2.11: S12. Particle Kinematics in an Expanding Universe- Newtonian Analysis - SOLUTIONS
Exercise 12.1.1
- Answer
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If the peculiar velocity is negative then the particle, over time, ends up with a smaller distance from the origin then it would have had if its peculiar velocity had been zero. In other words, it starts falling behind. Falling behind naturally brings its motion to be more similar to the surrounding fluid, as the fluid closer to the origin is moving more slowly. This improved agreement between its velocity and that of the surrounding fluid flow can be interpreted as a reduction in the magnitude of the peculiar velocity.
Exercise 12.1.2
- Answer
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Let’s look at just one of the three spatial components of the velocity. Notationally, we’ll drop the subscript “pec” to make
room for the subscript “\(x\)” indicating we are talking about the \(x\)-component of the velocity. For the other components it is the same solution.
From \(dv_x/dt = -(da/dt)/a v_x\) we can rearrange and cancel terms to get \(dv_x/v_x = -da/a\), which we can integrate up:\[\begin{equation*}
\begin{aligned}
\int_{v_{x,i}}^{v_x} \frac{dv'_x}{v'_x} = - \int_{a_i}^{a} \frac{da’}{a’}
\end{aligned}
\end{equation*}\]where the \(i\) subscript indicates “initial” and the integrals are easily done to get
\[\begin{equation*}
\begin{aligned}
\ln(v_x/v_{x,i}) = - \ln(a/a_i) = \ln(a_i/a)
\end{aligned}
\end{equation*}\]which can be solved to find
\[\begin{equation*}
\begin{aligned}
v_x = v_{x,i} (a_i/a).
\end{aligned}
\end{equation*}\]
The same goes for the other components of the peculiar velocity. So we have what we wanted to show, that the peculiar velocity reduces with expansion as \(1/a\).