S07. Distances as Determined by Standard Candles - SOLUTIONS

Exercise 7.1.1

Answer

Constant \(t\), \(r\), and \(\phi\) so we have

\[\begin{equation*}
\begin{aligned}
\int ds = \int_{0}^{d\theta} a(t)r \,d\theta ' = a(t)r\,d\theta
\end{aligned}
\end{equation*}\]

Exercise 7.1.2

Answer

Similar to 7.1.1 above, we now have constant \(t\), \(r\), and \(\theta\), which gives

\[\begin{equation*}
\begin{aligned}
\int ds = \int_{\phi}^{\phi+d\phi} a(t)r\sin\theta \,d\phi ' = a(t)r\sin\theta \,d\phi
\end{aligned}
\end{equation*}\]

Exercise 7.1.3

Answer

This is simply \(a^2(t)r^2 \sin \theta \, d\theta d\phi\).

Exercise 7.1.4

Answer

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

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

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

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

Substituting in Equation 7.3 to our result above we get

\[\begin{equation*}
\begin{aligned}
d^2_{\rm lum} = \frac{L}{4\pi}\frac{4\pi d^2(1 + z)^2}{L} \quad \Longrightarrow \quad d_{\rm lum} = d(1 + z)
\end{aligned}
\end{equation*}\]