26 Standing waves, radiation pressure

Transcription

1 26 Standing waves, radiation pressure We continue in this lecture with our studies of wave reflection and transmission at a plane boundary between two homogeneous media. In case of total reflection from a perfectly conducting mirror placed at =surface, Γ=, andtheincidentandreflectedwavesin < region combine to produce standing waves of electric and magnetic field: H i y H t = E i E t = x Region Region 2 E r H r Incidentwave(a traveling wave going in -direction): Ẽ i =ˆxE o e jβ and H i =ŷ E o e jβ, Reflectedwave(a traveling wave going in -direction): Standingwave: Ẽ r = ˆxE o e jβ and H r =ŷ E o e jβ, Ẽ = Ẽi+Ẽr =ˆxE o (e jβ e jβ ) and which simplify as H = H i + H r =ŷ E o (e jβ +e jβ ) Ẽ = j ˆx2E o sin(β ) and H =ŷ 2E o cos(β ). Standing waves

2 These are called standing wave phasors because when we go to the time-domain (by multiplying with e jωt and taking the real time as usual) we obtain: E x (, t) sin(β)sin(ωt) E(, t) =ˆx2E o sin(β )sin(ωt) and H(, t) =ŷ 2E o cos(β )cos(ωt); these, unlike d Alembert solutions of the format f(t v ),describeoscillations in time t, with different amplitudes at different positions (see margin and the animation linked in class calendar). λ = 2π β H y (, t) cos(β)cos(ωt) Standing waves carry no net energy, that is, with standing wave fields we have E H =, because of the cancellation of the power transported by its traveling wave components in opposite directions. Verification: Using the phasors we have Ẽ = jˆx2e o sin(β ) and H =ŷ 2E o cos(β ), E H = 2 Re{Ẽ H } = 2 Re{ jˆx2e o sin(β ) ŷ 2E o = ẑ 2E2 o sin(β ) cos(β )Re{ j} =. cos(β )} λ 2 Note: Nulls in Ex and Hy are separated by half wavelength. Adjacent nulls of Ex and Hy are separated by quarter wavelength. It is useful to think of nulls of Ex as "shorts" in analogy to shorts in circuits where v=. Conductor shorts Ex on its surface where a current flows. Also useful to think of nulls of Hy as "opens" in analogy to opens in circuits where i=. 2

3 Note that E(,t)=on =surface satisfying the tangential electric boundary condition as expected (recall that the fields are ero within the perfect conducting mirror). Also note that H(,t)=ŷ 2E o cos(ωt) on =surface. Since this tangential magnetic field is not ero, boundary condition equations imply that there must be an oscillating surface current E x (, t) sin(β)sin(ωt) λ = 2π β H y (, t) cos(β)cos(ωt) satisfying J s =ˆx 2E o cos(ωt) A m, ẑ H(,t)=J s. J s on mirror surfaces is really a convenient idealiation of volume currents flowing in thin layers just a few skin depths near goodconductor surfaces (real-life mirrors are good but not perfect conductors!). Radiation due to J s in effect causes the reflected wave and also cancels out the incident wave field in >. Next we examine reflections from a good conductor and see of how the limiting case of a perfect conductor is naturally reached. λ 2 Note: Nulls in Ex and Hy are separated by half wavelength. Adjacent nulls of Ex and Hy are separated by quarter wavelength. It is useful to think of nulls of Ex as "shorts" in analogy to shorts in circuits where v=. Conductor shorts Ex on its surface where a current flows. Also useful to think of nulls of Hy as "opens" in analogy to opens in circuits where i=. 3

4 Going back to the partial reflection case, consider the transmitted fields Ẽ t and H t in Region 2 shown in the margin. Also shown in the margin are the phasors for current density J t and magnetic flux density B t. In the box below we integrate the volumetric current density J t of a good conductor from =to and find out that this depth integral matches the surface current density found above for the case of perfect conductor. In this calculation we assume that Region is vacuum, and also take µ 2 = µ o : Ẽ t =ˆxτE o e γ 2 H t =ŷ τe o e γ 2 J t = σ 2 Ẽ t =ˆxσ 2 τe o e γ 2 B t = µ 2 Ht =ŷ µ 2τE o e γ 2 Effective surface current: Assuming that Region 2 is a good conductor, jωµ 2 jωµ2 = and γ 2 = jωµ 2 (σ 2 + jωɛ 2 ) jωµ 2 σ 2 σ 2 + jωɛ 2 σ 2 and therefore τ = 2 2 and σ 2 τ 2σ 2 = 2 jωµ2 σ 2 = 2γ 2. + The depth integral of the volumetric current density in Region 2, that is, the effective surface current of the region is then J t d =ˆx σ 2 τe o e γ 2 d =ˆxE o γ 2 (σ 2 τ)=ˆx 2E o in phasor form, matching the phasor of the time-domain result from above, namely J s =ˆx 2E o cos(ωt) A m representing the surface current on an idealied perfect conductor surface. 4

5 Surface resistance: Let J s stand for the effective surface current of a good conductor with a propagation constant γ jωµσ = α + jβ = α + jα and a volumetric current density J() such that J s = = J()d = = J()e γ d = J() γ. In that case J() = J s γe γ and E() = J s γ σ e γ inside the good conductor in terms of the effective surface current J s,andtheaverage power dissipated per unit volume (Joule heating) is J() E() = 2 J s γ 2e 2α = σ 2 J s 22α2 e 2α. σ The depth integral of the same quantity, that is the power dissipated per unit area, is then J() E() d = 2 R s J s 2, with R s α πfµσ πfµ σ = = σ σ called the surface resistance. The surface resistance concept is useful to model loss effects in waveguides and cavity resonators as studied in ECE 45. Also, we can make use of surface resistance when modeling lossy transmission lines (see Lecture 39). (Ω) 5

6 Let s finally calculate the magnetic component of the Lorent force on charge carriers of a good conductor due to the penetrating fields: Radiation pressure proportional to Radiation pressure: If there are N free charge carriers per unit volume inside a reflecting mirror, then NF = Nqv B t = J t B t will be the force per unit volume of the mirror, expressed in terms of current density J t = Nqv and the magnetic flux density B t. Its integral over all can be interpreted as the total force per unit area of the mirror, P rad = J t B t d, having a magnitude known as radiation pressure of the reflecting wave. This is a time-varying quantity, with a time-average P rad = 2 Re{ J B }d = ẑ 2 Re{(σ 2τE o )( µ 2τE o )}e 2α2 d where = ẑ E o 2 2 Re{(2γ 2 )( µ 2 2 )} 2α 2 =ẑ2 E o 2 2 Re{γ 2 } α 2 µ 2 = ẑ2 E o 2 2 µ o =2 S i /c, S i ẑ E o 2 2 is the time-average Poynting vector of the incident wave reflected from the mirror (factor of 2 in P rad is due to the recoil of the wave off the mirror; see Rothman and Boughn, Am. J. Phys., 77,22,977). S i /c shows that plane-tem waves not only carry and transport energy, but also momentum. TEM waves not only heat, butalsopush! ((It can also be shown that the momentum density of the wave is S i /c 2 N.s/m 3 and (spin) angular momentum density ± S i /ωc N.s/m 2 for right- and left-circular waves. Momentum per photon of energy ω can be obtained by dividing the above expressions by S i /ωc, the number density of photons in the wave field.)) 6