Differential equation for motion within stars

  • Thread starter leonmate
  • Start date
  • #1
84
1
I know:

f = ma
f = - GMm/r^2

a = -GM/r^2 can be easily derived,

But, we've been given the differential equation of motion of gas within a star as:

a = -GM/r^2 - 1/p * dP(r)/dr

I was wondering where the - 1/p * dP(r)/dr term is derived from? I can't find it in my text books.

Cheers
 

Answers and Replies

  • #2
84
1
p is density btw
 
  • #3
Matterwave
Science Advisor
Gold Member
3,965
326
That's the term that deals with pressure from the surrounding gas. Pressure is force/area, so certainly one should expect it to show up in our acceleration equation. There's a derivative in there because there's pressure from both sides and only the difference in pressure will contribute.
 
  • #4
84
1
Right,

So the first term is the grav force towards the centre and the second term represents the radiation pressure outwards (hence the negative sign), yes?
 
  • #5
Ken G
Gold Member
4,438
333
No, the second term does not need to be radiation pressure, it is pressure of any kind. That includes radiation pressure, but in most cases that is negligible compared to regular old gas pressure. If the temperature is regarded as known (which is the standard way of visualizing the situation, though it leads to some awkward misconceptions that I won't go into here), then we typically use the ideal gas law to get the gas pressure, and that is usually the P in that equation.

By the way, I think I understand where the questioner is coming from when they wonder about "what force is represented by that last term." Which of the fundamental forces is that, gravity or electromagnetic or what? The answer is, it is not any of the fundamental forces, because it is only a "force" in the sense that we have already adopted what is called the "fluid picture". In the fluid picture, we average over a small volume, and get coarse-grained quantities like temperature and pressure and density. But something else happens when we do that averaging-- any process that transports net momentum into our averaging box is going to get interpreted as a force on that box, even though there is no fundamental force there at all. Gas pressure is usually just the momentum that is transported into the box by virtue of the fact that the particles entering the box have more momentum in a given direction (say, outward, as in that equation), than the particles leaving the box. A great but fascinating subtlety arises here when we recognize that the sign of the outward momentum is of great importance-- particles that exit the box downward will remove negative outward momentum, while particles that enter the box upward will bring in positive outward momentum, so the "bottom side" of the box is responsible entirely for increasing the outward momentum in the box, even though there is no net transport of particles. That's an outward force, called "pressure" when you divide by the area of the bottom side of the box. Similarly, the top side of the box will be responsible entirely for decreasing the outward momentum in the box. That's also pressure, when divided by the area of the top of the box, but that points downward, so to get the net force, you need to subtract the two, and that's why it is the gradient in the pressure that gets interpreted as a force in the fluid model of a gas.

Note also that no collisions are required-- the equation holds as soon as you adopt the fluid model. Collisions just help that model hold true, because the model generally assumes the particle distribution function is essentially isotropic in the fluid frame, which gives us the necessary concept of "isotropic pressure", such that P is just a number. This can also cause confusion-- note that pressure is isotropic, but is still responsible for an upward force if you consider the bottom face of the fluid cell, and a downward force if you consider the top face. So you have an isotropic effect that induces forces by virtue of its spatial gradient, and it is not a fundamental force, it is an effective force that appears in the fluid description.
 
  • #6
84
1
Ah, that's interesting, thanks for explaining that in such good detail.

I've always wondered this with stars. Gravity is obvious, but as you said the opposing is a little more complicated. I always wanted to know what force causes degeneracy pressure in white dwarfs and neutron stars. I always thought it was the electromagnetic working in some strange way. And the fact that it can be overcome seems to give it a precise amount of force it can withstand/counter
 
  • #7
Ken G
Gold Member
4,438
333
I always wanted to know what force causes degeneracy pressure in white dwarfs and neutron stars. I always thought it was the electromagnetic working in some strange way. And the fact that it can be overcome seems to give it a precise amount of force it can withstand/counter
Hopefully you can now see that degeneracy pressure is just the effect I was describing, it is ordinary kinetic pressure. Things get a little weird in quantum mechanics in that you have a wave function instead of a velocity for every particle, but the answer comes out just the same, the pressure gradient gives the net rate that particles transport momentum into the cell, per cell volume-- even degeneracy pressure works just like that. The only place where the quantum mechanics there becomes important is if you are thinking thermodynamically, and so wish to associate a temperature to the situation. Complete degeneracy forces you to associate a zero temperature with the gas, but the pressure in the gas still just comes from the momentum flux, and that still ends up coming from the kinetic energy density in the usual way, which is called "kinetic pressure." As I mentioned above, this is no kind of fundamental force, it's just an effective force that only appears in the fluid description of the gas. Individual particles don't feel any such force, even in complete degeneracy.
 
  • Like
Likes leonmate

Related Threads on Differential equation for motion within stars

  • Last Post
Replies
15
Views
5K
Replies
9
Views
13K
Replies
5
Views
3K
Replies
1
Views
3K
Replies
3
Views
3K
Replies
7
Views
1K
Replies
4
Views
8K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
9
Views
2K
  • Last Post
Replies
2
Views
4K
Top