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CHAPTER VII

RAY ACOUSTICS

7.1 GENERAL INTRODUCTION

Ray based models have been used for many years in underwater acoustics. In the early

1960s, virtually all modelling was done using either normal modes or ray tracing and

primarily the latter. Today, however, ray tracing codes have fallen somewhat out of favor

in the research community, the problem being the inherent (high frequency)

approximation of the method which leads to somewhat coarse accuracy in the results. On

the other hand, ray methods are still used extensively in the operational environment

where speed is a critical factor and environmental uncertainty poses much more sever

constraints on the attainable accuracy. Furthermore, much of the insights derived from

studying ray theory are important in interpreting the results of other models (Jensen et al,

2000).

From a historical point of view, it is interesting that the behavior of ray paths was

understood long before ray theory was mathematically formalized. Ray theory originally

emerged from optics where it was used to understand the propagation of light even before

more fundamental equations for light propagation (Maxwell’s equation) were known.

Indeed, propagation and reflection of rays was originally studied by Euclid, while Snell’s

law governing the refraction of the rays dates back to 1626.

This development is analogous to the manner in which classical mechanics has come to

be understood as an approximation to more complete and complicated equations of

quantum mechanics. In the next we shall see how ray theory is formally derived from full

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wave equation. Mathematically the theory is somewhat complicated, however, the

resulting algorithm is both intuitively appealing and simple to describe.

7.2 MATHEMATICAL DERIVATION

Our starting point is the Helmholtz equation in Cartesian coordinates x̂ ( , , )x y z

22

s2ˆ ˆ ˆ ˆ(x) (x) (x x )

ˆ(x)p p

c

(7.1)

where ˆ(x)c is the sound speed and ω is the angular frequency of the source located at sx̂ .

To obtain the ray equation, we seek the solution of the Helmholtz equation in the form of

ˆ(x)

0

ˆ(x)ˆ(x)

( )

ji

jj

Ap e

i

(7.2)

This is called the ray series. It is generally divergent, but in certain cases it can be shown

to be an asymptotic approximation to exact solution. Taking derivatives of the ray series,

we obtain

,

0 0

ˆ(x)

( ) ( )

j j xi

x j jj j

A Ape i

x i i

(7.3)

2, ,2 2

20 0 0

ˆ(x)( ) 2

( ) ( ) ( )

j j x j xxi

x xx xj j jj j j

A A Ape i i

x i i i

(7.4)

2 2

, , 2 2

ˆ ˆ(x) (x) ˆ ˆ(x) (x)where , , , and .

j j

j x j xx x xx

A AA A

x x x x

Similarly

2, ,2 2

20 0 0

ˆ(x)( ) 2

( ) ( ) ( )

j j y j yyi

y yy yj j jj j j

A A Ape i i

y i i i

(7.5)

2, ,2 2

20 0 0

ˆ(x)( ) 2

( ) ( ) ( )

j j z j zzi

z zz zj j jj j j

A A Ape i i

z i i i

(7.6)

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Therefore, adding Eq 7.4, 7.5 and 7.6 we have

2

2 2 2 2

0 0 0

ˆ(x) | | 2 .( ) ( ) ( )

j j ji

j j jj j j

A A Ap e i i

i i i

(7.7)

Substituting this result into the Helmholtz equation we have

2

2 2 2

0 0 0

2

s20

| | 2 .( ) ( ) ( )

ˆ ˆ(x x )ˆ(x) ( )

j j ji

j j jj j j

ji

jj

A A Ae i i

i i i

Ae

c i

(7.8)

Equating terms of like order in ω, we obtain the following infinite sequence of equations

for functions (x) and (x)jA ,

2 2 2

2

0 0

1 2 2

1

ˆ ˆO( ) : | (x) | (x)

ˆ ˆ ˆ ˆO( ) : 2 (x). (x) ( (x)) (x) 0

:

ˆ ˆ ˆ ˆ ˆO( ) : 2 (x). (x) ( (x)) (x) (x) 1,2,.......j

j j j

c

A A

A A A j

(7.9)

The 2O( ) equation for (x) is known as the eikonal equation and the remaining

equation for (x)jA are known as the transport equation.

7.3 SOLUTION OF THE EIKONAL EQUATION (Jensen et al, 2000)

The Eikonal equation

2 2 2ˆ ˆO( ) : | (x) | (x)c (7.10)

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is a first order nonlinear PDE amenable to solution by the method of characteristics. In

brief, we introduce a family of curves (rays) which are perpendicular to the level curves

(wave fronts) of ˆ(x) as shown in the Fig. 7.1. This family of rays defines a new

coordinate system, and it turns out that in ray coordinates the Eikonal equation reduces to

a far simpler, linear, ordinary differential equation.

Fig. 7.1 Rays and Wave fronts

Since ˆ(x) is a vector perpendicular to the wave fronts, we can define the ray trajectory

x̂( )s by the following differential equation,

x̂ˆ(x)

dc

ds (7.11)

The factor of c is introduced so that the tangent vector x̂d

ds has unit length. This is easily

verified since

2

2 2x̂ˆ| (x) |

dc

ds (7.12)

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Now, from the Eikonal equation Eq. 7.11, the term on the right is found to be unity. Since

x̂1

d

ds , the parameter s is simply the arc length along the ray. The rays can also be

conveniently parameterized with respect to travel time or any other quantity which

increases monotonically along the ray.

Our definition for the rays is based on their being perpendicular to level curves of ˆ(x) , a

function which for the moment is still unknown. However, with some manipulations we

can write the ray equations in a form involving only ˆ(x)c . We consider first just the x -

component of Eq. 7.11. Differentiation with respect to s yields

2 2

2

1d dx d d x y

ds c ds ds dx x s x y s

(7.13)

Using Eq. 7.11 this can be written as

2 2

2

1d dxc

ds c ds x x x y y

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2

d dx c

ds c ds x x y

(7.14)

Then, using the Eikonal equation, Eq. 7.10, to replace the term in square brackets, we

obtain

2

1 1

ˆ2 (x)

d dx c

ds c ds x c

(7.15)

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2

ˆ1 1 (x)d dx c

ds c ds c x

(7.16)

By applying this process to each of the coordinates, we obtain the following vector

equation for the ray trajectories

2

1 x 1ˆ(x)

d dc

ds c ds c

(7.17)

In cylindrical coordinates ( , )r z these ray equations may be written in the first order form

2

1( )

dr d dcc s

ds ds c dr

(7.18)

2

1( )

dz d dcc s

ds ds c dz

(7.19)

where ( ( ), ( ))r s z s is the trajectory of the ray in the range-depth plane. We have

introduced the auxiliary variables ( )s and ( )s in order to write the equations in first

order form. Recall that the tangent vector to a curve [r(s), z(s)] is given by ( , )dr dz

ds ds Thus

from the above equations the tangent vector to the ray is ( ( ), ( ))c s s .

This set of ordinary differential equations is solved numerically as discussed in

subsequent sections. However, to complete the specifications of the rays we also need

initial conditions. As indicated in Fig. 7.2, the initial conditions are that the ray starts at

the source position ( , )s sr z with a specified take off angle . Thus we have

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cos,

(0)sr r

c

(7.20)

sin,

(0)sz z

c

(7.21)

The source coordinate is of course a given quantity whereas the take off angle for the

moment is an unknown variable.

If the index of refraction is independent of frequency then the ray paths are also

independent of frequency. However, the phase is frequency dependent. In practice there

is usually a frequency-dependent loss which adds an imaginary term to the index of

refraction. This loss introduces additional frequency dependence in the ray calculation.

To obtain the pressure field we need to associate a phase and amplitude with each ray.

The phase is obtained by solving the eikonal equation in the coordinate system of the

rays. Recall that,

),( ss zr

r

z Fig. 7.2 Schematic of 2-D ray geometry

130

2

1ˆ ˆ(x). (x)

ˆ(x)c (7.22)

Therefore,

2

1 x 1.

d

c ds c (7.23)

1or,

d

ds c

(7.24)

This is the eikonal equation written in terms of the ray coordinate s .Note that the original

nonlinear partial differential equation has been reduced to a linear ordinary equation

which is readily solve to yield,

0

1( ) (0) '

( ')

s

s dsc s

(7.25)

The integral term n this equation is the travel time along the ray. So, from a physical

point of view the phase of the wave is simply delayed in accordance with its travel time.

7.4 SOLUTION OF THE TRANSPORT EQUATION

The final step in computing the pressure is to associate amplitude with each ray, that is, t

solve the transport equation. Let us recall the form of the transport equation

2

0 0ˆ ˆ ˆ ˆ2 (x). (x) ( (x)) (x) 0A A (7.26)

The rays are defined in Eq. 7.11 as being perpendicular to the waterfronts. This allows us

rewrite the above equation as

131

2

0 0

ˆ2 xˆ ˆ ˆ. (x) ( (x)) (x) 0

dA A

c ds (7.27)

In principal, we could calculate 2 ˆ(x) and then integrate this equation along a ray path.

On the other hand,

2 x. .

d

ds (7.28)

So 2 ˆ(x) is given by the divergence of the rays. Loosely Eq. 7.27, states that the

amplitude along a ray changes in relation to the separating of the ray tube .To make this

statement piecewise, we introduce the Jacobian. For a three–dimensional problem we

have

x

( , , )

x x x

s

y y zJ

s s

y y z

s

(7.29)

when and are respectively the declination and azimuthal take off angles of the ray.

For a cylindrically symmetric problem this reduces to

r z z rJ r

s s

(7.30)

2 2

,z r

or J r

(7.31)

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Referring to Fig. 7.3 we see geometrically that the following are equivalent forms,

cos sinr r

r z r rJ

(7.32)

where r is the angle of the ray at the receiver. Also the Jacobian satisfies

2 1 d J

J ds c

(7.33)

So we can write Eq. 7.26 as

00

2( ) 0

dA c d JA

c ds J ds c

(7.34)

Integrating this equation we obtain our final result for the solution of the transport

equation,

dz

dr

cos sinr r

dz dr

r

( , )s sr z

Fig. 7.3 The ray tube cross section

133

1 2

0 0

( ) (0)( ) (0)

(0) ( )

c s JA s A

c J S (7.35)

7.5 INITIAL CONDITION

In order to complete the solution of the eikonal and transport equation we must specify

the initial conditions. This is done using the method of canonical problems: One

constructs a simpler problem for which the exact solution is known and the initial

condition are chosen so that the ray result agrees with the exact solution. In ocean

acoustics we are normally interested in a point source in an infinite homogeneous

medium as the canonical problem. We take the sound speed of the homogeneous medium

to be that of the original inhomogeneous problem evaluated at the source, ie, 0 0|sc c .

The solution to this problem is given by

0

0 ( )4

i s ce

p ss

(7.36)

where s is the distance from the source. Thus the amplitude and phase associated with

the solution are

0

0

0

1( )

4

( )

A ss

ss

c

(7.37)

Now, taking 0s we obtain (0) 0 as the initial condition for the eikonal. The

amplitude 0 ( )A s goes to infinity as 0.s posing a slight complication. This difficulty is

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resolved by consideration initial conditions for the product 0 (0) (0)A J which turns out to

be a bounded quantity. To compute the Jacobian we first need the solution of the ray

equation in a homogeneous medium. When the sound speed is constant the ray equations

16 are easily solved, yielding

x( ) x (cos cos ,cos sin ,sin )ss s (7.38)

Thus the rays are simply straight lines fanning out from the source point as indicated in

Fig. 1.From the rays the Jacobian determinant is calculate to be

2( ) cosJ s s (7.39)

Thus 1 2 1 2

0

1lim ( ) | ( ) | | cos |

4sA s J s

(7.40)

Substituting this Eq. 7.34 we have

1 2

0

1 ( )cos( )

4 (0) ( )

c sA s

c J S

(7.41)

Combining this result with that of the eikonal equation we obtained the pressure field as

0

11 2 '( ')1 ( )cos

( )4 (0) ( )

s

i dsc sc s

p s ec J S

(7.42)

Thus we obtain the pressure field obtained by dividing the energy of the point source

among each of the ray tubes (the tubes formed by pairs of adjacent rays).The change of

intensity along a ray tube is then inversely proportional to the cross section of that of that

135

tube. Finally the phase calculated from the travel time along a ray. The transmission loss

is given by

0

( )TL 20log

( 1)

p s

p s

(7.44)

where 0 ( 1)p s is given by Eq. 7.36.

The pressure and the transmission loss are basically calculated by the two MatLab

function ‘Eikonal_Solver’ and ‘Singleray_Tracing’ and they ar given in Table 7.1 and

Table 7.2 respectively.

Table 7.1: Eikonal_Solver

%function to solve the eikonal equation function [dx] = Eikonal_Solver(s,x) global ds theta c r z r0 z0 c0 dr=abs(ds*cos(theta)); dz=abs(ds*sin(theta)); c_r=( interp2(r0,z0,c0,r+dr,z)-interp2(r0,z0,c0,r,z) )/(dr);

%derivative of "c" w.r.t "r" c_z=( interp2(r0,z0,c0,r,z+dz)-interp2(r0,z0,c0,r,z) )/(dz);

%derivative of "c" w.r.t "z" dx(1)=c*x(2); dx(2)=(-1/(c.^2))*c_r; dx(3)=c*x(4); dx(4)=(-1/(c.^2))*c_z; [dx]=[dx(1),dx(2),dx(3),dx(4)]';

%----------------------End of the Function-------------------------

Table 7.2: Singleray_Tracing

function[R,Z,ray_length,pressure,TL] =

single_ray_tracing(finput,frequency,sdepth,stheta,... ds,depth,range,fout,n_radial,n_depth,m) %finput ---> Sound Profile %frequency ---> Source Frequency %sdepth ---> Source Depth %stheta ---> Initial Angle of Propagation %ds ---> Range Increment %depth ---> Recevier Depth %range ---> Maximum Range

136

%fout ---> Output File Name; global c r z r0 z0 c0 theta stheta=(pi/180)*stheta; %conversion of initial angle of propagation from degree to radians %======================Initial conditions============================== r=0;z=sdepth; co=interp2(r0,z0,c0,r,z); c=co;theta=stheta; phase(1)=0; R(1)=r;Z(1)=z;J(1)=0; eps=cos(theta)/c; ki=sin(theta)/c; angle=stheta; s=0; k=2; ray_length=s; while r<=range [t,x] = ode23s('Eikonal_Solver',[s s+ds],[r eps z ki]); %solving the differential equation if z<0 z=-z; ki=-ki; elseif z>depth z=2*depth-z; ki=-ki; else n=length(x(:,1)); r=x(n,1); %radial coordinate eps=x(n,2); %epsilon z=x(n,3); %depth ki=x(n,4); %ki c=1/sqrt(eps^2+ki^2); %speed of sound theta=acos(c*eps); %angle of propagation R(k)=r; %radial dimension for the ray trace Z(k)=z; %depth for ray trace phase(k)=phase(k-1)+ds/c; %the time dependent phase of wave J(k)=(-s^2)*cos(theta); %the jacobian value pressure(k)=1/(4*pi)*sqrt(abs(c*cos(theta)/(co*J(k))))*...

cos(frequency*phase(k)); %the pressure as given by transport equation TL(k)=-20*log10( abs(4*pi*(pressure(k))) ); k=k+1; angle=[angle,theta]; s=s+ds; ray_length=[ray_length,s]; end end

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7. 6 NUMERICAL EXAMPLE

Consider a harmonic point source located 50 m below the free surface with frequency f =

20 Hz, and initial angle of propagation = 7° and the sound speed profile given in the

text file ‘Norwegian Sea Profile.txt’ shown in Table 7.3. While the receiver is fixed at a

constant depth 6000 m then the corresponding Ray path, Transmission Loss and Pressure

in the range 0 m to 9000 m is shown in the Fig. 7.4 , Fig. 7.5 and Fig. 7.6

Table 7.3: Norwegian Sea Profile.txt

6 5

0.000

0 1522.0

300 1501.0

1200 1514.0

2000 1496.0

5000 1545.0

3000.00

0 1522.0

300 1501.0

1200 1514.0

2000 1496.0

5000 1545.0

6000.00

0 1522.0

300 1501.0

1200 1514.0

2000 1496.0

5000 1545.0

9000.00

0 1522.0

300 1501.0

1200 1514.0

2000 1496.0

5000 1545.0

16000.00

0 1522.0

300 1501.0

1200 1514.0

2000 1496.0

5000 1545.0

20000.00

0 1522.0

300 1501.0

1200 1514.0

2000 1496.0

5000 1545.0

Fig. 7.4 Ray Path (Sound speed: Norwegian Sea Profile)

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Fig. 7.5 Transmission Loss (Sound speed: Norwegian Sea Profile)

Fig. 7.6 Pressure Plot (Sound speed: Norwegian Sea Profile)

139

Consider another harmonic point source located 500 m below the free surface with

frequency f = 20 Hz, and initial angle of propagation = 10° and the sound speed

profile given in the text file ‘North Atlantic Profile.txt’ shown in Table 7.4. While the

receiver is fixed at a constant depth 5000 m then the corresponding Ray path,

Transmission Loss and Pressure in the range 0 m to 250000 m is shown in the Fig. 7.7,

Fig. 7.8 and Fig. 7.9

Table 7.4 North Atlantic Profile.txt

6 5 0.000

0 1522.0

300 1501.0

1200 1514.0

2000 1496.0

5000 1545.0

50000.000

0 1522.0

300 1501.0

1200 1514.0

2000 1496.0

5000 1545.0

100000.000

0 1522.0

300 1501.0

1200 1514.0

2000 1496.0

5000 1545.0

150000.000

0 1522.0

300 1501.0

1200 1514.0

2000 1496.0

5000 1545.0

200000.000

0 1522.0

300 1501.0

1200 1514.0

2000 1496.0

5000 1545.0

250000.000

0 1522.0

300 1501.0

1200 1514.0

2000 1496.0

5000 1545.0

Fig. 7.7 Ray Path (Sound speed: North Atlantic Profile)

140

Fig. 7.8 Transmission Loss (Sound speed: North Atlantic Profile)

Fig. 7.9 Pressure plot (Sound speed: North Atlantic Profile)