Equation involving H and a re Accerlerating Universe

  • Thread starter Thread starter Buzz Bloom
  • Start date Start date
  • Tags Tags
    Universe
Buzz Bloom
Gold Member
Messages
2,517
Reaction score
465
In the Wikipedia article https://en.wikipedia.org/wiki/Accelerating_universe#Evidence_for_acceleration the following equation
(1)
Equation2.png

[where the four currently hypothesized contributors to the energy density of the universe are curvature, matter, radiation and dark energy]

is given without any derivation from the previous equations
(2)
Equation.png

(3)
Equation3.png

(4)
Equation4.png
I have several question about these equations.

(a) I assume that in (2) K is -1, 0, +1 respectively for the space being hyperbolic, flat, or spherical. Then R would be the radius of curvature of the universe at time t. Is that correct?
(b) How is (1) derived form (2), (3), and (4)?
(c) Is the sum of the four Omegas in (1) supposed to equal the Omega in (4)?
(d) In the Einstein equations, isn't Lambda a constant density, independent of a?
(e) Isn't Omega[DE] = Lambda/rho[sub-c] and Lambda a constant independent of a?
(f) If e is correct, then
Omega[DE] = Lambda * (8*pi*G/3) * (a/adot)^2
OK, if that's right, where does the exponent with w in the coefficient of Omega[DE] come from?
 
Space news on Phys.org
I found this document very useful as a thorough introduction to flrw/lcdm equations:
http://casa.colorado.edu/~ajsh/phys5770_08/frw.pdf
 
Last edited by a moderator:
Buzz Bloom said:
I assume that in (2) K is -1, 0, +1 respectively for the space being hyperbolic, flat, or spherical.

Yes.

Buzz Bloom said:
Then R would be the radius of curvature of the universe at time t

If K = 1, yes. If K = 0 or -1, R doesn't have that simple interpretation; it's just a "scale factor" than can be used to track the universe's expansion.

Buzz Bloom said:
Is the sum of the four Omegas in (1) supposed to equal the Omega in (4)?

No. (4) is just a definition; each individual ##\Omega## in (1) is obtained from the corresponding energy density by dividing by the critical density.

Buzz Bloom said:
In the Einstein equations, isn't Lambda a constant density, independent of a?

If it's a cosmological constant, yes. But the term "dark energy" is also used more generally to denote anything that can cause accelerating expansion; for example, a scalar field.

Buzz Bloom said:
where does the exponent with w in the coefficient of Omega[DE] come from?

##w## is the parameter in the equation of state that relates pressure to density; in general, ##p = w \rho##. For non-relativistic matter, ##w = 0##; for radiation or highly relativistic matter, ##w = 1/3##; for a cosmological constant, ##w = -1##; and for other more general types of "dark energy" such as a scalar field, ##w## can vary.

If you look at the second Friedmann equation, and note the factor ##\rho + 3p## on the RHS (where we include a cosmological constant and dark energy in ##\rho## and ##p##), you can see that, if we know ##w##, the factor becomes ##\rho \left( 1 + 3 w \right)##, so any ##w < - 1/3## will make the factor negative and therefore the acceleration ##\ddot{a} / a## will be positive. Combining the second Friedmann equation with the first leads to the factor ##- 3 \left( 1 + w \right)## in (1); note that for a cosmological constant, ##1 + w = 0## and the dark energy density term is constant, as desired. But any ##w < - 1/3## can in principle appear in that term, so it is not necessarily constant in general.
 
Last edited:
Buzz Bloom said:
(a) I assume that in (2) K is -1, 0, +1 respectively for the space being hyperbolic, flat, or spherical. Then R would be the radius of curvature of the universe at time t. Is that correct?
Not quite. The way the equations above are written, R is a constant (the current radius of curvature), and the radius of curvature at any given time would be Ra.

(PeterDonis' answer above is for the more common notation where R(t) replaces a(t).)

Buzz Bloom said:
(b) How is (1) derived form (2), (3), and (4)?
By using stress-energy conservation. Stress-energy conservation determines how the energy density of each component scales with the expansion.

Buzz Bloom said:
(c) Is the sum of the four Omegas in (1) supposed to equal the Omega in (4)?
The sum of the four ##\Omega## parameters is equal to one, by definition. You can see this by considering the situation where ##a = 1## (now), as by definition ##H(a=1) = H_0##.

Buzz Bloom said:
(d) In the Einstein equations, isn't Lambda a constant density, independent of a?
Yes.

Buzz Bloom said:
(e) Isn't Omega[DE] = Lambda/rho[sub-c] and Lambda a constant independent of a?
As PeterDonis mentioned, Lambda is the special case for dark energy where ##w = -1##. You should be able to show that if ##w = -1##, the dark energy in equation (1) is a constant.

Buzz Bloom said:
(f) If e is correct, then
Omega[DE] = Lambda * (8*pi*G/3) * (a/adot)^2
OK, if that's right, where does the exponent with w in the coefficient of Omega[DE] come from?
The way that equation (1) is written, ##\Omega_{DE}## is the current density fraction of dark energy. That is to say, equation (4) should be written as ##\Omega_x = \rho_x(now) / \rho_c(now)##, where ##x## is whatever component of the energy density we're referring to.
 
Last edited:
wabbit said:
I found this document very useful as a thorough introduction to flrw/lcdm equations:
http://casa.colorado.edu/~ajsh/phys5770_08/frw.pdf

Thanks wabbit. The document looks like I will find a lot of useful stuff in it. I will probably have more questions after I study it.
 
Last edited by a moderator:
Buzz Bloom said:
(a) I assume that in (2) K is -1, 0, +1 respectively for the space being hyperbolic, flat, or spherical. Then R would be the radius of curvature of the universe at time t. Is that correct?

PeterDonis said:
.
If K = 1, yes. If K = 0 or -1, R doesn't have that simple interpretation; it's just a "scale factor" than can be used to track the universe's expansion.

Chalnoth said:
Not quite. The way the equations above are written, R is a constant (the current radius of curvature), and the radius of curvature at any given time would be Ra.

Thanks PeterDonis and Chalnoth.

It seems like the notation is a bit confusing. The clarification in Chalnoth's post helps to clarify it.
 
PeterDonis said:
No. (4) is just a definition; each individual Ω\Omega in (1) is obtained from the corresponding energy density by dividing by the critical density.

I think what confused me was seeing in several places that the sum of the contributing Omegas added to 1.

Ah, I see now that Chalnoth explained that the sum always equals 1 by definition.
 
Last edited:
Thanks once again to PeterDonis. Your answers to my questions (e) and (f) were very clear and helpful.answers to my questions (e) and (f) were very clear and helpful.
 
I have been thinking some more about the 1st equation of my original question.

equation2-png.81011.png


(a) What would be the value of Omega[k] be if |radius of curvature| = R[0], and k = +1 or -1?

(b) Also, since a is a relative distance with a[0] arbitraily set = 1, as I now undestand it, the equation would need to be modified, replaceing a with R/R[0]. Would an alternative to this be to modify H[0] by replacing its standard distance 1 Mpc with R[0], that is, dividing H[0] by R[0] and/or by reinterperting the unit of time from 1 sec to R[0]/c?
 
Back
Top