When a star enters a red giant phase, does the pressure in its core increase or decrease?

Question

When a star becomes a red giant, does the pressure in its core increase or decrease?

From one side, the only source of pressure in a star is gravity. A red giant is much larger than an ordinary star, thus the same mass spread inside a larger volume must produce a lower interior pressure.

But from the other side, pressure is needed for fusion. Since a red giant is much more luminous than an ordinary star, a higher core pressure would be needed to cause more intense fusion (and to fuse other elements apart from H).

It seems like a contradiction - so what is true?

Answer 1

If you heat up a balloon, the balloon will expand. Is the balloon's internal pressure higher? Yes. Is the balloon less dense? Also yes. Is this a contradiction?

I think what you aren't seeing is that a star is in equilibrium: The inward pressure of gravity is exactly balanced by the outward pressure of the star's material, which is primarily due to thermal energy.

When a star starts to run low on fuel, it undergoes a lot of complicated evolution between the core and a series of "shells" that are fusing different elements. The higher heat output from that shell fusion increases the outward thermal pressure in the star, which makes the upper layers spread out and become less dense to return to equilibrium.

So the core is more dense, but the outer layers are much less dense, so the star overall is less dense despite the core being under more pressure.

Answer 2

Newton's shell theorem tells you that what is outside the core has no gravitational influence on what is at the core (as long as the envelope is spherically symmetric).

The equation of hydrostatic equilibrium tells us that the pressure gradient must balance the inward gravity $$ \frac{dP}{dr} = - \rho g\ $$ where $P$ is the pressure, $\rho$ is the density and $g$ is the gravitational acceleration, which is only governed by what mass is interior to radius $r$ and thus $g = GM(r)/r^2$, where $M(r) = 4\pi r^3 \rho/3$.

The next part is the crucial bit.

If the structure of the star is that of a core and an envelope, and we define the core to be say that region over which the presssure drops by an order of magnitude from its central value, then we can approximate the hydrostatic equilibrium equation for the core as $$\frac{P_c}{R_c} \simeq \frac{3GM^2(r)}{4\pi R_c^5}\, , $$ $$ P_c \simeq \frac{3GM^2(r)}{4\pi R_c^4}\ , $$ where $P_c$ is the central pressure and $R_c$ is the radius of the core.

In a first-ascent red giant star, the central core is a compact ball of partially degenerate helium, with a mass of $\sim 0.5 M_\odot$ and a core radius of just $R_c \simeq 10^4$ km. Putting these values into the equation above we get $P_c \simeq 1.6\times 10^{21}$ Nm$^{-2}$.

In contrast, the core of the Sun has a density of $1.6\times 10^{5}$ kg/m$^3$ and a temperature of $1.5\times 10^7$ K. Using the perfect gas law with a mean particle mass of $0.5 \times 1.67 \times 10^{-27}$ kg (ionised hydrogen), gives $P_c \simeq 4 \times 10^{16}$ Nm$^{-2}$.

So you can see that the core pressure of a giant star is way bigger than that of an equivalently massed main sequence star. The main sequence star obeys the same equation of hydrostatic equilibrium, but the difference is that for something like the Sun, the core is much less compact and much less dense so that $M_c^2/R_c^4$ is several orders of magnitude lower.

Answer 3

A red giant is really, really different than the sun. A red giant is basically two stars, one inside of the other- and the inner one is tiny and has a substantial fraction of the mass.

Here is the density profile of the sun. You can see that the core is 1000 times denser than the surface. Note that the radius axis is linear.

Here is the density profile of a Red Giant. The core is 1000,000,000,000 times denser than the surface. Note that the radius axis is logarithmic.