1 / 16 \pi$ is for convenience, to make certain formulas look the way physicists were used to having them look.

 Quote by Mentz114 In this paper, Padmanabhan proposes an action for the gravitational field equations such that the cosmological constant arises as a constant of integration. It's here http://uk.arxiv.org/abs/gr-qc/0609012v2
I had not seen this, looks interesting.

This paper made me think quite a bit, oldie but a goodie:http://ajp.aapt.org/resource/1/ajpias/v39/i8/p901_s1

 Quote by PeterDonis I don't have a handy online reference, but MTW discusses it. See below for a quick summary. No, there isn't. They have nothing to do with each other, other than appearing in the same Lagrangian. The idea behind the Lagrangian is to construct the most general Lorentz scalar that is composed of no higher than second derivatives of the metric. There are only two such scalars possible: the scalar curvature $R$, composed of second derivatives of the metric, and a constant $\Lambda$, which is a "zeroth derivative" of the metric. So the most general Lagrangian is just the sum of those two terms (the factor of 2 in front of $\Lambda$ is there so that the field equation will just contain $\Lambda$ instead of $\Lambda / 2$). But there's no other connection between them; they are completely independent of each other as far as GR is concerned. The $\sqrt{-g}$ is there to make a Lorentz invariant integration measure (since we're going to integrate the Lagrangian over all spacetime and then vary it with respect to the metric to obtain the field equation). The factor of $1 / 16 \pi$ is for convenience, to make certain formulas look the way physicists were used to having them look.
Gives me quite a bit to look into and think about. I really do appreciate you taking your time on this board Peter. Thanks.

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 Quote by Messenger I really do appreciate you taking your time on this board Peter. Thanks.
You're welcome! Thanks for the appreciation.

 Quote by PeterDonis There are only two such scalars possible: the scalar curvature $R$, composed of second derivatives of the metric, and a constant $\Lambda$, which is a "zeroth derivative" of the metric. So the most general Lagrangian is just the sum of those two terms (the factor of 2 in front of $\Lambda$ is there so that the field equation will just contain $\Lambda$ instead of $\Lambda / 2$). But there's no other connection between them; they are completely independent of each other as far as GR is concerned.
Hi Peter,

Looking into the "zeroth derivative". I take it from the Ricci scalar being composed of second derivatives, it is never possible for the action principle to = 0 with $2\Lambda=-R$ since this would mean that either the Ricci scalar or the cosmological constant would have to be the same order derivative of the metric. My inquiry stems from unimodular descriptions of the cosmological constant. If this were possible, and the definition of the Ricci scalar is $R=g^{\mu\nu}R_{\mu\nu}$, then this would imply that $\Lambda=g^{\mu\nu}\Lambda_{\mu\nu}$. I ask because for zero curvature it would seem feasible to then write
$\frac{1}{2}Rg_{\mu\nu}-R_{\mu\nu}=\frac{3}{4}\Lambda g_{\mu\nu}-\Lambda_{\mu\nu}=0$.
If I set $\frac{3}{4}\Lambda g_{\mu\nu}=0$, I don't see any way to tell that $\Lambda_{\mu\nu}=G_{\mu\nu}$ isn't true.
I would like to understand more in depth the proof that $\Lambda$ is only a zeroth derivative.