2.7: S07. Distances as Determined by Standard Candles - SOLUTIONS
- Page ID
- 7668
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Exercise 7.1.1
- Answer
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Constant \(t\), \(r\), and \(\phi\) so we have
\[\begin{equation*}
\begin{aligned}
ds = a(t)r\,d\theta
\end{aligned}
\end{equation*}\]
Exercise 7.1.2
- Answer
-
Similar to 7.1.1 above, we now have constant \(t\), \(r\), and \(\theta\), which gives
\[\begin{equation*}
\begin{aligned}
ds = a(t)r\sin\theta \,d\phi
\end{aligned}
\end{equation*}\]
Exercise 7.1.3
- Answer
-
This is simply \(a^2(t)r^2 \sin \theta \, d\theta d\phi\).
Exercise 7.1.4
- Answer
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Using the result from 7.1.3 above, we now just integrate over \(\theta\) and \(\phi\), so the area is
\[\begin{equation*}
\begin{aligned}
A &= \int_{0}^{2\pi}\int_{0}^{\pi} a(t)rsin\theta d\theta d\phi \\ \\ &= \int_{0}^{2\pi} 2a^2(t)r^2d\phi \\ \\ &= 4\pi a^2(t)r^2
\end{aligned}
\end{equation*}\]
Exercise 7.1.5
- Answer
-
We know that Luminosity = (Flux)\(\times\)(Surface Area). Making the appropriate substitutions we do indeed find that \(F = L/(4\pi d^2 a^2)\).
Exercise 7.2.1
- Answer
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As we found in Chapter 6, the rate of arrival of the wave crests will be slower than the rate of emission by the factor of \(a(t_r)/a(t_e)\). The same argument applies to the rate of arrival of photons. We also saw that wavelength would be stretched out by a factor \(1+z \equiv \frac{\lambda_r}{\lambda_e} = a(t_r)/a(t_e)\). Therefore the rate of arrival of photons will be slowed down by a factor of \(1+z\).
Exercise 7.2.2
- Answer
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The relationship between photon energy and wavelength is \(E = \frac{hc}{\lambda}\). Substituting this into the definition of redshift, we find that
\[\begin{equation*}
\begin{aligned}
E_r/E_e = \frac{1}{1 + z}
\end{aligned}
\end{equation*}\]
Exercise 7.3.1
- Answer
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Solving Equation 7.4 for \(d_{\rm lum}\) we get
\[\begin{equation*}
\begin{aligned}
d_{\rm lum} = \sqrt{\frac{L}{4\pi F}}
\end{aligned}
\end{equation*}\]
Exercise 7.3.2
- Answer
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Substituting in Equation 7.3 to our result above we get
\[\begin{equation*}
\begin{aligned}
d^2_{\rm lum} = \frac{L}{4\pi}\frac{4\pi d^2(1 + z)^2}{L} \quad \Longrightarrow \quad d_{\rm lum} = d(1 + z)
\end{aligned}
\end{equation*}\]