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In Landau & Lifshitz volume 2, they argue that the eikonal equation can be derived by direct transition to the limit $\lambda \to 0$ in the wave equation. For a field $f$ satisfying the wave equation \begin{equation} \frac{\partial^2 f}{\partial x_i \partial x^i} = 0, \end{equation} we substitute $f = a e^{i \psi}$, where $\psi$ is the eikonal, to obtain \begin{equation} \frac{\partial^2 a}{\partial x_i \partial x^i} e^{i\psi} + 2i \frac{\partial a}{\partial x_i} \frac{\partial \psi}{\partial x^i} e^{i\psi} + if \frac{\partial^2 \psi}{\partial x_i \partial x^i} - \frac{\partial \psi}{\partial x_i}\frac{\partial \psi}{\partial x^i}f = 0. \end{equation} The procedure is then to argue that the eikonal $\psi$ is a large quantity (which makes sense), and as a result we may neglect the first three terms leaving only the last; the eikonal equation is recovered.

My problem is that I do not understand why the first three terms may be neglected in this limit relative to the fourth. What makes the fourth so much larger relative to the others as $\psi$ becomes large?

Qmechanic
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Jack Hughes
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1 Answers1

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It is important that not just $\psi$ is large, but also its' gradient is large: eiconal changes by $2\pi$ over a wavelength which is small in this limit, so $\dfrac{\partial \psi}{\partial x^i} \sim\dfrac{1}{\lambda}$. On the other hand, amplitude in the geometric optics approximation does not change much over the wavelength, so $\dfrac{\partial a}{\partial x^i}\ll \dfrac{a}{\lambda}$. It is also reasonable to say that second derivative of the eiconal is much smaller than $\dfrac{1}{\lambda^2}$ because otherwise the wavevector $k_i = \dfrac{\partial \psi}{\partial x^i}$ would change a lot over a wavelength which would be beyond the scope of geometric optics. So summarize, the first three terms are certainly $o(\dfrac{a}{\lambda^2})$ while the last is $O(\dfrac{a}{\lambda^2})$ so for $\lambda\rightarrow 0$ we leave only the last one.

Viking
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