Luminosity

Luminosity is the intrinsic brightness of an object. It is measured in the units of energy per time or watts (W)

For example, the Sun radiates \(4\times10^{26}\)W. Solar parameters are given special units. This amount of luminosity is called solar luminosity and is denoted by \(L_\odot\). $$L_\odot=4\times10^{26}\text{W}$$

In astronomy, it is common to use cgs units. \(1\text{W} = 10^7\text{erg s}^{-1}\). $$L_\odot=4\times10^{33}\text{erg s}^{-1}$$ The luminosity of our galaxy is about \(10^{11}L_\odot\)

Flux

The amount of radiation received by an observer who is at some distance away from the source of luminosity is the flux received by them.

Naturally, this follows an inverse square law (see diagram below).

Flux and luminosity are hence related according to the following relation $$F=\frac{L}{4\pi r^2}$$ Flux tells us how bright an object appears. The farther we go, the fainter a source appears.
Units of flux: energy per unit area.

Apparent & Absolute Magnitudes

Apparent Magnitude

Human eyes are actually designed to percieve brightness logarithmically. It is therefore more convenient to have a logarithmic scale to measure brightness instead of a linear scale.

Hipparchus was a Greek astronomer in second century who catalogued stars visible to the naked eye. He divided stars into 6 categories based on their apparent magnitudes (m). The apparent magnitude of \(m=1\) would be the brightest and \(m=6\) would be the faintest star.

Astronomers today use a similar scale even today! \(m=6\) is the limit of the naked eye. This is the faintest we can see without any aid. However, today, much fainter stars have been observed with telescopes.

Relation between the flux (brightness) of two objects and their apparent magnitudes is: $$m_1 -m_2 = 2.5 \log_{10} \frac{F_2}{F_1}$$ This scale is defined in such a way that a brightness difference of 100 corresponds to 5 magnitudes.

So, a star of magnitude 1 is
\(2.512\) times brighter than a star of magnitude 2
\(2.512^2 = 6.3\) times brighter than a star of magnitude 3
\(2.512^3 = 16\) times brighter than a star of magnitude 4
\(2.512^4 = 40\) times brighter than a star of magnitude 5
\(2.512^5 = 100\) times brighter than a star of magnitude 6

As expected, the difference of 5 magnitude means a \(100\times\) difference in brightness.
Note: Smaller magnitude => brighter object

Example: If we place another Sun along with our Sun, the brightness (flux) would be (nearly) 2 times.
How much will be the apparent magnitude if the current apparent magnitude is \(m_\odot=4.83\)?
Solution:
Since the flux is now twice the value, we have \(m_1 = m_\odot, \ m_2 = ? F_2 = 2F, \ F_1 = F\) we solve the following:
\[ \begin{align*} m_1 - m_2 & = 2.5 \log_{10} \frac{F_2}{F_1} \\ m_\odot - m_2 & = 2.5 \log_{10} \frac{2F}{F} \\ 4.83 - m_2 & = 2.5 \log_{10} 2 \\ m_2 & = 4.83 - 2.5 \log_{10} 2 \\ m_2 & = 4.07 \end{align*} \] ∴ the apparent magnitude of two Suns would be 4.07. A smaller value since it would be brighter.
The magnitude decrease of roughly 0.76 (i.e., 4.83-4.07) corresponds to doubling of the flux or brightness.

Absolute Magnitude

Since apparent magnitude is related to the flux received, it does not tell us anything about the true brightness (luminosity). A faint star very close to us could easily appear brighter than a much brighter star which is far-far away.
Wouldn't it be convenient if we could place all the stars on the equal footing?
That's why we define the absolute magnitude (M) of a star. This is the apparent magnitude of a star when it is placed \(10\) pc away from us.
This definition makes provides the formula: $$m-M=2.5 \log{\frac{\frac{L}{4\pi r^2}}{\frac{L}{4\pi 10^2}}}$$ Note: \(r\) is in pc and L dosen't change (it's intrinsic, dosen't depend on the distance). Simplifying, $$m-M=5\log{\frac{r}{10}}$$

Here, \(r\) is measured in pc.

The interesting this to note is that this formula can be used to find the absolute magnitude of a source, when its distance is known. Distances can be known using the parallax of a source which is relatively easy to measure.
Conversely, if we know the absolute magnitude of a source (say it is a known type star), we can get a measure of its distance by simple comparing the apparent and absolute magnitudes using the above formula.

Distance Modulus

Having known the apparent and absolute magnitude of a star, we can infer how far it is from us.
$$m-M=5\log{\frac{r}{10}}$$ $$r = 10\times10^{(m-M)/5} \text{ pc}$$
Use the following calculators to find the distance to any star given its \(m\) and \(M\).

r (distance) using m (apparent) and M (absolute) magnitudes

Enter the apparent magnitude \(m\) and absolute magnitude \(m\) to calculate \(r\)

Tip: 💡
The apparent magnitude of the Sun is \(m_\odot = -26.74\) and the absolute magnitude is \(M_\odot = 4.83\)
Try substituting these values in the input fields. You should get 1 AU for the distance 😉


Distance \(r\)