Spectral Line Broadening

An A-type star at 9600 K spins fast enough that v sin i = 150 km/s. Take its H-gamma line, 434.05 nm in air, and work out the full width at half maximum you would measure. Then do what a spectrum quietly invites you to do: read that whole width as thermal motion, and see what temperature comes back.

Open this problem in the tool

1. Thermal broadening starts with one atom's speed along the line of sight. A Maxwell-Boltzmann gas gives the line a Gaussian profile of width λ√(2kT/(Auc²)), and that width is the 1/e half-width. It is neither the standard deviation, smaller by √2, nor the FWHM, larger by 1.6651, and the row is labelled Thermal half-width (1/e) so you cannot mistake which one is on offer. For hydrogen at 9600 K the radicand is 1.76211 × 10⁻⁹, its square root 4.19775 × 10⁻⁵, and 434.05 nm times that is 1.8220 × 10⁻² nm.

   $$\Delta {\lambda }_{\mathrm{th}}=\lambda \sqrt{\frac{2kT}{Au{c}^2}}=434.05\mathrm{nm}\times \sqrt{1.76211\times 10^{-9}}=1.8220\times 10^{-2}\mathrm{nm}$$

2. Rotation widens the same line by an entirely different mechanism. The approaching limb is blueshifted by λ v sin i / c and the receding limb is redshifted by as much, so the profile smears over that whole range. 150 km/s divided by c = 2.99792458 × 10⁵ km/s is 5.00346 × 10⁻⁴, and 434.05 nm times that is 2.1718 × 10⁻¹ nm. Note what the tool assumed for you: i = 90°, edge-on, sin i = 1. A star seen closer to pole-on hides part of its spin, and no measurement of one line can tell you which case you have.

   $$\Delta {\lambda }_{\mathrm{rot}}=\frac{\lambda v\sin i}{c}=\frac{434.05\mathrm{nm}\times 150km/s\times \sin 90^{\circ }}{2.99792458\times 10^{5}km/s}=2.1718\times 10^{-1}\mathrm{nm}$$

3. Add the two in quadrature. That is exact for a pair of Gaussians, and the rotation kernel is not a Gaussian, so the tool stands a Gaussian of the same 1/e half-width in for it and is honest about the few percent that costs. √((1.8220 × 10⁻²)² + (2.1718 × 10⁻¹)²) = 2.1794 × 10⁻¹ nm. Look at what the thermal term contributed: the combined width is 0.35% wider than rotation alone.

   $$\Delta \lambda =\sqrt{\Delta {\lambda }_{\mathrm{th}}^2+\Delta {\lambda }_{\mathrm{rot}}^2}=\sqrt{{\left(1.8220\times 10^{-2}\right)}^2+{\left(2.1718\times 10^{-1}\right)}^2}=2.1794\times 10^{-1}\mathrm{nm}$$

4. The number a spectroscopist quotes is the FWHM, and for a profile written as exp(−(x/Δλ)²) that is 2√(ln 2) times the half-width. 1.6651 × 2.1794 × 10⁻¹ = 3.6289 × 10⁻¹ nm, printed in the Combined FWHM row. Drag the rotation speed to zero and the same row falls to 3.0339 × 10⁻² nm, twelve times narrower, which is the line this star would show if it stopped turning.

   $$\mathrm{FWHM}=2\sqrt{\ln 2}\Delta \lambda =1.6651\times \left(2.1794\times 10^{-1}\right)=3.6289\times 10^{-1}\mathrm{nm}$$

5. The Thermal width ratio row holds at 7.443 whatever you do to the temperature or the wavelength, because √(A_Fe/A_H) is all that survives once the rest cancels. Now press Iron with the rotation still at 150 km/s. Iron's thermal width really is 7.443 times narrower, 2.4479 × 10⁻³ nm, and its combined width is 2.1719 × 10⁻¹ nm against hydrogen's 2.1794 × 10⁻¹. Two atoms 55 times apart in mass, and the lines you would actually measure differ by a third of a percent.

   $$\frac{\Delta {\lambda }_{\mathrm{H}}}{\Delta {\lambda }_{\mathrm{Fe}}}=\sqrt{\frac{A_{\mathrm{Fe}}}{A_{\mathrm{H}}}}=\sqrt{\frac{55.845}{1.008}}=7.443$$

6. Put the thermal width back into speed units to see where the crossover sits. Δλ_th/λ is v_th/c with v_th = √(2kT/(Au)), which for hydrogen at 9600 K is 12.58 km/s. The star's surface is moving 11.92 times faster than its own atoms move along the line of sight. Below roughly 12.6 km/s the thermal term is the larger of the two and the width still measures temperature; above it, rotation takes over and keeps going.

   $$v_{\mathrm{th}}=\sqrt{\frac{2kT}{Au}}=12.58km/s,\frac{v\sin i}{v_{\mathrm{th}}}=\frac{150}{12.58}=11.92$$