Physics 504, Lecture 7 Feb 14, Energy Flow, Density and Attenuation. 1.1 Energy flow and density. Last Latexed: February 11, 2011 at 10:54 1

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1 Last Latexe: February, at :54 Physics 54, Lecture 7 Feb 4, Energy Flow, Density an ttenuation We have seen that there are iscrete moes for electromagnetic waves with E, B e ikz it corresponing, for real an k, to waves travelling in the z irection, with a ispersion relation k = µɛ γ, with iscrete values of γ for TE or TM moes, or γ = for TEM moes. This relation for TE an TM moes are of the same form as for the high-frequency behavior of ielectrics, (J 7.6) with γ/ µɛ playing the roll of the plasma frequency. The phase velocity is v p = /k = µɛ > µɛ, greater than the velocity in an infinite meium (R 3 ), as for all travelling waves we have >. On the other han, as k µɛ = constant, we have kk = µɛ, so the group velocity v g = k = k µɛ = <, µɛ v p µɛ less than the infinite meium velocity.. Energy flow an ensity These calculations assume the walls ha infinite conuctivity, but generally the walls are goo but not perfect conuctors,. Then there will be ey currents an energy loss in the walls, an a right-moving wave will be attenuate in the z irection, so k will evelop a small positive imaginary part. We may estimate this attenuation constant by comparing the energy travelling past z to the energy per unit length lost at z. The energy flux is given by the z component of the Poynting vector, so the power transmitte own the pipe is P = ẑ S. Up to now we have worke with quantities linear in E or H so the proviso that the real fiels are just the real parts of our complex fiels coul be swept 54: Lectures 7 Last Latexe: February, at :54 uner the rug. But the Poynting vector S = E phys H phys is quaratic. Our wave propagating in the z irection is actually E phys (x, y, z, t) = ( E(x, y, k, )e ikz it + E (x, y, k, )e ikz+it). So when we talk in terms of a wave with a single k an, we have E phys (x, y, z, t) = Re E(x, y, z, t) = Re ( E(x, y, k, )e ikz it ), S phys = E phys H phys = 4( ( E(x, y, k, )e ikz it + E (x, y, k, )e ikz+it) = 4 ( H(x, y, k, )e ikz it + H (x, y, k, )e ikz+it)) ( E(x, y, k, ) H(x, y, k, )e ikz it + E (x, y, k, ) H(x, y, k, )+ E(x, y, k, ) H (x, y, k, ) + E (x, y, k, ) H (x, y, k, )e ikz+it ) The first an last term are rapily oscillating, so if we are intereste in the average of S,wehave S = ( E (x, y, k, ) H(x, 4 y, k, )+ E(x, y, k, ) H (x, y, k, ) ) = Re ( E(x, y, k, ) H (x, y, k, ) ) The power transmitte own the waveguie is the integral of the z component of this, so only the transverse components of E an H are neee, an these are given by TM: E z = ψ, Et = i k γ t ψ, TE: H z = ψ, Ht = i k γ Ht = i ɛ ẑ γ t ψ, ψ S = t ψ, Et = i µ ẑ γ t ψ, n S =

2 54: Lectures 7 Last Latexe: February, at : : Lectures 7 Last Latexe: February, at :54 4 with ( t + γ )ψ =. Thus the average transmitte power is P =ẑ Re S = k { γ ɛ (for TM) 4 t ψ µ (for TE) The integral t ψ = ψ S n ψ t ψ =+γ ψ ψ. With := γ / µɛ, k = µɛ /,thisis P = { ɛ (for TM) ψ ψ µɛ µ (for TE) We might also calculate the energy per unit length in the waveguie, U = u = ( Ephys D phys + B phys H ) phys,so U = ɛ E 4 + µ H Here the longituinal (ẑ) components enter as well as the transverse ones. For the TM moe, we have H = H t = E = E t + E z = ɛ t ψ γ ( k γ ) t ψ + Ez s E z = ψ, U = ( k ɛ 4 γ ) ɛ 4 t ψ + ψ + µ γ t ψ = ɛ µɛ ψ = ɛ ψ γ (TM moe). Similarly for the TE moe, we nee H t = ikγ t ψ an H z = ψ H = k γ 4 ψ k + ψ = + ψ = ψ γ while so Note that in either case, E = µ γ 4 U = µ ψ = µ ɛ ψ (TE moe). P U = ɛµ = v g. ψ, s we might expect, the energy flux is the energy ensity times the group velocity.. ttenuation We now turn to the major effect of the finiteness of the conuctivity σ in the walls of the waveguie. We saw in section 8. that the power loss per unit area of interface is given by P loss = H δσ, where δ = /µ c σ is the skin epth. This gives a loss of power per unit length proportional to the square of the fiels an therefore to the power, so the power transmitte along the waveguie will fall off exponentially. This can be escribe by giving k a small imaginary part, Im k = β, giving a factor e βz to each of the fiels, an e βz to the power flow. Thus P z = β P (z) = σδ For the TM moe, so β = 4σδ ˆn H ˆn H =ˆn H t = iɛ ˆn (ẑ γ t ψ)= iɛ γ ɛ γ / kɛ n γ 4 ψ = ɛ kσδ l. (ˆn t ψ ) ẑ / n ψ

3 54: Lectures 7 Last Latexe: February, at : : Lectures 7 Last Latexe: February, at :54 6 Though we on t know the ratio of the integrals, they iffer only in that numerator takes only the square of one of the two components of ψ, an integrates it over the circumference rather than the area. Furthermore the ratio of these integrals epens only on the moe an not on the frequency of the transmitte wave. Let us write / n ψ = ξ, where an are the circumference an the area of the wave guie s cross section, an ξ is a imensionless number expecte to be of orer-ofmagnitue. Let us isplay the frequency epenance explicitly by writing δ = δ /, k = µɛ /.Thus β = ɛ µ σδ / For the TE moe, ˆn H =ˆn H t +ˆn ẑh z so ˆn H = ˆn H k t + Hz = ˆn t ψ γ + ψ. gain let us write ˆn t ψ / ψ / = ξ, ψ ψ = ζ. where ζ is another imensionless number of orer one, an ξ is somewhat ifferently efine from the TM case, but still of orer one. Then ˆn t ψ / ψ = γ ξ. The attenuation coefficient is ɛ / ] β = [ξ + ζ µ σδ ξ. Then for both moes we can write the attenuation coefficient as [ ɛ / ] β = ξ + η, µ σδ where η = ζ ξ for the TE moe. Note that for the TM moe η =. The attenuation coefficient calculate in this approximation iverges as, an is it is proportional to for large, soithasaminimumfor some small multiple of, 3 for the TM moes, but epenent on η /ξ, an hence the shape of the waveguie, for TE moes. We will skip section ttenuation for ircular yliner We have seen that the TE an TM moes in a circular wave guie are etermine by ψ TE mn = J m(x mn ρ/r)cosmφ, ψtm mn = J m(x mn ρ/r)cosmφ, where x mn an x mn are the n th zeros of J m(x) anj m (x) respectively. The cutoff frequencies are given in terms of γ TE mn = x mn /r, γtm mn = x mn/r. To evaluate the attenuation coeffieients, we nee ψ,an π = r φγ J n m (γr)cos φ = πrγ J m (γr)( + δ m), which we nee only for TM moes. For TE moes we nee π ψ = rjm(x mn) cos mφ φ = πrjm(x mn)( + δ m ), π ˆn t ψ = r φ = π r φ r J m (x mn ) (m sin mφ) = π r J m(x mn)( + δ m ),

4 54: Lectures 7 Last Latexe: February, at : : Lectures 7 Last Latexe: February, at :54 8 The only integral that requires more than table look-up is r π r ψ = ρρjm(γρ) φ cos (mφ) =π ρρjm(γρ)( + δ m ). The integral is connecte to the orthnormalization properties of Bessel functions, an is : [J m (x mn u)] uu = J m+(x mn ) [J m (x mnu)] uu = ( ) m J (x m(x mn) mn ) Thus for the TM moes, we have / ξtm mn = (γ TM n mn) ψ = πrj m (x mn) πr J m+(x mn ) = J m (x mn) r Jm+(x mn ) In fact, there is an ientity (see footnote again) J m(x) = m J x m(x) J m+ (x), which means, as J m (x mn ) =, that J m (x mn) = J m+ (x mn ), ξtm mn =/r, an ɛ βmn TM = / µ rσδ for all TM moes. For the TE moes, / ξte mn = ˆn t ψ (γmn TE ) ψ m πj = m(x mn)/r = m r(x mn m ). π(γ TE mn ) r ( (m/x mn ) ) J m (x mn ) / ζ mn TE = ψ ψ πrjm = (x mn ) π ( (m/(x mn) )) Jm(x mn) = x mn r (x mn m ). So the attenuation coefficient is [ ɛ βmn TE = / ] µ rσδ (x mn m ) +. rfken.5, problems.. an..3 (3r E.) or see For TM moes, mn TM = x mnc/r, wherec =/ µɛ is the spee of light, For copper, the resistivity is ρ = σ =.7 8 Ω m, an we may take the permeability to be essentially µ. lso = γ c. δ = /µ c σ ɛ =8.854 /N m,so ɛ cɛ γ m = µ σδ σ = γ s N m Ωm = m / xmn r. The units combine to m / as Ω = V/ =(J/)/(/s) = Nms/. In comparison to the TM moe for a square of sie a, weseethatβ TM = a r β TM. s the cutoff frequencies are.448c/r an 5πc/a respectively, we see that the comparable imensions are r =(.448/ 5π)a =.34a, much smaller, an then a/r =.46, so the smaller pipe oes have faster attenuation. For TE moes, there is an extra factor of (x mn m ) +. whichforthelowestmoeis ( /) compare to.5 +( /) for the square. But the cutoff frequencies are now.84c/r an πc/a, so comparable imensions have r =.84a/ π =.44a..4 Resonant avities We have consiere wave guies uniformly extene in the z irection, infinite in both irections, an foun that there are moes of propagation with arbitrary efinite wavenumber k an frequency given by µɛ = k + γ. Thus for a particular an >, there are two possible waves, a rightmoving an a left-moving one. Superposition will then give us staning waves suitable to escribe a resonant cavity mae by placing flat conucting en-caps on the wave guie, say at z =anz =. Thus quite generally each fiel will be a superposition of wave with k = k an one with k = k. For the TM case, the etermining fiel is E z = ( ψ (k) e ikz + ψ ( k) e ikz) e it, In calculating the transverse fiel, the piece coming from ψ ( k) nees the minus in 8.33, so E t = i k ( t ψ (k) e ikz γ t ψ ( k) e ikz)

5 54: Lectures 7 Last Latexe: February, at : : Lectures 7 Last Latexe: February, at :54 This is a fiel parallel to the conuctor surface at z =anz =, so must vanish (or be very small) there for a perfect (goo) conuctor encap. Vanishing of E t at z = implies ψ (k) = ψ ( k),anthenψ (k) (i sin k) = from z =. sweon twantψ =,weseethatk = pπ/ with p an integer, pπz E z =cos ψ(x, y) E t = pπ γ H t = i ɛ cos γ pπz sin t ψ pπz ẑ t ψ { for TM moes with p Z where in using 8.6, we recall that k takes a minus sign for the part of the fiel flowing leftwar. For TE moes, the etermining fiel is H z, which must vanish at z = an z = if the encaps are perfect conuctors an exclue magnetic fiels, as ˆn B is continuous at the bounary. Thus pπz H z =sin ψ(x, y) H t = pπ γ E t = i µ γ pπz cos t ψ pπz sin ẑ t ψ { for TE moes with p Z,p.. s for the waveguie, the values of γ epen on the moe (TE or TM) an the cross section, which often means two inices. For example, for a circular guie, there is an angular inex m an another inex n specifying which root of J m (fortm)orofj(x)/x (for TE). With x mn the n th zero of J m (x) (notcountingx = ) an x mn the n th zero of J (x), we have x γ mn = x mn /R (TM moes) or γ mn = x mn /R (TE moes). s µɛ = k + γ we have mnp = mnp = x mn µɛ R + p π x mn µɛ R + p π with p fortmmoes, with p>fortemoes. The lowest TM moe is thus = x / µɛ R =.45/ µɛ R, whichis inepenent of the length of the cavity. s p cannot be zero for TE moes, (or the etermining fiel H z = ) the lowest TE moe is = R µɛr, (π/.84 =.9). This moe has the avantage that its frequency can be tune by moving a piston back an forth, changing. This calculation has assume no power losses, but of course a real cavity will generally have walls of finite conuctivity, an power will be lost as we iscusse earlier in the walls, not only along the z irection but also in the encaps. gain the power lost will be proportional to the energy store in the cavity. Let Q be the π times the energy store U ivie by the energy lost in one cycle U (in time t =π/), so Q =π U. ssuming Q, U the energy loss per cycle will be small compare to U, with U π U, t an the energy will ecay exponentially, with U(t) =U()e t/q, Q = U/ U/t. This means that if at time zero something excites an electromagnetic fiel in the cavity, the fiels will have a time epenence E(t) =E e i( i/q)t Θ(t), where Θ(t) = fort an zero earlier. Thus the frequency response to what is essentially a elta-function excitation (an therefore equal for all frequencies) is E() = π = ie π i/, E(t)e it t = E e i( i/)t t π My is what Jackson calls +, with his the resonant frequency of the unampe cavity. The change in resonant frequency ue to amping is generally small, a fractional change of the orer /Q, an I will ignore that effect.

6 54: Lectures 7 Last Latexe: February, at :54 where := /Q. This response etermines how the cavity will respon to excitations of any frequency, with the energy absorbe proportional to E() ( ) + /4. This resonance shape is the form of response the 3 simplest resonant structures have in response to a stimulus of frequency, as for example the ratio of the energy of ampe harmonic oscillator to the riving force, or the resonant absorption of light by an atomic transition. In nuclear physics this is calle the Breit-Wigner amplitue., mistakenly calle the half-with, is actually the full with of the region with a response at least half the maximum value, which is [ /, +/]. The value of for a resonant cavity can be calculate as for the attenuation of a waveguie. That is, we compare the power lost in the walls to the energy in the electromagnetic fiels in the cavity. This is one in Jackson, pp , base on the same tools as use in calculating the attenuation of the waveguie, but I will skip it. Earth an Ionosphere n interesting resonant cavity is forme by the surface of the Earth an the ionosphere, a layer of ionize gas starting about km up, which provies sufficient conuctivity to reflect raio waves. But this cavity is not a cyliner with encaps, but clearly calls out for spherical coorinates. s you also nee an introuction to other curvilinear coorinate systems to o your homework, an as you have tol me you have not learne about these, let us igress to iscuss curvilinear coorinates in general, orthogonal coorinates more particularly, an finally spherical coorinates.