Vacuum energy properties

  • #1
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Main Question or Discussion Point

I was wondering, since vacuum energy is supposed to have some energy density,according to standard comology equating it to dark energy and cosmological constant , does this energy gravitate? I mean it looks like according to GR its stress-energy tensor with negative pressure should generate some curvature in the metric tensor, is this what is meant when is said to be gravitationally repulsive?
 

Answers and Replies

  • #2
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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.
 
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  • #3
Chalnoth
Science Advisor
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I was wondering, since vacuum energy is supposed to have some energy density,according to standard comology equating it to dark energy and cosmological constant , does this energy gravitate? I mean it looks like according to GR its stress-energy tensor with negative pressure should generate some curvature in the metric tensor, is this what is meant when is said to be gravitationally repulsive?
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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Hi, stevedb1
I follow your reasoning up to this last statement
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.
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?
 
  • #5
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Hi TrickyDicky

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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Hi TrickyDicky
I'm saying that a flat universe can be an accelerating universe if you take into account pressure.
Ok, I see.

Do you agree with chalnoth that the vacuum energy generates gravitational field?
 
  • #7
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Do you agree with chalnoth that the vacuum energy generates gravitational field?
A negative one, hence the acceleration.
 
  • #8
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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.
..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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