sinγ(h y > ) exp(iωt iqx)dωdq

Transcription

1 Lectue 9/28/5 Can we ecove a ay pictue fom the above G fo a membane stip? Such a pictue would be complementay to the above expansion in a seies of integals along the many banches of the dispesion elation. Recall, ou G fo the membane stip was (now with the width h einseted and γ ω 2 q 2 ) G(x, y, y s,t)= 4π 2 sinγy < sinγ(h y > ) exp(iωt iqx)dωdq γ sinγh Last time I evaluated this by esidues and picked up an infinite numbe of poles wheeve sinγh =, one fo each banch of the dispesion elation. The esponse theeby being witten as a sum ove all the diffeent modes of popagation, each tem of which was an integal along the banch ( in eithe q o in ω). This gives a convenient epesentation fo the esponse when the eceive is in the fa field and/o when inteest is confined to a naow fequency o low fequency band such that only a few of the banches contibute. Howeve, we also know that the esponse ought be epesentable as a sum of ays. Had we eplaced the edges with image souces ( and we'd need an infinite numbe of them because we have two edges ), the esponse would be a sum of ays, one fom each image souce, each of which looked like the G fo an unbounded membane, each epesenting some set of eflections of the two edges and containing factos of eflection coefficients off the edges. How can this expession fo G be made to exhibit that kind of behavio? The tick is to ealize that we need a sum of tems that look like exp(iωt iq x x-iq y d y ) whee d y is a distance in the y diection. We can get that by expanding the sin γh in the denominato sinγh = 2i 2i exp( iγh) = exp(iγh) exp( iγh) exp( 2iγh) = 2i exp( iγh) exp( 2inγh) n= This is now a sum ove an index n that epesents a numbe of double eflections off the edges, each eflection incuing an exta phase -2iγh fo one ound tip. G is then witten as a sum of an infinite numbe of tems, a typical one being of the fom G(x, y, y s,t)~ exp(iγy < )exp(iγh iγy > )exp( iγh)exp( 2inγh)exp(iωt iqx) dωdq γ This is the desied ay pictue. Each tem of this is intepetable as a popagation ove some distance in y that may involve seveal bounces off the edges, and a distance x. Each tem of this integand is simila to that seen when we discussed the unbounded membane by doing ou fist integal ove q y : ~exp(iωt iq x x-iη d y )/η. The facto of exp(-2inγh) epesents a ound tip bounce off both edges of the stip, a distance of 2h. 83

2 At shot times we would need only a few of the tems. As each tem is integable exactly (ecall the Cagniad tick), one ends up with an analytic expession fo G as a sum ove ay aivals. This is efficient at shot times t and distances x, with no estiction in fequency. We'll see the same complementay desciptions of esponses in elastic wavguides Begin 3-d Scala waves in unbounded 3-d. Befoe examining the elastic wave eqn in 3-d, it is helpful to take a look at the simple poblem of a scala wave in an unbounded 3-d medium. c 2 2 ψ(x, y, z,t) + t 2 ψ(x, y, z,t)= f (x, y, z;t) ( In acoustics in a fluid, ψ epesents pessue (diffeence fom ambient) and f is a souce tem elated to the density of injected volume, and c 2 is bulk modulus ove mass density. But such intepetations ae unnecessay fo what we will do hee. ) It is not had to show that we may define quantities E 2 c2 ψ ψ + 2 ( tψ) 2 ; F - &ψ ψ that obey an enegy-like continuity equation (with an enegy souce tem on the ight side) d dt E + F = &ψ f The G fo such a medium satisfies c 2 2 G(x, y,z,t) + t 2 G(x, y,z,t)= δ(t)δ(x)δ(y)δ(z) To find it we take a quaduple FT Thus [c 2 q 2 ω 2 ] % G( q,ω)= %G(,ω) = (2π) 3 d 3 q exp( iq ) / [c 2 q 2 ω 2 ] Let us fist examine this at =. Switching to spheical coodinates fo the vecto q: %G( =,ω) = (2π) 3 q 2 dqdφ q sinθ q dθ q / [c 2 q 2 ω 2 ] 84

3 The q integation uns fom to infinity. The φ q fom to 2π the θ q fom to π. %G( =,ω) = 2π 2 q 2 dq / [c 2 q 2 ω 2 ] This diveges. But that is OK. As in the membane, a point foce leads to an infinite esponse at the point of application. One could egulaize this by intoducing a shot distance cutoff ( equivalent to a lage q cutoff) due to a finite footpint of the agent applying the foce. If we do not cae to lose geneality that way, we can still examine the imaginay pat. i Im %G( =,ω) = q 2 dq / [c 2 q 2 (ω iε) 2 ] 2π 2 whee the integal now just uns clockwise ove the semicicle above the pole. The esult is i Im G( % =,ω) = 2π ( iπ) q 2 2 * / [2c 2 q * ] whee q * = ω/c is the value of q at the pole. Thus Im G( % =,ω) = ω 4πc 3 A concentated hamonic souce geneates field ψ that satisfies c 2 2 ψ(x, y, z,t) + t 2 ψ(x, y,z,t)= δ(x)δ(y)δ(z)cos(ωt) The steady state esponse is ψ(x, y, z,t) = Re[ % G(x, y,z;ω)exp(iωt)] The souce is doing wok at a ate Π = &ψ(,,,t) f (t) = Re[iω % G(x, y,z;ω)exp(iωt)]cos(ωt) whose time aveage is < Π >= (ω / 2)Im % G(,,;ω) = ω2 8πc 3 ======= If, the pocess of finding G is moe complicated. But not as had as it was in 2-d. We eplace the dot poduct in the exponent with q cosθ, and we do the easy integal ove φ, to get. %G(,ω) = (2π) 2 We change vaiables ζ = cosθ, to get π q 2 dq sinθ dθ exp( iq cosθ) / [c 2 q 2 ω 2 ] 85

4 %G(,ω) = (2π) 2 = (2π) 2 q 2 dq = 4iπ 2 q 2 dq dζ exp(iqζ) / [c 2 q 2 ω 2 ] [ exp(iq) exp( iq)] iq [c 2 q 2 ω 2 ] q dq exp(iq) / [c 2 q 2 ω 2 ] Thee ae poles at q = ±(ω iε) / c. We close in uppe half plane ( because > ) and pick up the esidue fom q = -ω/c +iε, and obtain = 4iπ 2 (2iπ)( ω / c)exp( iω / c) / [2c2 ( ω / c)] = This is an outgoing spheical wave. exp( iω / c) 2 4πc ( as a useful execise, confim that this has dimensions consistent with Gtilde as defined fom its PDE. c 2 2 % G ω 2 % G = δ(x)δ(y)δ(z) ) ( as anothe useful execise, confim that it has negative imaginay pat at the oigin) G in the time domain is now easy to constuct G(t,) = 2π exp(iωt) %G(ω,) = δ(t / c) 2 4πc Unlike in 2-d, G in the time domain has no tail. G(t,) is a simple geometically attenuating delta function, a shap pulse with amplitude that dops like /. In the HW6. you ae asked to calculate the esponse to a simila looking souce, but with an the x-deivative of a delta function instead of δ(x) and a step function Θ in time instead of a δ(t). The pocedue you ae asked to follow is vey simila to that used above Distibuted souces, on beyond the concentated point souce If a hamonic souce is distibuted in space, c 2 2 ψ(x, y,z,t) + t 2 ψ(x, y,z,t)= f ( )exp(iωt) 86

5 (whee we mean to take the eal pat so that the focing and the esponse ae eal) Then we seek a solution by i) looking fo a steady state esponse ψ = Ψ(x,y,z)exp(iωt) and ii) taking a tiple spatial FT. The esult is c 2 q 2 Ψ( q) ω 2 Ψ( q)= f ( q); Ψ( q) = f ( q) / [c 2 q 2 ω 2 ] = f ( q) % G( q,ω) Let us ask fo the ate at which wok is done by such a souce tem f(x,y,z) exp(iωt). Fo simplicity let us assume that f() is eal. Ou continuity equation above shows what the spatial density of powe is, f ψ/ t. Thus the total powe applied by f is, instantaneously, (notice I einset the eal pat befoe multiplying hamonically vaying quantities) Π(t) = Re &ψ( ;t) Re f ( ;t) d 3 = Re{iωΨ( )e iωt } f ( )cosωt d 3 = (2π) 3 Re d 3 q d 3 {iω f ( q) G( % q,ω)e iωt e i q f ( )cosωt} The integation is easy, it is meely the definition of the spatial FT of f()*. Π(t) = Re &ψ( ;t) Re f ( ;t) d 3 = Re{iωΨ( )e iωt } f ( )cosωt d 3 = (2π) 3 Re d 3 q {iω f ( q) 2 G( % q,ω)e iωt cosωt} If we take a time aveage, this simplifies futhe: < Π >= (2π) 3 ω 2 d 3 q f ( q) 2 Im G( % q,ω) As a check, we confim this gives the familia esult in the case that f() =δδδ. f ( q) = In that case ============= Elastic waves in unbounded 3-d. If the medium is isotopic, ou govening PDE is ρ && u(,t) = (λ + µ) ( u(,t)) + µ 2 u(,t) + f (,t) It diffes fom the fields we have been looking at in that the dependent vaiable is now a vecto field u(,t). And the souce is a vecto field f(,t). We aleady know the dispesion elation; thee ae two types of plane waves, each nondispesive. 87

6 The Geens function is now a 3x3 tenso, and is the solution to the tenso PDE ρ && G ij (,t) = (λ + µ) i k G kj (,t) + µ k 2 G ij (,t) + δ ij δ (3) ( )δ(t) such that the solution u(,t) to the souce f(,t) is given by u i (,t) = dt ' d 3 ' ρ G ij ( ',t t ') f j ( ',t ') G ij (,t) epesents the esponse at in the i diection to a souce at the oigin that is a concentated impulse in the j diection. As befoe, we solve by quaduple FT ρω 2 % G ij ( q,ω) +(λ + µ)q i q k % Gij ( q,ω) + µ q k q k % Gij ( q,ω) = δ ij This is a 3x3 matix algebaic equation fo the tansfomed G. in dyadic notation ( {µ q 2 ρω 2 } [ I ] + (λ + µ) [ q q ] ) [ % G] = [ I ] fom which we see that the matix [ % G] is the invese of a matix. [ % G] = [ (µ q 2 ρω 2 ) [ I ] + (λ + µ)q 2 [ ˆq ˆq ] ] We can ewite the above ( ˆq i being the i th component of the unit vecto in the diection of q ) Thee ae closed fom expessions fo the invese of abitay 3x3 matices, but they ae not paticulaly simple. Howeve, we may take advantage of the isotopy to speculate that the solution, being some function of the vecto q, and noticing that thee ae no othe quantities with which to expess G's tenso chaacte othe than q and the Konecke delta, G must take the fom % G ij ( q,ω) = ˆq i ˆq j g L ( q,ω) +(δ ij ˆq i ˆq j )g T ( q,ω) [ % G] = [ ˆq ˆq] g L ( q,ω) + ([I] [ ˆq ˆq])g T ( q,ω) fo some yet to be detemined scala functions g of the magnitude of q and of ω. let us solve fo the g. ( {µ q 2 ρω 2 } [ I ] + (λ + µ)q 2 [ ˆq ˆq] ) ( [ ˆq ˆq] g L + ([ I] [ ˆq ˆq])g T )= [ I ] 88

7 Notice that the matices [P L ] = [ ˆq ˆq] and [P T ] = [ I] [ ˆq ˆq] ae disjoint pojection opeatos. They have popeties [P L ][P T ] = [P T ][P L ] = ; [P L ] 2 = [P L ]; [P T ] 2 = [P T ] and [P L ]+[P T ] = [ I]. Rewiting the above in tems of the [ P ], we get ( {µ q 2 ρω 2 } [P L +P T ] ] + (λ + µ)q 2 [P L ] ) ( [P L ] g L + ([ P T ])g T )= [ I ] o, {µ q 2 ρω 2 }g T [P T ] + {(λ + 2µ)q 2 ρω 2 }g L [P L ] = [ I ] Clealy g T = {µ q 2 ρω 2 } and g L = {(λ + 2µ)q 2 ρω 2 } and % G ij ( q,ω) = %G ij (,ω) = 8π 3 ρ [ ˆq i ˆq j (λ + 2µ)q 2 ρω 2 + δ ij ˆq i ˆq j µ q 2 ρω 2 d 3 q ˆq i ˆq j exp( iq ) + d 3 q q 2 ω 2 δ ij ˆq i ˆq j c T 2 q 2 ω 2 exp( i q )] As eve, the fist thing we do with this is evaluate it at the oigin. It is clea that the integals do not convege ( displacement at the position of a point foce is infinite). But we can ask fo the imaginay pat that we'd need to use to evaluate the powe flow by consideing the velocity (in say the z diection due to a foce in the z diection ) Im % G 33 (,ω) = 8π 3 ρ Im But ˆq 3 is meely cosθ q so Im %G 33 (,ω) = 8π 3 ρ Im = 4πρ Im π q2 dq q2 dqsinθ q dθ q dφ q [ q2 dq ˆq 3 ˆq 3 q 2 ω + ˆq 3 ˆq 3 2 c 2 T q 2 ω ] 2 sinθ q dθ q dφ q [ cos2 θ q q 2 ω + sin2 θ q 2 c 2 T q 2 ω ] 2 sinθ q dθ q [ cos2 θ q q 2 ω + sin2 θ q 2 c 2 T q 2 ω ] 2 = 4πρ Im 2 / 3 q2 dq[ q 2 ω + 4 / 3 2 c 2 T q 2 ω ] 2 As peviously, the imaginay pat comes fom the half esidue at the pole, 89

8 i Im % G 33 (,ω) = 4πρ ( iπ)[ ω 3c L 3 + 2ω 3c T 3 ] Im %G 33 = ω 2ρ [ c L c T 3 ] A hamonic point foce appeas to do much geate amounts of wok on the shea waves than on the P-waves, by a facto of 2 c L 3 /c T 3. This numbe is typically to 6, depending on Poisson atio. 9