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Are Einstein's field equations without the cosmological constant scale invariant?

If so does the addition of the cosmological constant break the scale invariance?

John

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- #1

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Are Einstein's field equations without the cosmological constant scale invariant?

If so does the addition of the cosmological constant break the scale invariance?

John

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[itex]\rho_{crit} = \frac{3c^2H^2}{8\pi G}[/itex]

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timmdeeg

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Hi Mordred, I rather think the universe would be spatially flat in that case.

[itex]\rho_{crit} = \frac{3c^2H^2}{8\pi G}[/itex]

[itex]\rho_{crit} = \frac{3c^2H^2}{8\pi G}[/itex] combined with the first Friedmann equation yields k = 0.

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timmdeeg

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Hi Mordred, I rather think the universe would be spatially flat in that case.

[itex]\rho_{crit} = \frac{3c^2H^2}{8\pi G}[/itex] combined with the first Friedmann equation yields k = 0.

as far as I know the definition of critical density is the density at which stops expansion granted its also related to the curvature. Mind you dark energy complicates this definition.

http://www.astro.virginia.edu/~jh8h/glossary/criticaldensity.htm

http://www.collinsdictionary.com/dictionary/english/critical-density

Its also the definition in my textbooks which I can't post one of them being Barbera Ryden's introductory to cosmology

more accurately its the density at which expansion stops without the cosmological constant. Ignoring the cosmological constant it would describe the fate of the universe a flat is static, positive curvature is open and expanding forever, negative would be collapsing. However the cosmological constant complicates this older reasoning.

http://map.gsfc.nasa.gov/universe/uni_fate.html

see the link above for further details on the effect of geometry has on expansion

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cepheid

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as far as I know the definition of critical density is the density at which stops expansion granted its also related to the curvature. Mind you dark energy complicates this definition.

http://www.astro.virginia.edu/~jh8h/glossary/criticaldensity.htm

Mordred

I'm afraid that timdeeg is right in post #4, and you are wrong. The definition of the critical density is the density at which the universe is neither spatially open nor closed, but flat (zero curvature). In the link you posted above, you seem to have missed a very important part of the description:

Indeed, (for the zero lambda case), if the density is critical, the universe expandsMordred's link above said:The mass density of the universe which just stops the expansion of space,after infinite cosmic time has elapsed.

EDIT: also this whole discussion is kind of off topic, it seems unrelated to what the OP was asking.

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fair enough, good to know and yes I did miss that aspect thanks for the clarification

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Chronos

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Nabeshin

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The EFE do not single out a length scale, hence are 'scale invariant' in that sense. Simply put, gravity on its own knows only about c and G, out of which a length scale cannot be formed. Given matter fields in the form of a stress energy tensor, one can again construct a length scale: [itex] \ell \sim \left(G c^{-4} < T > \right)^{-1/2}[/itex]

The addition of Lambda does also introduce a length scale (one can after all think of it as a stress energy). The dimensions of Lambda are L^-2, so obviously 1/sqrt(Lambda) is a preferred length.

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Also to answer #2, it is kind of true that in the absence of matter-energy scale invariance should be recovered, however in practice this is not so at least in the Schwarzschild case as it is usually interpreted physically (weak field), as mass-energy manages to sneak in thru the boundary condition(central mass) that makes it lose the scale invariance.

But the equation by itself is trivially scale invariant.

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