Blog #14: Worksheet 5.1 - Extragalactic Distance Ladder

(a) Suppose you are observing two stars, Star A and Star B. Star A is 3 magnitudes fainter than Star B. How much longer do you need to observe Star A to collect the same amount of energy in your detector as you do for Star B? 

In order to relate the magnitudes of a star to their observation time, we need to look at how much light we are receiving per unit of time for each star. Flux is the measure of energy per time per area, and in this case, the "per area" is the same, since we are looking at the perceived flux reaching our eyes (or telescope if we are observing with a telescope). Since we are looking to see how much longer it will take for the light of Star A to reach us, we can use the proportion of fluxes of stars A and B using the relation:

\begin{align}
\frac{F_{B}}{F_{A}} \approx 2.5^{(m_A - m_B)}
\end{align}

where in this case, the factor $(m_A - m_B)$ is the difference in magnitudes between Star A and Star B, which we know to be 3. Therefore, the ratio of fluxes of Star A and Star B is:

\begin{align}
\frac{F_{B}}{F_{A}} \approx 2.5^{(m_A - m_B)} \approx 16
\end{align}

Since flux is related linearly with time, since the flux of Star A is 16 times less than the flux of Star B, it will take 16 times longer for Star A to collect the same amount of energy as Star B. 

(b) Stars have both an apparent magnitude, $m$, which is how bright they appear from Earth. They also have an absolute magnitude, $\mathcal{M}$, which is the apparent magnitude a star would have at $d = 10$ pc. How does the apparent magnitude, $m$, of a star with absolute magnitude $\mathcal{M}$, depend on its distance, $d$ away from you? 

Let's consider a star with an apparent magnitude, $m$, and an absolute magnitude, $\mathcal{M}$. The flux of the star is determined by the equation :

\begin{align}
F = \frac{\text{Luminosity}}{\text{Area}} = \frac{L}{\pi d^2}
\end{align}

where, L is the inherent luminosity of the star, and $d$ is the distance the star is away from you.


Knowing this, we can use the equation for the ratio of fluxes to determine the relationship between the apparent magnitude of a star, $m$ at a distance $d$, and the apparent magnitude of the same star if it were located $d = 10$ pc away.

The flux of the star at any given distance is:

\begin{align}
F_m = \frac{L}{4\pi d ^2}
\end{align}

The flux of a star at exactly $d = 10$ pc is as follows:

\begin{align}
F_M = \frac{L}{4\pi (10 \text{ pc}) ^2}
\end{align}

To solve for the relationship between the the apparent magnitude, $m$, and the absolute magnitude, $\mathcal{M}$ the ratio of the fluxes of these two values are as follows:

\begin{align}
\frac{F_{m}}{F_{M}} &= 10^{0.4(M - m)}\\
\frac{\frac{L}{4\pi d ^2}}{\frac{L}{4\pi (10 \text{ pc}) ^2}} &= 10^{0.4(M - m)}\\
\left(\frac{10 \text{ pc}}{d}\right)^2 &= 10^{0.4(M - m)}\\
\log{\left(\frac{10 \text{ pc}}{d}\right)^2} &= 0.4(M - m)\\
\frac{\log{\left(\frac{10 \text{ pc}}{d}\right)^2}}{0.4} &= M - m\\
m &= M - \frac{2 \log{\left(\frac{10 \text{ pc}}{d}\right)}}{0.4} \\
m &= M - 5 \log{\left(\frac{10 \text{ pc}}{d}\right)}
\end{align}

This equation is known as the distance module, and the distance needs to be measured in parsecs.

(c) What is the star's parallax in terms of its apparent and absolute magnitudes? 

The star's parallax with respect to its distance is given by the equation:

\begin{align}
\theta  = \frac{1 \text{AU}}{d}
\end{align},

where, $theta$ is the parallax angle, and $d$ is the distance to the star, measured in parsecs.

We can use the equation derived in part (b) to solve for the distance, $d$, as follows:

\begin{align}
m &= M - 5 \log{\left(\frac{10 \text{ pc}}{d}\right)}\\
M - m &= 5 \log{\left(\frac{10 \text{ pc}}{d}\right)}\\
\frac{M - m}{5} &= \log{\left(\frac{10 \text{ pc}}{d}\right)}\\
10^{\frac{M - m}{5}} &= \frac{10 \text{ pc}}{d}\\
d &= \frac{10^{0.2(M - m)}}{10 \text{ pc}}\\
\end{align}

Now that we have $d$ in terms of magnitudes, we can substitute it into the parallax equation above as follows:

\begin{align}
\theta  &= \frac{1 \text{AU}}{d}\\
\theta  &= \frac{1 \text{AU}}{\frac{10^{0.2(M - m)}}{10 \text{ pc}}}\\
\theta  &= \frac{1 \text{AU}}{10^{0.2(M - m)}} \times 10 \text{ pc}

\end{align},