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- Thread starter TrickyDicky
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- #2

stevebd1

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This is a good question. When considering if the universe is flat, hyperbolic or hyperspherical you only need to consider the density, the critical density being expressed as-

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

where for a flat universe [itex]\rho/\rho_c=1[/itex], less and the universe is hyperbolic, more and the universe is hyperspherical.

Normally, based on regular matter with an equation of state of 0, [itex]\rho/\rho_c=1[/itex] would halt the expansion of the universe but if we include for dark energy with an eos of -1, then the critical density can imply a flat but accelerating universe, the algebraic form for gravity in GR expressed as-

[tex]\rho+3P/c^2=\rho_m+\rho_\Lambda+3(-\rho_\Lambda c^2)/c^2=\rho_m-2 \rho_\Lambda[/tex]

where [itex]\rho_m[/itex] is baryonic & dark matter, [itex]\rho_\Lambda[/itex] is dark energy and [itex]P=w\rho c^2[/itex] where w is eos.

Based on the above, for a static universe, the following should apply-

[tex]\rho_m=2\rho_\Lambda[/tex]

which based on current observations isn't the case ([itex]\rho_\Lambda[/itex] being estimated as being considerably greater than [itex]\rho_m[/itex]), so it's my understanding that while the EFE, taking into account pressure in the [itex]T_{\mu\nu}[/itex] component would imply an accelerating universe, the critical density (not taking into account pressure) equation implies a flat one.

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

where for a flat universe [itex]\rho/\rho_c=1[/itex], less and the universe is hyperbolic, more and the universe is hyperspherical.

Normally, based on regular matter with an equation of state of 0, [itex]\rho/\rho_c=1[/itex] would halt the expansion of the universe but if we include for dark energy with an eos of -1, then the critical density can imply a flat but accelerating universe, the algebraic form for gravity in GR expressed as-

[tex]\rho+3P/c^2=\rho_m+\rho_\Lambda+3(-\rho_\Lambda c^2)/c^2=\rho_m-2 \rho_\Lambda[/tex]

where [itex]\rho_m[/itex] is baryonic & dark matter, [itex]\rho_\Lambda[/itex] is dark energy and [itex]P=w\rho c^2[/itex] where w is eos.

Based on the above, for a static universe, the following should apply-

[tex]\rho_m=2\rho_\Lambda[/tex]

which based on current observations isn't the case ([itex]\rho_\Lambda[/itex] being estimated as being considerably greater than [itex]\rho_m[/itex]), so it's my understanding that while the EFE, taking into account pressure in the [itex]T_{\mu\nu}[/itex] component would imply an accelerating universe, the critical density (not taking into account pressure) equation implies a flat one.

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- #3

Chalnoth

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Yes, vacuum energy absolutely gravitates, and contributes to the space-time curvature. If we had the same expansion rate, and the same amount of normal/dark matter, but no dark energy, then we would have negative spatial curvature at present.

- #4

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I follow your reasoning up to this last statement

Do you mean that in your opinion,taking into account pressure, a flat universe shouldn't be an accelerating universe? or maybe I misunderstood this last part?it's my understanding that while the EFE, taking into account pressure in the [itex]T_{\mu\nu}[/itex] component would imply an accelerating universe, the critical density (not taking into account pressure) equation implies a flat one.

- #5

stevebd1

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Do you mean that in your opinion,taking into account pressure, a flat universe shouldn't be an accelerating universe? or maybe I misunderstood this last part?

I'm saying that a flat universe can be an accelerating universe if you take into account pressure.

- #6

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Ok, I see.Hi TrickyDicky

I'm saying that a flat universe can be an accelerating universe if you take into account pressure.

Do you agree with chalnoth that the vacuum energy generates gravitational field?

- #7

stevebd1

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Do you agree with chalnoth that the vacuum energy generates gravitational field?

A negative one, hence the acceleration.

- #8

stevebd1

Gold Member

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Here's a few useful equations that demonstrate a flat and accelerating universe-

Friedmann equation-

[tex]H^2=\left(\frac{\dot{a}}{a}\right)^2=\frac{8\pi G}{3}\rho-\frac{kc^2}{a^2}+\frac{\Lambda c^2}{3}[/tex]

Friedmann acceleration equation-

[tex]\dot{H}+H^2=\frac{\ddot{a}}{a}=-\frac{4\pi G}{3}\left(\rho+\frac{3P}{c^2}\right)+\frac{\Lambda c^2}{3}[/tex]

where [itex]P[/itex] is pressure.

Both equations can be rewritten where-

[tex]\rho'\Rightarrow \rho_m+\frac{\Lambda c^2}{8\pi G}=(\rho_m+\rho_\Lambda)[/tex]

[tex]P'\Rightarrow P_m-\frac{\Lambda c^4}{8\pi G}=(P_m-P_\Lambda)[/tex]

where the equation of state for dark energy is [itex]w=-1[/itex] and for ordinary and dark matter, [itex]w=0[/itex].

Friedmann equation becomes-

[tex]H^2=\frac{8\pi G}{3}\rho'-\frac{kc^2}{a^2}[/tex]

which reduces to the critical density equation when [itex]k=0[/itex].

Friedmann acceleration equation becomes-

[tex]\dot{H}+H^2=-\frac{4\pi G}{3}\left(\rho'+\frac{3P'}{c^2}\right)[/tex]

which would normally show that both energy density and pressure would cause a deceleration in the expansion of the universe though the inclusion of the cosmological constant (or dark energy or vacuum energy) means the universe accelerates.

[tex]\frac{\dot{H}}{H^2}=-(1+q)[/tex]

where q is the deceleration parameter-

[tex]q=\frac{1}{2\rho_c}\left(\rho'+\frac{3P'}{c^2}\right)=\frac{1}{2}(1+3w)[/tex]

where [itex]w=P'/(\rho'c^2)[/itex], the equation of state of the universe.

http://en.wikipedia.org/wiki/FLRW#Solutions

http://en.wikipedia.org/wiki/Deceleration_parameter

http://scienceworld.wolfram.com/physics/DecelerationParameter.html

http://www.answers.com/topic/deceleration-parameter

Friedmann equation-

[tex]H^2=\left(\frac{\dot{a}}{a}\right)^2=\frac{8\pi G}{3}\rho-\frac{kc^2}{a^2}+\frac{\Lambda c^2}{3}[/tex]

Friedmann acceleration equation-

[tex]\dot{H}+H^2=\frac{\ddot{a}}{a}=-\frac{4\pi G}{3}\left(\rho+\frac{3P}{c^2}\right)+\frac{\Lambda c^2}{3}[/tex]

where [itex]P[/itex] is pressure.

Both equations can be rewritten where-

[tex]\rho'\Rightarrow \rho_m+\frac{\Lambda c^2}{8\pi G}=(\rho_m+\rho_\Lambda)[/tex]

[tex]P'\Rightarrow P_m-\frac{\Lambda c^4}{8\pi G}=(P_m-P_\Lambda)[/tex]

where the equation of state for dark energy is [itex]w=-1[/itex] and for ordinary and dark matter, [itex]w=0[/itex].

Friedmann equation becomes-

[tex]H^2=\frac{8\pi G}{3}\rho'-\frac{kc^2}{a^2}[/tex]

which reduces to the critical density equation when [itex]k=0[/itex].

Friedmann acceleration equation becomes-

[tex]\dot{H}+H^2=-\frac{4\pi G}{3}\left(\rho'+\frac{3P'}{c^2}\right)[/tex]

which would normally show that both energy density and pressure would cause a deceleration in the expansion of the universe though the inclusion of the cosmological constant (or dark energy or vacuum energy) means the universe accelerates.

[tex]\frac{\dot{H}}{H^2}=-(1+q)[/tex]

where q is the deceleration parameter-

[tex]q=\frac{1}{2\rho_c}\left(\rho'+\frac{3P'}{c^2}\right)=\frac{1}{2}(1+3w)[/tex]

where [itex]w=P'/(\rho'c^2)[/itex], the equation of state of the universe.

..A value of q greater than 0.5 indicates that the expansion is decelerating quickly enough for the universe eventually to collapse. A value less than 0.5 indicates that the expansion will continue for ever. In models with a cosmological constant, q can even be negative, indicating an accelerated expansion, as in the inflationary universe.

http://en.wikipedia.org/wiki/FLRW#Solutions

http://en.wikipedia.org/wiki/Deceleration_parameter

http://scienceworld.wolfram.com/physics/DecelerationParameter.html

http://www.answers.com/topic/deceleration-parameter

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