HW #6

1. Inflaton

(a) slow-roll regime

In the slow-roll regime, we neglect the kinetic energy as well as (Overscript[ϕ, ··]) term in the equation of motion.  Then
\(TraditionalForm\`H\^2 = \(\(8  π\)\/3\) \(G\_N\) \(m\^2\/2\) ϕ\^2\), 3  H (Overscript[ϕ, ·]) + (m^2) ϕ = 
        0.
We use 8  π (Subscript[G, N]) = ((Subscript[M, Pl])^(-2)).  Putting them together,
3 ((m ϕ)/((Sqrt[6]) 
                  (Subscript[M, Pl]))) (Overscript[ϕ, ·]) + (m^2) ϕ = 
        0,
and hence
d ϕ = -(Sqrt[(2/3)]) m (Subscript[M, Pl]) d t.
The solution is simply
ϕ(
          t) = ϕ(0) - (Sqrt[(2/3)]) m (Subscript[M, Pl]) t.
Using this solution, the kinetic energy is
\(TraditionalForm\`\(1\/2\) \(ϕ\&·\)\^2 = \(1\/3\) \(m\^2\) M\_Pl\^2\),
while the potential energy is
((m^2)/2) (ϕ^2).
Therefore, if ϕ ≫ (Subscript[M, Pl]), we see that the kinetic energy is indeed smaller than the potential energy.  At the same time, (Overscript[ϕ, ··]) = 
        0 for the above solution and indeed is negligible compared other non-vanishg terms.  The scale factor can be obtained from the Friedmann equation
(H^2) = ((((Overscript[R, ·]))^2)/(R^2)) = (1/(3 
                   ((Subscript[M, Pl])^2))) (1/2) (m^2) (ϕ^2),
(((Overscript[R, ·]))/R) = (1/((Sqrt[6]) (Subscript[M, Pl]))) 
          m ϕ,
d log R = (1/((Sqrt[6]) (Subscript[M, Pl]))) 
          m ϕ d t
and hence
log ((R(t))/(R(0))) = (1/((Sqrt[6]) 
                (Subscript[M, Pl]))) m(ϕ(0) t - (1/(Sqrt[6])) m (Subscript[M, Pl]) 
                (t^2)) 
When ϕ ≫ (Subscript[M, Pl]), the first term in the parentheses dominates and the scale factor grows exponentially with time.

(b) oscillating regime

In the oscillating regime ϕ ≪ (Subscript[M, Pl]) and m t ≫ 1, we study the equation of motion
(Overscript[ϕ, ··]) + 
          3  H (Overscript[ϕ, ·]) + (m^2) ϕ = 0,
\(TraditionalForm\`H\^2 = \(1\/\(3 
                 M\_Pl\^2\)\) \(1\/2\) \((\(ϕ\&·\)\^2 + \(m\^2\) ϕ\^2)\)\)
with the ansatz
ϕ(t) = ϕ((Subscript[t, 1])) ((Subscript[t, 1])/t) 
          cos m(t - (Subscript[t, 1])).
\(TraditionalForm\`ϕ\&\(\(·\)\(·\)\) = \(ϕ(t\_1)\) \(t\_1\/t\) \((\(-m\^2\) cos\ \(m(t - t\_1)\) + 
              2 \( m\/t\) sin\ \(m(t - t\_1)\) + 2\/t\^2)\)\)
The leading term in 1/(m t) is the first term in the parantheses which is canceled by (m^2) ϕ term.  To see if the next-leading term is also cancelled in the equation of motion, we work out
(Overscript[ϕ, ·]) = ϕ(
            (Subscript[t, 1])) ((Subscript[t, 1])/t) (-m sin m(t - (Subscript[t, 1])) - (1/t) 
                cos m(t - (Subscript[t, 1]))),
(H^2) = (1/(3  ((Subscript[M, Pl])^2))) (1/2) ((ϕ(
                (Subscript[t, 1])))^2) (((Subscript[t, 1])^2)/(t^2)) ((m^2) + 
              2  (m/t) sin m(t - (Subscript[t, 1])) 
                cos m(t - (Subscript[t, 1])) + (1/(t^2)) (cos^2) m(
                  t - (Subscript[t, 1]))).
The equation of motion is then
ϕ(
              (Subscript[t, 1])) ((Subscript[t, 1])/t) (2  (m/t) sin m(t - (Subscript[t, 1])) + 
                (2/(t^2))) + 
          3  (1/((Sqrt[6]) (Subscript[M, Pl]))) ((Subscript[t, 1])/t) (Sqrt[(m^2) + 
                  2  (m/t) sin m(t - (Subscript[t, 1])) 
                    cos m(t - (Subscript[t, 1])) + (1/(t^2)) (cos^2) m(
                      t - (Subscript[t, 1]))]) ((ϕ(
                  (Subscript[t, 1])))^2) ((Subscript[t, 1])/t) (-m sin m(
                    t - (Subscript[t, 1])) - (1/t) cos m(t - (Subscript[t, 1]))) = 0. 
The next leading terms are
ϕ(
              (Subscript[t, 1])) ((Subscript[t, 1])/t) (2  (m/t) sin m(t - (Subscript[t, 1]))) - 
          3  (1/((Sqrt[6]) (Subscript[M, Pl]))) ((Subscript[t, 1])/t) 
            m ((ϕ((Subscript[t, 1])))^2) ((Subscript[t, 1])/t) m sin m(t - (Subscript[t, 1])) = 
        0. 
This is satisfied if
2 - 
          3  (1/((Sqrt[6]) (Subscript[M, Pl]))) (Subscript[t, 1]) m ϕ((Subscript[t, 1])) = 0. 
(Subscript[t, 1]) = 
        2 (Sqrt[ (2/3)]) ((Subscript[M, Pl])/(m ϕ((Subscript[t, 1])))).
With this choice, the equation of motion is satisfied both for the leading and the next-leading terms in the expansion in 1/m t.

The expansion rate to the leading order is
(H^2) = (1/(3  ((Subscript[M, Pl])^2))) (1/2) ((ϕ(
                  (Subscript[t, 1])))^2) (((Subscript[t, 1])^2)/(t^2)) 
            (m^2) = (1/(3  ((Subscript[M, Pl])^2))) (1/2) 
              4  (2/3) (((Subscript[M, Pl])^2)/(m^2)) (1/(t^2)) (m^2) = (4/9) 
              (1/(t^2))
This differential equation can be integrated easily and we find
R=R((Subscript[t, 1]))((((t/(Subscript[t, 1]))))^(2/3)).
Therefore the energy density scales as
\(TraditionalForm\`ρ = \(3 \( M\_Pl\^2\) 
            H\^2 = \(3 \( M\_Pl\^2\) \(4\/9\) 
              1\/t\^2 = \(4\/3\) \(1\/t\_1\^2\) \((\(R(t\_1)\)\/R)\)\^3\)\)\)
Indeed, the energy density is that of matter-domnated universe.

(c) numerical solution

We solve the differential equation numerically.  A word of caution is that this is actually a not very safe thing to do if the solution oscillates crazy like this one.  The numerical errors may build up.  Mathematica seems to handle it fairly well, though.

I didn't specify the boundary conditions.  Normally, we take the initial time derivative to vanish, but of course we don't really know what the right boundary conditions are.

sol = 
      NDSolve[{ϕ''[t] + 
                  3  H ϕ'[t] + (m^2) ϕ[t] == 
                0 /. {H -> (Sqrt[(1/(3 (MPl^2))) (1/2) (ϕ'[t]^2 + (m^2) ϕ[t]^2)])} /. {MPl -> 1, 
              m -> (10^(-2))}, ϕ[0] == 
            100, ϕ'[0] == 0}, ϕ, {t, 0, 
          100000}]

Plot[Evaluate[ϕ[t] /. sol[[1]]], {t, 0, 20000}, 
      PlotRange -> {-1, 100}, 
      PlotStyle -> RGBColor[0, 1, 0]]

The slow-roll solution does not have vanishing time derivative at the initial time.

Plot[(Subscript[ϕ, 0]) - (Sqrt[(2/3)]) m MPl t /. {(Subscript[ϕ, 0]) -> 100, 
          m -> (10^(-2)), MPl -> 1}, {t, 0, 13000}, 
      PlotRange -> {-1, 100}, 
      PlotStyle -> RGBColor[1, 0, 0]]

Show[%, %%]

The fact that they agree so well with each other is a demonstration that the inflation is quite insensitive to the initial values; the solution quickly approaches the slow-roll solution.

Plot[Evaluate[ϕ[t] /. sol[[1]]], {t, 10000, 100000}, 
      PlotRange -> {-1, 1}]

Plot[Evaluate[ϕ[t] /. sol[[1]]], {t, 10000, 100000}, 
      PlotRange -> {-0.1, 0.1}]

Note that the time t in this solution is not the same as the time t in the analytic solution in the oscillating regime; they are offset by the contribution from the slow-roll regime.  But the offset is quickly forgotten as time goes on m t ≫ 1.  By multiplying the amplitude by time, we can see that the oscillation is more or less 1/t, consistent with the analytic approximate solution.

Plot[
      Evaluate[t ϕ[t] /. sol[[1]]], {t, 10000, 100000}]

To compare the numerical and anlaytic solutions head-to-head, we need to do some more work.  We fix the parameters by looking at one period in the numerical solution

Plot[Evaluate[ϕ[t] /. sol[[1]]], {t, 40000, 41000}]

(Subscript[t, 0]) = 
      t /. FindRoot[ϕ[t] == 0 /. sol[[1]], {t, 
              40300}][[1]]

Evaluate[ϕ[t] /. 
            sol[[1]] /. {t -> 
              (Subscript[t, 0]) + (π/2) (1/m)} /. {m -> 
            (10^(-2))}]

Then the approximate analytic

Plot[((Subscript[t, 1])/(t + (Subscript[t, 1]) - 
                    (Subscript[t, 0]) - (1/m) (π/2))) (Subscript[ϕ, 1]) 
              Cos[m (t - (Subscript[t, 0])) - (π/2)] /. {(Subscript[t, 1]) -> 
                2 (Sqrt[ (2/3)]) (MPl/(m (Subscript[ϕ, 1])))} /. {m -> 
              (10^(-2)), MPl -> 1} /. {(Subscript[ϕ, 1]) -> 
            0.005730285458006179`, (Subscript[t, 0]) -> 40284.31696953443`}, {t, 40000, 
        41000}]

Plot[Evaluate[t ϕ[t] /. sol[[1]]], {t, 40000, 50000}, 
      PlotStyle -> RGBColor[1, 0, 0]]

Plot[t  ((Subscript[t, 1])/(t + (Subscript[t, 1]) - 
                    (Subscript[t, 0]) - (1/m) (π/2))) (Subscript[ϕ, 1]) 
              Cos[m (t - (Subscript[t, 0])) - (π/2)] /. {(Subscript[t, 1]) -> 
                2 (Sqrt[ (2/3)]) (MPl/(m (Subscript[ϕ, 1])))} /. {m -> 
              (10^(-2)), MPl -> 1} /. {(Subscript[ϕ, 1]) -> 
            0.005730285458006179`, (Subscript[t, 0]) -> 40284.31696953443`}, {t, 40000, 
        50000}, PlotStyle -> RGBColor[0, 1, 0]]

Show[%, %%]

They agree very well with each other.

Optional

It is too cumbersome to write many equations in Mathematica.  It will be provided as a separate PDF file.