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S16. Distance and Magnitude - SOLUTIONS

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    Exercise 16.1.1

    Answer

    \(D_A = d\) and (d_L = d (1+z)\) so \(D_A = d_L/(1+z) \)

    Exercise 16.2.1

    Answer

    The path is at constant \(\theta\) and \(\phi\) so only \(r\) is varying. Therefore \(\sqrt{ds^2} = a(t) dr/\sqrt{1-kr^2}\). The comoving length is given by integrating this up, while setting the scale factor equal to unity so \(\ell = \int_0^r dr/\sqrt{1-kr^2}\). For \(kr^2 << 1\), \(\ell \simeq \int_0^r dr(1+k r^2/2) = r + kr^3/6\).

    Exercise 16.3.1

    Answer

    Curvature affects the history of the expansion rate, \(H(a)\), via the Friedmann equation. It also affects the time a photon has to travel to come from coordinate distance \(r\) due to the \(dr^2/(1-kr^2)\) term in the invariant distance equation.

    Exercise 16.4.1

    Answer

    The Friedmann equation is \(H^2 = 8\pi G \rho/3\ - k/a^2\). If \(\rho = \rho_c \equiv 3H^2/(8\pi G)\) then the Friedmann equation becomes \(H^2 = 8 \pi G/3 \times 3H^2/(8\pi G) - k/a^2\) which simplifies to \(H^2 = H^2 - k/a^2\) from which we can infer that \(k = 0\).

    Exercise 16.4.2

    Answer

    If \(\rho = \rho_m + \rho_\Lambda\) and we multiply the \(8\pi G \rho/3\) term in the Friedmann equation by \(\rho_{c,0}/\rho_{c,0}\) we get

    \[H^2 = 3 H_0^2/(8\pi G) \times 8\pi G/3 (\rho_m + \rho_\Lambda)/\rho_{c,0}-k/a^2 = H_0^2 \left(\rho_{m,0}a^{-3}/\rho_{c,0}+\rho_\Lambda/\rho_{c,0}-k/(H_0^2 a^2)\right).\]

    From here the definitions of the \(\Omega\)s gets us to the desired result.

    Exercise 16.4.3

    Answer

    This is now a simple matter of setting \(a=1\) in the equation we just derived and recognizing that \(H(a=1)\) is \(H_0\).

    This page titled S16. Distance and Magnitude - SOLUTIONS is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Lloyd Knox.