How Do We Calculate the Kinetic Energy of Gases in Stars?

[Piano man]

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?

(Note to mods: I posted this question in the Astrophysics section earlier today, but after 130 views with no replies, I think I may get a better response in this subforum. I know spread-posting is frowned upon but I do not intend to troll!)

 

[ehild]

We see the photons which arrive from the stars, and these photons are radiated out by the atoms (molecules, ions) of the chromosphere. The particles have rotational, vibrational and electronic degrees of freedom. The visible, UV and higher frequency radiation originates from the transition between the electronic states of the particles. We get information about the temperature of the stars from the frequency distribution of their radiation.
Read about "black body radiation". http://www.egglescliffe.org.uk/physics/astronomy/blackbody/bbody.html
The wavelength of maximum intensity of radiation is proportional to the absolute temperature of the star, according to Wien's Law.

ehild

 

[Piano man]

Thank you.
I should have said that I was working with velocities and temperatures of chromospheric winds, not just the temperature of the chromosphere. The wind would be anisotropic, as it is being ejected from the star, but the turbulent velocity of the particles within the wind would be isotropic.
So how does that effect my original question?

 

[ehild]

I see. I do not know if those stellar winds are isotropic or not. The star can have magnetic field and I think it has its effect. I guess the velocity distribution in the chromosphere is described by the Maxwell-Boltzmann distribution, but the particles lose energy when escaping the star. ehild

 

[paranormal]

Hi there,

Thank you for your question. I would like to clarify that the kinetic energy of gases in stars is a complex and multi-dimensional concept, and there is no one definitive answer to your question. However, I will do my best to provide a response based on current scientific understanding.

First, it is important to note that the motion of particles in stars is not solely limited to one dimension. While we may only observe the motion along our line of sight, the particles are actually moving in three dimensions. Therefore, it would be more accurate to use the three-dimensional kinetic energy equation (E=3/2kT) to calculate the temperature of the star's chromosphere.

However, it is also important to consider the concept of equipartition of energy, which states that in a system in thermal equilibrium, the average kinetic energy per degree of freedom is equal to 1/2kT. In this case, the average kinetic energy (E=kT) would also be a valid equation to use in calculating the temperature of the star's chromosphere.

In conclusion, both the three-dimensional kinetic energy and the average kinetic energy equations are valid for calculating the temperature of the star's chromosphere. Which one to use would depend on the specific context and assumptions of the situation. It is always important to carefully consider the assumptions and limitations of any mathematical equation used in scientific calculations. I hope this helps clarify your question.