Judging a Star’s Actual Brightness

We are bounded in a nutshell of Infinite Space: Worksheet # 7, Problem #2: Judging a Star’s actual Brightness

2. Okay, let’s use your new-found knowledge of magnitudes.

(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?

(b) Stars have both an apparent magnitude, m, which is how bright they appear from the Earth. They also have an absolute magnitude, M, which is the apparent magnitude a star would have at \( d = 10 ~parsec\) (One parsec = 3.26 light years. We’ll learn more about this quantity in the next question). How does the apparent magnitude, m, of a star with absolute magnitude M, depend on its distance, d away from you?

(c) Assume that the part of the Galaxy that we live in is uniformly filled with stars with a number density \(n_\star \). Suppose you plan to measure the spectrum of every star down to an apparent magnitude m. How many more stars do you have to observe if you observe to a limit that is one magnitude fainter? In other words, what is the fractional change (to one digit) in the number of targets for each magnitude fainter you can observe? (Hint: Don’t overthink this problem. We don’t have enough information about the distribution of stars to consider specifics.)

a. Ok, so we’ve worked with magnitudes before, an ancient system of measuring a stellar object’s brightness. But in this case we have to prove the equations we have been using thus far. First off, let us remember how magnitudes work: the higher the number, the fainter the star is, and this scale works in a logarithmic relationship where each increase in magnitude corresponds to an increase in flux as according to the equation:  \[ \frac{F_1}{F_2} = 10^{0.4(m_2 - m_1)}.\] Now then, we can use this equation to work through the current problem we are tasked with.

We know from the problem that \[ m_A = m_B - 3\] \[\Delta m = 3,\] and so this difference in magnitude corresponds to a “precise” (not the best word considering most of these systems are approximations half the time) flux ratio, as seen in the equation above. Furthermore, we can rewrite the same equation in the following way: \[ \frac{F_1}{F_2} = 10^{0.4(m_2 - m_1)}\] \[ m_1 - m_2 = -2.5 \log_{10} \left(\frac{F_1}{F_2} \right) ,\] by changing the way we use the logarithmic bases. Now, we just plug in the right definitions within the context of this problem and get: \[ m_ A - m_B = -2.5 \log_{10} \left(\frac{F_A}{F_B} \right) ,\] and now plugging in real numbers: \[ m_ B - 3 - m_B = -2.5 \log_{10} \left(\frac{F_A}{F_B} \right) \]  \[ -3 = -2.5 \log_{10} \left(\frac{F_A}{F_B} \right) \] \[\frac{6}{5} = \log_{10} \left(\frac{F_A}{F_B} \right) ,\]  which finally yields: \[ \frac{F_A}{F_B} = 10^(6/5) = 15.8,\] which describes the ratio of time necessary to collect the same amount of light for each star, meaning Star A takes 15.8 times longer than Star B.

b. First, we shall define in a broader sense what the Flux ratio is. From the simple definition of Flux we know it relates surface area and Luminosity, which can be simplified once we assume several things are the same, which is what we do here: \[ \frac{F_A}{F_a} = \frac{\frac{L_A}{4\pi d_A ^2}}{\frac{L_a}{4\pi d_a ^2}} = \frac{d_a ^2}{d_A ^2 }, \] and now, knowing this star has an Absolute Magnitude M, which we know is the brightness at 10 pc, therefore we can establish what is the relationship of a star’s absolute and apparent magnitude because of its distance. So if the Absolute magnitude is at 10 pc: \[\frac{d_a ^2}{d_A ^2 } = \frac{d_a ^2}{ 100 ~pc^2}, \] which we will plug back into the original equation.  

Knowing the equation from the last problem, we change it so it represents the star we are looking at now: \[ m_1 - m_2 = -2.5 \log_{10} \left(\frac{F_1}{F_2} \right) \]  \[ m_A - m_a = -2.5 \log_{10} \left(\frac{F_A}{F_a} \right), \] which is also just a way of describing absolute and apparent magnitudes (we now plug in our definition of distance in this case): \[ M - m = -2.5 \log_{10} \left(\frac{d_a ^2}{ 100} \right) ,\] and we can start simplifying down with the properties of logarithms:  \[ M - m = -2.5 \log_{10} (d_a ^2) - \log_{10} (100) \]  \[ M - m = -2.5 (2 \log_{10} (d) - 2) ,\] which yields the equation describing the relationship of apparent and absolute magnitude, also known as a use for the distance modulus: \[ M = m - 5 \log (d) - 5.\]

c. In this problem, we simply want to understand how many more stars, of they were uniformly arranged, we would see if we looked one magnitude fainter than we previously did. Therefore, we start setting up with the standard equation for relating two  magnitudes: \[ m_x - m_{x_0} = -2.5 \log_{10} \left(\frac{F_x}{F_{x_0} } \right) ,\]  and we know the new stars we want to see are of 1 magnitude greater than our base stars: \[ 1  = -2.5 \log_{10} \left(\frac{F_x}{F_{x_0} } \right) ,\] Now just a matter of simplifying for a bit (and adding the broader definition of flux we saw earlier since the stars are uniform), we see that: \[ -\frac{2}{5} = \log_{10} \left(\frac{\frac{L_x}{4\pi d_x ^2}}{\frac{L_{x_0} }{4\pi d_{x_0}  ^2}}  \right) \] \[ -\frac{2}{5} = \log_{10} \left(\frac{d_{x_0}  ^2 } {d_x ^2}\right)\] \[ \frac{d_{x_0} ^2 }{ d_x ^2 } = 10^{-0.4},\] which means we can find a direct relationship between the distances we can see out two now by increasing our observation to one more magnitude: \[ \frac{d_{x_0} }{ d_x} = 0.63 \] \[ d_{x_0} = 0.63 ~d_x.\]

This increased radius of observation can then be used in a density equation where we substituted out the mass for number of stars: \[N_\star = V \times n_\star\] \[ \frac{N_{\star~x}}{N_{\star~ {x_0}}} =\frac{V_{x} \times n_\star}{V_{x_0}  \times n_\star},\] where we define the spherical volume by: \[ V = \frac{4}{3} ~\pi ~r^3,\] and placing it in the equation we get: \[ \frac{N_{\star~x}}{N_{\star~ {x_0}}} = \frac{\frac{4}{3} ~\pi ~d_x ^3 \times n_\star}{\frac{4}{3} ~\pi ~d_{x_0}^3 \times n_\star},\] which can be simplified down to: \[ \frac{N_{\star~x}}{N_{\star~ {x_0}}} = \frac{d_x ^3}{ d_{x_0} ^3},\] which in turn can be plugged in with the extension of the radius of observation we calculated earlier: \[ \frac{N_{\star~x}}{N_{\star~ {x_0}}} = \frac{d_x ^3}{ (0.63~ d_x) ^3} \] \[ \frac{N_{\star~x}}{N_{\star~ {x_0}}} = \frac{1}{0.25},\] which means by gaining a magnitude of observation sensitivity, we increase the amount of stars we can see by a factor of: \[ \frac{N_{\star~x}}{N_{\star~ {x_0}}} = 4.\]