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?


3 Answers 3


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.

  • $\begingroup$ The pressure inside the balloon will be equal to atmospheric pressure, or else the balloon will be expanding (or contracting). $\endgroup$ Commented May 17 at 0:45
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    $\begingroup$ @ElementsInSpace That would only be true if the balloon had no elastic force pressing inward. $\endgroup$ Commented May 17 at 2:57
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    $\begingroup$ This is a good answer, but the balloon analogy only works for balloons with very little air in them, as balloons generally decrease in pressure with increasing volume. See This answer, or this video showing two connected balloons, with one losing its air to the other. $\endgroup$
    – quant
    Commented May 17 at 23:45

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.

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    $\begingroup$ Pressure is the integral of local gravity * local density out to vacuum. So, at each distance the gravity only depends on whats inside the shell, but the pressure definitely depends on whats outside the shell $\endgroup$ Commented May 17 at 13:30
  • $\begingroup$ @QuadmasterXLII I think you are right and it maybe that the back-of-the-envelope approach really isn't quite sufficient here. $\endgroup$
    – ProfRob
    Commented May 17 at 15:18
  • $\begingroup$ I second @QuadmasterXLII. Sure, with gravity yes -- but for pressure it's both: It depends on what's outside the shell and the gravity at that point (i.e., what's inside). The dependency on the outer shell can be seen with the two extremes: On the surface, pressure is zero, at the center it is maximal. The gradient does depend on what's inside, as your formula states -- because that's what will be outside and increase pressure accordingly when you move past it inward ;-). The gradient's limit (but not the pressure's ;-) ) approaching the center is zero because gravity becomes zero... $\endgroup$ Commented May 18 at 13:27
  • $\begingroup$ Addressed. The pressure in the core is essentially determined by the properties of the core and not what lies outside, because the pressure outside the core is much lower than inside. Therefore one can still use $P_c/R_c$ to approximate the pressure gradient. $\endgroup$
    – ProfRob
    Commented May 18 at 15:32

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. enter image description here

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. enter image description here


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