Pressure support and gravity

Support depends on a pressure difference

Pressure is force per area, measured in pascals, where 1 Pa = 1 N m⁻². In a fluid with isotropic pressure, it acts normally on each face of a small element. Pressure is not intrinsically a one-way outward push.

If opposite faces experience the same pressure and have the same area, their pressure forces cancel. If the inward face of a radial element experiences greater pressure than the outward face, the difference supplies a net outward force. That difference per distance is a pressure gradient.

Gravity acts on the element’s mass. Hydrostatic balance describes a state in which the pressure-gradient force balances gravity, so the element has no net acceleration in the simplified model. The fluid may have a large pressure even where the local pressure gradient is small.

Much stellar material is plasma: matter containing ions and free electrons, generally close to electrically neutral on the large scales considered here. Its density is mass per volume, in kilograms per cubic metre. Pressure support and energy production are separate questions about that material.

Balance the forces on one declared small cell

Consider an original locally flat fluid cell with face area A = 2 m², radial thickness h = 10 m and density ρ = 100 kg m⁻³. Stipulate approximately constant inward gravitational acceleration g = 200 m s⁻² across the cell. This is a teaching calculation, not a measured solar layer.

Its mass is ρAh = 2000 kg, so its inward weight is ρAhg = 400000 N. Choose inner-face pressure P_in = 1000000 Pa and outer-face pressure P_out = 800000 Pa. The pressure difference is 200000 Pa.

The outward net pressure force is (P_in − P_out)A = 400000 N. It balances the stipulated inward weight. Each face force is larger than the net pressure force; the cancellation between them is essential to the calculation.

Dividing the balance by A gives P_in − P_out = ρgh for this locally constant-density, constant-gravity approximation. A complete star is spherical and stratified, so its pressure changes must be followed through the varying enclosed mass, density and radius.

Only the appropriate enclosed mass enters spherical gravity

For a spherical Newtonian model, let r be distance from the centre, m(r) the mass enclosed inside that radius, ρ(r) the local density and P(r) the local pressure. Mass conservation gives dm/dr = 4πr²ρ. Hydrostatic balance gives dP/dr = −Gm(r)ρ/r².

Here G is the gravitational constant. The minus sign says pressure decreases outward when density and enclosed mass are positive. Replacing m(r) by the entire stellar mass at every interior radius would generally give the wrong gravitational field.

In a regular finite-density centre, enclosed mass approaches zero proportionally to r³. The local gravitational acceleration Gm(r)/r² therefore tends to zero proportionally to r. The apparent division by r² is not a physical central infinity in that regular model.

The spherical equations omit rapid acceleration, rotational support, magnetic stresses and relativistic effects. They are a controlled description for suitable quasistatic conditions, not a complete equation for every stellar collapse, distorted binary or compact remnant.

Central pressure can be high while central gravity is zero

To see the distinction clearly, construct an original uniform-density sphere of total mass M and radius R, with prescribed surface pressure P_s. Its density is 3M/(4πR³). This is a mechanical toy, with no stellar atmosphere, equation of state or energy-transport solution fitted to the Sun.

Define x = r/R, between 0 and 1, and surface gravity g_s = GM/R². Then m(r)/M = x³ and g(r)/g_s = x. Integrating the pressure equation inward from the supplied boundary gives P(r) − P_s = [3GM²/(8πR⁴)](1 − x²).

The central pressure excess is P_c − P_s = 3GM²/(8πR⁴), so its normalized profile is (P − P_s)/(P_c − P_s) = 1 − x². The complete table gives five radii on the same x scale. The three plotted quantities have separate dimensionless definitions rather than the same physical units.

At x = 0, the local gravitational acceleration and pressure gradient vanish, while pressure reaches its maximum. At x = 1, the enclosed mass is M and pressure equals its prescribed surface value. These endpoint properties are consequences of this stated model, not a measured solar profile.

An equation of state and an energy budget close different parts of the problem

An equation of state relates quantities such as pressure, density, temperature and composition. For a sufficiently dilute, nondegenerate classical gas at one common temperature, particle contributions can be described by P_gas = nk_BT, where n is the number of pressure-contributing particles per volume and k_B is Boltzmann’s constant.

In an ionized mixture, both ions and free electrons contribute to that particle count under the stated gas approximation. Thermal radiation can contribute pressure too. Dense, degenerate or strongly interacting material requires an appropriate different treatment; the ideal-gas formula is not universal.

A mechanically supported star still radiates energy. Maintaining its thermal state requires an energy budget describing nuclear production, losses, transport and possible changes of internal and gravitational energy. Fusion does not add a separate direct mechanical thrust to the hydrostatic equation.

Quasistatic means the mechanical structure can adjust through a sequence of near-equilibrium states on a sufficiently slow evolutionary time scale. It does not require every quantity to remain exactly constant forever. The next lessons examine the energy source and transport that accompany this pressure–gravity balance.

Worked examples

Try the reasoning

1. If equal pressure acts on both equal-area radial faces of the small cell, what is its net radial pressure force?

   Show the answer and reasoning

   Zero.

   The opposite pressure forces cancel even if each pressure is large.

2. Does zero gravity at the regular centre require zero central pressure in the stated sphere?

   Show the answer and reasoning

   No.

   The pressure gradient vanishes at the centre, while pressure itself can be maximal.

3. Within the uniform-density family, double M while keeping R and P_s fixed. How does P_c − P_s change?

   Show the answer and reasoning

   It increases by a factor of 4.

   The derived central pressure excess is proportional to M²/R⁴. This compares two stipulated mechanical models, not a full stellar-evolution prediction.

4. Should a separate “fusion thrust” be added to the pressure–gravity force balance?

   Show the answer and reasoning

   No.

   Fusion contributes to the energy budget. Pressure and its gradient provide the mechanical fluid force within the hydrostatic description.

What to carry forward

• Pressure support comes from a gradient, not the magnitude of pressure alone.
• Spherical interior gravity depends on enclosed mass, while the pressure profile also depends on boundary conditions.
• Mechanical equilibrium, an equation of state and thermal energy balance are related but distinct parts of a stellar model.