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Understanding double-peaked HI profiles

  1. Oct 28, 2013 #1
    Hi everyone. This isn't a specific homework problem, I'm just trying to understand a concept.

    I've been studying the HI profiles (from the 21cm emission line). Almost all of them look like the attached profile. The velocity on the x-axis corresponds to the amount that the line has been redshifted, and the flux on the y-axis tells how much gas has that particular velocity.

    With a simple model of an observer viewing a galaxy that is a solid, uniform disk edge-on, could someone help me understand why one would expect a double-peak HI profile. I've drawn tons of circles and tried to determine the flux at each velocity and the velocity at each angle, but I'm not getting it.

    Ideally, could someone show me how to derive a function for flux as a function of velocity for the simplest case? I understand that the velocity of the center of the peak gives the total recession velocity of the galaxy and the width of the peak gives rotation velocity, but I don't see why there's a double peak rather than a single one.

    Thanks so much!
     

    Attached Files:

  2. jcsd
  3. Oct 28, 2013 #2

    mfb

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    Staff: Mentor

    Consider a uniform ring, seen edge-on, where particles rotate with a velocity of v. Consider a point with the angle θ between our line of sight and the point on the ring (so θ=0 is the closest point, θ=pi/2 is at one side and so on). If we consider the radial component of the velocity only, we get ##v(\theta)=\sin(\theta)## and therefore $$\frac{dv}{d\theta}=v\cos(\theta)$$ and $$\frac{d\theta}{dv}=\frac{1}{v\cos(\theta)}$$. But ##\frac{d\theta}{dv}## just quantifies your intensity - how much material you have per velocity bin. As ##\cos(\theta)## is small for θ close to pi/2 and -pi/2, you get a large intensity there, the two peaks at the edge of the spectrum.

    A disk is like a collection of many rings. Many galaxy rotation curves have nearly constant velocities (as a function of radius), so the peaks can be seen for the whole galaxy.
     
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