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Chapter 34: Problem 29

Prove, for a plane electromagnetic wave that is normally incident on a plane surface, that the radiation pressure on the surface is equal to the energy density in the incident beam. (This relation between pressure and energy density holds no. matter what fraction of the incident energy is reflected.)

Short Answer

Expert verified
The radiation pressure on the surface is equal to the energy density in the incident beam if the surface absorbs all the energy.

Step by step solution

01

Understand Radiation Pressure

Radiation pressure, denoted as \( P \), is the pressure exerted by electromagnetic radiation on a surface. It is caused by the momentum transfer from the photons in the radiation to the surface.
02

Energy Density of Electromagnetic Wave

The energy density \( u \) of an electromagnetic wave refers to the amount of energy stored in a given volume by the wave. For a plane electromagnetic wave, it is given by the formula: \[ u = \frac{E^2 + B^2}{2\mu_0} \]where \( E \) is the electric field strength, \( B \) is the magnetic field strength, and \( \mu_0 \) is the permeability of free space.
03

Relation Between Radiation Pressure and Momentum

The pressure exerted on the surface by the wave is related to the change in momentum of the photons. For a normally incident wave, the radiation pressure can be expressed using the momentum flux and the reflection coefficient.
04

Calculate Momentum Flux

The momentum flux \( \frac{I}{c} \) is the measure of the rate at which momentum is transferred per unit area. Here, \( I \) is the intensity of the incident wave and \( c \) is the speed of light. Since energy density \( u \) is related to intensity \( I \) by \( u = \frac{I}{c} \), we can write:\[ \text{Momentum flux} = \frac{I}{c} = u \]
05

Determine Radiation Pressure

Consider the reflection coefficient \( R \) which determines the fraction of the incident wave that is reflected. For complete reflection, \( R = 1 \). The radiation pressure considering both the incident and reflected waves is:\[ P = \left( 1 + R \right) u \] In the scenario of complete reflection, \( P = 2u \). For partial reflection where \(R\) lies between 0 and 1, the radiation pressure adapts accordingly.
06

Equalize Radiation Pressure and Energy Density

If the surface absorbs all the incident energy (no reflection), \( R = 0 \). Hence, by substituting in the formula, we get:\[ P = u \] Therefore, the radiation pressure is equal to the energy density in this case.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electromagnetic Wave
Electromagnetic waves are waves of electric and magnetic fields oscillating perpendicular to each other and the direction of propagation. These waves travel through space and can carry energy, momentum, and thus exert pressure. This type of wave includes visible light, radio waves, microwaves, and X-rays. The waves can interact with surfaces, such as reflecting off or being absorbed. When dealing with radiation pressure, the behavior of these waves upon hitting a surface is crucial.
Energy Density
Energy density is the amount of energy stored in a given system or region of space per unit volume. For an electromagnetic wave, energy density is given by the formula: \[ u = \frac{E^2 + B^2}{2\mu_0} \]where:
  • \(E\) is the electric field strength
  • \(B\) is the magnetic field strength
  • \(\mu_0\) is the permeability of free space
Energy density tells us how much energy the wave carries through space. Due to the oscillatory nature of electromagnetic waves, both electric and magnetic fields contribute to the energy density.
Momentum Transfer
When an electromagnetic wave hits a surface, it transfers momentum to that surface. This process is the basis for radiation pressure. The momentum transfer involves a change in the wave鈥檚 momentum, which exerts force over the area of incidence. The relationship between radiation pressure and momentum transfer can be encapsulated by the formula for momentum flux:\[ \text{Momentum flux} = \frac{I}{c} \]where:
  • \(I\) is the intensity of the incident wave
  • \(c\) is the speed of light
This flux represents the rate of momentum being transferred per unit area.
Reflection Coefficient
The reflection coefficient \(R\) measures the fraction of the incident wave reflected by the surface. It ranges from 0 to 1.
  • \(R = 1\) means total reflection, where all the wave鈥檚 energy is reflected
  • \(R = 0\) implies perfect absorption, with no energy reflected
For partial reflection, \(R\) lies between 0 and 1. The radiation pressure \(P\) on the surface can be affected by this coefficient according to:\[ P = \left( 1 + R \right) u \]If no energy is reflected, radiation pressure equals the energy density. If full reflection occurs, the pressure doubles.
Intensity of Wave
The intensity \(I\) of an electromagnetic wave is the power carried per unit area by the wave. It is directly related to energy density and is essential for understanding radiation pressure. The energy density \(u\) can be expressed using intensity:\[ u = \frac{I}{c} \]Since intensity represents how much energy the wave carries into a surface per time per area, this relationship allows us to calculate pressure based on wave intensity. As intensity increases, the energy density, thus the radiation pressure, also increases.

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Most popular questions from this chapter

Radiation from the Sun reaching the Earth (just outside the atmosphere) has an intensity of \(1.4 \mathrm{~kW} / \mathrm{m}^{2}\). (a) Assuming that the Earth (and its atmosphere) behaves like a flat disk perpendicular to the Sun's rays and that all the incident energy is absorbed, calculate the force on the Earth due to radiation pressure. (b) Compare it with the force due to the Sun's gravitational attraction.

The intensity of direct solar radiation th?t is not absorbed by the atmosphere on a particular summer day is \(100 \mathrm{~W} / \mathrm{m}^{2}\). How close would you have to stand to a \(1.0 \mathrm{~kW}\) electric heater to feel the same intensity? Assume that the heater radiates uniformly in all directions.

A particle in the solar system in under the combined influence of the Sun's gravitational attraction and the radiation force due to the Sun's rays. Assume that the particle is a sphere of density \(1.0 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\) and that all the incident light is absorbed. (a) Show that, if its radius is less than some critical radius \(R\), the particle will be blown out of the solar system. (b) Calculate the critical radius.

A black, totally absorbing piece of cardboard of area \(A=2.0 \mathrm{~cm}^{2}\) intercepts light with an intensity of \(10 \mathrm{~W} / \mathrm{m}^{2}\) from a camera strobe light. What radiation pressure is produced on the cardboard by the light?

Frank D. Drake, an investigator in the SETI (Search for Extra-Terrestrial Intelligence) program, once said that the large radio telescope in Arecibo, Puerto Rico, "can detect a signal which lays down on the entire surface of the earth a power of only one picowatt." (a) What is the power that would be received by the Arecibo antenna for such a signal? The antenna diameter is \(300 \mathrm{~m} .\) (b) What would be the power of a source at the center of our galaxy that could provide such a signal? The galactic center is \(2.2 \times\) \(10^{4}\) ly away. Take the source as radiating uniformly in all directions.
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