Einstein tensor with the cosmological constant present.by Messenger Tags: constant, cosmological, einstein, present, tensor 

#1
Dec312, 07:58 PM

P: 68

I can fairly well grasp the trace relationship between the Einstein tensor and the Ricci tensor, and see that Ricci tensor is a multiple of the metric. If the cosmological constant is included I don't get why the Einstein tensor shouldn't become a multiple of the metric (leaving out physical considerations) to still achieve a Ricci flat manifold. Any text suggestions to help me explore this appreciated.
[tex]G_{\mu\nu}=R_{\mu\nu}\frac{1}{2}R g_{\mu\nu}=0[/tex] [tex]g^{\mu\nu} G_{\mu\nu}=g^{\mu\nu} R_{\mu\nu}\frac{1}{2}R g^{\mu\nu} g_{\mu\nu}=0[/tex] [tex]G=R2R=0[/tex] [tex]G=R=0[/tex] or [tex]G_{\mu\nu}=R_{\mu\nu}\frac{1}{2}R g_{\mu\nu}=0[/tex] [tex]G_{\mu\nu}\Lambda g_{\mu\nu}=R_{\mu\nu}\frac{1}{2}R g_{\mu\nu}=0[/tex] [tex]\Lambda g_{\mu\nu}\Lambda g_{\mu\nu}=R_{\mu\nu}\frac{1}{2}R g_{\mu\nu}=0[/tex] [tex]\Lambda g^{\mu\nu} g_{\mu\nu}\Lambda g^{\mu\nu} g_{\mu\nu}=g^{\mu\nu} R_{\mu\nu}\frac{1}{2}R g^{\mu\nu}g_{\mu\nu}=0[/tex] [tex]4\Lambda  4\Lambda =R 2R=R=0[/tex] 



#2
Dec312, 08:05 PM

PF Gold
P: 4,081

As far as I know the cosmological constant adds curvature so that if R=0, then [itex]G^a_a=g^a_a\Lambda[/itex]. This is your equation
[tex] G_{\mu\nu}\Lambda g_{\mu\nu}=R_{\mu\nu}\frac{1}{2}R g_{\mu\nu}=0 [/tex] with R=0 



#3
Dec312, 08:20 PM

Physics
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PF Gold
P: 5,506

[tex]G_{ab} + \Lambda g_{ab} = R_{ab}  \frac{1}{2} g_{ab} R + \Lambda g_{ab} = 8 \pi T_{ab}[/tex] If we have a spacetime with no matter or radiation present, then [itex]T_{ab} = 0[/itex] and we have [tex]G_{ab} + \Lambda g_{ab} = R_{ab}  \frac{1}{2} g_{ab} R + \Lambda g_{ab} = 0[/tex] In other words, the presence of a cosmological constant doesn't change the definition of the Einstein tensor; it's still [itex]G_{ab} = R_{ab}  1/2 g_{ab} R[/itex]. What the cosmological constant does is give more possible solutions for [itex]G_{ab}[/itex] when [itex]T_{ab} = 0[/itex]. One way of interpreting this is that the cosmological constant provides another form of "stressenergy" which isn't ordinary matter or radiation. (Sometimes the term "dark energy" is used for this, although that term is also used to refer to other ways for nonzero stressenergy to be present without any normal matter or radiation.) In other words, we rearrange the second equation above to [tex]G_{ab} = R_{ab}  \frac{1}{2} g_{ab} R =  \Lambda g_{ab}[/tex] So the cosmological constant now looks like a form of "stressenergy tensor" that is proportional to the metric; we just write [itex](T_{ab})_{\Lambda} =  ( \Lambda / 8 \pi ) g_{ab}[/itex]. 



#4
Dec312, 08:28 PM

P: 68

Einstein tensor with the cosmological constant present. 



#5
Dec312, 08:35 PM

P: 68

[tex]"G_{\alpha\beta}"=R_{ab}  \frac{1}{2} g_{ab} R + \Lambda g_{ab}=G_{ab} + \Lambda g_{ab}[/tex] 



#6
Dec312, 08:54 PM

Physics
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PF Gold
P: 5,506

[tex]\frac{1}{16 \pi} R \sqrt{g}[/tex] Hilbert and Einstein claimed that this was the right Lagrangian because it was the most general one using no greater than second derivatives of the metric. But they hadn't realized that that was, strictly speaking, false: the most general such Lagrangian also includes a constant term: [tex]\frac{1}{16 \pi} \left( R + 2 \Lambda \right) \sqrt{g}[/tex] Varying this with respect to the metric gives the EFE with a cosmological constant, as I wrote it down. So there is certainly "proof" of that version. So no, I'm not bothered. (Technical note: I left out the Lagrangian due to ordinary matter and radiation in the above, so strictly speaking, varying what I wrote above gives the vacuum EFE, with or without a cosmological constant. Adding the matter action term doesn't change anything I said, it just adds the ordinary stressenergy tensor [itex]T_{ab}[/itex] on the RHS of the EFE.) 



#7
Dec312, 08:55 PM

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PF Gold
P: 5,506





#8
Dec312, 09:01 PM

P: 68

Peter,
Thanks for some thought provoking comments, I will be back later with more questions after researching more in depth what you have stated. 



#9
Dec312, 09:34 PM

Physics
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PF Gold
P: 5,506

Saw one other item on rereading the OP:
http://en.wikipedia.org/wiki/Einstein_manifold 



#10
Dec312, 09:55 PM

P: 68

it states that g22 and g33 are the same as flat spacetime, and that g00 and g11 are subject to the boundary conditions that spacetime becomes flat at infinity. Isn't the [tex]1\frac{2M}{r}[/tex] the same derivation for Newtonian gravity, as in a small perturbation away from flat spacetime? Isn't this solution for Ruv=0 Schwarzschild solution only applicable since it is so near Minkowski space? This is also what I don't understand. The cosmological constant for Minkowski space would seem to be added to the "1" with the perturbation becoming the Newtonian phi. How does one exactly derive the cosmological constant into the formula for Newtonian gravity? I have seen the result that Hobbs et. al. arrived at General Relatvity: An Introduction for Physicists but not the actual logic in the derivation. 



#11
Dec412, 12:10 AM

Physics
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PF Gold
P: 5,506

http://en.wikipedia.org/wiki/De_Sitter_space The static coordinates on this spacetime do indeed make the cosmological constant look like a Newtonian "potential". However, that doesn't mean you have to be able to "derive" de Sitter spacetime as a perturbation on Minkowski spacetime. Once again, there's nothing that requires the "correction" term in the de Sitter metric in static coordinates to be small compared to 1. 



#12
Dec612, 02:43 PM

P: 68

If for [tex]\frac{1}{16 \pi} R \sqrt{g}[/tex] [tex]R \rightarrow 0 [/tex] does that mean for [tex]\frac{1}{16 \pi} \left( R + 2 \Lambda \right) \sqrt{g}[/tex] [tex]R \rightarrow 2\Lambda[/tex]? 



#13
Dec612, 02:47 PM

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PF Gold
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#14
Dec612, 03:13 PM

P: 68

so to let it also "vanish" with a constant present in Euclidean space (Minkowski), wouldn't R have to become equal to [itex]2\Lambda[/itex]? 



#15
Dec612, 03:39 PM

Physics
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PF Gold
P: 5,506

http://en.wikipedia.org/wiki/De_Sitter_space http://en.wikipedia.org/wiki/Anti_de_Sitter_space 



#16
Dec612, 03:54 PM

P: 68

I need to read up more on the derivation of the LaGrangian that you showed. I would think there would be some correlation between R and [itex]\Lambda[/itex] for them to show up together in it, similar to potential and kinetic energy.




#17
Dec612, 04:07 PM

PF Gold
P: 4,081





#18
Dec612, 04:13 PM

Physics
Sci Advisor
PF Gold
P: 5,506

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 [itex]R[/itex], composed of second derivatives of the metric, and a constant [itex]\Lambda[/itex], 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 [itex]\Lambda[/itex] is there so that the field equation will just contain [itex]\Lambda[/itex] instead of [itex]\Lambda / 2[/itex]). But there's no other connection between them; they are completely independent of each other as far as GR is concerned. The [itex]\sqrt{g}[/itex] 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 [itex]1 / 16 \pi[/itex] is for convenience, to make certain formulas look the way physicists were used to having them look. 


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