Solution to Binney and Merrifield Problem 2.1

Surface brightness is defined as the radiative flux per unit solid angle of the image. If we are given a surface brightness $I_{1} = 19 \, \text{mag}\,\mathrm{arcsec}^{-2}$ at a given point in a galaxy, we know that point emits as much as light as a star of magnitude 19. We know that apparent magnitudes are related by,

\begin{equation} m - m_{\mathrm{ref}} = -2.5\log_{10}\left(\frac{F}{F_{\mathrm{ref}}}\right) \end{equation}

and we wish to determine the combined surface brightness of $I_1$ which corresponds to the surface brightness from the point in M31 and from $I_2$ from the center of a globular cluster. Let's choose Vega as our reference.

\begin{equation} m_{\mathrm{M31}}-m_{\mathrm{Vega}} = -2.5\log_{10}\left(\frac{F_{\mathrm{M31}}}{F_{\mathrm{Vega}}}\right) \end{equation}

Since Vega is defined to have a magnitude of 0 in all bands,

\begin{equation} m_{\mathrm{M31}}= -2.5\log_{10}\left(\frac{F_{\mathrm{M31}}}{F_{\mathrm{Vega}}}\right) \end{equation}

We can do the same for the globular cluster:

\begin{equation} m_{\mathrm{GC}}= -2.5\log_{10}\left(\frac{F_{\mathrm{GC}}}{F_{\mathrm{Vega}}}\right) \end{equation}

We cannot add magnitudes to get a total magnitude but we can add fluxes to get a total flux. If we solve for the flux of both M31 and the globular cluster andd add them together, we get:

\begin{equation} F_{\mathrm{tot}} = F_{\mathrm{M31}} + F_{\mathrm{GC}} = F_{\mathrm{Vega}}(10^{-\frac{m_{\mathrm{M31}}}{2.5}} + 10^{-\frac{m_{\mathrm{GC}}}{2.5}}) \end{equation}

Now we apply (1) with our $F_{\mathrm{tot}}$ and $F_{\mathrm{Vega}}$.

\begin{equation} m_{\mathrm{tot}} - m_{\mathrm{Vega}} = -2.5\log_{10}\left(\frac{F_{\mathrm{Vega}}(10^{-\frac{m_{\mathrm{M31}}}{2.5}} + 10^{-\frac{m_{\mathrm{GC}}}{2.5}})}{F_{\mathrm{Vega}}}\right) \end{equation}

Now, because $m_{\mathrm{Vega}} = 0$ and the $F_{\mathrm{Vega}}$ will cancel from the numerator and denominator, we can substitute $m_{\mathrm{M31}} = 19$ and $m_{\mathrm{GC}} = 20$ and get:

\begin{equation} \boxed{m_{\mathrm{tot}} = 18.6\, \text{mag}\,\mathrm{arcsec}^{-2}} \end{equation}

This makes sense, as the combined light should be brighter than either of the two light sources individually.

Solution to Binney and Merrifield Problem 2.2

First, we will convert from $L_{\odot}\, \mathrm{pc}^{-2}$ to $ \mathrm{mag}\,\mathrm{arcsec}^{-2}$. Let's suppose we have a square 1 pc by 1 pc patch on the sky a distance $d$ away from us, and the angle it subtends from our point of view is given by $\theta$. Using the small-angle approximation, these variables are related by,

\begin{equation} \theta = \frac{D}{d} \end{equation}

where $D$ is a side length of the square patch, which is 1 pc. Now we will compare the absolute magnitude of the square 1 pc by 1 pc patch with the absolute magnitude of the sun in the $\textit{B}$-band.

\begin{equation} M_{\mathrm{patch}} - M_{\odot} = -2.5\log_{10}\left(\frac{F_{\mathrm{patch}}}{F_{\odot}}\right) \end{equation}

We recall that $F = \frac{L}{4\pi d^2} $, and when comparing absolute magnitudes, both objects are considered at a distance of $d = 10 \, \mathrm{pc}$. Given this, the ratio of their fluxes will now become the ratio of their luminosities.

\begin{equation} M_{\mathrm{patch}} - M_{\odot} = -2.5\log_{10}\left(\frac{L_{\mathrm{patch}}}{L_{\odot}}\right) \end{equation}

We know that the $\textit{B}$-band surface brightness of the galactic disk is:

\begin{equation} I_B\frac{L_{\odot}}{\mathrm{pc}^2} \end{equation}

We also know that from the angular diameter equation,

\begin{equation} \mathrm{Square \, Patch \, Area} = D^2 = \theta^2 d^2 \end{equation}

If we multiply the area of the square patch, $D^2 = \theta^2 d^2 $, with the surface brightness of the patch, $I_B \frac{L_\odot}{\mathrm{pc}^2} $, we will get the luminosity of the patch:

\begin{equation} L_{\mathrm{patch}} = \left(\theta^2 d^2\right)\left(I_B \frac{L_\odot}{\mathrm{pc}^2}\right) \end{equation}

However, we must convert $\theta = 1''$ to radians.

\begin{equation} L_{\mathrm{patch}} = (1'')^2d^2I_B \frac{L_{\odot}}{\mathrm{pc}^2}\times \frac{1\, \mathrm{rad}^2}{(206,265'')^2} \end{equation}

\begin{equation} L_{\mathrm{patch}} = d^2I_B \frac{L_{\odot}}{\mathrm{pc}^2}\times \frac{1 }{(206,265)^2} \end{equation}

We also know that,

\begin{equation} \mu_B = m_{\mathrm{patch}} = M_{\mathrm{patch}} + 5\log_{10}(d) - 5 \end{equation}

\begin{equation} M_{\mathrm{patch}} = M_{\odot} -2.5\log_{10}\left(\frac{L_{\mathrm{patch}}}{L_{\odot}}\right) \end{equation}

so this means,

\begin{equation} \mu_B = M_{\odot} - 2.5\log_{10}\left(d^2 I_B \frac{L_{\odot}}{\mathrm{pc}^2 L_{\odot}} \times \frac{1}{206,265}\right) + 5\log_{10}(d) - 5\end{equation}

Using property of logs, we can simplify this to:

\begin{equation} \mu_B = M_{\odot} - 2.5\log_{10}(d^2) - 2.5\log_{10}(I_B) - \log_{10}\left(\frac{1}{206,265}\right) + 5\log_{10}(d) - 5 \end{equation}

The 2nd and 5th terms will cancel each other, and the constants combine to give,

\begin{equation} \mu_B = M_{\odot} - 2.5\log_{10}(I_B) + 21.572 \end{equation}

with $M_{\odot} = 5.48$,

\begin{equation} \boxed{\mu_B = 27.05 - 2.5\log_{10}(I_B)} \end{equation}