Electromagnetic Wave Energy and Intensity

Hello! Welcome to our third lesson on electromagnetic waves.

In our previous session, we established the fundamental properties of plane electromagnetic waves in a vacuum. We learned that they are transverse, with the electric field $E$ , magnetic field $B$ , and propagation vector $k$ forming a mutually orthogonal, right-handed system. We also derived the crucial relationship between the field magnitudes, $E = c B$ .

Today, we will address a critical question: what do these propagating fields carry with them? The answer is energy and momentum. This lesson focuses on quantifying the energy transport by an electromagnetic wave. Our learning outcome is to calculate the Poynting vector, energy density, and intensity of electromagnetic waves. Understanding these concepts is essential for analyzing almost any phenomenon involving light and other EM radiation, from the power received by a radio antenna to the energy balance of a planet.

1. Energy Density in Electromagnetic Waves

Electromagnetic fields store energy. From electrostatics and magnetostatics, you may recall the expressions for the energy stored per unit volume (energy density) in an electric field and a magnetic field:

Electric Energy Density: $u_{E} = \frac{1}{2} ϵ_{0} E^{2}$
Magnetic Energy Density: $u_{B} = \frac{1}{2 μ _{0}} B^{2}$

An electromagnetic wave consists of both electric and magnetic fields, so the total energy density $u$ at any point in space and time is the sum of these two:

u = u_{E} + u_{B} = \frac{1}{2} ϵ_{0} E^{2} + \frac{1}{2 μ _{0}} B^{2}

A remarkable property of plane electromagnetic waves is that the energy is shared equally between the electric and magnetic fields at all times. We can prove this using the relationship $B = E / c$ and the formula for the speed of light, $c^{2} = 1/ (μ_{0} ϵ_{0})$ .

Let's substitute these into the expression for magnetic energy density:

u_{B} = \frac{1}{2 μ _{0}} (\frac{E}{c})^{2} = \frac{1}{2 μ _{0} c ^{2}} E^{2}

Since $1/ c^{2} = μ_{0} ϵ_{0}$ , we have:

u_{B} = \frac{1}{2 μ _{0}} (μ_{0} ϵ_{0}) E^{2} = \frac{1}{2} ϵ_{0} E^{2} = u_{E}

Thus, for a plane electromagnetic wave, the energy is perfectly partitioned between the electric and magnetic fields: $u_{E} = u_{B}$ .

The total instantaneous energy density can therefore be written in several equivalent ways:

u = u_{E} + u_{B} = 2 u_{E} = ϵ_{0} E^{2} = 2 u_{B} = \frac{1}{μ _{0}} B^{2}

2. The Poynting Vector and Poynting's Theorem

Now that we know how much energy is stored in a volume of space, we need a way to describe the flow of this energy. This is precisely what the Poynting vector, $S$ , does. It represents the rate of energy flow per unit area. Its direction is the direction of energy transport, and its magnitude is the power per unit area (units of W/m²).

Given your preference for rigorous derivations, we will derive the Poynting vector from the principle of energy conservation in electromagnetism. This leads to a powerful statement known as Poynting's Theorem.

The following reading from a set of university lecture notes provides a formal derivation. It starts with the work done by fields on charges ( $J \cdot E$ ) and uses Maxwell's equations and a vector identity to arrive at a continuity equation for energy.

PHYS 352 Electromagnetic Waves

This section rigorously derives the energy conservation law for electromagnetic fields. It introduces the Poynting vector as a natural consequence of this law.

Please read section '1.6.1 Conservation of energy and the Poynting flux' (pages 21-22). Focus on understanding the steps that transform the expression for work done on charges, -\mathbf{J}\cdot\mathbf{E}, into the final energy conservation equation. Pay close attention to the physical interpretation of each term.

The key result from the derivation is Poynting's theorem:

\frac{\partial u}{\partial t} + \nabla \cdot S = - J \cdot E

where $u = \frac{1}{2} ϵ_{0} E^{2} + \frac{1}{2 μ _{0}} B^{2}$ is the total energy density.

This equation is a statement of energy conservation in differential form:

$- J \cdot E$ is the rate at which the fields do work on charges per unit volume (i.e., the power transferred from the fields to the matter).
$\frac{\partial u}{\partial t}$ is the rate of change of the energy density stored in the electromagnetic fields.
$\nabla \cdot S$ is the divergence of the energy flux, representing the energy flowing out of a differential volume.

From this theorem, we identify the Poynting vector as:

S = \frac{1}{μ _{0}} (E \times B)

For a plane wave where $E$ and $B$ are perpendicular, the magnitude is $S = \frac{1}{μ _{0}} E B$ . Since $E$ , $B$ , and $k$ form a right-handed system, the direction of $S$ is, as expected, the direction of wave propagation $k$ .

Poynting Vector and Intensity — Caption: A summary of the key definitions. The Poynting vector $\mathbf{S}$ is defined by the cross product of the electric and magnetic fields, indicating the direction of energy flow. Intensity $I$ is its time-average.

3. Wave Intensity

The electric and magnetic fields in a wave oscillate in time, so the Poynting vector $S (t)$ also oscillates. For a sinusoidal wave propagating in the z-direction, $E (z, t) = E_{0} cos (k z - ω t)$ and $B (z, t) = B_{0} cos (k z - ω t)$ . The magnitude of the Poynting vector is:

S (t) = \frac{1}{μ _{0}} E (t) B (t) = \frac{1}{μ _{0}} E_{0} B_{0} cos^{2} (k z - ω t)

In most practical situations, we are interested in the average power delivered over time, not this instantaneous value. This time-averaged power per unit area is called the intensity ( $I$ ) of the wave.

I = ⟨ S (t)⟩

To find the average, we use the fact that the time average of $cos^{2} (θ)$ over one or more full cycles is $\frac{1}{2}$ .

I = \frac{1}{μ _{0}} E_{0} B_{0} ⟨ cos^{2} (k z - ω t)⟩ = \frac{E _{0} B _{0}}{2 μ _{0}}

Using the relationships $E_{0} = c B_{0}$ and $c^{2} = 1/ (μ_{0} ϵ_{0})$ , we can write the intensity in several useful forms:

I = \frac{E _{0} B _{0}}{2 μ _{0}} = \frac{E _{0}^{2}}{2 μ _{0} c} = \frac{c B _{0}^{2}}{2 μ _{0}} = \frac{1}{2} c ϵ_{0} E_{0}^{2}

Notice also that the average total energy density is $⟨ u ⟩ = ⟨ ϵ_{0} E^{2} ⟩ = ϵ_{0} E_{0}^{2} ⟨ cos^{2} ⟩ = \frac{1}{2} ϵ_{0} E_{0}^{2}$ . Comparing this with the last expression for intensity, we find a simple and intuitive relationship:

I = c ⟨ u ⟩

This states that the intensity (energy flux) is simply the average energy density flowing past a point at the speed of light.

The following text provides a clear walkthrough of these derivations and includes a practical example.

Chapter 13 Maxwell's Equations and Electromagnetic Waves

This reading from the MIT TEAL course notes provides a concise summary of the Poynting vector and intensity, showing the derivations for the various forms of the intensity formula and its relation to energy density.

Read the subsections 'Poynting Vector' and the parts defining 'Intensity' on pages 13-16 and 13-17. Follow equations 13.6.8 through 13.6.12. Also, review 'Example 13.1: Solar Constant', as it is a classic application of these concepts.

Test your understanding!

A 100 W light bulb radiates light uniformly in all directions. Assume it converts electrical power to light with 5% efficiency.

What is the intensity of the light at a distance of 2.0 m from the bulb?
What are the amplitudes of the electric field ( $E_{0}$ ) and magnetic field ( $B_{0}$ ) at this distance?

(Use $ϵ_{0} \approx 8.85 \times 1 0^{- 12} F/m$ and $c \approx 3.0 \times 1 0^{8} m/s$ ).

Show answer

Intensity: The bulb radiates light power $P_{light}$ uniformly over the surface of a sphere. The area of a sphere of radius $r$ is $A = 4 π r^{2}$ .
- Light power: $P_{light} = 100 W \times 0.05 = 5.0 W$ .
- Intensity $I = \frac{P _{light}}{A} = \frac{5.0 W}{4 π ( 2.0 m ) ^{2}} = \frac{5.0}{16 π} \approx 0.0995 W/m^{2}$ .
Field Amplitudes: We can find $E_{0}$ from the intensity formula $I = \frac{1}{2} c ϵ_{0} E_{0}^{2}$ .
- Rearranging for $E_{0}$ :
  $E_{0} = \frac{2 I}{c ϵ _{0}} = \frac{2 \times 0.0995}{( 3.0 \times 1 0 ^{8} ) \times ( 8.85 \times 1 0 ^{- 12} )} \approx 75.0 \approx 8.66 V/m$
- We can then find $B_{0}$ using $E_{0} = c B_{0}$ :
  $B_{0} = \frac{E _{0}}{c} = \frac{8.66 V/m}{3.0 \times 1 0 ^{8} m/s} \approx 2.89 \times 1 0^{- 8} T$

4. An Engineering Application: Energy Flow in a Wire

A fascinating application of the Poynting vector is understanding how energy is delivered to a simple resistor. We usually think of energy flowing along the wire with the electrons. Poynting's theorem provides a different, field-centric perspective.

Consider a cylindrical wire of radius $a$ and length $L$ carrying a steady current $I$ . It has a resistance $R$ .

There is a uniform electric field $E$ inside the wire parallel to its axis, pointing in the direction of the current, with magnitude $E = V / L = (I R) / L$ .
The current creates a circular magnetic field $B$ around the wire. At the surface of the wire, Ampere's Law gives its magnitude as $B = μ_{0} I / (2 π a)$ .

The Poynting vector $S = \frac{1}{μ _{0}} (E \times B)$ is directed radially inward at the surface of the wire. The fields are supplying energy from the space outside the wire, flowing into the wire, where it is dissipated as heat.

The magnitude of $S$ at the surface is:
$S = \frac{1}{μ _{0}} E B = \frac{1}{μ _{0}} (\frac{I R}{L}) (\frac{μ _{0} I}{2 π a}) = \frac{I ^{2} R}{2 π a L}$
The total power flowing into the wire is this flux multiplied by the surface area of the cylinder ( $A_{s u r f} = 2 π a L$ ):
$P = S \times A_{s u r f} = (\frac{I ^{2} R}{2 π a L}) (2 π a L) = I^{2} R$
This is exactly the familiar formula for Joule heating! The energy dissipated as heat in the resistor is precisely accounted for by the energy flux from the electromagnetic field entering the conductor from its sides.

Conclusion

In this lesson, we have developed the tools to describe the energy carried by electromagnetic waves. This bridges the gap between the abstract wave solutions and their physical effects.

Key Takeaways:

Energy in an EM wave is stored in both the electric and magnetic fields, with the total energy density given by $u = ϵ_{0} E^{2}$ . The energy is shared equally between the E and B fields.
The Poynting vector, $S = \frac{1}{μ _{0}} (E \times B)$ , represents the instantaneous rate and direction of energy flow per unit area.
The intensity $I$ is the time-averaged magnitude of the Poynting vector. For a sinusoidal wave, it is given by $I = ⟨ S ⟩ = \frac{1}{2} c ϵ_{0} E_{0}^{2}$ , where $E_{0}$ is the electric field amplitude.
Poynting's theorem provides a rigorous statement of energy conservation, showing how field energy changes, flows, and is converted to other forms (e.g., heat in a resistor).

Preview of the Next Lesson:
Now that we can quantify the energy flux of an electromagnetic wave, we are ready to analyze what happens when this wave encounters a boundary between two media. In the next lesson, we will analyze the reflection and transmission of electromagnetic waves at normal incidence on a dielectric interface. We will use the concepts of energy flux and intensity to calculate how much of the wave's energy is reflected and how much is transmitted.

Recommended resources

1. Energy Density in Electromagnetic Waves

2. The Poynting Vector and Poynting's Theorem

3. Wave Intensity

4. An Engineering Application: Energy Flow in a Wire

Conclusion