GR modification to Newtonian Friedmann Equation:
In Question 1, you have derived the Friedmann Equation in a matter-only universe in the Newtonian
approach. That is, you now have an equation that describes the rate of change of the size of
the universe, should the universe be made of matter (this includes stars, gas, and dark matter) and
nothing else. Of course, the universe is not quite so simple. In this question we’ll introduce the full
Friedmann equation which describes a universe that contains matter, radiation and/or dark energy.
We will also see some correction terms to the Newtonian derivation.
A) The full Friedmann equations follow from Einstein’s GR, which we will not go through in
this course. Analogous to the equations that we derived in Question 1, the full Friedmann
equations express the expansion/contraction rate of the scale factor of the universe in terms
of the properties of the content in the universe, such as the density, pressure and cosmological
constant. We will directly quote the equations below and study some important consequences.
Use the first Friedmann equation: \[ \left( \frac{\dot{a}}{a} \right)^2 = \frac{8 \pi}{3} G \rho + \frac{ \kappa c^2 }{a^2} + \frac{ \Lambda }{3} \] and the second Friedmann equation: \[ \frac{\ddot{a}}{a} = - \frac{4 \pi G}{3c^2} ( \rho c^2 + 3P) + \frac{ \Lambda }{3} \] To find the third equation in a flat universe \[ \dot{ \rho} c^2 = -3 \frac{\dot{a}}{a}( \rho c^2 + P) \]
With the universe being flat and multiplying the first equation by acceleration squared gives us: \[ \ddot{a}^2 = \frac{8 \pi}{3} G \rho a^2+ \frac{ \Lambda }{3} a^2 \] Taking the time derivative gives us \[ 2 \dot{a} \ddot{a} = \frac{8 \pi}{3} G \dot{\rho} a^2 + \frac{16 \pi}{3} G \rho \dot{a} a + \frac{2 \Lambda }{3} \dot{a} a \] Rearranging this allows us to set it equal to the second equation: \[ \frac{ \ddot{a} }{a} = \frac{4 \pi a}{3 \dot{a} } G \dot{\rho} + \frac{ 8 \pi}{3} G \rho + \frac{ \Lambda }{3} = - \frac{4 \pi G}{3c^2} ( \rho c^2 + 3P) + \frac{ \Lambda }{3} \] Which simplifies with a few cancellations to: \[ \dot{ \rho} c^2 = -3 \frac{\dot{a}}{a}( \rho c^2 + P) \]
B) Cold matter dominated universe. If the matter is cold, its pressure P = 0, and the
cosmological constant Λ = 0. Use the third Friedmann equation to derive the evolution of the
density of the matter ρ as a function of the scale factor of the universe a. You can leave this
equation in terms of ρ, ρ0, a and a0, where ρ0 and a0 are current values of the mass density
and scale factor.
The result you got has the following simple interpretation. The cold matter behaves like
“cosmological dust” and it is pressureless (not to be confused with warm/hot dust in the
interstellar medium!). As the universe expands, the mass of each dust particle is fixed, but the
number density of the dust is diluted - inversely proportional to the volume.
Using the relation between ρ and a that you just derived and the first Friedmann equation,
derive the differential equation for the scale factor a for the matter dominated universe. Solve
2
Astronomy 17 - Galactic and Extragalactic Astronomy Fall 2015
this differentiation equation to show that \(a(t) \propto t^{\frac{3}{2}}\) . This is the characteristic expansion
history of the universe if it is dominated by matter.
Without pressure, the third equation can be written as \[ \frac{\dot{\rho}}{\rho} = -3 \frac{\dot{a}}{a}\] Taking the derivative of both sides with respect to time starting at \( \rho_0 \: \text{and} \: a_0 \) produces: \[ \frac{\rho}{\rho_0} = \left(\frac{a}{a_0} \right)^{-3}\] Plugging this relationship into the first equation gives: \[ \left( \frac{\dot{a}}{a} \right)^2 = \frac{8 \pi}{3} G \left( \frac{a_0^3 \rho_0}{a^3} \right) \] Taking the derivative of each side gives \[ a^{\frac{1}{2}} = c \: dt \] Where c is a constant (because we just want a relationship between a and t). Taking the integral of each side gives \[ \frac{2}{3} a^{\frac{3}{2} } = c t \] So \[ a(t) \propto t^{\frac{3}{2}} \]
C) Radiation dominated universe. Let us repeat the above exercise for a universe filled with
radiation only. For radiation, \(P = \frac{1}{ 3} ρc^2 \)and Λ = 0. Again, use the third Friedmann equation
to see how the density of the radiation changes as a function of scale factor.
The result also has a simple interpretation. Imagine the radiation being a collection of photons.
Similar to the matter case, the number density of the photon is diluted, inversely proportional
to the volume. Now the difference is that, in contrast to the dust particle, each photon can be
thought of as wave. As you learned last week, the wavelength of the photon is also stretched as
the universe expands, proportional to the scale factor of the universe. According to quantum
mechanics, the energy of each photon is inversely proportional to its wavelength: E = hν.
Unlike the dust case where each particle has a fixed energy. So in an expanding universe, the
energy of each photon is decreasing inversely proportional to the scale factor. Check that this
understanding is consistent with the result you got.
Again using the relation between ρ and a and the first Friedmann equation to show that \(a(t) \propto t^{\frac{1}{2}}\) for the radiation only universe.
With the new pressure, the third equation can be written as \[ \frac{\dot{\rho}}{\rho} = -4 \frac{\dot{a}}{a}\] Taking the derivative of both sides with respect to time starting at \( \rho_0 \: \text{and} \: a_0 \) produces: \[ \frac{\rho}{\rho_0} = \left(\frac{a}{a_0} \right)^{-4}\] Plugging this relationship into the first equation gives: \[ \left( \frac{\dot{a}}{a} \right)^2 = \frac{8 \pi}{3} G \left( \frac{a_0^4 \rho_0}{a^4} \right) \] Taking the derivative of each side gives \[ a = c \: dt \] Where c is a constant (because we just want a relationship between a and t). Taking the integral of each side gives \[ \frac{1}{2} a^{2 } = c t \] So \[ a(t) \propto t^{\frac{1}{2}} \]
D) Cosmological constant/dark energy dominated universe. Imagine a universe dominated
by the cosmological-constant-like term. Namely in the Friedmann equation, we can set ρ = 0
and P = 0 and only keep Λ nonzero.
As a digression, notice that we said “cosmological-constant-like” term. This is because the
effect of the cosmological constant may be mimicked by a special content of the universe which
has a negative pressure \(P = - ρc^2 \). Check that the effect of this content on the right-hand-side
of third Friedmann equation is exactly like that of the cosmological constant. To be general we
call this content the Dark Energy. How does the energy density of the dark energy change
in time?
Show that the scale factor of the cosmological-constant-dominated universe expands exponentially
in time. What is the Hubble parameter of this universe?
From the first equation we have: \[ \left( \frac{\dot{a}}{a} \right)^2 = \frac{ \Lambda }{3} \] Taking the derivative gives: \[ \frac{da}{a} = \sqrt{ \frac{ \Lambda }{3} } dt \] Ignoring constants, the integral of this relationship turns into \[ ln(a) \propto t \] \[ a \propto e^t \] And because \[ H= \frac{\dot{a}}{a} \] The Hubble constant in this universe is \[ H = \sqrt{ \frac{ \Lambda }{3} } \]
E) Suppose the energy density of a universe at its very early time is dominated by half matter
and half radiation. (This is in fact the case for our universe 13.7 billion years ago and only 60
thousand years after the Big Bang.) As the universe keeps expanding, which content, radiation
or matter, will become the dominant component? Why?
Radiation will dominate because it falls off at a slower rate ( \(a(t) \propto t^{\frac{1}{2}}\) ) than matter ( \(a(t) \propto t^{\frac{2}{3}}\) ).
F) Suppose the energy density of a universe is dominated by similar amount of matter and dark
energy. (This is the case for our universe today. Today our universe is roughly 68% in dark
energy and 32% in matter, including 28% dark matter and 5% usual matter, which is why it is
acceleratedly expanding today.) As the universe keeps expanding, which content, matter or the
dark energy, will become the dominant component? Why? What is the fate of our universe?
Dark energy will be dominant as the universe expands and matter is scattered. Energy, and therefore normal matter, cannot be created or destroyed, by the law of conservation of energy. We have also shown that dark energy will increase at a faster rate ( \(a \propto e^t\) ) than radiation ( \(a(t) \propto t^{\frac{2}{3}}\) ). Eventually, our universe will be a very sparse, vast expanse of dark energy with few pockets of matter spread throughout.
Great job Danielle! 5/5
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