Help Deriving Hydrostatic Equilibrium


by jumi
Tags: deriving, equilibrium, hydrostatic
jumi
jumi is offline
#1
Apr2-12, 11:24 PM
P: 28
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]."

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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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NascentOxygen
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#2
Apr3-12, 09:18 AM
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Quote Quote by jumi View Post
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
jumi
jumi is offline
#3
Apr3-12, 12:57 PM
P: 28
By where does the (dP/dr)*dr term come from?



Why wouldn't it just be P(r) = P(r+dr)?

NascentOxygen
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#4
Apr6-12, 07:38 PM
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P: 4,710

Help Deriving Hydrostatic Equilibrium


Quote Quote by jumi View Post
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.


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