Published by Patrick Mutisya · 14 days ago
Describe redshift as an increase in the observed wavelength of electromagnetic radiation emitted from receding stars and galaxies.
When a star or galaxy moves away from Earth, the waves of light it emits are stretched. This stretching increases the wavelength (\$\lambda\$) and decreases the frequency (\$f\$). The change in wavelength is expressed as the redshift \$z\$:
\$z = \frac{\Delta\lambda}{\lambda0} = \frac{\lambda{\text{observed}} - \lambda0}{\lambda0}\$
where \$\lambda0\$ is the wavelength emitted at the source and \$\lambda{\text{observed}}\$ is the wavelength measured on Earth.
For speeds much less than the speed of light (\$v \ll c\$), the redshift is approximately proportional to the recessional velocity \$v\$:
\$z \approx \frac{v}{c}\$
Thus, measuring \$z\$ allows us to calculate the velocity at which a galaxy is receding.
Edwin Hubble discovered that more distant galaxies have larger redshifts, leading to the empirical relationship:
\$v = H_0 d\$
where \$H_0\$ is Hubble’s constant (≈ 70 km s⁻¹ Mpc⁻¹) and \$d\$ is the distance to the galaxy. Combining this with the redshift–velocity relation gives:
\$z \approx \frac{H_0 d}{c}\$
Suppose a spectral line that is normally at \$\lambda0 = 500\ \text{nm}\$ is observed at \$\lambda{\text{observed}} = 510\ \text{nm}\$.
\$z = \frac{510 - 500}{500} = 0.02\$
\$v = z c = 0.02 \times 3.00 \times 10^8\ \text{m s}^{-1} = 6.0 \times 10^6\ \text{m s}^{-1} \approx 6000\ \text{km s}^{-1}\$
\$d = \frac{v}{H_0} = \frac{6000\ \text{km s}^{-1}}{70\ \text{km s}^{-1}\text{Mpc}^{-1}} \approx 86\ \text{Mpc}\$
| Quantity | Symbol | Formula / Relationship | Typical Units |
|---|---|---|---|
| Redshift | \$z\$ | \$z = \dfrac{\lambda{\text{obs}} - \lambda0}{\lambda_0}\$ | dimensionless |
| Recessional \cdot elocity | \$v\$ | \$v \approx zc\$ (for \$v \ll c\$) | km s⁻¹ |
| Hubble’s Constant | \$H_0\$ | Empirical constant linking \$v\$ and \$d\$ | km s⁻¹ Mpc⁻¹ |
| Distance to Galaxy | \$d\$ | \$d = \dfrac{v}{H_0}\$ | Mpc |