---
title: "How acoustic radiation force pushes beads to the nodes of a standing wave"
description: "Derive the radiation force on a small particle from the Gor’kov potential, find the stable point, and work out force, stiffness and return time for a 5 µm bead."
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language: "en"
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  zh-CN: "https://sandboxphysics.com/zh/stories/acoustics-ultrasound/acoustic-radiation-force/"
last_modified: "2026-10-29"
---

[Acoustics and ultrasound](https://sandboxphysics.com/stories/acoustics-ultrasound/)

L3 · Chapter 5

# How acoustic radiation force pushes beads to the nodes of a standing wave

Sandbox Physics · 2026-10-29

Over one period the sound pressure is positive half the time and negative the other half, so it averages to zero. Yet beads are pushed firmly to the middle of the channel. The reason is a second-order effect: the particle scatters the sound, and the product of the scattered and incident waves leaves a nonzero average force. This chapter writes that force as a potential and asks where its valley lies, how deep it is and how stiff.

## Before you begin

The standing-wave pressure and velocity of Chapter 3, and how a potential energy sets a force: the force points down the steepest slope of the potential. Complex numbers appear only in the viscous correction, which can be skipped at first.

- Use the Gor’kov potential and the acoustic contrast factor to find the radiation force in a one-dimensional standing wave and decide whether particles go to pressure nodes or antinodes.
- Find the trap stiffness and overdamped return time near the stable point, and check a numerical integration against the exact solution.

## How a field that averages to zero leaves a force

The first-order fields $p_1$ and $\mathbf v_1$ average to zero over a period. A particle whose density and compressibility differ from water’s scatters the incident wave, though, and the product of incident and scattered fields is second order. Its time average is not zero: that is the acoustic radiation force.

When the particle is much smaller than the wavelength ($ka\ll1$) the force is minus the gradient of a potential with two terms. A particle stiffer than water has higher energy where the pressure swings strongly (the monopole term, coefficient $f_1$). A particle denser than water has lower energy where the liquid sloshes strongly (the dipole term, coefficient $f_2$).

$$
U=\frac{4\pi}{3}a^3\left[f_1\frac{\kappa_0}{2}\langle p_1^2\rangle-f_2\frac{3}{4}\rho_0\langle v_1^2\rangle\right],\qquad \mathbf F=-\nabla U
$$

Settnes and Bruus, Eqs. 48a–b. $a$ is the particle radius and angle brackets denote time averages.

$$
f_1=1-\frac{\kappa_p}{\kappa_0},\qquad f_2=\frac{2(\tilde\rho-1)}{2\tilde\rho+1},\quad \tilde\rho=\frac{\rho_p}{\rho_0}
$$

This is the form without viscosity. With the viscous boundary layer, $f_2$ becomes the complex expression of Eqs. 48d–e, of which the real part is taken.

Source: M. Settnes and H. Bruus · R. Barnkob

## Force and contrast factor in a one-dimensional standing wave

Insert the standing wave of Chapter 3 into the potential and the two terms combine into a sinusoidal force. All the material properties fold into one number, the acoustic contrast factor $\Phi$, whose sign decides where the particle goes.

Particles with $\Phi>0$ are pushed to pressure nodes: polystyrene beads have $\Phi=0.165$, Pyrex beads about 0.54, and a cell with the parameters tabulated by Settnes and Bruus about 0.068. Particles with $\Phi<0$, more compressible than water or much lighter, are pushed to pressure antinodes; in an $n=1$ channel those are the side walls.

$$
F_y=4\pi\Phi\,ka^3E_{\mathrm{ac}}\sin(2ky),\qquad \Phi=\frac{f_1}{3}+\frac{f_2}{2}
$$

Eq. 2 of Barnkob and co-workers with the origin at the left wall. The force scales with $a^3$, the particle volume.

Source: R. Barnkob · M. Settnes and H. Bruus

## Stable point, stiffness and return time

The force vanishes where $\sin(2ky)=0$. Only zeros where the force has a negative slope are stable: a particle displaced to the right is pushed back left. For $\Phi>0$ this needs $\cos(2ky)<0$, which is exactly a pressure node; the same point is unstable for $\Phi<0$. Near the stable point the force grows in proportion to the displacement, like a spring.

A bead in water is so small and slow that its inertia dies out within tens of microseconds, and the motion is overdamped: the radiation force balances the Stokes drag $6\pi\eta a$. On the centre plane this equation can be solved exactly, with no numerical integration. The experiment uses that solution as an independent reference for the integrator’s release curve, and the two agree to better than $4\times10^{-13}\,\mathrm{m}$.

$$
\kappa_{\mathrm{trap}}=8\pi|\Phi|k^2a^3E_{\mathrm{ac}},\qquad \tau=\frac{6\pi\eta a}{\kappa_{\mathrm{trap}}}
$$

The stiffness is the magnitude of the force slope at the stable point, and $\tau$ is the time constant with which a small displacement decays.

$$
\dot y=A\sin(2ky)\;\Rightarrow\;\tan(ky)=\tan(ky_0)\,e^{2kAt},\qquad A=\frac{2\Phi ka^2E_{\mathrm{ac}}}{3\eta}
$$

Separate the variables and integrate. $A$ is the amplitude of the radiation drift speed; streaming on the centre plane has the same sinusoidal shape, so including it only adds to $A$.

Half-wave resonance across a channel with hard side walls. Teal: pressure amplitude, largest at the walls and zero at the centre. Amber: velocity amplitude, zero at the walls and largest at the centre. Beads with $\Phi>0$ collect on the pressure node. Original schematic.

Source: R. Barnkob · M. Settnes and H. Bruus

## Worked example: a 5 µm polystyrene bead

The experiment’s defaults: a channel $w=377\,\mu\mathrm{m}$ wide at $n=1$, so $k=\pi/w=8.33\times10^3\,\mathrm{m^{-1}}$; $E_{\mathrm{ac}}=20\,\mathrm{J/m^3}$; a polystyrene bead 5 µm across with $\Phi=0.165$; water viscosity $\eta=0.890\,\mathrm{mPa\,s}$.

The largest radiation force is $4\pi\Phi ka^3E_{\mathrm{ac}}=5.40\,\mathrm{pN}$, halfway between the node and a wall. Dividing by the Stokes coefficient $6\pi\eta a=4.19\times10^{-8}\,\mathrm{N\,s/m}$ gives a top drift speed of 129 µm/s.

The stable point is the channel centre. The stiffness is $\kappa_{\mathrm{trap}}=9.00\times10^{-8}\,\mathrm{N/m}$ and the return time constant $\tau=0.466\,\mathrm{s}$, the value the experiment shows. Released from $y_0=w/8$, the closed-form solution of the drift equation takes 1.45 s to bring the bead within 10 µm of the node (with streaming included).

The well depth is $F_{\max}/k=6.5\times10^{-16}\,\mathrm{J}$, about $1.6\times10^5$ times the thermal energy at room temperature, and thermal motion jiggles the bead about the node by about 0.21 µm root mean square. The experiment leaves out Brownian motion, which is negligible for a bead this size and matters for sub-micrometre particles.

Source: R. Barnkob · M. Settnes and H. Bruus

## Try it in the Lab

### [Drag a bead and watch it return to the node](https://sandboxphysics.com/labs/acoustic-radiation-trap/)

1. In the default preset, drag the probe bead to a quarter of the way across the channel and let go. Compare the simulated curve with the exact solution in the “Probe return after release” chart.
2. Turn off “Include acoustic streaming” and repeat. How much does the return time change?
3. Choose “Soft particles flee to the walls” and see where the particles go once the contrast factor is negative. Drag one to the channel centre and let go: does it stay?

## Exercises with explained answers

1. In the same field, what are the largest radiation force and the return time for a 5 µm Pyrex bead?

   Show the answer

   With the viscous correction $\Phi=0.562$ (0.539 without it), so the largest force is $5.40\times0.562/0.165=18.4\,\mathrm{pN}$ and the return time $0.466\times0.165/0.562=0.137\,\mathrm{s}$, 3.4 times faster.

2. Halve the bead diameter to 2.5 µm. What is the return time now, and why?

   Show the answer

   The force scales with $a^3$ and the drag with $a$, so the drift speed scales with $a^2$ and the time constant grows four times, to about 1.86 s. Small particles are hard to focus, which is why streaming starts to win in the next chapter.

3. Why can a particle with $\Phi<0$ not rest at a pressure node?

   Show the answer

   At the node $\cos(2ky)=-1$, and the force slope $8\pi\Phi k^2a^3E_{\mathrm{ac}}\cos(2ky)$ is positive when $\Phi<0$: a particle nudged right is pushed further right. It slides to the pressure antinodes, where the slope is negative; in an $n=1$ channel those are the two side walls.

[Previous: Why a point comes out as a blob when echo times become an image](https://sandboxphysics.com/stories/acoustics-ultrasound/pulse-echo-imaging/) [Next: How acoustic streaming in a standing wave stirs the liquid](https://sandboxphysics.com/stories/acoustics-ultrasound/acoustic-streaming/)

## Sources and what we recomputed

1. [R. Barnkob, P. Augustsson, T. Laurell and H. Bruus, Acoustic radiation- and streaming-induced microparticle velocities determined by micro-PIV in an ultrasound symmetry plane, arXiv:1208.6534v1; Physical Review E 86, 056307 (2012)](https://arxiv.org/abs/1208.6534v1)

   Version 1, 31 August 2012, read in full. Eq. 1 (standing-wave field), Eq. 2 (radiation force), Eqs. 5–7 (streaming scale and Rayleigh streaming), Eqs. 9–10 (geometry and thermoviscous corrections), Eqs. 12–14 (velocity amplitude, velocity ratio, critical diameter), Eqs. 16–18 and Table I (wall drag), Table II (material data), Tables IV–V and Figs. 8–9 (measured energy densities, streaming coefficients and velocity ratios). We recompute the theory values; the measured data are quoted, not reproduced.

2. [M. Settnes and H. Bruus, Forces acting on a small particle in an acoustical field in a viscous fluid, arXiv:1110.6037v1; Physical Review E 85, 016327 (2012)](https://arxiv.org/abs/1110.6037v1)

   Version 1, 27 October 2011, read in full. Eq. 1 (viscous penetration depth), Eqs. 4–5 (first-order pressure and compressibility), Eqs. 48a–e (Gor’kov potential with the viscous dipole coefficient) and Table I (Pyrex and cell parameters).