---
title: "Standing waves inside a sphere"
description: "Derive radial acoustic modes from a regular center and a pressure-release boundary, then calculate frequencies and node positions."
canonical_url: "https://sandboxphysics.com/stories/stellar-seismology/radial-standing-waves/"
language: "en"
page_type: "story"
alternate_languages:
  zh-CN: "https://sandboxphysics.com/zh/stories/stellar-seismology/radial-standing-waves/"
last_modified: "2026-11-03"
---

[Read the interior of a star](https://sandboxphysics.com/stories/stellar-seismology/)

AC26 · L1

# Standing waves inside a sphere

Sandbox Physics · 2026-11-03

A sphere permits a discrete set of radial sound modes because a single wave shape must satisfy both the center and the outer boundary. In the uniform acoustic model, the frequencies form an even ladder. We can derive the ladder and its nodes without treating the sphere as a real stellar evolution model.

## Before you begin

Read Why stars ring. You need sine and cosine, derivatives as rates of change, and the idea that compression raises a gas’s pressure. Each use of the wave equation is explained below.

## Specify the gas and its motion

Take a spherical acoustic cavity of radius $R$, measured in meters. Its equilibrium density $\rho$ is constant, in kilograms per cubic meter, and its sound speed $c$ 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 $r$ increases outward from the center. The pressure perturbation $p(r,t)$ is the small changing part of the pressure, in pascals. It is not the much larger equilibrium pressure. The radial gas velocity $u(r,t)$ 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.

$$
\frac{\partial u}{\partial t}=-\frac{1}{\rho}\frac{\partial p}{\partial r},\qquad \frac{\partial p}{\partial t}=-\rho c^2\frac{1}{r^2}\frac{\partial(r^2u)}{\partial r}
$$

The first equation is radial force per unit mass. In the second, the divergence of velocity measures the fractional volume expansion rate.

Reading: MESA School 2026, Lab 3: Asteroseismology

## 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 $P(r)$ multiplied by $\cos(\omega t)$. The angular frequency $\omega$ is measured in radians per second. One full cycle contains $2\pi$ radians, so the ordinary frequency is $\nu=\omega/(2\pi)$ in hertz. Differentiating the cosine twice contributes a factor $-\omega^2$.

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.

$$
\frac{\partial^2p}{\partial t^2}=c^2\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial p}{\partial r}\right)
$$

This follows by eliminating velocity from the two linear acoustic equations.

$$
p(r,t)=P(r)\cos(\omega t),\qquad P^{\prime\prime}+\frac{2}{r}P^\prime+\frac{\omega^2}{c^2}P=0
$$

A prime denotes a derivative with respect to outward radius. The radial shape satisfies an ordinary differential equation.

Reading: MESA School 2026, Lab 3: Asteroseismology

## The center removes one possible solution

The apparent singularity at the center comes from spherical coordinates. The gas pressure itself must remain finite. Define $\psi=rP$. 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 $A$.

This is also why a drawing should not place a pressure node at the center of the fundamental mode. The transformed quantity $\psi$ is zero there, but pressure need not be. Changing variables simplified the equation; it did not turn the central gas into a fixed endpoint.

$$
\psi=rP,\qquad \psi^{\prime\prime}+k^2\psi=0,\qquad k=\omega/c
$$

The wavenumber $k$ has units of inverse meters.

$$
P(r)=A\frac{\sin(kr)}{kr},\qquad \lim_{r\to0}P(r)=A
$$

The regular radial pressure shape is the spherical Bessel function of degree zero, written explicitly here so no special-function knowledge is needed.

Reading: MESA School 2026, Lab 3: Asteroseismology

## 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 $n$ 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.

$$
P(R)=0\quad\Longrightarrow\quad k_nR=n\pi,\qquad \nu_n=\frac{nc}{2R},\qquad n=1, 2,\ldots
$$

The angular frequency has been divided by $2\pi$ to obtain hertz.

$$
r_j=\frac{jR}{n},\qquad j=1,\ldots,n-1
$$

These positions count the pressure zeros strictly inside the uniform cavity, excluding the outer boundary.

Reading: MESA School 2026, Lab 3: Asteroseismology · GYRE v7.1: Boundary conditions

## Zero pressure does not mean a motionless surface

Let $\xi_r(r,t)$ 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.

$$
\Xi_r(r)=\frac{P^\prime(r)}{\rho\omega^2}
$$

$\Xi_r$ is the displacement amplitude of the deterministic harmonic mode. Its overall normalization is freely chosen within the small-amplitude approximation.

Reading: MESA School 2026, Lab 3: Asteroseismology

## 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.

Reading: MESA School 2026, Lab 3: Asteroseismology

## Exercises and explanations

1. Draw the pressure signs and internal nodes for the first three modes. Does the center count as a pressure node?

   Read the explanation

   The first mode keeps one sign inside the sphere and has no internal node. The second changes sign at half the radius. The third changes sign at one-third and two-thirds of the radius. Central pressure is finite and nonzero for these normalized modes; it is the transformed function $\psi$ and the radial displacement that vanish at the center.

2. A uniform cavity has twice the original radius and half the original sound speed. What happens to every frequency?

   Read the explanation

   Frequency is proportional to sound speed and inversely proportional to radius. The two changes multiply, giving one-quarter of every original frequency. This uses the same boundary condition and uniform model.

3. Why does a boundary-pressure detector fail in this model even though the boundary moves?

   Read the explanation

   The imposed pressure-release condition sets the pressure perturbation to zero at the boundary. Its gradient still accelerates the gas, so displacement can be nonzero. The boundary-pressure signal stays zero, while a displacement-sensitive measurement can see the mode.

## Try the prediction

### [Change a layer and read its modes](https://sandboxphysics.com/labs/stellar-oscillation-modes/)

1. Select the uniform preset and predict the internal node count before changing radial order.
2. Compare the selected mode’s dashed node rings with its order. Keep transformed-coordinate zeros separate from pressure nodes.
3. Increase core sound speed slightly. Explain which uniform-speed assumption has changed and how to join the solutions across the new interface.

[Previous: Why stars ring](https://sandboxphysics.com/stories/stellar-seismology/why-stars-ring/) [Next: Sound crossing a layered star](https://sandboxphysics.com/stories/stellar-seismology/layered-acoustic-interiors/)

## Sources and calculation scope

1. [GYRE v7.1: Boundary conditions](https://gyre.readthedocs.io/en/v7.1/ref-guide/osc-equations/bound-conditions.html)

   Central regularity and the Lagrangian free-surface pressure condition in stellar models; distinguish these from the fixed-radius Eulerian pressure-release condition of this gravity-free cavity.

2. [MESA School 2026, Lab 3: Asteroseismology](https://mesastar.org/summer-school-2026-labs/monday/lab-3/)

   Official classroom treatment of oscillation frequencies, sound-travel integrals and model evolution. The controllable constant-density cavity here is a separate declared teaching model.