A Summing Up — Question 6

Section 44-14 · Physics pages 7562–7563

Question 6 Matter and vacuum energy affect acceleration with opposite signs For a flat universe with pressureless matter and a cosmological constant, use q0 = Ωm/2 −ΩΛ. Let Ωm = .30 and ΩΛ = .70, with densities scaling as a to the power -3 and a to the power 0. t a Schematic; labels, not drawing lengths, specify the data. (a) Find the present deceleration parameter and identify acceleration or deceleration. (b) Find the scale factor and redshift when acceleration begins, using ρm = 2ρΛ.
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Question 6 - Solution Matter and vacuum energy affect acceleration with opposite signs t a (a) Method and calculation q0 = .30/2 −.70 = −.55 . q = −a¨a/˙a2 < 0 ⇒¨a > 0 . (b) Interpretation and further calculation .30a−3 = 2(.70) ⇒aacc = (.30/1.40)1/3 ≃.5984 . zacc = a−1 acc −1 ≃.6711 . Physical check The transition to acceleration differs from matter-vacuum equality because pressure contributes to the acceleration equation.

Question 6

Matter and vacuum energy affect acceleration with opposite signs

For a flat universe with pressureless matter and a cosmological constant, use q0=Ωm/2−ΩΛq_0=\Omega_m/2-\Omega_\Lambda. Let Ωm=.30\Omega_m=.30 and ΩΛ=.70\Omega_\Lambda=.70, with densities scaling as a to the power -3 and a to the power 0.

Diagram shown in the original SVG worksheet.

  1. Find the present deceleration parameter and identify acceleration or deceleration.

  2. Find the scale factor and redshift when acceleration begins, using ρm=2ρΛ\rho_m=2\rho_\Lambda.

Question 6 - Solution

Matter and vacuum energy affect acceleration with opposite signs

Diagram shown in the original SVG worksheet.

(a) Method and calculation

q0=.30/2−.70=−.55.\boxed{q_0=.30/2-.70=-.55}.q=−aä/ȧ2<0⇒ä>0.q=-a\ddot a/\dot a^2<0\Rightarrow\boxed{\ddot a>0}.

(b) Interpretation and further calculation

.30a−3=2(.70)⇒aacc=(.30/1.40)1/3≃.5984..30a^{-3}=2(.70)\Rightarrow\boxed{a_{\mathrm{acc}}=(.30/1.40)^{1/3}\simeq.5984}.zacc=aacc−1−1≃.6711.\boxed{z_{\mathrm{acc}}=a_{\mathrm{acc}}^{-1}-1\simeq.6711}.

Physical check

The transition to acceleration differs from matter-vacuum equality because pressure contributes to the acceleration equation.