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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                    4R

                                                    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)
                  8Zr      r0  r
                                                 C         8Zr          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)
                  8Zr      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 ) 
                  8Zr         ' (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     )
                  8Zr                                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
language:English
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