Magnitude-Limited surveys and Volume-Limited Surveys

Suppose the galaxy has two types of stars, Q-type stars with twice the Sun's luminosity, and R-type stars with a third the Sun's luminosity. R-type stars are four times as numerous in the galaxy as Q-type stars, and both types are uniformly distributed in space.

Note

A magnitude limited survey is a survey where any star is accepted as long it is not too faint (magnitude has to be below a certain number)

A volume limited survey is a survey where any star is accepted as long as it is within a certain distance and is bright enough to be picked up with the instruments you are using (max distance is used)

A. If you are conducting a magnitude-limited survey of stars, compare the number of Q-type stars you'll observe to the number of R-type stars.

Defining Some Variables

$L_o$ = the suns luminosity

$L_Q$= $2L_o$ = luminosity of Q type stars

$L_R$= $\frac{L_o}{3}$ = luminosity of Q type stars

$n_Q$= number of Q type stars

$n_R$= number of R type stars

$m_{max}$= faintest observable magnitude

$m_Q$ = magnitude of Q stars

$m_R$ = magnitude of Q stars

N= number density

n=number

Equations

We are given the luminosity so here is the equation for luminosity

$L$= $ \frac {Energy}{time}$ 

Given the luminosity we can figure out flux because we know area of a sphere, although now we have a d variable which we do not know. 

$\frac {2L_o}{d_Q^2}$=$\frac {\frac {L_o}{3}}{d_R^2}$ $\times 2.5^{m_R-m_Q}$

Luminosities cancel out

$\frac {2}{d_Q^2}$=$\frac {\frac {1}{3}}{d_R^2}$ $\times 2.5^{m_R-m_Q}$

$\frac {6}{d_Q^2}$=$\frac {1}{d_R^2}$ $\times 2.5^{m_R-m_Q}$

Since this is magnitude limited, the magnitudes of the stars are the same

$\frac {6}{d_Q^2}$=$\frac {1}{d_R^2}$

There is a d variable in number density which we can solve for

$d^3$=$ \frac {N}{n}$

$d$= $\sqrt [3] { \frac {N}{n}}$

We can solve for the ratio of R stars to Q stars

$ {( \frac {d_Q}{d_R})}^3$= $\frac {\frac {N_Q}{n_Q}}{\frac {N_R}{n_R}}$

$ {( \frac {d_Q}{d_R})}^3$= $\frac {N_Q}{n_Q} \times {\frac {n_R}{N_R}}$

Which gives us the ratio of Q stars to R stars

$ {( \frac {d_Q}{d_R})}^3 \times \frac {N_R}{N_Q}$= $\frac {n_R}{n_Q}$

B. If you decide instead to conduct a volume-limited survey but still have the same magnitude limit, what distance corresponds to your volume limit?

Defining Some Variables

$L_o$ = the suns luminosity

$L_Q$= $2L_o$ = luminosity of Q type stars

$L_R$= $\frac{L_o}{3}$ = luminosity of Q type stars

$n_Q$= number of Q type stars

$n_R$= number of R type stars

$d_{max}$= furthest distance

$m_{max}$= faintest observable magnitude

Equations

This equation is taken from part A. Since this is a volume limited survey, the volumes will be equal so distances will be equal.

$\frac {6}{d_Q^2}$=$\frac {1}{d_R^2}$ $\times 2.5^{m_R-m_Q}$

We can also solve for magnitude difference between Q and R type stars

$F$= $ \frac {Energy}{time \times area}$ = $ \frac {L}{4 \pi d^2}$

$ \frac {F_Q}{F_R}$ = $2.5^{m_R-m_Q}$

$m_R-m_Q$= $\log{2.5} \frac {F_Q}{F_R}$ 

$d_Q =d_R= d_{max}$ 

This means volumes of Q and R stars are equal

$2L_o$=$\frac {L_o}{3} \times 2.5^{m_R-m_Q}$