# S07. Distances as Determined by Standard Candles - SOLUTIONS

- Page ID
- 7668

# 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*}\]