Help Deriving Hydrostatic Equilibrium

In summary, the poster is seeking help with understanding hydrostatic equilibrium and the derivation of the equation of motion for a volume element. They are specifically confused about the last term in the equation, which comes from the linear approximation of a complex curve, and the significance of (dP/dr)*dr in relation to pressure and radial distance.
  • #1
jumi
28
0
Hey all,

I posted this in the Astrophysics sub-forum, but traffic here seems higher (if a mod sees this, can you close the other thread please?)

Long story short, for my Modern Physics course, we have to do a research paper on a physics topic we didn't cover in class. Since I've always been interested in astronomy and the cosmos, I figured I'd do star formation / life cycle of stars. The paper has to have mathematical and physical reasoning for everything we present.

Anyway, I found some cool books that have helped me out thus far, but I'm having trouble following an explanation for hydrostatic equilibrium.

The book in question "Evolution of Stars and Stellar Populations" by Maurizio Salaris and Santi Cassisi.

They start the derivation by "finding the equation of motion of a generic infinitesimal cylindrical volume element with axis along the radial direction, located between radii r and r+dr", with a base (perpendicular to the radial direction) of area dA and density ρ.

They then obtain the mass, dm, contained in the element: dm = ρdrdA.

"[They] neglect rotation and consider self-gravity and internal pressure as the only forces in action. The mass enclosed within the radius r acts as a gravitational mass located at the center of the star; this generates an inward gravitational acceleration: g(r) = G[itex]m_{r}[/itex] / [itex]r^{2}[/itex]."

"Due to spherical symmetry, the pressure forces acting on both sides perpendicular to the radial direction are balanced, and only the pressure acting along the radial direction has to be determined. The force acting on the top of the cylinder is P(r+dr)dA, whereas the force acting on the base of the element is P(r)dA. By writing: P(r+dr) = P(r) + [itex]\frac{dP}{dr}[/itex]dr and remembering that drdA = dm/ρ, the equation of motion for the volume element can be written as: [itex]\frac{d^{2}r}{dt^{2}}[/itex]dm = -g(r)dm - [itex]\frac{dP}{dr}\frac{dm}{ρ}[/itex]."

---

They go further to get the final equation, but I get lost at this last paragraph. I don't understand where the last term of this equation comes from: P(r+dr) = P(r) + [itex]\frac{dP}{dr}[/itex]dr. Why wouldn't it just be P(r+dr) = P(r)?

And then once they get that equation, how do they get to this equation: [itex]\frac{d^{2}r}{dt^{2}}[/itex]dm = -g(r)dm - [itex]\frac{dP}{dr}\frac{dm}{ρ}[/itex]? I know they skipped a bunch of steps... Maybe I'm just not thinking right.

Any help would be appreciated.

Thanks in advance.
 
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  • #2
jumi said:
I get lost at this last paragraph. I don't understand where the last term of this equation comes from: P(r+dr) = P(r) + [itex]\frac{dP}{dr}[/itex]dr.
That looks like the standard linear approximation to a complex curve. For small excursions about x, we can linearize the complex curve using the gradient at x, viz., f(x+∆x) ≈ f(x) + (df(x)/dx) * ∆x
 
  • #3
By where does the (dP/dr)*dr term come from?

http://img69.imageshack.us/img69/884/physforums.png [Broken]

Why wouldn't it just be P(r) = P(r+dr)?
 
Last edited by a moderator:
  • #4
jumi said:
By where does the (dP/dr)*dr term come from?
That relates the change of pressure with radial distance.
Why wouldn't it just be P(r) = P(r+dr)?
That would be saying there is no change of pressure with distance, and the mass enclosed within the volume dA·dr must be zero.
 
  • #5


Hi there,

Hydrostatic equilibrium is an important concept in astrophysics, particularly in understanding the structure and evolution of stars. It refers to the balance between the inward force of gravity and the outward force of pressure, which keeps a star from collapsing or expanding.

In order to derive the equation for hydrostatic equilibrium, the authors of the book you mentioned are using the equation of motion for a cylindrical volume element. This element has a base with area dA and a height of dr, and they are considering it to be located between radii r and r+dr. The mass of this element is given by dm = ρdrdA, where ρ is the density.

The first term in the final equation, \frac{d^{2}r}{dt^{2}}dm, represents the acceleration of the element due to the forces acting on it. The second term, -g(r)dm, represents the gravitational force pulling the element inward. The third term, - \frac{dP}{dr}\frac{dm}{ρ}, represents the pressure force acting on the element.

Now, to answer your questions about the derivation:

1. Why wouldn't P(r+dr) just be equal to P(r)?

The reason P(r+dr) is not equal to P(r) is because the pressure at a certain radius is not constant throughout the volume element. The pressure at the top of the element, P(r+dr), will be slightly higher than the pressure at the base, P(r), due to the weight of the material above it. This is why the authors have added the term \frac{dP}{dr}dr to account for this change in pressure.

2. How do they get to the final equation?

To get to the final equation, they are essentially applying Newton's second law of motion (F=ma) to the cylindrical volume element. They are considering the forces acting on the element and setting them equal to the mass of the element multiplied by its acceleration. The gravitational force is given by -g(r)dm, and the pressure force is given by \frac{dP}{dr}\frac{dm}{ρ}. By rearranging the terms, they arrive at the final equation.

I hope this helps to clarify the derivation for you. If you have any further questions, feel free to ask. Good luck with your research paper!
 

1. What is hydrostatic equilibrium?

Hydrostatic equilibrium is a state in which the pressure exerted by a fluid at any given point is equal in all directions. This means that there is no net force acting on the fluid, and it is in a state of balance.

2. What is the importance of hydrostatic equilibrium?

Hydrostatic equilibrium is important in many scientific fields, including astronomy, meteorology, and fluid mechanics. It helps us understand the forces that act on fluids and how they affect the behavior of objects in the fluid.

3. How is hydrostatic equilibrium derived?

Hydrostatic equilibrium is derived from the fundamental principles of fluid mechanics, namely the equations of motion and the equation of continuity. These equations are solved to determine the pressure distribution within the fluid.

4. What are the key assumptions made when deriving hydrostatic equilibrium?

The key assumptions made when deriving hydrostatic equilibrium include that the fluid is incompressible, that there are no external forces acting on the fluid, and that the fluid is at rest or moving at a constant velocity.

5. How is hydrostatic equilibrium related to atmospheric pressure?

In the Earth's atmosphere, hydrostatic equilibrium is responsible for the decrease in pressure with increasing altitude. This is because as you go higher in the atmosphere, there is less air above you, resulting in a decrease in the weight of the air pushing down on you and therefore a decrease in atmospheric pressure.

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