The chain rule and the high frequency limit

The familiar partial-differential equations of physics come to us in (t,x,z)-space. The chain rule for partial differentiation will convert the partial derivatives to (t' ,x' ,z' )-space. For example, differentiating (30) with respect to z gives  

$${\partial P \over \partial z} \eq {\partial P' \over \partia... ...artial P' \over \partial z' }\ {\partial z' \over \partial z\,}$$ (31)

Using (27), (28) and (29) to evaluate the coordinate derivatives gives  

$${\partial P \over \partial z} \eq -\ {1 \over \bar v}\ {\pa... ...' \over \partial t' }\ \ +\ \ {\partial P' \over \partial z' }$$ (32)

There is nothing special about the variable P in (31) and (32). We could as well write  

$${\partial\ \over \partial z} \eq -\ {1 \over \bar v }\ {\partial\ \ \over \partial t' }\ +\ {\partial\ \ \over \partial z' }$$ (33)

where the left side is for operation on functions that depend on (t,x,z) and the right side is for functions of (t' ,x' ,z' ). Differentiating twice gives  

$${\partial^2\ \over \partial z^2} \eq \left( {- 1 \over \bar... ... \partial t' } \ +\ {\partial\ \ \over \partial z' } \ \right)$$ (34)

Using the fact that the velocity is always time-independent results in  

$${\partial^2\ \over \partial z^2} \eq {1 \over \bar v^2 }\ ... ... {d \bar v \, \over d z' } \ {\partial\ \ \over \partial t' }$$ (35)

Except for the rightmost term with the square brackets it could be said that ``squaring'' the operator (33) gives the second derivative. This last term is almost always neglected in data processing. The reason is that its effect is similar to the effect of other first-derivative terms with material gradients for coefficients. Such terms, cause amplitudes to be more carefully computed. that all such terms should be included, from the beginning.