Know that the distance d of a far galaxy can be determined using the brightness of a supernova in that galaxy

Published by Patrick Mutisya · 14 days ago

IGCSE Physics 0625 – Topic 6.2.3 The Universe: Determining Galaxy Distances with Supernovae

6.2.3 The Universe – Determining the Distance to a Far Galaxy

Learning Objective

Know that the distance d of a far galaxy can be determined using the brightness of a supernova in that galaxy.

Key Concepts

  • Standard candle: An astronomical object whose absolute luminosity L is known.
  • Type Ia supernova: Exploding white dwarfs that have a very uniform peak absolute magnitude, making them excellent standard candles.
  • Inverse‑square law for light: The apparent brightness b of a source falls off as the square of the distance:

    \$b = \frac{L}{4\pi d^{2}}\$

  • Distance modulus: Relates apparent magnitude m, absolute magnitude M and distance d (in parsecs):

    \$m - M = 5\log_{10} d - 5\$

Why Type Ia Supernovae?

When a carbon‑oxygen white dwarf in a binary system accretes matter and reaches the Chandrasekhar limit (\overline{1}.4 M☉), it undergoes a thermonuclear explosion. The peak luminosity is almost the same for all such events, giving a typical absolute magnitude:

M ≈ –19.3 (in the visual band).

Step‑by‑Step Procedure to Find the Distance d

  1. Observe the supernova in the target galaxy and record its apparent magnitude m at peak brightness.
  2. Assume the absolute magnitude M = –19.3 (or use a calibrated value if provided).
  3. Insert m and M into the distance modulus equation and solve for d:

    \$d = 10^{\frac{m - M + 5}{5}} \text{ parsecs}\$

  4. If required, convert parsecs to light‑years (1 pc ≈ 3.26 ly) or to megaparsecs (1 Mpc = 10⁶ pc).

Example Calculation

Suppose a Type Ia supernova in a distant galaxy is measured to have an apparent magnitude m = 22.0.

Using M = –19.3, the distance modulus is:

\$m - M = 22.0 - (-19.3) = 41.3\$

Now solve for d:

\$d = 10^{\frac{41.3 + 5}{5}} = 10^{9.26} \text{ pc} \approx 1.8 \times 10^{9} \text{ pc}\$

Converting to megaparsecs:

\$d \approx 1.8 \times 10^{3} \text{ Mpc}\$

Thus the galaxy is about 1.8 Gpc away.

Table – Typical \cdot alues for a Type Ia Supernova

QuantitySymbolTypical \cdot alueUnits
Absolute magnitude (peak)M–19.3mag
Apparent magnitude (example)m22.0mag
Distance (calculated)d1.8 × 10⁹pc

Important Points for Exam Answers

  • State that a Type Ia supernova is a standard candle because its absolute magnitude is known.
  • Write down the inverse‑square law or distance modulus as the governing equation.
  • Show the substitution of the given apparent magnitude and the standard absolute magnitude.
  • Calculate d step‑by‑step, keeping track of units.
  • Convert the final answer to the required unit (pc, Mpc, or ly) if the question asks.

Common Misconceptions

  • Confusing apparent and absolute magnitude: Remember that apparent magnitude depends on distance, while absolute magnitude is intrinsic.
  • Using the wrong form of the distance equation: The distance modulus is the most convenient form for magnitudes; the inverse‑square law uses luminosity and brightness.
  • Neglecting extinction: In real observations dust can dim the supernova, making the galaxy appear farther away. For IGCSE level, extinction is usually ignored.

Suggested Diagram

Suggested diagram: A schematic showing a distant galaxy, a Type Ia supernova within it, and the observer on Earth. Include labels for absolute magnitude (M), apparent magnitude (m), and the distance d.

Summary

By treating a Type Ia supernova as a standard candle with a known absolute magnitude, the distance to its host galaxy can be found using the distance modulus formula. This method underpins modern cosmology and allows astronomers to map the scale of the Universe.

Practice Question

A Type Ia supernova in a remote galaxy is observed with an apparent magnitude of 24.5. Calculate the distance to the galaxy in megaparsecs. (Use M = –19.3.)