# Ray equation for velocity linear with depth

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Series Geophysical References Series Digital Imaging and Deconvolution: The ABCs of Seismic Exploration and Processing Enders A. Robinson and Sven Treitel 2 http://dx.doi.org/10.1190/1.9781560801610 9781560801481 SEG Online Store

Suppose we have a 2D stratified medium with horizontal coordinate x and depth coordinate y. We work in the first quadrant, so the surface of the earth is the line $y{\rm {=0}}$ . The y axis in the upward direction indicates depth. We assume that $v_{0}$ , a are two known positive constants. Let the velocity $v\left(x,y\right){\rm {=}}v_{0}{\ +}\ a\ y$ increase linearly with depth y but not vary with horizontal coordinate x. It follows that slowness $n\left(x,y\right){\rm {=}}\ {\rm {1}}/v\left(x,y\right)$ decreases with depth. Because

 {\begin{aligned}{\frac {\partial n}{\partial y}}&{\rm {=}}{\frac {\partial v^{-{\rm {1}}}}{\partial y}}{\rm {=}}-v^{-{\rm {2}}}{\frac {\partial v}{\partial y}}{\rm {=}}-{\frac {a}{v^{\rm {2}}}},\end{aligned}} (26)

it follows that the gradient of the slowness is

 {\begin{aligned}\mathrm {grad} \ {\textit {n}}&{\rm {=}}\left({\frac {\partial n}{\partial x}}{\rm {,\ }}{\frac {\partial n}{\partial y}}\right){\rm {=}}\left({\rm {0,\ }}{\frac {\partial n}{\partial y}}\right){\rm {=}}\left({\rm {0,\ }}-{\frac {a}{v^{\rm {2}}}}\right).\end{aligned}} (27)

We see that in this case the gradient of slowness is a vector that always points in the negative direction on the depth axis. In other words, the gradient of slowness points vertically toward the surface of the earth.

In ray theory, it is conventional to let $\theta$ be the angle between the positive y direction and the raypath. Thus, the unit tangent vector to the ray is

 {\begin{aligned}\mathbf {u=} {\frac {d\mathbf {r} }{ds}}&{\rm {=}}\left({\frac {dx}{ds}},{\frac {dy}{ds}}\right){\rm {=}}\left({\rm {\ sin\ }}\theta {\rm {,\ cos\ }}\theta \right).\end{aligned}} (28)

It seems odd to see the sine and cosine in this order, but it is a built-in pitfall of ray theory that one must get used to. The ray equation

 {\begin{aligned}{\frac {d}{ds}}\left(n\mathbf {u} \right)&{\rm {=}}\mathrm {grad} \ {\textit {n}}\end{aligned}} (29)

becomes

 {\begin{aligned}{\frac {d}{ds}}\left(n\left({\rm {\ sin\ }}\theta {\rm {,\ \ cos\ }}\theta \right)\right)&{\rm {=}}\left({\frac {\partial n}{\partial x}}{\rm {\ ,\ }}{\frac {\partial n}{\partial y}}\right).\end{aligned}} (30)

Because $\partial n/\partial x{\rm {=0}}$ , the first component of the ray equation 30 is

 {\begin{aligned}{\frac {d}{ds}}&\left(n{\rm {\ sin\ }}\theta \right){\rm {=0}},\end{aligned}} (31)

which says that $n{\rm {\ sin\ }}\theta$ is a constant. This constant is designated by p and is called the Snell parameter. Thus, the first component 31 of the ray equation gives the so-called Snell equation,

 {\begin{aligned}p{\rm {=}}n{\rm {\ sin\ }}\theta {\rm {=}}\mathrm {constant} .\end{aligned}} (32)

The sine refers to the horizontal component. Equation 32 says that the horizontal component of the raypath vector nu is the same at all points along the raypath. At the shot point, the ray makes the angle ${\theta }_{0}$ with the vertical. Choose a specific point (x,y) on the raypath. Designate the angle at this point by $\theta$ . The Snell equation says that

 {\begin{aligned}p&={\frac {\mathrm {sin} \theta _{0}}{v_{0}}}={\frac {\mathrm {sin} \theta }{v}}\end{aligned}} (33)

The second component of ray equation 30 is

 {\begin{aligned}{\frac {d}{ds}}&\left(n{\rm {\ cos\ }}\theta \right){\rm {=}}{\frac {\partial n}{\partial y}}.\end{aligned}} (34)

The cosine refers to the vertical component. If we carry out the differentiation on the left side of equation 34, we obtain

 {\begin{aligned}&-n{\rm {\ sin\ }}\theta {\frac {d\theta }{ds}}{\rm {+\ cos\ }}\theta {\frac {dn}{dy}}{\frac {dy}{ds}}{\rm {=}}{\frac {\partial n}{\partial y}},\end{aligned}} (35)

which gives

 {\begin{aligned}n\mathrm {sin} \theta {\frac {d\theta }{ds}}&{\rm {=}}\left({\rm {cos}}^{\rm {2}}\theta -{\rm {1}}\right){\frac {\partial n}{\partial y}}\end{aligned}} (36)

or

 {\begin{aligned}&n{\frac {d\theta }{ds}}{\rm {=}}-{\rm {\ sin\ }}\theta {\frac {\partial n}{\partial y}}\end{aligned}} (37)

or

 {\begin{aligned}{\frac {d\theta }{ds}}&{\rm {=}}-v{\rm {\ sin\ }}\theta \left(-{\frac {a}{v^{\rm {2}}}}\right).\end{aligned}} (38)

It follows that the second component 34 of the ray equation gives

 {\begin{aligned}{\frac {d\theta }{ds}}&{\rm {=}}{\frac {a{\rm {\ sin\ }}\theta }{v}}{\rm {=}}ap{\rm {=}}{\frac {a{\rm {\ sin\ }}{\theta }_{0}}{v_{0}}}.\end{aligned}} (39)

The first component 31 of the ray equation (Snell’s law) says that p is constant. Thus, the second component 34 of the ray equation gives

 {\begin{aligned}{\frac {ds}{d\theta }}&={\frac {1}{pa}}=\mathrm {constant} \ \Xi \ \rho \end{aligned}} (40)

This equation is the differential equation of a circle with radius

 {\begin{aligned}\rho {\ =\ }{\frac {\rm {1}}{ap}}{\ =\ }{\frac {v_{0}}{a{\rm {\ sin\ }}{\theta }_{0}}}{\ =\ }\mathrm {constant} .\end{aligned}} (41)

In conclusion, the solution of the ray equation tells us that the raypath is an arc of a circle of radius $\rho .$ 