# Kinetic energy of gases in stars

Hi,

I have a conceptual and mathematical question about gases in stars.
The information we have from stars is due to the motion of particles in one dimension: along our line of sight.
We assume that this motion is isotropic and that regardless of where on the star we look, we'll get the same motion.
So, when calculating the temperature of the star's chromosphere, do we use the one dimensional kinetic energy E=1/2kT, the three-dimensional kinetic energy E=3/2kT or the average kinetic energy E=kT to equate to 1/2mv^2, where v is the one dimensional velocity?

Chronos
Gold Member
The sun is the only star whose chromosphere is observationally accessible. We are still trying to figure out its heating mechanism. The photosphere temperature is relatively simple to calculate. The most widely used method is based on Wien's law, which only requires measurement of the peak emission frequency. This can be done both with the sun and any star bright enough for spectroscopic measurements. Another, more accurate, method relies on the strength of different absorption lines in the spectrum.

Drakkith
Staff Emeritus
Hi,
The information we have from stars is due to the motion of particles in one dimension: along our line of sight.

Could you elaborate on this? I'm not sure I understand it correctly.

Ken G
Gold Member
So, when calculating the temperature of the star's chromosphere, do we use the one dimensional kinetic energy E=1/2kT, the three-dimensional kinetic energy E=3/2kT or the average kinetic energy E=kT to equate to 1/2mv^2, where v is the one dimensional velocity?
Sort of the middle one, E=kT, but I wouldn't call that "average kinetic energy" because I don't know what you mean by that (the average kinetic energy is the 3D kinetic energy, 3/2kT). First I should point out that we must be clear what v is-- it isn't actually the "one dimensional velocity," because if we have a T, we have many different velocities there. So v either has to be the standard deviation of the one-dimensional velocity, or more commonly, it is the "Doppler width" of the line in v units, which is something different. What you actually do is observe the line broadening, which gives you the spread in velocity along the line of sight (let's use the "Doppler width" meaning for v). You then assume the velocities are distributed isotropically and like a Maxwellian (assumptions we have good reason to expect in high densities), and reverse engineer what the T has to be to produce the spread in line-of-sight velocities that you see. It comes out the same as equating 1/2mv^2 to kT, but that's just how it comes out-- I don't see any way to assert that in advance, because neither 1/2mv^2 nor kT have any particular physical meaning in this situation.

Getting back to the tricky difference between what we call the "Doppler width" of a line, and what is the "standard deviation" of the line-- if we define v to be the standard deviation (the root-mean-square line-of-sight velocity), then it would be appropriate to call 1/2mv^2 to be the kinetic energy in the line-of-sight direction, and equate it with 1/2kT, which is the energy at T in that same direction. But that v isn't what is meant by the Doppler width v, which is the v appearing directly in the Gaussian shape of the line. The latter v, the Doppler width, is larger, so that v would use the expression 1/4mv^2 as the kinetic energy from the component of motion along that direction. You'd equate that to 1/2kT, and get the same result as what we were talking about above.