33-2 Energy Transport and the Poynting Vector

33-2 Energy Transport and the Poynting Vector#

Prompts

  • What is the Poynting vector \(\vec{S}\)? How is it related to \(\vec{E}\) and \(\vec{B}\)? What does its direction tell you?

  • For an EM wave, how do you calculate the instantaneous rate of energy flow per unit area? How does it relate to \(E\) alone?

  • Define the intensity \(I\) of an EM wave. How is it related to \(E_{\text{rms}}\) and \(E_m\)?

  • What is an isotropic point source? Why does intensity from such a source fall off as \(1/r^2\)? Use energy conservation.

  • The electric and magnetic fields in an EM wave carry energy. Are the energy densities \(u_E\) and \(u_B\) equal or different? Why?

Lecture Notes#

Overview#

  • An EM wave transports energy; the rate per unit area is given by the Poynting vector \(\vec{S}\).

  • \(\vec{S} \propto \vec{E} \times \vec{B}\) gives both direction (energy flow) and magnitude (power per area).

  • The intensity \(I\) is the time-averaged power per unit area; \(I \propto E_{\text{rms}}^2\).

  • For an isotropic point source, \(I = P_s/(4\pi r^2)\)—intensity falls as \(1/r^2\) by energy conservation.

The Poynting vector#

The Poynting vector \(\vec{S}\) (after John Henry Poynting) describes the rate of energy transport per unit area in an EM wave:

(230)#\[ \vec{S} = \frac{1}{\mu_0} \vec{E} \times \vec{B} \]

  • Direction: \(\vec{S}\) points in the direction of wave travel and energy flow (perpendicular to both \(\vec{E}\) and \(\vec{B}\)).

  • Magnitude: \(S = \frac{1}{\mu_0} EB\) (since \(\vec{E} \perp \vec{B}\)).

  • SI unit: W/m² (power per area).

Using \(B = E/c\) (from 33-1):

(231)#\[ S = \frac{1}{c\mu_0} E^2 \]

So the instantaneous energy flow rate depends only on \(E\) (and constants).

Important

The electric and magnetic fields in an EM wave carry equal energy densities: \(u_E = u_B\). Although \(E = cB\) and \(c\) is large, the energy density \(u_B = B^2/(2\mu_0)\) equals \(u_E = \varepsilon_0 E^2/2\) when \(B = E/c\) and \(c = 1/\sqrt{\mu_0\varepsilon_0}\).

Intensity#

The intensity \(I\) is the time-averaged power per unit area:

(232)#\[ I = S_{\text{avg}} = \left(\frac{\text{power}}{\text{area}}\right)_{\text{avg}} \]

For a sinusoidal wave \(E = E_m \sin(kx - \omega t)\), the average of \(E^2\) over a cycle gives \(E_m^2/2\). Defining the root-mean-square electric field:

(233)#\[ E_{\text{rms}} = \frac{E_m}{\sqrt{2}} \]

we obtain

(234)#\[ I = \frac{1}{c\mu_0} E_{\text{rms}}^2 = \frac{1}{2c\mu_0} E_m^2 \]

  • Intensity \(\propto\) amplitude squared—typical for waves.

  • Most detectors (e.g., eyes, solar cells) respond to the electric field; \(E_{\text{rms}}\) and \(E_m\) are the practical measures.

Isotropic point source#

An isotropic point source emits EM waves equally in all directions—spherical wavefronts.

Energy conservation: All power \(P_s\) emitted must pass through any sphere of radius \(r\) centered on the source. The sphere area is \(4\pi r^2\), so

(235)#\[ I = \frac{P_s}{4\pi r^2} \]

  • \(1/r^2\) falloff: Intensity decreases as the inverse square of distance—same power spread over a larger area.

  • Applies to stars, small bulbs, and other sources that radiate roughly isotropically.

Beamed vs isotropic

A searchlight or laser beam does not emit isotropically; intensity falls off differently (e.g., slowly within the beam). Eq. (235) applies only when the source radiates equally in all directions.

Poll: Electric field amplitude vs distance

The amplitude of the oscillating electric field at your cell phone is 4.0 μV/m when you are 10 km east of the broadcast antenna. What is the electric field amplitude when you are 20 km east of the antenna?

  • (A) 4.0 μV/m

  • (B) 2.0 μV/m

  • (C) 1.0 μV/m

  • (D) 0.5 μV/m

Example: Order-of-magnitude energy and force from EM wave

An EM wave with known \(\vec{B}\)-field amplitude is incident on a person. Estimate (a) the energy delivered to the person in 1 second, and (b) the force on the person from the wave.

Solution: (a) From \(B_0\), get \(E_0 = cB_0\); intensity \(I = E_0^2/(2c\mu_0)\). Estimate person’s cross-sectional area \(A \sim 1\) m². Energy \(\approx I \cdot A \cdot \Delta t\). (b) Force \(F \approx IA/c\) for absorption.

[FIGURE: EM wave incident on person; indicate propagation direction and typical \(E\), \(B\) orientation]

Summary#

  • Poynting vector \(\vec{S} = (1/\mu_0)\vec{E} \times \vec{B}\): direction and rate of energy transport per unit area.

  • Instantaneous: \(S = (1/(c\mu_0)) E^2\).

  • Intensity \(I = S_{\text{avg}} = (1/(c\mu_0)) E_{\text{rms}}^2\); \(E_{\text{rms}} = E_m/\sqrt{2}\).

  • Isotropic point source: \(I = P_s/(4\pi r^2)\); \(1/r^2\) from energy conservation.

  • Electric and magnetic energy densities in an EM wave are equal: \(u_E = u_B\).

Discussion#

Energy and momentum in fields#

We usually think of electric (\(\vec{E}\)) and magnetic (\(\vec{B}\)) fields as just math on a chalkboard. They aren’t. They are real, physical entities, like matter.

  • The electromagnetic (EM) field itself can store energy and momentum.

  • It is not just a “messenger” between charges; it is a physical system with its own “weight”.

  • During interactions, matter and field can exchange energy/momentum, while total energy and total momentum still follow conservation laws.

Energy and momentum density#

We use the density language to characterize the distribution of energy and momentum in the field. For EM fields in vacuum,

  • Energy density \(u\): energy per unit volume (J/m\(^3\)).

\[ u = \frac{1}{2}\varepsilon_0 E^2 + \frac{1}{2\mu_0}B^2 \]

  • Momentum density \(\vec g\): momentum per unit volume (kg/(m\(^2\)·s), equivalently N·s/m\(^3\)).

\[ \vec g = \varepsilon_0\,(\vec E\times\vec B) \]

Thought experiment: building the field in a box#

Use one small rectangular box in vacuum, volume

\[ V=\Delta x\,\Delta y\,\Delta z. \]

Turn fields on from 0 to final values over a short time.

  • For energy: accumulate work/power into the box.

  • For momentum: accumulate impulse into the box.

Energy#

You have to do work in order to establish an EM field, and that work is where the energy comes from.

Electric part of energy density \(u_E\)

Let \(\vec E= E \hat z\).

  • Electric potential drop across height: \(V_{\text{pot}}=E\,\Delta z\).

  • Charge on top/bottom faces from Gauss law: \(Q=\varepsilon_0E(\Delta x\Delta y)\).

  • Charging current while \(E\) builds: \(I=dQ/dt=\varepsilon_0(\Delta x\Delta y)\,dE/dt\).

  • Power input to the box to build \(E\): \(P=V_{\text{pot}}I\).

Energy for \(\vec E\) field accumulates over time by the power input:

\[ U_E=\int P\,dt =\int (E\Delta z)\,\varepsilon_0(\Delta x\Delta y)\frac{dE}{dt}\,dt =V\,\varepsilon_0\int_0^E E'\,dE' =V\left(\frac12\varepsilon_0E^2\right). \]

Therefore

\[ u_E=\frac{U_E}{V}=\frac12\varepsilon_0E^2. \]

Magnetic part of energy density \(u_B\)

Let \(\vec B= B \hat z\).

  • Changing flux through area \(\Delta x\Delta y\) gives electromotive force (EMF): \(V_{\text{emf}}=d\Phi_B/dt=(\Delta x\Delta y)\,dB/dt.\)

  • Circular current to sustain magnetic field, by Ampere law along length \(\Delta z\): \(B\,\Delta z=\mu_0 I\;\Rightarrow\;I=\frac{B\Delta z}{\mu_0}.\)

  • Power input to the box to build \(B\): \(P=V_{\text{emf}}I\).

Energy for \(\vec B\) field accumulates over time by the power input:

\[ U_B=\int P\,dt =\int \left(\Delta x\Delta y\frac{dB}{dt}\right)\left(\frac{B\Delta z}{\mu_0}\right)dt =V\,\frac1{\mu_0}\int_0^B B'\,dB' =V\left(\frac1{2\mu_0}B^2\right). \]

Therefore

\[ u_B=\frac{U_B}{V}=\frac1{2\mu_0}B^2. \]

Put together, \(u=u_E + u_B\).

Momentum#

In physics, momentum isn’t just “mass times velocity”—it’s a record of impulse. If a force pushes on the EM field for a certain amount of time, it will give the field a momentum.

Momentum density \(\vec g\)

Choose directions \(\vec E= E \hat x\), \(\vec B= B \hat y\).

  • As \(E\) grows, displacement current through area \(\Delta y\Delta z\) is \(I_d=\varepsilon_0\frac{d\Phi_E}{dt}=\varepsilon_0(\Delta y\Delta z)\frac{dE}{dt}\).

  • This current in magnetic field \(B\) feels Lorentz force \(F=I_d\,\Delta x\,B=\varepsilon_0(\Delta x\Delta y\Delta z)B\frac{dE}{dt}\).

  • Impulse received over time: \(dp = F\,dt\).

Lorentz force constantly pushing on the EM field creats the momentum:

\[ p=\int F\,dt=V\,\varepsilon_0B\int_0^E dE'=V(\varepsilon_0EB). \]

Therefore, the scalar density in this geometry:

\[ g=\frac{p}{V}=\varepsilon_0EB. \]

Direction is \(\hat x\times\hat y=\hat z\), so in vector form:

\[ \vec g=\varepsilon_0(\vec E\times\vec B). \]

The moment the EM field hit an object—like a mirror or a solar sail—the field “collapses,” and pays back that debt, delivering that stored momentum to the object, creating what we measure as radiation pressure.

Connect back to Poynting vector#

Now let us compare two definitions:

(236)#\[ \vec S=\frac1{\mu_0}(\vec E\times\vec B), \qquad \vec g=\varepsilon_0(\vec E\times\vec B). \]

Both vectors point along \(\vec E\times\vec B\), but they describe different things:

  • \(\vec S\): energy flow rate (How much power power transmitted through each scqure).

  • \(\vec g\): momentum density (How much “kink” is sored in each cube).

Since \(c^2=1/(\mu_0\varepsilon_0)\), (236) implies a universal relation:

\[ \vec S=c^2\vec g, \qquad \vec g=\frac{\vec S}{c^2}. \]

Why \(c^2\)?

Imagine energy and momentum of EM field is carried by a particle call photon.

  • Setup: suppose a photon carries energy \(E_\gamma\) and moving with velocity \(\vec{c}\).

  • Momentum: the photon momentum is \(\vec{p}_\gamma = E_\gamma / c^2 \vec{c}\) (relativity \(E=cp\) for massless particle)

  • Energy flow: It will contribute to energy flow of \(E_\gamma \vec{c}\) (energy \(\times\) flow velocity)

  • Density perspective: Suppose the photon density is \(n\): \(\vec{g}=n \vec{p}_\gamma = n E_\gamma /c^2 \vec{c}\), \(\vec{S}=n E_\gamma \vec{c}\)

Therefore \(\vec{S}=c^2 \vec{g}\)