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Lecture 10 9/28/15

Can we recover a ray picture from the above G for a membrane strip ? Such a picturewould be complementary to the above expansion in a series of integrals along the manybranches of the dispersion relation.Recall, our G for the membrane strip was (now with the width h reinserted and γ ≡ ω2 − q2 )

G(x, y, ys ,t)=14π 2 ∫∫

sinγy< sinγ(h − y> )γ sinγh

exp(iωt − iqx)dωdq

Last time I evaluated this by residues and picked up an infinite number of poles whereversinγh = 0, one for each branch of the dispersion relation. The response thereby beingwritten as a sum over all the different modes of propagation, each term of which was anintegral along the branch ( in either q or in ω). This gives a convenient representation for theresponse when the receiver is in the far field and/or when interest is confined to a narrowfrequency or low frequency band such that only a few of the branches contribute.

However, we also know that the response ought be representable as a sum of rays. Had wereplaced the edges with image sources ( and we'd need an infinite number of them becausewe have two edges ), the response would be a sum of rays, one from each image source, eachof which looked like the G for an unbounded membrane, each representing some set ofreflections of the two edges and containing factors of reflection coefficients off the edges.How can this expression for G be made to exhibit that kind of behavior?

The trick is to realize that we need a sum of terms that look like exp(iωt – iqxx-iqy dy ) wheredy is a distance in the y direction. We can get that by expanding the sin γh in thedenominator1

sinγh=

2iexp(iγh)− exp(−iγh)

=2i exp(−iγh)1− exp(−2iγh)

= 2i exp(−iγh) exp(−2inγh)n=0

∞

∑

This is now a sum over an index n that represents a number of double reflections off theedges, each reflection incurring an extra phase -2iγh for one round trip. G is then written asa sum of an infinite number of terms, a typical one being of the form

G(x, y, ys ,t)~ ∫∫exp(iγy< )exp(iγh − iγy> )exp(−iγh)exp(−2inγh)exp(iωt − iqx)

γdωdq

This is the desired ray picture. Each term of this is interpretable as a propagation over somedistance in y that may involve several bounces off the edges, and a distance x. Each term ofthis integrand is similar to that seen when we discussed the unbounded membrane by doingour first integral over qy: ~exp(iωt – iqxx-iη dy )/η. The factor of exp(-2inγh) represents around trip bounce off both edges of the strip, a distance of 2h.

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At short times we would need only a few of the terms. As each term is integrable exactly(recall the Cagniard trick) , one ends up with an analytic expression for G as a sum over rayarrivals. This is efficient at short times t and distances x, with no restriction in frequency.

We'll see the same complementary descriptions of responses in elastic wavguides.--------

Begin 3-d

Scalar waves in unbounded 3-d.

Before examining the elastic wave eqn in 3-d, it is helpful to take a look at the simplerproblem of a scalar wave in an unbounded 3-d medium.

−c2∇2ψ(x, y, z,t)+∂t2ψ(x, y, z,t)= f (x, y, z;t)

( In acoustics in a fluid, ψ represents pressure (difference from ambient) and f is a sourceterm related to the density of injected volume, and c2 is bulk modulus over mass density. Butsuch interpretations are unnecessary for what we will do here. ) It is not hard to show that wemay define quantities

E ≡ 12 c

2∇ψ ⋅∇ψ + 12 (∂tψ)

2;r

F ≡ - &ψr∇ψ

that obey an energy-like continuity equation (with an energy source term on the right side)

ddt

E +r∇ ⋅

rF = &ψ f

The G for such a medium satisfies

−c2∇2G(x, y, z,t)+∂t2G(x, y, z,t)= δ(t)δ(x)δ(y)δ(z)

To find it we take a quadruple FT

[c2q2 −ω2 ] %G(rq,ω)=1

Thus

%G(rr ,ω) = (2π )−3 d 3rq∫∫∫ exp(−irq ⋅ rr ) / [c2q2 −ω2 ]

Let us first examine this at r =0. Switching to spherical coordinates for the vector q:

%G(rr = 0,ω) = (2π )−3 q2 dq∫∫∫ dφq sinθq dθq / [c

2q2 −ω2 ]

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The q integration runs from 0 to infinity. The φq from 0 to 2π the θq from 0 to π.

%G(rr = 0,ω) = 12π 2 0

∞

∫ q2dq / [c2q2 −ω2 ]

This diverges. But that is OK. As in the membrane, a point force leads to an infiniteresponse at the point of application. One could regularize this by introducing a shortdistance cutoff ( equivalent to a large q cutoff) due to a finite footprint of the agent applyingthe force. If we do not care to lose generality that way, we can still examine the imaginarypart.

i Im %G(rr = 0,ω) = 1

2π 2 ∫ q2dq / [c2q2 − (ω − iε)2 ]where the integral now just runs clockwise over the semicircle above the pole. The result is

i Im %G(rr = 0,ω) = 1

2π 2(−iπ ) q*

2 / [2c2q*]where q* = ω/c is the value of q at the pole. Thus

Im %G(rr = 0,ω) = − ω

4πc3

A concentrated harmonic source generates field ψ that satisfies−c2∇2ψ(x, y, z,t)+∂t

2ψ(x, y, z,t)= δ(x)δ(y)δ(z)cos(ωt)The steady state response is

ψ(x, y, z,t) = Re[ %G(x, y, z;ω)exp(iωt)]

The source is doing work at a rate

Π = &ψ(0,0,0,t) f (t) = Re[iω %G(x, y, z;ω)exp(iωt)]cos(ωt)whose time average is

<Π >= −(ω / 2)Im %G(0,0,0;ω) = ω2

8πc3

=======If r ≠ 0, the process of finding G is more complicated. But not as hard as it was in 2-d. Wereplace the dot product in the exponent with q r cosθ, and we do the easy integral over φ, toget….

%G(rr ,ω) = (2π )−2 q2 dq0

∞

∫ sinθ dθ0

π

∫   exp(−iqr cosθ) / [c2q2 −ω2 ]

We change variables ζ = −cosθ, to get

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%G(rr ,ω) = (2π )−2 q2 dq0

∞

∫ dζ−1

1

∫   exp(iqrζ ) / [c2q2 −ω2 ]

= (2π )−2 q2 dq0

∞

∫ [ exp(iqr)− exp(−iqr)]iqr [c2q2 −ω2 ]

=1

4iπ 2rqdq

−∞

∞

∫ exp(iqr) / [c2q2 −ω2 ]

There are poles at q = ±(ω − iε) / c . We close in upper half plane ( because r > 0) and pickup the residue from q = -ω/c +iε, and obtain…

=1

4iπ 2r(2iπ )(−ω / c)exp(−iωr / c) / [2c2 (−ω / c)] = 1

4πrc2exp(−iωr / c)

This is an outgoing spherical wave.

( as a useful exercise, confirm that this has dimensions consistent with Gtilde as defined from

its PDE. −c2∇2 %G −ω2 %G= δ(x)δ(y)δ(z) )

( as another useful exercise, confirm that it has negative imaginary part at the origin)

G in the time domain is now easy to construct

G(t,r) = 1

2πexp(iωt)

−∞

∞

∫ %G(ω,r) = 14πrc2

δ(t − r / c)

Unlike in 2-d, G in the time domain has no tail. G(t,r) is a simple geometrically attenuatingdelta function, a sharp pulse with amplitude that drops like 1/r.

In the HW6.1 you are asked to calculate the response to a similar looking source, but with anthe x-derivative of a delta function instead of δ(x) and a step function Θ in time instead of aδ(t). The procedure you are asked to follow is very similar to that used above.

------Distributed sources, on beyond the concentrated point source

If a harmonic source is distributed in space,

−c2∇2ψ(x, y, z,t)+∂t

2ψ(x, y, z,t)= f (rr )exp(iωt)

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(where we mean to take the real part so that the forcing and the response are real) Then weseek a solution by i) looking for a steady state response ψ = Ψ(x,y,z)exp(iωt) and ii) takinga triple spatial FT. The result is

c2 | rq |2 Ψ(rq)−ω2Ψ(rq)= f (rq); Ψ(rq) = f (rq) / [c2 | rq |2 −ω2 ] = f (rq) %G(rq,ω)

Let us ask for the rate at which work is done by such a source term f(x,y,z) exp(iωt). Forsimplicity let us assume that f(r) is real. Our continuity equation above shows what thespatial density of power is, f ∂ψ/∂t. Thus the total power applied by f is, instantaneously,(notice I reinsert the real part before multiplying harmonically varying quantities)

Π(t) = ∫∫∫ Re &ψ(rr;t) Re f (rr;t) d 3rr = ∫∫∫ Re{iωΨ(rr )eiωt} f (rr )cosωt d 3rr

= (2π )−3 Re d 3rq∫∫∫ d 3rr∫∫∫ {iω f (rq) %G(rq,ω)eiωte−irq ⋅rr f (rr )cosωt}

The r integration is easy, it is merely the definition of the spatial FT of f(r)*.

Π(t) = ∫∫∫ Re &ψ(rr;t) Re f (rr;t) d 3rr = ∫∫∫ Re{iωΨ(rr )eiωt} f (rr )cosωt d 3rr

= (2π )−3 Re d 3rq∫∫∫ {iω | f (rq) |2 %G(rq,ω)eiωt cosωt}

If we take a time average, this simplifies further:

<Π >= −(2π )−3ω

2d 3rq∫∫∫ | f (rq) |2 Im %G(rq,ω)

As a check, we confirm this gives the familiar result in the case that f(r) =δδδ. In that case

f (rq) =1

=============

Elastic waves in unbounded 3-d.

If the medium is isotropic, our governing PDE is

ρ  r&&u(rr ,t) = (λ +µ)

r∇(

r∇ ⋅

ru(rr ,t))+µ∇2 ru(rr ,t)+rf (rr ,t)

It differs from the fields we have been looking at in that the dependent variable is now avector field u(r,t). And the source is a vector field f(r,t).

We already know the dispersion relation; there are two types of plane waves, eachnondispersive.

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The Greens function is now a 3x3 tensor, and is the solution to the tensor PDE

ρ   &&Gij (

rr ,t) = (λ +µ)∂i∂kGkj (rr ,t)+µ ∂k

2Gij (rr ,t) + δij δ

(3) (rr )δ(t)

such that the solution u(r,t) to the source f(r,t) is given by

ui (

rr ,t) = dt ' ∫∫∫ d 3rr '∫ ρ  Gij (rr − rr ',t − t ')  f j (

rr ',t ')

Gij (

rr ,t) represents the response at r in the i direction to a source at the origin that is a

concentrated impulse in the j direction.

As before, we solve by quadruple FT

−ρω2   %Gij (

rq,ω) +(λ +µ)qiqk%Gij (

rq,ω)+µ qkqk%Gij (

rq,ω) = δij

This is a 3x3 matrix algebraic equation for the transformed G. We can rewrite the abovein dyadic notation

( {µ q2 − ρω2} [ I ] + (λ +µ) [ rqrq ] ) ⋅ [ %G] = [ I ]

from which we see that the matrix [%G] is the inverse of a matrix .

[%G] = [ (µ q2 − ρω2 ) [ I ] + (λ +µ)q2 [ q̂q̂ ] ]−1

( q̂i being the ith component of the unit vector in the direction of rq ) There are closed form

expressions for the inverse of arbitrary 3x3 matrices, but they are not particularly simple.However, we may take advantage of the isotropy to speculate that the solution, being somefunction of the vector q, and noticing that there are no other quantities with which to expressG's tensor character other than q and the Kronecker delta, G must take the form

%Gij (rq,ω) = q̂iq̂ jgL (|

rq |,ω)+(δij − q̂iq̂ j )gT (|rq |,ω)

[ %G] = [q̂q̂] gL (|rq |,ω) + ([I ]− [q̂q̂])gT (|

rq |,ω)for some yet to be determined scalar functions g of the magnitude of q and of ω.

let us solve for the g….

( {µ q2 − ρω2} [ I ] + (λ +µ)q2 [q̂q̂] ) ⋅ ( [q̂q̂] gL + ([ I]− [q̂q̂])gT )= [ I ]

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Notice that the matrices [PL ] = [q̂q̂] and [PT ] = [ I]− [q̂q̂] are disjoint projection

operators. They have properties [PL ][PT ] = [PT ][PL ] = 0; [PL ]2= [PL ]; [PT ]

2 = [PT ] and[PL ]+ [PT ] = [ I] . Rewriting the above in terms of the [ P ], we get

( {µ q2 − ρω2} [PL +PT ] ] + (λ +µ)q2 [PL ] ) ⋅ ( [PL ] gL + ([PT ])gT )= [ I ]

or,

{µ q2 − ρω2}gT [PT ] + {(λ + 2µ)q2 − ρω2 }gL [PL ] = [ I ]

ClearlygT = {µ q

2 − ρω2}−1 and gL = {(λ + 2µ)q2 − ρω2 }−1

and

%Gij (rq,ω) =

q̂iq̂ j(λ + 2µ)q2 − ρω2 +

δij − q̂iq̂ jµ q2 − ρω2

%Gij (rr ,ω) = 1

8π 3ρ[ d 3rq∫∫∫

q̂iq̂ j exp(−irq ⋅ rr )

cL2q2 −ω2 + d 3rq∫∫∫

δij − q̂iq̂ jcT2 q2 −ω2 exp(−i

rq ⋅ rr )]

As ever, the first thing we do with this is evaluate it at the origin. It is clear that the integralsdo not converge ( displacement at the position of a point force is infinite). But we can ask forthe imaginary part that we'd need to use to evaluate the power flow by considering thevelocity (in say the z direction due to a force in the z direction )

Im %G33(0,ω) =

18π 3ρ

Im q2 dq∫∫∫ sinθqdθqdφq[q̂3q̂3

cL2q2 −ω2 +

1− q̂3q̂3cT2 q2 −ω2 ]

But q̂3 is merely cosθq so

Im %G33(0,ω) =1

8π 3ρIm q2 dq∫∫∫ sinθqdθqdφq[

cos2θqcL2q2 −ω2 +

sin2θqcT2 q2 −ω2 ]

=14πρ

Im q2 dq0

∞

∫ 0

π

∫ sinθqdθq[cos2θqcL2q2 −ω2 +

sin2θqcT2 q2 −ω2 ]

=14πρ

Im q2 dq0

∞

∫ [ 2 / 3cL2q2 −ω2 +

4 / 3cT2 q2 −ω2 ]

As previously, the imaginary part comes from the half residue at the pole,

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i Im %G33(0,ω) =14πρ

(−iπ )[ ω3cL

3 +2ω3cT

3 ]

Im %G33 = −ω12ρ

[ 1cL3 +

2cT3 ]

A harmonic point force appears to do much greater amounts of work on the shear waves thanon the P-waves, by a factor of 2 cL

3/cT3 . This number is typically 10 to 16, depending on

Poisson ratio.