Specify the gas and its motion
Take a spherical acoustic cavity of radius , measured in meters. Its equilibrium density is constant, in kilograms per cubic meter, and its sound speed is constant, in meters per second. We omit gravity and buoyancy. This is a useful idealization for learning about waves and boundaries; a real star’s equilibrium requires additional equations.
The radial coordinate increases outward from the center. The pressure perturbation is the small changing part of the pressure, in pascals. It is not the much larger equilibrium pressure. The radial gas velocity is positive outward. A radial mode has the same pressure at every point on a sphere of a given radius.
Two physical statements govern the motion. A pressure gradient accelerates the gas. Expansion of a small volume reduces its pressure. Keeping only terms linear in the small disturbance gives the equations below. The factor involving the square of the radius accounts for the changing area of spherical shells.
Separate the shape from the rhythm
Differentiate the pressure equation once more in time and substitute the acceleration equation. Because density and sound speed are constant, they can be taken outside radial derivatives. The result is the spherical wave equation.
Now ask for one repeating pattern. Write pressure as a radial shape multiplied by . The angular frequency is measured in radians per second. One full cycle contains radians, so the ordinary frequency is in hertz. Differentiating the cosine twice contributes a factor .
The separation is what makes this a mode calculation. All radii share one frequency, while the radial function says which layers compress together and where the pressure stays zero.
The center removes one possible solution
The apparent singularity at the center comes from spherical coordinates. The gas pressure itself must remain finite. Define . Substituting it into the radial equation cancels the first-derivative term, leaving the same sinusoidal equation used for a one-dimensional string.
Its general solution contains both sine and cosine. Dividing a cosine term by radius would make pressure diverge at the center, so that term must vanish. The remaining sine divided by radius is finite: near zero, sine is proportional to its argument. It is convenient to normalize the pressure shape by its finite central value .
This is also why a drawing should not place a pressure node at the center of the fundamental mode. The transformed quantity is zero there, but pressure need not be. Changing variables simplified the equation; it did not turn the central gas into a fixed endpoint.
The outer boundary selects the frequencies
Choose a pressure-release outer boundary: the pressure perturbation is zero at the fixed equilibrium radius. Applying that condition to the regular solution requires the sine to vanish at the boundary. Its positive roots give the allowed modes.
The integer is the radial order in this cavity, starting at one. The fundamental mode has no internal pressure node. The second mode has one, and the third has two. An internal node is a radius where this mode’s pressure perturbation vanishes at all phases. Neighboring regions on its two sides compress with opposite signs.
There is no nontrivial zero-frequency member of this ladder: a constant nonzero pressure would fail the pressure-release condition. At nonzero frequency, the boundary and the center jointly select a discrete spectrum.
A stellar free surface needs a different distinction. GYRE states its simplest pressure condition for a moving gas element: the Lagrangian perturbation vanishes. That includes the equilibrium pressure gradient sampled by the displacement. Our gravity-free cavity instead prescribes zero Eulerian pressure perturbation at a fixed radius. Its evenly spaced frequencies should not be carried unchanged into a stratified star with an atmosphere.
Reading: MESA School 2026, Lab 3: Asteroseismology · GYRE v7.1: Boundary conditions
Zero pressure does not mean a motionless surface
Let be the radial displacement of a gas layer. Velocity is its time derivative. For a sinusoidal mode, the acceleration is minus angular frequency squared times displacement. Substituting into the force equation relates the displacement shape to the pressure gradient.
At the outer boundary, pressure is zero but its radial gradient generally is not. The gas can move there. Conversely, the displacement vanishes at the center by symmetry, although central pressure oscillates. Pressure nodes and displacement nodes should therefore be distinguished in a diagram.
This distinction matters when generating an observable. Sampling boundary pressure in this model would give a flat zero signal. A nonzero boundary-displacement proxy can carry the modes. Turning that proxy into a real stellar brightness calculation would require the response of temperature, radiation and the atmosphere.
Work a complete one-meter example
Suppose the cavity radius is one meter and sound speed is 300 meters per second. First identify the model: uniform density, uniform sound speed and the pressure-release boundary just derived. Divide speed by twice the radius to obtain a fundamental frequency of 150 hertz. The next two modes are 300 and 450 hertz.
For the third mode, the internal pressure nodes are one-third and two-thirds of a meter from the center. The wave travel time from center to boundary is one three-hundredth of a second. A round trip takes twice as long, equal to the fundamental period. Higher modes fit multiple oscillations into that same travel time.
If the radius doubles while sound speed stays fixed, every frequency halves. If sound speed instead increases by ten percent at fixed radius, every frequency rises by ten percent. A change confined to one layer has a richer effect because different mode shapes sample the layer differently. That is the next chapter’s question.