<p>If you have a gas of non-degenerate ions and degenerate electrons, then by definition you have an upper limit to the temperature, since if it were higher the electrons would not be degenerate.</p>

<p>Below this temperature, the electron degeneracy pressure will always greatly exceed any perfect gas pressure from the ions. Since degeneracy pressure does not depend on temperature, then the central temperature could be <em>anything</em> lower than the temperature below which the electrons can be assumed degenerate. For example, the mass, radius and central density of white dwarfs of a given mass will be very similar, although their central temperatures will change by more than an order of magnitude as they cool over billions of years.</p>

<p>If you wish to make the small corrections, either for partial degeneracy (the pressure is then slightly temperature dependent) or to include the ion pressure, then the only way to proceed is add the pressures and then solve the equations of stellar structure numerically.</p>

CC BY-SA 3.0
2018-02-17T08:11:34.953Z
2018-02-17T08:11:34.953Z
ProfRob
<p>If we have a stellar density model $\rho(r) = \rho_c(1-r/R)$, where the star is composed of ions behaving as perfect gas and electron with non-relativistic degeneracy. The central pressure is due to both gas and degeneracy pressure. My final goal is to find central temperature of this model.</p>

<p>Now for the linear density model, I calculated the central pressure only for gas to be $P_c = \frac{5GM^2}{4 \pi R^4}$. And central temperature for only gas is $T_c = \frac{5G \mu M_{H} M}{12KR}$</p>

<p>I know the Pressure for non-relativistic degeneracy $P_{nrd} = C \rho^{5/3}$. I was wondering to find central pressure for degeneracy, may I use the Lane-Emden equations for polytrops and simply find central pressure and temperature from there? Then add the two temperatures to find the maximum central temperature for a gas+degenerate star.</p>

<p>For Lane-Emden model, they also consider density dependence on radial distance, but they express the density function as $\rho = \rho_c \theta^n$. Is it equivalent to $\rho(r) = \rho_c(1-r/R)$ for gas model?</p>

<stellar-astrophysics><degenerate-matter>
Central temperature of a stellar model with combination of gas and degeneracy
bhjghjh
# Central temperature of a stellar model with combination of gas and degeneracy

(Asked by: username_0 on 2018-02-17)

If we have a stellar density model $\rho(r) = \rho\_c(1-r/R)$, where the star is composed of ions behaving as perfect gas and electron with non-relativistic degeneracy. The central pressure is due to both gas and degeneracy pressure. My final goal is to find central temperature of this model.

Now for the linear density model, I calculated the central pressure only for gas to be $P\_c = \frac{5GM^2}{4 \pi R^4}$. And central temperature for only gas is $T\_c = \frac{5G \mu M\_{H} M}{12KR}$

I know the Pressure for non-relativistic degeneracy $P\_{nrd} = C \rho^{5/3}$. I was wondering to find central pressure for degeneracy, may I use the Lane-Emden equations for polytrops and simply find central pressure and temperature from there? Then add the two temperatures to find the maximum central temperature for a gas+degenerate star.

For Lane-Emden model, they also consider density dependence on radial distance, but they express the density function as $\rho = \rho\_c \theta^n$. Is it equivalent to $\rho(r) = \rho\_c(1-r/R)$ for gas model?

**Tags:** stellar-astrophysics, degenerate-matter


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## 1 answers

### Answer by username_1 on 2018-02-17. Score: 2 | Accepted answer

If you have a gas of non-degenerate ions and degenerate electrons, then by definition you have an upper limit to the temperature, since if it were higher the electrons would not be degenerate.

Below this temperature, the electron degeneracy pressure will always greatly exceed any perfect gas pressure from the ions. Since degeneracy pressure does not depend on temperature, then the central temperature could be *anything* lower than the temperature below which the electrons can be assumed degenerate. For example, the mass, radius and central density of white dwarfs of a given mass will be very similar, although their central temperatures will change by more than an order of magnitude as they cool over billions of years.

If you wish to make the small corrections, either for partial degeneracy (the pressure is then slightly temperature dependent) or to include the ion pressure, then the only way to proceed is add the pressures and then solve the equations of stellar structure numerically.


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