Balance momentum carried by radiation
Compute vacuum radiation pressure from incoming and outgoing momentum, including absorption, reflection, transmission, and finite pulses.
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The question for today
What you’ll understand
- Relate vacuum electromagnetic energy flux to momentum density and flux.
- Compute pressure using both incident and outgoing beam momentum.
- Distinguish normal force, absorbed heating, and net recoil from an angular emission pattern.
Radiation can exert force without depositing all its energy as heat. A perfect stationary mirror reverses the normal momentum of incident light while absorbing no energy in the ideal limit. An absorber takes both energy and momentum, and a transparent object can pass both onward. Momentum balance therefore requires knowing what leaves as well as what arrives. We will work in vacuum with a stationary planar target, so material momentum conventions do not enter the calculation.
Momentum density and momentum flux have different units
The vacuum Poynting vector is energy flow per unit area per unit time. Divide it by speed squared to obtain field momentum per unit volume. For a traveling plane wave, momentum moves at c, so its flux in the propagation direction is intensity divided by c. One quantity is a volume density and the other is a force per area. Confusing them leaves a missing speed factor and turns pressure into the wrong physical dimensions.
For a finite object, the electromagnetic momentum balance includes the change of field momentum stored inside a surrounding volume and momentum crossing its surface. In a steady beam, the time-averaged storage term vanishes and the difference between incoming and outgoing momentum flux gives the force. For a switched pulse, integrate over the complete interaction and include any residual stored fields. A local intensity alone cannot determine force without the target's scattering and absorption behavior.
Absorption and reflection give different impulses
At normal incidence, label energy fractions absorbed, specularly reflected, and transmitted as A, R, and T. They sum to one for a passive target in the stated steady situation. Each absorbed unit of incident energy transfers momentum equal to that energy divided by c. A reflected unit reverses direction and transfers twice as much momentum. A transmitted unit continuing forward in vacuum leaves the same momentum as it brought and transfers none in this idealization.
The resulting pressure is the sum of absorption and twice reflection times incident intensity divided by c. The formula assumes normal incidence, an effectively stationary target, and forward transmission in the same exterior vacuum. Oblique incidence requires projections of both crossing rate and momentum, while diffraction or diffuse scattering requires integrating outgoing directions. For a moving mirror, Doppler shifts alter the energy balance. Do not transplant the stationary normal-incidence expression into those different geometries.
Anisotropic emission can also recoil
Radiation emitted by a source carries momentum direction by direction. Integrate the angular power weighted by its outward unit vector and divide by c to obtain the momentum-loss rate. A stationary fixed-axis electric dipole pattern is symmetric under reversal of direction, so opposite angular contributions cancel. It radiates energy while carrying zero net vector momentum in that frame. Nonzero total power alone therefore does not determine a recoil force or a preferred recoil direction.
For a normally incident pulse, total energy replaces intensity times area times duration in the impulse balance. Pressure during the pulse depends on its temporal shape, while the integrated impulse depends on the total momentum difference. The vector lab can illustrate addition and reversal of a directed momentum component, but it contains only one vector. Reflection coefficients, heating, forces, and the integration of an angular emission pattern must be calculated on paper rather than inferred from nonexistent readouts.
Work it through
Separate pressure from heating
A stationary opaque plate in vacuum receives a normally incident 900 W/m² beam over 0.020 m². It specularly reflects 75% of the energy and absorbs the remainder. With c=3.0×10⁸ m/s, find normal force and absorbed power.
Transmission is zero, so A=0.25 and R=0.75. The momentum factor is A+2R=1.75, while the heating fraction remains only 0.25.
Pressure is 1.75×900/(3.0×10⁸)=5.25×10⁻⁶ Pa. Multiply by 0.020 m² to obtain force 1.05×10⁻⁷ N along the incident beam.
The incident optical power is 900×0.020=18 W. Only one quarter is absorbed, giving 4.5 W of heating in this ideal optical balance. Thermal reradiation and mechanical motion would require additional bookkeeping if included.
The beam produces 1.05×10⁻⁷ N normal force and 4.5 W absorbed power. Its reflected energy contributes strongly to force despite not being absorbed.
The living notebook
Explore the model
Interactive lab · vectors
Reverse a momentum component
For fresh runs Reset and set BOTH angle 0°, magnitude 1; then Reset and set angle 180°, magnitude 1. Predict horizontal momenta +1 and −1 in normalized units. Their difference is two units, illustrating the momentum change on reflection. The lab does not model a mirror, intensity, or force.
- x component
- 1.393
- y component
- 0.975
- Magnitude
- 1.7
Playback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forReversing a beam momentum component transfers twice that component to the target.
Pause & reason
Check your understanding
Your turn
Try it yourself
A 6.0 mJ pulse hits a stationary planar target normally in vacuum. Fractions R=0.40 and T=0.10 leave as backward specular and forward transmitted beams. Find absorbed energy and normal impulse using c=3.0×10⁸ m/s.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
First find A from energy conservation, then distinguish its role in heating from the reflection contribution to impulse.
Compare with the solution
A=1−0.40−0.10=0.50, so absorbed energy is 3.0 mJ. The momentum factor is A+2R=1.30. Impulse is 1.30×0.0060/(3.0×10⁸)=2.6×10⁻¹¹ N·s along the incident direction. The pulse duration is unnecessary for impulse but would be needed to determine average force during illumination.
Keep the idea
Take it with you
- Vacuum momentum density is S/c², while traveling-wave momentum flux is I/c.
- Normal reflection transfers twice the incident normal momentum per reflected energy.
- Energy emission and net vector momentum emission require different angular integrals.
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