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Background

Light propagating in an anisotropic medium does not (in general) take a straight-line path between two points. The propagation time between those points, then, is dependent on the total path length and the velocity of light at each point along that path. The path may be determined by the Fermat principle (or principle of least time).


Problem

I'm working with a medium whose index of refraction is a function of depth. More specifically, it has the form:

$$A - Be^{Cz}$$

...where $A,B$ and $C$ are positive constants, and $z$ is depth. I'm given that the time taken for propagation between a point $\langle x_0, y_0, z_0 \rangle$ and a point $\langle x_1, y_1, z_1 \rangle$ (in Cartesian coordinates) is given by the integral:

$$T = \frac{1}{c}\int_{z_0}^{z_1} \sqrt{1 + \bigg(\frac{dx}{dz}\bigg)^2}\times n(z) dz$$

And, hence, may be written in code (here, in Python) as:

from scipy.integrate import quad

c = 3e8 # Speed of light

def ray_trace(n, pos0, pos1):
    path = quad(lambda z: np.sqrt(1 + ((pos1[2]-pos0[2])/(pos1[0]-pos0[0]))**2)
                    * n(z), pos1[2], pos0[2])

    return path[0] / c

It is unclear to me why the integral depends only upon $x$ and $z$. For fixed $x$ and $z$, if $y$ is increased (and thus the distance between the points increased), why is it that the total propagation time does not increase?

I suspect that some error has been made here.

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  • $\begingroup$ Are you integrating $\mathrm{d}\,z$? If so, the expression for the arclength looks wrong. Also, since $n(z)$ is independent of $x$ and $y$ (cylindrical symmetry), we can choose a coordinate system where $y_0=y_1$. $\endgroup$
    – DanDan面
    Commented May 27, 2023 at 2:17
  • $\begingroup$ @DanDan0101 Yes, my apologies, I dropped the $dz$. So, the $dx$ here is really the distance between the two points in the $xy$-plane? $\endgroup$ Commented May 27, 2023 at 2:19

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After staring at it for some time, and @DanDan0101 pointing out what now seems obvious, it looks like we're implicitly assuming that $y_0 = y_1$ (i.e. choosing a coordinate system where $y_0 = y_1$), hence $dx$ is really the "$xy$-distance" between the two points. Hence, in general $T$ does depend on both $x$ and $y$.

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  • $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Commented May 27, 2023 at 3:42

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