Show that 1/2mv^2 = 3/2kT for a star cluster

[warfreak131]

Homework Statement

Demonstrate that the temperature applicable to a stellar system
is
T = m < v² (x) >
...3k
where it is assumed that the stars have equal mass m, v(x) is the
velocity of a star relative to the center-of-mass, and <> represents
an average.

Homework Equations

The Attempt at a Solution

I don't know how to approach this problem. The only thing I could guess is that the spherically symmetric stars can be considered the particles of a system, so it's just a much larger system, but the physics is the same overall.

 

[Physics Enemy]

Isn't it something like: Assume the stars move about with a mean thermal energy, which is their KE. So KE = 1/2mv^2 = (3/2)KT
Then mv^2 = 3kT => T = mv^2/3k?

 

[warfreak131]

Isn't it something like: Assume the stars move about with a mean thermal energy, which is their KE. So KE = 1/2mv^2 = (3/2)KT
Then mv^2 = 3kT => T = mv^2/3k?

I emailed him about it, citing this argument, and he said that we have to demonstrate it for a cluster of stars, not particles.

 

[Physics Enemy]

I emailed him about it, citing this argument, and he said that we have to demonstrate it for a cluster of stars, not particles.

But isn't the idea that they are behaving like particles in a gas?

 

[warfreak131]

But isn't the idea that they are behaving like particles in a gas?

tell that to him, not me :P

i don't know what the hell he wants us to do

 

[Physics Enemy]

tell that to him, not me :P

i don't know what the hell he wants us to do

I think he wants you to justify why they behave like particles in a gas, before you write down the equations. Perhaps Wiki it? I think you need to explain the assumptions.

Hey, would you mind checking my thread? It's a basic numeracy one lol.

 

[warfreak131]

thats what i was thinking, but then again, you don't know my teacher, or how horrible he is.

and i checked your numeracy question, and I am not quite sure what its asking for :P sorry

i don't want to hijack my own thread and turn it into another question, so i won't discuss it here