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Find all elements of the Maxwell stress tensor for a monochromatic plane wave traveling in the z direction and linearly polarized in the x direction (Eq. 9.48). Does your answer make sense? (Remember that -Trepresents the momentum flux density.) How is the momentum flux density related to the energy density, in this case?

Short Answer

Expert verified

Answer

The all the elements Tzzare 0, where Tzz-ε0E02cos2(kz-ωt+δ), the answer make sense as the direction of the field is in the z direction. The relation between momentum flux density and the energy is 1cuz^.

Step by step solution

01

Expression for the electric field, and magnetic field:

Write the expression for the electric field.

Ez,t=E0coskz-ωtx^ ……. (1)

Write the expression for the magnetic field.

Bz,t=1cE0coskz-ωty^ ……. (2)

02

Determine the required relation:

The momentum flux density Tijis given by,

Tij=ε0(EiEj-12δijE2)+1μ0(BiBj-12δijB2)

With the fields in Eq. 9.48, E has only an x component, and B only has a y component. So, all the “off-diagonal” (ij)terms will be zero.

As for the “diagonal” elements:

Txx=ε0(ExEx-12E2)+1μ0(-12B2)=12(ε0E2-1μ0B2)=0

Solve for second diagonal element.

Tyy=ε0(-12E2)+1μ0(ByBy-12B2)=12(-ε0E2+1μ0B2)=0

Solve for third diagonal element.

Tu=ε0(-12E2)+1μ0(-12B2)=-u

So, Tzz=-ε0E02cos2(kz-ωt+δ)(all other elements zero).

The momentum of these fields is in the z direction, and it is being transported in the z direction, so yes, it does make sense that Tzzshould be the only nonzero element in Tij.

It is known that localid="1658405424229" -T·dais the rate at which momentum crosses an area da. Here we have no momentum crossing areas oriented in the x or y-direction.

The momentum per unit time per unit area flowing across a surface oriented in thez-direction is,

-Tzz=u=gc

Therefore,

p=gcATpT=gcA=momentumperunittimecrossingareaA

It is known that momentum flux density is equal to energy density. Therefore,

g=1cε0E02cos2(kz-ωt+δ)z^=1cuz^

Therefore, the all the elements Tzzare 0, where Tzz-ε0E02cos2(kz-ωt+δ), the answer make sense as the direction of the field is in the z direction. The relation between momentum flux density and the energy is 1cuz^.

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Most popular questions from this chapter

In writing Eqs. 9.76 and 9.77, I tacitly assumed that the reflected and transmitted waves have the same polarization as the incident wave—along the x direction. Prove that this must be so. [Hint: Let the polarization vectors of the transmitted and reflected waves be

n^T=cosθTx^+sinθTy^,n^R=cosθRx^+sinθRy^prove from the boundary conditions that θT=θR=0.]

Question:According to Snell's law, when light passes from an optically dense medium into a less dense one the propagation vector bends away from the normal (Fig. 9.28). In particular, if the light is incident at the critical angle

θc=sin-(n2n1)

Then , and the transmitted ray just grazes the surface. If exceeds , there is no refracted ray at all, only a reflected one (this is the phenomenon of total internal reflection, on which light pipes and fiber optics are based). But the fields are not zero in medium ; what we get is a so-called evanescent wave, which is rapidly attenuated and transports no energy into medium 2.26

Figure 9.28

A quick way to construct the evanescent wave is simply to quote the results of Sect. 9.3.3, with and

kT=kTsinθTx^+cosθTz^

the only change is that

sinθT=n1n2sinθI

is now greater than, and

cosθT=1-sin2θT

is imaginary. (Obviously, can no longer be interpreted as an angle!)

(a) Show that

ET(r,t)=E0Te-kzeI(kx-ωt)

Where

kωc(n1sinθ1)2-n22

This is a wave propagating in the direction (parallel to the interface!), and attenuated in the direction.

(b) Noting that (Eq. 9.108) is now imaginary, use Eq. 9.109 to calculate theirreflection coefficient for polarization parallel to the plane of incidence. [Notice that you get reflection, which is better than at a conducting surface (see, for example, Prob. 9.22).]

(c) Do the same for polarization perpendicular to the plane of incidence (use the results of Prob. 9.17).

(d) In the case of polarization perpendicular to the plane of incidence, show that the (real) evanescent fields are

Er,t=E0e-kzcoskx-ωty^Br,t=E0ωe-kzksinkx-ωtx^+kcoskx-ωtz^

(e) Check that the fields in (d) satisfy all of Maxwell's equations (Eq. 9.67).

(f) For the fields in (d), construct the Poynting vector, and show that, on average, no energy is transmitted in the z direction.

The "inversion theorem" for Fourier transforms states that

ϕ(Z)=-ϕ(k)eikzdkϕ(k)=12π-ϕ(z)e-ikzdz

Use this to determine A(k), in Eq. 9.20, in terms of f(z,0)andf*(z,0)

Question: Use Eq. 9.19 to determineA3andδ3in terms ofrole="math" localid="1653473428327" A1,A2,δ1, andδ2.

A microwave antenna radiating at 10GHzis to be protected from the environment by a plastic shield of dielectric constant2.5. . What is the minimum thickness of this shielding that will allow perfect transmission (assuming normal incidence)? [Hint: Use Eq. 9.199.]

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