Emission line width units

Question

What do these units mean: the large velocity widths of emission lines (in AGN) are 2,000 - 10,000 km s^-1? I've looked for the answer but keep getting swamped in myriads of details. I want to know what the term velocity means in large velocity widths and also why km s^-1. Thanks

Answer 1

Velocity and wavelength shifts are connected by the Doppler effect, with which you may already be familiar. Basically, objects that are approaching appear bluer and those receding appear redder.

Now, suppose body of material emits some spectral line and that the particles of that material have line-of-sight velocities that range from $-10^4\text{km.s}^{-1}$ to $+10^4\text{km.s}^{-1}$. The particles moving away from us emit a line that is red-shifted. Those moving towards us emit a blue-shifted line. If you add up all the lines, you'll get a line that is broadened across the rest-frame line. i.e. the wavelength of the line in the absence of any motion.

To convert between the two, you can re-arrange the formula on the Wikipedia page for the Doppler effect to get

$v/c=1-\lambda_0/\lambda \Leftrightarrow\lambda=\lambda_0/(1-v/c)$

where $v$ is the line-of-sight velocity, $\lambda_0$ is the rest wavelength, $\lambda$ is the observed wavelength and $c$ is the speed of light. To get the wavelength shift, use

$\Delta\lambda=\lambda-\lambda_0=\lambda_0v/(c-v)$

So a velocity of $10^4\text{km.s}^{-1}$ gives a wavelength shift of about 3.4%.

The units are chosen just because that's what the velocities end up being. The velocity can itself imply a few things. The main implication is that the emitting material is a superposition of particles orbiting with a spread of velocities. If you imagine a ring of gas orbiting something, the gas on one side is receding and on the other side approaching. There's more gas between these points, so the peak should still be on the rest wavelength but there is a finite spread that is related to the orbital velocities and therefore to the gravitational potential.