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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

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1 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.

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Hi, thanks for the help. I still feel like a dunce, sorry. So does "large velocity widths of AGN" refer to "fast" radial velocity of accretion disk? What is this sentence (the one I quoted in my question) really saying about AGN? –  Gigi Giles Feb 25 '12 at 0:02
    
It pretty much means the velocities of the gas or stars or whatever is emitting the line is fast. Bear in mind it doesn't have to be a disk or gas. It could be stars. I'm not an expert so I don't want to make an authoritatize statement about what the high velocities mean. For AGN, I think it supports the presence of a substantial concentrated mass, like a black hole, but I'm not sure that big star clusters can't have large velocity widths. –  Warrick Feb 27 '12 at 9:19

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