Published by Patrick Mutisya · 14 days ago
By the end of this lesson you should be able to explain that an object of known luminosity is called a standard candle and describe how it is used to determine astronomical distances.
\$\mu = m - M = 5\log_{10}\!\left(\frac{d}{10\ \text{pc}}\right)\$
where \$m\$ is the apparent magnitude, \$M\$ the absolute magnitude, and \$d\$ the distance in parsecs.
Because the inverse‑square law relates luminosity and flux,
\$F = \frac{L}{4\pi d^{2}},\$
knowing \$L\$ allows us to solve for the distance \$d\$ from a measured flux \$F\$. This provides a fundamental rung on the cosmic distance ladder.
| Type | Typical Luminosity (Absolute Magnitude \$M\$) | Distance Range | Key Characteristics |
|---|---|---|---|
| Cepheid \cdot ariable | \$M_V \approx -5\$ to \$-7\$ | \overline{0}.1 – 30 Mpc | Period–luminosity relation; pulsating supergiants. |
| RR Lyrae | \$M_V \approx +0.6\$ | \overline{0}.01 – 1 Mpc | Old, low‑mass stars; used for globular clusters. |
| Type Ia Supernova | \$M_B \approx -19.3\$ | \overline{10} Mpc – 10 Gpc | Thermonuclear explosion of a white dwarf; very uniform peak brightness. |
| Tip of the Red Giant Branch (TRGB) | \$M_I \approx -4.0\$ | \overline{0}.1 – 20 Mpc | Sharp cutoff in luminosity of red giants. |
\$d = 10^{\frac{\mu+5}{5}}\ \text{pc}.\$
Suppose a Type Ia supernova is observed with an apparent magnitude \$m = 15.2\$. The calibrated absolute magnitude for Type Ia supernovae is \$M = -19.3\$.
Distance modulus:
\$\mu = 15.2 - (-19.3) = 34.5.\$
Distance:
\$d = 10^{\frac{34.5+5}{5}} = 10^{7.9}\ \text{pc} \approx 7.9 \times 10^{7}\ \text{pc} \approx 257\ \text{Mpc}.\$
A standard candle is an astronomical object whose intrinsic luminosity is known. By comparing this known luminosity with the observed brightness, we can determine the object's distance using the inverse‑square law or the distance modulus formula. Standard candles are essential tools for constructing the cosmic distance ladder and for probing the scale of the universe.