<p>Because the total mass of the star has to be conserved.</p>

<p>If you consider a linear density model, for a star of fixed mass and radius, 
you have to make the central density higher than the average density in a constant density model.</p>

<p>Furthermore, because the mass of each shell goes as $\rho r^2$, there needs to more shells with a. density higher than average. This means the average density of a mass shell is higher than in the constant density model and more of the mass is at smaller radii.</p>

<p>Hydrostatic equilibrium demands that the pressure gradient is proportional to $-\rho M(r)/r^2$, where $M(r)$ is the mass inside radius $r$. At any radius, this will be larger on average in the linear density model because more of the mass is at smaller radii.</p>

<p>If the pressure gradient is higher, then the central pressure will be higher. The value of the central temperature then just comes from gas laws.</p>

CC BY-SA 3.0
2018-02-15T07:48:04.480Z
2018-02-15T07:48:04.480Z
ProfRob
<p>I just derived the pressure and temperature profile for a linear density model of a star. If I compare the profiles of pressure and temperature for a constant density model of stars, I see that the linear density model central pressure is higher, but the central temperature is lower.</p>

<p>What is the physical reason behind this?</p>

<star>
Central temperature and pressure for constant density vs linear density model of stars
bhjghjh
# Central temperature and pressure for constant density vs linear density model of stars

(Asked by: username_0 on 2018-02-15)

I just derived the pressure and temperature profile for a linear density model of a star. If I compare the profiles of pressure and temperature for a constant density model of stars, I see that the linear density model central pressure is higher, but the central temperature is lower.

What is the physical reason behind this?

**Tags:** star


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## 1 answers

### Answer by username_1 on 2018-02-15. Score: 1 | Accepted answer

Because the total mass of the star has to be conserved.

If you consider a linear density model, for a star of fixed mass and radius,
you have to make the central density higher than the average density in a constant density model.

Furthermore, because the mass of each shell goes as $\rho r^2$, there needs to more shells with a. density higher than average. This means the average density of a mass shell is higher than in the constant density model and more of the mass is at smaller radii.

Hydrostatic equilibrium demands that the pressure gradient is proportional to $-\rho M(r)/r^2$, where $M(r)$ is the mass inside radius $r$. At any radius, this will be larger on average in the linear density model because more of the mass is at smaller radii.

If the pressure gradient is higher, then the central pressure will be higher. The value of the central temperature then just comes from gas laws.


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