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Annexe I: Radiation pattern of a point body force in solid A monochromatic point body-force in a s solid have the following displacement field (Aki and Richard, p.89 Quantitative seismology): U 0 ( )e ik P r 1 1 2 u ij [ i j (3 i j ij )( ) (3 i j ij )( ) ] 4 C P r 2 ik P r ik P r U 0 ( )e ik S r 1 1 2 [ i j (3 i j ij )( ) (3 i j ij )( ) ] 4 C S r 2 ik S r ik S r For a source orientation parallel to the displacement component arbitrarily chosen along the z- axis: U ( )e ik P r 1 1 2 u zz ( ) 0 [sin 2 (3 sin 2 1)( ) (3 sin 2 1)( ) ] 4 C P r 2 ik P r ik P r U 0 ( )e ik S r 1 1 2 [sin 2 (3 sin 2 1)( ) (3 sin 2 1)( ) ] 4 C S r 2 ik S r ik S r The observation is done along the x-axis that is to say for =0: U ( ) e ik P r 1 1 2 e ik S r 1 1 2 u zz (0) 0 2 [1 ( ) ] 2 [1 ( ) ] 4 r C P ik P r ik P r CS ik S r ik S r It follows that: U 0 ( ) 1 cos k P r 1 sin k P r 1 cos k S r 1 sin k S r Re[u zz (0)] 3[ (1 2 2 ) ] 3 [ (1 2 2 ) ] 4 r C P k P r kPr kPr CS k S r kS r kS r 1 Now, for soft incompressible solids one can approximate cos k P r 1 and neglect 1 2 2 : kPr U 0 ( ) cos k S r 1 cos k S r sin k S r Re[ u zz (0)] [ ) 2 2 ] (1) 4 rC S 3 3 kS r 3 kS r kS r If the observation is done along the z-axis or in other words =/2: U 0 ( ) 2e ik P r 1 1 2 2e ik S r 1 1 2 u zz ( ) 2 [ ( ) ] [ ( ) ] 2 4 r C P ik P r ik P r 2 C S ik S r ik S r It follows that: U 0 ( ) 1 cos k P r sin k P r 1 cos k S r sin k S r Re[u zz ( )] 3 [ 3 3 2 2 ] 3 [ 3 3 2 2 ] 2 4 r C P k P r kPr CS k S r kS r 1 Now, one can approximate cos k P r 1 and neglect 1 : 2 k r2 P U 0 ( ) cos k S r 1 sin k S r Re[ u zz ( )] [ 2 2 ] (2) 2 4 rC S 3 3 kS r 3 kS r Characterizing the non-axisymetric nature of the point-body force can be achieved from the computation of the zero-crossing abscissa of (1) and (2). A graphical resolution using f=100Hz as the emission frequency, Cs=3.2m.s-1 as the shear wave speed gives rz=1.60cm and rx=1.08cm for the longitudinal and transverse dimension of the radiation pattern. This result clearly shows the orientation dependence of a monochromatic point body force pattern in solids. Annexe II: Time reversal cavity in fluid The time-reversal (TR) cavity is a sphere centered on O with a radius r0. Let S be a point of its surface associated with a position vector r0 . The spherical symmetry of the problem allows to select arbitrarily the orientation of the spatial coordinate system. The z-axis is chosen in the direction of an observing point M located on r . The distance vector between the observing point M and S is defined as R r r0 . z R S M r rO O y x FIG.1: The spherical time reversal cavity Standard notations of spherical coordinates were used for the following computation. Basic relations are first recall: x r0 sin cos y r0 sin sin , z r cos 0 R 2 r02 r 2 2rr0 cos , r0 dS r02 sin d d R dR d . r a) Monopole source In a first step, a pulsing sphere located in O emits a pulse and the pressure field on the surface of the TR cavity is: p(r0 , t ) (0,0) G(r0 , t;0,0)dV G(r0 , t;0,0) (3) V The particle velocity measured on sensors are time reversed and reemitted, creating the TR pressure field: 1 G (r0 , T t ;0,0) p (r , t ) S G (r , t ; r0 , t 0 )dS TR (4) Z t r r0 R (t t 0 ) (t ) C C , one has to solve: Using the green’s function G(r , t; r0 , t 0 ) 4 r r0 4R r0 R ' (t ) 1 p TR (r , t ) C dS , (5) 16 2 Zr0 S R where ’ stands for the time derivative of a Dirac delta function. Using an adequate element 2 r0 RdR surface dS , it results: r r0 r 1 r0 R 1 2r r 2r r p (r , t ) ' (t )dR [ (t 0 ) (t 0 TR )] (6) 8Zr r0 r C 8Zr C C From (6) it appears that the recreated field is antisymetric as regard to the collapse time 2r0/C. b) Dipolar source The general TR field leads to lengthy computations. The aim of the following part is to show that when a dipole source is used during the first step of a TR experiment, the field recreated by a TR cavity is no more time symmetric as regard to the collapse time. For sake of simplicity, the demonstration is made for an observing point located along the z-axis which happens to be a symmetry axis of the dipolar source. Starting from Eq.4 and using the green’s function of a dipolar source in the far field r r0 (t t 0 ) C G(r , t; r0 , t 0 ) sin(r , r0 ) , one has to solve: 4 r r0 r0 R ' (t ) 1 p TR (r , t ) C sin dS (7) 16 2 Zr0 S R 2 r0 RdR Using the element surface dS , it results: r r0 r 1 r0 R p (r , t ) ' (t ) sin dR TR (8) 8Zr r0 r C A gate function R12 , equals to one between R1=r0-r and R2=r0+r and null everywhere else is R now introduced. 1 r0 R r 2 r 2 R 2 2 R2 p TR (r , t ) 8Zr ' (t C ) 1 ( 0 2rr0 ) R1 dR r02 r 2 (Ct r0 ) 2 2 2r r 2r r p TR (r , t ) 1 1 ( ) ( (t 0 ) (t 0 ) 8Zr 2rr0 C C d r 2 r 2 (Ct r0 ) 2 2 R2 / C 1 ( 0 ) R1 / C dt 2rr0 The first term is antisymetric as respect to the collapse time 2r0/C, but the total field is not symmetric: changing Ct-r0 in r0-Ct change the result of the derivative. A simple interpretation follows from this observation: the TR field that first reaches the point M, comes from one side of the TR cavity and the TR field that last reaches the point M comes from the other side. These two sides are perfectly identical for a monopolar source and the TR field is symmetric. Whereas in the case of a dipolar source, the two sides act differently, giving an apparent “aperture” bigger for one side than for the other resulting in the non symmetric diffraction effects around the focal point.

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posted: | 11/29/2011 |

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