- #1
johne1618
- 371
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As I understand it the de Sitter model is a model of the Universe with:
rho = matter density = rho = 0
p = pressure = 0
k = spatial curvature = 0
cosmological constant = Lambda = non zero
Putting these values in the Friedmann equations one finds the solution for the scale factor a(t) is:
a(t) = exp( sqrt(Lambda c^2/3) * t)
This describes an accelerating empty universe with a non-zero cosmological constant.
Although this model has the right deceleration parameter q = -1 it is contrary to observations as we know there is matter in the Universe.
Now consider the following model:
p = - rho c^2
k = 0
Plugging these values into the Friedmann equations we find we are left with the following equation for the scale factor a:
a'^2 = a a''
This also has the solution:
a(t) = exp(H * t)
where
H^2 = 8 Pi G rho' / 3
where rho' = rho + Lambda c^2
Now this model describes a matter-filled accelerating Universe with no explicit cosmological constant provided that the equation of state of the matter is:
p = -rho c^2
Is this right?
Does this latter model describe the present Universe provided that p = -rho c^2 holds for present day matter?
In this model the negative pressure is associated with the particles of matter themselves rather than having a cosmological constant that is associated with the background space.
Perhaps the negative pressure is a zero-point energy phenomenon holding the individual particles of matter together (in the same manner as the Casimir effect pushes conducting plates together).
rho = matter density = rho = 0
p = pressure = 0
k = spatial curvature = 0
cosmological constant = Lambda = non zero
Putting these values in the Friedmann equations one finds the solution for the scale factor a(t) is:
a(t) = exp( sqrt(Lambda c^2/3) * t)
This describes an accelerating empty universe with a non-zero cosmological constant.
Although this model has the right deceleration parameter q = -1 it is contrary to observations as we know there is matter in the Universe.
Now consider the following model:
p = - rho c^2
k = 0
Plugging these values into the Friedmann equations we find we are left with the following equation for the scale factor a:
a'^2 = a a''
This also has the solution:
a(t) = exp(H * t)
where
H^2 = 8 Pi G rho' / 3
where rho' = rho + Lambda c^2
Now this model describes a matter-filled accelerating Universe with no explicit cosmological constant provided that the equation of state of the matter is:
p = -rho c^2
Is this right?
Does this latter model describe the present Universe provided that p = -rho c^2 holds for present day matter?
In this model the negative pressure is associated with the particles of matter themselves rather than having a cosmological constant that is associated with the background space.
Perhaps the negative pressure is a zero-point energy phenomenon holding the individual particles of matter together (in the same manner as the Casimir effect pushes conducting plates together).
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