2.10: S11. The Friedmann Equation - SOLUTIONS
( \newcommand{\kernel}{\mathrm{null}\,}\)
Exercise 11.1.1
- Answer
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K.E. is 12mv2, where v=˙aℓ, so the test particle's kinetic energy is
12m˙a2ℓ2
Exercise 11.1.2
- Answer
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P.E. is −GM(<d)md, where by M(<d) we mean the mass contained in the sphere of radius d, so M(<d)=43πd3ρ.
Therefore the test particle's potential energy is
−G43πρd2m
Exercise 11.1.3
- Answer
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12m˙a2ℓ2−G43πρd2m=κ
Exercise 11.1.4
- Answer
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Recall that d=aℓ⟹ℓ=da, substituting this in and rearranging our equation we get
12md2˙a2a2−12md28πGρ3=κ
dividing through by 12md2 gives
˙a2a2−8πGρ3=2κm1d2
Then we substitute back in d=aℓ and solve for (˙aa)2:
(˙aa)2=8πGρ3+2κml2×1a2