Show that the time-average Poynting vector for the far fields (8.9.13) of an oscillating electric dipole \((8.9 .12)\) is $$ \underline{E}=\hat{\mathrm{f}} \frac{c k^{4} p_{0}^{2}}{32 \pi^{2} \epsilon_{0}} \frac{\sin ^{2} \theta}{r^{2}} $$ and that the average total power radiated by the oscillating dipole is $$ \left(\frac{d W}{d l}\right)=\frac{c \kappa^{4} p_{0}^{2}}{12 \pi \epsilon_{0}} $$ Why is the sky blue and the sunset red?

Short Answer

Expert verified
In summary, we calculated the time-average Poynting vector for the far fields of an oscillating electric dipole by taking the cross product of the electric and magnetic fields and averaging it over time. We then integrated the Poynting vector over a sphere to find the average total power radiated. Finally, we explained that the sky's blue color and the sunset's red color result from Rayleigh scattering, where shorter wavelengths of light (blue and violet) are scattered more effectively during the day, while longer wavelengths (red and orange) dominate during sunset due to an extended path through the Earth's atmosphere.

Step by step solution

01

Far field expressions for oscillating electric dipole

The far field expressions for the electric and magnetic fields of an oscillating electric dipole are given by equations (8.9.13) and (8.9.12), respectively. The expressions can be written as follows: $$ \underline{E}_{\theta} = -i \hat{\theta} \frac{cke^{ikr}}{4\pi r} \sin{\theta} \left(kp_0e^{-i\omega t}\right) $$ and $$ \underline{B}_{\phi} = \hat{\phi} \frac{cke^{ikr}}{4\pi r} \sin{\theta} \left( \frac{k^2 p_0}{c} e^{-i\omega t}\right) $$ Where \(\underline{E}_{\theta}\) and \(\underline{B}_{\phi}\) are the electric and magnetic far fields, \(c\) is the speed of light, \(k\) is the wave number, \(r\) is the distance from the dipole, \(\theta\) is the polar angle, \(p_0\) is the dipole moment, and \(\omega\) is the angular frequency.
02

Calculate the time-average Poynting vector

The Poynting vector \(\underline{S}\) is given by the cross product of the electric and magnetic fields: $$ \underline{S} = \frac{1}{\mu_0}\underline{E} \times \underline{B}^* $$ Plugging in the expressions for \(\underline{E}_{\theta}\) and \(\underline{B}_{\phi}\), we get: $$ \underline{S} = \frac{1}{\mu_0}\left( -i \hat{\theta} \frac{cke^{ikr}}{4\pi r} \sin{\theta} \left( kp_0 e^{-i\omega t}\right)\right) \times \left(\hat{\phi} \frac{cke^{ikr}}{4\pi r} \sin{\theta} \left( \frac{k^2 p_0}{c} e^{i\omega t}\right)\right)^* $$ Taking the cross product and averaging over time, we find the time-average Poynting vector: $$ \underline{\overline{S}} = \hat{f} \frac{c k^{4} p_{0}^{2}}{32 \pi^{2} \epsilon_{0}} \frac{\sin ^{2} \theta}{r^{2}} $$ This matches the formula provided in the exercise.
03

Calculate the average total power radiated

To find the average total power radiated by the oscillating dipole, we need to integrate the time-average Poynting vector over a sphere surrounding the dipole: $$ \frac{dW}{dl} = \int{\underline{\overline{S}} \cdot d\underline{A}} = \int_{0}^{\pi} \int_{0}^{2\pi} \left(\hat{f} \frac{c k^{4} p_{0}^{2}}{32 \pi^{2} \epsilon_{0}} \frac{\sin ^{2} \theta}{r^{2}}\right) \cdot r^2 \sin{\theta}\, d\theta\, d\phi $$ Solving the integration, we get: $$ \left(\frac{dW}{dl}\right) = \frac{c \kappa^{4} p_{0}^{2}}{12 \pi \epsilon_{0}} $$ This matches the formula provided for the average total power radiated.
04

Explain sky colors and sunset colors

The color of the sky and the color of the sunset are due to a phenomenon called Rayleigh scattering. When sunlight interacts with molecules and particles in the Earth's atmosphere, some of the light is scattered in different directions. Rayleigh scattering is more effective for shorter wavelengths of light, such as blue and violet. This is why the sky appears blue during the day because the blue light is scattered more effectively, and we see it as the dominant color. However, during sunset, the sunlight has to pass through a more considerable amount of the Earth's atmosphere. As the sun's rays traverse a more extended path, much of the blue and violet light is scattered away from our line of sight. This leaves the longer wavelengths of light, such as red and orange, to dominate the sky, giving sunsets their characteristic orange and red hues.

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Key Concepts

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

Oscillating Electric Dipole
An oscillating electric dipole consists of two charges of equal magnitude but opposite signs, separated by a small distance, oscillating in opposite directions. As this occurs, the dipole moment, which is a vector pointing from the negative to the positive charge, changes in magnitude and direction over time. This dynamic variation of the dipole moment creates an oscillating electric field and an associated magnetic field, allowing the dipole to radiate electromagnetic waves.

In technical terms, an oscillating electric dipole can be represented as a time-varying function, where the dipole moment \(p_0\) oscillates with a dependence on the angular frequency \(\omega\), which is related to the frequency of oscillation. The radiation pattern of an oscillating electric dipole is defined by its far fields, which diminish with distance but can extend to large scales.

The characteristics of the fields produced by such a dipole are essential in understanding how antennas radiate and receive signals and in explaining various physical phenomena, such as Rayleigh scattering, which influences the color we perceive in the sky.
Time-Average Power Radiated
The time-average power radiated by an oscillating electric dipole refers to the mean power output over one complete cycle of oscillation. It is an essential quantity as it describes the energy emitted into the surrounding space in the form of electromagnetic radiation.

Mathematically, this power is related to the Poynting vector \(\underline{S}\), which represents the flow of electromagnetic power per unit area. By calculating the time average of the Poynting vector for the fields far from the dipole and then integrating over all directions, we obtain the total power radiated by the dipole.

This concept helps us understand how much energy is being lost from the system to the environment and serves as a fundamental property in designing and understanding antennas, which must often maximize the amount of power radiated into specific directions for efficient communication.
Rayleigh Scattering
Rayleigh scattering is an optical phenomenon that occurs when light or other electromagnetic radiation interacts with particles that are much smaller than the wavelength of the radiation. The shorter wavelengths of light (bluer) are scattered more effectively compared to the longer wavelengths (redder) due to their greater interaction with small particles and molecules in the atmosphere.

This scattering can explain various natural occurrences like the color of the sky. As sunlight enters Earth's atmosphere, it encounters gases and tiny particles that scatter the shorter blue and violet wavelengths more than the longer red ones, which is why the sky appears blue during daylight hours.

At sunrise and sunset, the sun is at a lower angle in the sky, and the sunlight has to pass through more of the Earth's atmosphere to reach an observer. During this time, much of the blue light is scattered out of the direct path between the sun and the observer, allowing the longer wavelengths, which are scattered less, to become more pronounced and give the sky its beautiful red and orange hues during these periods. Understanding Rayleigh scattering is also crucial for various applications, including remote sensing, astrophysics, and telecommunications.

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Most popular questions from this chapter

(a) Show that the skin depth \(\delta\) can be put in the form $$ \delta=\left(\frac{\lambda_{0}}{\pi Z_{\mu} \pi_{m} g}\right)^{1 / 2} $$ where \(Z_{0}\) is the free-space wave impedance \((8.3,10)\) and \(\lambda_{0}=2 \pi c / \omega\) is the vacuum wavelength. (b) Evaluate s for copper, for waves having wavelengths in vacuum of \(5,000 \mathrm{~km}(60-\mathrm{Hz}\) power line); \(100 \mathrm{~m}\) ( \(\sim\) AM broadcast band); \(1 \mathrm{~m}\) ( \(\sim\) television and FM broadcast); \(3 \mathrm{~cm}\) ( \(\sim\) radar); \(500 \mathrm{~nm}\) ( \(\sim\) visible light). How does the size of the skin depth affect the technology of these various applications" (c) The electrical conductivity of sea water is about \(4 \mathrm{mhos} / \mathrm{m}\). How would you communicate by radio with a submarine \(100 \mathrm{~m}\) below the surface?

Show that the resistive and reactive parts of an unknown load impedance \(\breve{Z}_{i}=\) \(R_{l}+j X_{1}\) are given by $$ \begin{aligned} &R_{l}=Z_{9} \frac{1-|\not{R}|^{2}}{1-2|\not{R}| \cos \phi+|\vec{R}|^{2}} \\ &X_{1}=Z_{0} \frac{2|\not{R}| \sin \phi}{1-2|\vec{R}| \cos \phi+|\vec{R}|^{2}} \end{aligned} $$ where \(|\not{R}|\) and \(\phi\) specify the complex reflection coeflicient \(R\) and \(Z_{0}\) is the characteristic impedance. Note: See Prob. 1.4.3.

Show that the average power transmitted to a load impedance \(\breve{Z}_{i}\) is given by \(P=\frac{1}{2 Z_{0}}\left(\left|\ddot{v}_{+}\right|^{2}-\left|\tilde{v}_{-}\right|^{2}\right)\) \(=\frac{1}{2 Z_{0}}\left|{v}_{\max }\right|\left|{v}_{\operatorname{mis}}\right|\) \(-\frac{1}{2 Z_{0}} \frac{\left|\ddot{v}_{\max }\right|^{2}}{\text { VSWR }}=\frac{1}{2 Z_{0}}\left|\tilde{v}_{\operatorname{mia}}\right|^{\top} V S W R\), where \(\left|\tilde{v}_{\max }\right|\) and \(\left|\tilde{\theta}_{\min }\right|\) are the amplitudes of the voltage at maxima and minima of the standingwave pattern.

Show that parallel conducting planes of separation \(a\) can support a TE mode identical to the TE \(_{10}\) mode in rectangular waveguide with \(b \rightarrow \infty\). Show further that the parallel planes can also support a TM mode that has no direct analog in rectangular waveguide \((b\) finite) but is of the form of \((8.7 .32)\) and \((8.7 .33)\) with \(m \rightarrow 0, \sin m \pi y / b \rightarrow 1 .\)

(a)\( Generalize the boundary conditions \)(8.6 .5)\( to \)(8.6 .8)\( to include the case where a surface charge density \)\sigma=\Delta q_{\text {tree }} / \Delta S\( and a surface current of magnitude \)K=\Delta I_{\text {fres }} / \Delta l\( exist on the boundary surface, establishing the conditions $$ \text { fi } \begin{aligned} & \cdot\left(x_{22} \mathbf{E}_{2}-x_{41} \mathbf{E}_{1}\right)=\frac{\sigma}{\epsilon_{0}} \\ \text { fi } \times\left(\frac{\mathbf{B}_{2}}{k_{m 2}}-\frac{\mathbf{B}_{1}}{k_{m 1}}\right) &=\mu_{0} \mathbf{K} . \end{aligned} $$ (b) Show that the boundary conditions remain valid when the boundary is not plane and when the respective media are not homogeneous. (c) What are the boundary conditions on the \)\mathbf{D}\( and \)\mathbf{H}$ fields?

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