Effective weight of a star's layers

[Nabeshin]

I'm attempting to derive a function which relates the radius from the center of a star to the outward pressure which must be exerted to keep the star in equilibrium. So, I figure the pressure should be equal and opposite to the gravitational weight. My analysis is:

For some shell at a given radius away from the center of the star, we can write its mass as:
$$dm=4\pi r^2 \rho (r) dr$$
Since the entire shell is at the same radius from the center, we can essentially treat it as a sheet and write the equation for the differential force as:

$$dF=\frac{G M_{enc} dm}{r^2}$$

Where,
$$M_{enc}=\int_{0}^{r}dm=\int_{0}^{r}4\pi r^2 \rho (r) dr$$

If we take,
$$\rho (r)=\frac{\rho_0}{r}$$

Then we can rewrite this as:
$$dF=\frac{G (\int_{0}^{r}4\pi r \rho_0 dr) 4\pi r \rho_0 dr}{r^2}$$

$$dF=\frac{16 \pi ^{2} \rho_0^2(r^2/2)r dr}{r^2}$$
$$dF=8 \pi^2 \rho_0^2 r dr$$

If we call the radius to some shell in question r_s, and integrate from r_s to some larger radius, r_0,
$$F=\int_{r_s}^{r_{\odot}}8 \pi^2 \rho_0^2 r dr$$
$$F=4\pi^2\rho_0^2(r_{\odot}^2-r_s^2)$$

Which seemed all fine and good until I realized this gives a weight of zero for the outermost shell. My thoughts are that maybe there is some way to split the weight into the sum of the layers above the layer in question (effectively what I've found) and the weight of the layer itself, but when I write the layer as an infinitesimal it has a weight zero.

Anyone have any thoughts, or if I'm on the right track?

 

[Chronos]

That is the expected result and is technically correct. The point where radiation pressure equals gravitational attraction lies beyond the visible surface of a star.

 

[Widdekind]

Please ponder two Main Sequence stars, one more massive, and larger, than the other.

Now, please consider a single given radius (R) lying inside both said stars (R < R_small < R_big).

Is it not the case, that, at that given radius, the Enclosed Mass M_enc(R) is greater, for the larger & more massive star (M_enc,big(R) > M_enc,small(R)) ? Wouldn't that have to be the case, for the core of the bigger, brighter star to keep from exploding, from its much more prodigeous fusion rate (L_* ~ M_*^3.5-4), and resulting higher Temperature / Pressure ?

Thus, although bigger brighter O/B stars are, overall, on average, of lower Bulk Density... their cores are much denser than smaller dimmer M/K stars of the same radius (ie, at R_core,O/B = R_surface,M/K) ?

 

[Nabeshin]

Please ponder two Main Sequence stars, one more massive, and larger, than the other.

Now, please consider a single given radius (R) lying inside both said stars (R < R_small < R_big).

Is it not the case, that, at that given radius, the Enclosed Mass M_enc(R) is greater, for the larger & more massive star (M_enc,big(R) > M_enc,small(R)) ? Wouldn't that have to be the case, for the core of the bigger, brighter star to keep from exploding, from its much more prodigeous fusion rate (L_* ~ M_*^3.5-4), and resulting higher Temperature / Pressure ?

Thus, although bigger brighter O/B stars are, overall, on average, of lower Bulk Density... their cores are much denser than smaller dimmer M/K stars of the same radius (ie, at R_core,O/B = R_surface,M/K) ?

Right, but this is deviating from my assumption of the density function. In a more realistic case, density depends on other factors such as the mass of the star, rather than simply the distance from the center.

Edit: Instead of working with an ambiguous constant, we can write it in terms of observable parameters of a star, namely the radius and and mass:
$$M_{\odot}=\int_{0}^{r_{\odot}}\rho (r) dV = 4 \pi \rho_0 \int_{0}^{r_{\odot}} r dr = 2 \pi \rho_0 r_{\odot}^2$$
$$\rho_0=\frac{M_{\odot}}{2 \pi r_{\odot}^2}$$

This would seem to suggest, because of the squared radius factor in the denominator, that larger stars will in general be much less dense than smaller ones.

The disagreement on core density, like I said, probably arises from the simplicity of the density modeling.

 

[Widdekind]

Very roughly speaking, M_* ~ R_*^3/2. However, the exponent is closer to 1.3 for M thru A-Class stars. Then, the exponent is much higher for big, bright, O/B-Class stars. I understand, that this, too, is where the Main Sequence on the Hertzsprung-Russell Diagram begins to curve upwards. Apparently, O/B-Class stars are a "class unto themselves", being "overbright", for the same Temperature, b/c they puff up so much, from their prodigeous rates of fusion.

 

[qraal]

Try chasing up a few references on using the theory of polytropes to model stars - they're somewhat more accurate than your initial equations, but less intensive than full stellar modelling computer codes. Exercised my braincells trying to follow the derivation of the Lane-Embden equation and its various solutions, but is definitely worthwhile for more of an insight into how a star's interior behaves.

 

[Orion1]

Tolman-Oppenheimer-Volkoff equation...


If you are modeling stellar interiors, then you must base your model on General Relativity, specifically the Tolman-Oppenheimer-Volkoff (TOV) equation for your Equation of State.

In astrophysics, the Tolman-Oppenheimer-Volkoff (TOV) equation constrains the structure of a spherically symmetric body of isotropic material which is in static gravitational equilibrium, as modeled by General Relativity, the equation:

$$\frac{dP(r)}{dr} = - \frac{G}{r^2} \left[ \rho(r) + \frac{P(r)}{c^2} \right]\left[ M(r) + 4 \pi r^3 \frac{P(r)}{c^2} \right] \left[1 - \frac{2GM(r)}{c^2 r} \right]^{-1}$$

Here, r is a radial coordinate, and ρ(r0) and P(r0) are the density and pressure, respectively, of the material at r=r0. M(r0) is the total mass inside radius r=r0, as measured by the gravitational field felt by a distant observer and satisfies boundary condition M(0) = 0:
$$\frac{dm(r)}{dr} = 4 \pi \rho(r) r^2$$

Model density function:
$$\boxed{\rho(r) = \frac{\rho_0}{r}}$$

Mass function integration:
$$M(r) = 4 \pi \int_r^R \rho(r) r^2 dr = 4 \pi \rho_0 \int_r^R r dr = 2 \pi \rho_0 \left(R^2 - r^2)$$

Model mass equation solution:
$$\boxed{M(r) = 2 \pi \rho_0 \left(R^2 - r^2 \right)}$$

Total mass function integration:
$$M(r) = 4 \pi \int_0^R \rho(r) r^2 dr = 4 \pi \rho_0 \int_0^R r dr = 2 \pi \rho_0 R^2$$

Model total mass:
$$\boxed{M_0 = 2 \pi \rho_0 R^2}$$

Polytropic index:
$$\gamma = \frac{n + 1}{n}$$
$$n = 3$$ - main sequence star.
$$\boxed{\gamma = \frac{4}{3}}$$

Main sequence stars similar to the sun are usually modeled by a polytrope with index n = 3, corresponding to the Eddington standard model of stellar structure.

Polytropic constant:
$$\boxed{K = \left({ \frac{N_A k_B T_0}{\bar{\mu}}}\right) \rho_0^{-1/n}}$$

Polytropic pressure solution to the Lane-Emden equation:
$$\boxed{P(r) = K \rho(r)^{\gamma}}$$

Integration by substitution:
$$P(r) = K \rho(r)^{\gamma} = K \left( \frac{\rho_0}{r} \right)^{\gamma}$$

Model polytropic pressure equation solution:
$$\boxed{P(r) = K \left( \frac{\rho_0}{r} \right)^{\gamma}}$$

Model polytropic TOV Equation of State:
$$\boxed{\frac{dP(r)}{dr} = - \frac{G}{r^2} \left[ \frac{\rho_0}{r} + \frac{K}{c^2} \left( \frac{\rho_0}{r} \right)^{\gamma} \right]\left[ 2 \pi \rho_0 \left(R^2 - r^2 \right) + \frac{4 \pi K \rho_0^{\gamma} r^{3 - \gamma}}{c^2} \right] \left[1 - \frac{4 \pi G \rho_0}{c^2} \left( \frac{R^2 - r^2}{r} \right) \right]^{-1}}$$

Reference:
http://en.wikipedia.org/wiki/Tolman-Oppenheimer-Volkoff_equation"
http://en.wikipedia.org/wiki/Polytrope"
http://en.wikipedia.org/wiki/Standard_Solar_Model"

 

[protonchain]

Just out of curiosity, isn't this Hydrostatic Equilibrium? Aka, the balancing act between gravity and electron degeneracy pressure in stars.

 

[Orion1]

Polytropic index...

Just out of curiosity, isn't this Hydrostatic Equilibrium? Aka, the balancing act between gravity and electron degeneracy pressure in stars.

It is the General Relativity Equation of State for stellar Hydrostatic Equilibrium, however for white dwarf degenerate matter and electron degeneracy pressure, the polytropic index is n = 3/2.

Neutron stars are well modeled by polytropes with index about in the range between n = 1/2 and n = 1.

At relatively low densities, a polytrope with index n = 3/2 is a good model for degenerate star cores like those of red giants, for white dwarfs, brown dwarfs, giant gaseous planets (like Jupiter), or even for rocky planets.

At very high densities, where most of the particles are forced into quantum states with relativistic energies, the polytropic index n = 3.

Main sequence stars similar to the sun are usually modeled by a polytrope with index n = 3, corresponding to the Eddington standard model of stellar structure.

A polytrope with index n = 5 has an infinite radius. It corresponds to the simplest plausible model of a self-consistent stellar system, first studied by A. Schuster in 1883.

Polytropic index:
n = 1/2 to 1 - neutron stars
n = 3/2 - red giants, white dwarfs, brown dwarfs, giant gaseous planets (like Jupiter), rocky planets.
n = 3 - main sequence stars - Eddington standard model of stellar structure.
n = 5 - stellar system

Note that the higher the polytropic index, the density distribution is more condensed at the center.

Reference:
http://en.wikipedia.org/wiki/Polytrope"
http://en.wikipedia.org/wiki/Degenerate_matter#Concept"

 

[Vanadium 50]

If you are modeling stellar interiors, then you must base your model on General Relativity, specifically the Tolman-Oppenheimer-Volkoff (TOV) equation for your Equation of State.

I disagree. Unless you are talking about compact objects like white dwarfs, the GR effects are small.

Main sequence stars similar to the sun are usually modeled by a polytrope with index n = 3, corresponding to the Eddington standard model of stellar structure.

Polytropes? You're worried about GR effects and you're modeling with polytopes? Why are you willing to use a big approximation if you don't want to use a small one?

 

[Helios]

Instead of the Eddington ( n=3 ) model of the Sun ( for example ), a much better model is a composite polytrope with a n=3 radiative core connected to a n=3/2 convective zone shell. The radius of the radiative core is about 5/7 R. Note that the Sun is mostly convective by volume.

 

[Orion1]

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Resigned from thread.

 

[Vanadium 50]

You lectured us that we needed GR. I challenge that.

Helios made an excellent suggestion.