# Astronomy and the Relativistic Doppler Shift

1. Dec 10, 2006

### Fallen Seraph

1. The problem statement, all variables and given/known data

The hydrogen atoms in a star are also moving at high velocity because of the random motions caused by their high temperature. As a result, each atom is Doppler shifted a little bit differently, leading to a finite width of each spectral line, such as the 656.46-nm line we were just discussing. For a star like our sun, this leads to a finite width of the spectral lines of roughly %Delta lambda = 0.04 nm. If our instruments can only resolve to this accuracy, what is the lowest speed V, greater than 0, that we can measure a star to be moving?

2. Relevant equations
Lambda_o = Lamba_s*sqrt((c+v)/(c-v))

where lambda_o is the observed wavelength and lambda_s is the wavelength in the rest frame of the source.

3. The attempt at a solution

All I could try for a solution was to rearrange that equation to get an expression for the change in wavelength over the wavelength in the source's rest frame in terms of c and v, then set lambda_s=656.46-nm and the change=0.04nm and solve for v. But frankly this method seems daft and nonsensical to me. My main problem with the question is that I don't really understand what it means. I don't really see how the width of the spectral lines can be the same as the change in wavelength, I don't see how that relationship can hold or where it comes from. It seems to be a bunch of dissimilar things to me...

Also to clarify, when they say "finite lengths" it refers to the lines having a clear stopping and starting point, yes?

2. Dec 10, 2006

### OlderDan

I am not clear on what %Delta lambda = 0.04 nm means. If this is indeed a percent change in the wavelength, then the value should be dimensionless. So is it %Delta lambda = 0.04, or is it Delta lambda = 0.04 nm?

The finite width does not imply an abrupt cut-off. The velocity distribution of the hydrogen atoms relative to the center of mass of the star depends on speed and direction. For any given speed, if you reasonably assume an isotropic distribution you could calculate the distribution of the velocity component along the line from the star to the observer, but the speeds are distributed also. A reasonable, if not precise assumption is that the average velocity component relative to the center of the star peaks at zero, with a sort of Gaussian distribution. "Finite width" just means that the vast majority of hydrogen atoms have velocity components within a few standard deviations of the mean.

Your relativistic doppler equation seems to be unconventional. The usual expression involves the frequencies with f = fo*sqrt[(1+v/c)/(1-v/c)] where positive v means approaching (f is usually represented by the Greek letter nu). Since wavelength is proportional to the inverse of frequency, your ratio appears to be inverted.

You may be able to use the low speed approximation in your problem

http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/reldop3.html#c2

The variation of the wavelenths associated with the variation in doppler frequencies does result a broadening of the spectral line. If you look at a broadened line, you have an idea where the central wavelenth is. If your resolution is only as good as the width of the line, then unless the line moves as far as the width of the line, you will not know for sure that it has moved.

3. Jan 11, 2007

### matthew baird

Hey Dan, I have a similar issue that I am facing right now in a homework problem. I am using the correct equation though that you listed above where,
f=f_o*sqrt[(1-b)/(1+b)] where b=beta or v/c

here is my question, what if the relative speed of the source is c? That would make the entire expression undefined since 1-1=0....I have been searching in my book for some direction but have found nothing. Help is very much appreciated. :)

4. Jan 12, 2007

### Dick

'Sources' don't travel at c. They must have mass. Only 'signals' travel at c.