Hamiltonian formalism

Within the context of the Hamiltonian formalism the travel time can be written as

$$\tau = \displaystyle{ \int\limits_{A}^{B}} \left( \sigma_x \dot{x} + \sigma_y \dot{y} + \sigma_z \dot{z} - H \right) ds ,$$ (A.18)

where ${\cal H}$ represents the system's Hamiltonian:

$${\cal H}(x,y,z,\sigma_x,\sigma_y,\sigma_z) = \sigma_x \dot{x} + \sigma_y \dot{y} + \sigma_z \dot{z} - \sigma ,$$ (A.19)

or compactly, in vector notation:

$${\cal H}(x,y,z,\sigma_x,\sigma_y,\sigma_z) = \mbox{\boldmath$\sigma$}\cdot\mbox{$\mathbf{e}$}_s - \sigma .$$ (A.20)

The perturbation in travel time becomes

$$\delta\tau = \displaystyle{ \int\limits_{A}^{B}} \left[ \do... ...+ \sigma_z \delta\dot{z} - \left( \ldots \right) \right] ds$$

where

$$\left( \ldots \right) = \frac{\partial{{\cal H}}}{\partial{... ...sigma_z + \frac{\partial{{\cal H}}}{\partial{z}} \delta z .$$

It follows then that

$$\delta\tau = \displaystyle{ \int\limits_{A}^{B}} \left[ \le... ...gma_y \delta\dot{y} + \sigma_z \delta\dot{z} \right] ds .$$

where

$$\left( \ldots \right) =$$

$$= \left( \dot{x} - \frac{\partial{{\cal H}}}{\partial{\sigma... ...sigma_z - \frac{\partial{{\cal H}}}{\partial{z}} \delta z .$$

Let us notice now that

$$\displaystyle{ \int\limits_{A}^{B}} \left( \sigma_x \delta\do... ...+ \sigma_y \delta\dot{y} + \sigma_z \delta\dot{z} \right) ds =$$

$$= \underbrace{\left. \sigma_x \delta x + \sigma_y \delta y +... ...dot{\sigma_y}\delta y + \dot{\sigma_z}\delta z \right) ds ,$$

so the perturbation in travel time becomes

$$\delta\tau = \displaystyle{ \int\limits_{A}^{B}} \left[ \b... ...} \right) \delta z \phantom{+} \end{array} \right] ds .$$

According to Fermat's principle $\delta\tau = 0$, which allows to infer the following system of equations

$$\begin{array}{ccc} \displaystyle{ \frac{dx}{ds} = \frac{\par... ...-\frac{\partial{ {\cal H} }}{\partial{z}} } . \\ \end{array}$$ (A.21)

The Hamiltonian can equally be rewritten in order to proceed with an integration along travel time, and by substituting the components of sound slowness with the components of the wavenumber. In fact, by taking into account that

$$ds = d\tau/\sigma$$

and that

$$k_x = \omega \sigma_x , \hskip5mm k_y = \omega \sigma_y , \hskip5mm k_z = \omega \sigma_z ,$$

one can obtain the Hamiltonian[17]^A.1:

$${\cal H} = \omega^2 - c^2k^2$$ (A.22)

(this expression corresponds to the Hamiltonian given by Eq.(A.19), multiplied by the factor $-\omega^2/\sigma$); the system of equations becomes

$$\frac{dx}{d\tau} = \frac{\partial{ {\cal H} }}{\partial{k_x}}... ...c{dk_x}{d\tau} = -\frac{\partial{ {\cal H} }}{\partial{x}} ,$$

$$\frac{dy}{d\tau} = \frac{\partial{ {\cal H} }}{\partial{k_y}}... ...c{dk_y}{d\tau} = -\frac{\partial{ {\cal H} }}{\partial{y}} ,$$

$$\frac{dz}{d\tau} = \frac{\partial{ {\cal H} }}{\partial{k_z}}... ...c{dk_z}{d\tau} = -\frac{\partial{ {\cal H} }}{\partial{z}} .$$

As shown by the two-dimensional case with cylindrical symmetry follows automatically from this case as

$$\frac{dr}{d\tau} = \frac{\partial{ {\cal H} }}{\partial{k_r}}... ...c{dk_r}{d\tau} = -\frac{\partial{ {\cal H} }}{\partial{r}} ,$$

$$\frac{dz}{d\tau} = \frac{\partial{ {\cal H} }}{\partial{k_z}}... ...c{dk_z}{d\tau} = -\frac{\partial{ {\cal H} }}{\partial{z}} .$$

There are also alternativa approaches, which consider a Hamiltonian written in terms of $s$ or $r$. For the first case and for two-dimensional case with cylindrical symmetry one can obtain the Hamiltonian

$${\cal H} = \sigma_r \dot{r} + \sigma_z \dot{z} - \sigma ,$$ (A.23)

related to the system of equations

$$\frac{dr}{ds} = \frac{\partial{ {\cal H} }}{\partial{\sigma_r... ...d\sigma_r}{ds} = -\frac{\partial{ {\cal H} }}{\partial{r}} ,$$

$$\frac{dz}{ds} = \frac{\partial{ {\cal H} }}{\partial{\sigma_z... ...d\sigma_z}{ds} = -\frac{\partial{ {\cal H} }}{\partial{z}} ;$$

as for the second case the Hamiltonian and associated system of equations correspond to[18]

$${\cal H} = -\sqrt{ \frac{1}{c^2} - \sigma_z^2 } ,$$ (A.24)

and

$$\frac{dz}{dr} = \frac{\partial{ {\cal H} }}{\partial{\sigma_z... ...d\sigma_z}{dr} = -\frac{\partial{ {\cal H} }}{\partial{z}} .$$