Solution to Question 8
1. Condition for Circular Orbit of Light
For a light ray to travel in a circular path concentric with the planet, the bending of the ray due to the refractive index gradient must exactly match the curvature of the circle. In a spherically symmetric medium, the generalized Snell’s Law (Bouguer’s formula) is: $$ \mu(r) \cdot r \cdot \sin \theta = \text{constant} $$ For a circular path at a constant radius $r = R+h$, the angle of incidence with the radial direction is always $\theta = 90^\circ$. Thus, $\sin \theta = 1$, and the condition becomes: $$ \mu r = \text{constant} $$
2. Deriving the Height
For the trajectory to be stable at radius $r$, the derivative of the quantity $\mu r$ with respect to $r$ must be zero (or strictly, the gradient must provide the necessary curvature). Differentiating $\mu r$ with respect to $r$: $$ \frac{d}{dr} (\mu r) = \mu + r \frac{d\mu}{dr} = 0 $$ $$ \frac{d\mu}{dr} = -\frac{\mu}{r} \quad \text{…(1)} $$
We are given the refractive index profile as a function of altitude $h$: $$ \mu = \mu_0 – kh $$ Since $r = R + h$, we have $dr = dh$. Thus: $$ \frac{d\mu}{dr} = \frac{d\mu}{dh} = -k $$
Substituting this into equation (1): $$ -k = -\frac{\mu_0 – kh}{R + h} $$ $$ k(R + h) = \mu_0 – kh $$ $$ kR + kh = \mu_0 – kh $$ $$ 2kh = \mu_0 – kR $$
3. Alternative Approach: Huygens’ Principle (Wavefront Geometry)
We can bypass the calculus of ray trajectories entirely by visualizing the light flash as a series of physical wavefronts. For a beam traveling horizontally, its wavefronts are perfectly vertical planes.
When this light enters an atmosphere where the refractive index $\mu$ decreases with altitude, the speed of light $v = \frac{c}{\mu}$ increases with altitude. Let’s look at a tiny vertical cross-section of this wavefront with a height $dh$:
- The bottom edge is at radius $r$ and moves at speed $v$. In a tiny time $dt$, it covers an arc length of $s_{bottom} = v \, dt$.
- The top edge is at radius $r + dh$ and moves at a slightly faster speed, $v + dv$. In the same time $dt$, it covers an arc length of $s_{top} = (v + dv) \, dt$.
For the light to travel in a perfect circle, the wavefronts must always point directly toward the center of curvature. Therefore, the ratio of their arc lengths must match the ratio of their radii: $$ \frac{(v + dv)dt}{v \, dt} = \frac{r + dh}{r} $$ $$ 1 + \frac{dv}{v} = 1 + \frac{dh}{r} $$ $$ \frac{dv}{v} = \frac{dh}{r} \quad \text{…(2)} $$
Next, we connect speed to the refractive index ($v = c\mu^{-1}$). Differentiating this gives $dv = -c\mu^{-2}d\mu$. Dividing by $v$ gives the fractional change in speed: $$ \frac{dv}{v} = \frac{-c\mu^{-2}d\mu}{c\mu^{-1}} = -\frac{d\mu}{\mu} $$
Substituting this back into equation (2): $$ -\frac{d\mu}{\mu} = \frac{dh}{r} \implies r = -\frac{\mu}{\frac{d\mu}{dh}} $$
4. Derivation of the Trajectory Equation (Radius of Curvature)
This section details the formal calculus derivation of the radius of curvature $r$ for a ray moving in a medium with a varying refractive index, precisely as mapped out in the handwritten notes.
Step A: Analyzing the Ray Path
Applying Snell’s law at the starting horizontal point ($\mu_0$, where angle $\theta = 90^\circ$ with the vertical y-axis) and an arbitrary point along the curve:
$$ \mu_0 \times 1 = \mu \sin\theta $$From the geometry of the curve, if $\phi$ is the angle with the horizontal, then $\tan\phi = \frac{dy}{dx}$. Because $\theta$ is the angle with the vertical, $\tan\phi = \cot\theta$. Using trigonometric identities, $\sin\theta = \left(1 + \cot^2\theta\right)^{-1/2}$. Let $\dot{y} = \frac{dy}{dx}$. Substituting this in:
$$ \mu_0 = \frac{\mu}{\left[1 + \left(\frac{dy}{dx}\right)^2\right]^{1/2}} $$ $$ 1 + \dot{y}^2 = \frac{\mu^2}{\mu_0^2} $$ $$ \dot{y} = \sqrt{\frac{\mu^2}{\mu_0^2} – 1} $$Step B: Finding the Second Derivative
Differentiate $\dot{y}$ with respect to $x$ to find $\ddot{y} = \frac{d^2y}{dx^2}$:
$$ \ddot{y} = \frac{1}{2}\left(\frac{\mu^2}{\mu_0^2} – 1\right)^{-1/2} \times \frac{2\mu}{\mu_0^2} \frac{d\mu}{dx} $$Recognize that $\left(\frac{\mu^2}{\mu_0^2} – 1\right)^{1/2} = \dot{y} = \frac{dy}{dx} = \cot\theta$. Using the identity $\cot\theta = [\csc^2\theta – 1]^{1/2}$, we can substitute back:
$$ \ddot{y} = \frac{\mu}{\mu_0^2 \left[\csc^2\theta – 1\right]^{1/2}} \times \frac{d\mu}{dx} $$Since $\frac{d\mu}{dx} \div \frac{dy}{dx} = \frac{d\mu}{dy}$, the equation simplifies beautifully to:
$$ \ddot{y} = \frac{\mu}{\mu_0^2} \times \frac{d\mu}{dy} $$Step C: Applying the Radius of Curvature Formula
The standard mathematical formula for the radius of curvature $r$ is:
$$ r = \frac{\left[1 + \left(\frac{dy}{dx}\right)^2\right]^{3/2}}{\left|\frac{d^2y}{dx^2}\right|} = \frac{\left[1 + \dot{y}^2\right]^{3/2}}{|\ddot{y}|} $$Substitute $1 + \dot{y}^2 = \left(\frac{\mu}{\mu_0}\right)^2$ and our derived $\ddot{y}$ into this formula:
$$ r = \frac{\left[\left(\frac{\mu}{\mu_0}\right)^2\right]^{3/2}}{\frac{\mu}{\mu_0^2} \left|\frac{d\mu}{dy}\right|} = \frac{\left(\frac{\mu}{\mu_0}\right)^3}{\frac{\mu}{\mu_0^2} \left|\frac{d\mu}{dy}\right|} = \frac{\mu^3}{\mu_0^3} \times \frac{\mu_0^2}{\mu \left|\frac{d\mu}{dy}\right|} $$ $$ r = \frac{\mu^2}{\mu_0 \left|\frac{d\mu}{dy}\right|} $$Final Simplification: Because we are analyzing the specific point where the beam is horizontal, it remains in the layer where the local refractive index $\mu \approx \mu_0$. The $\mu^2$ and $\mu_0$ partially cancel out, yielding the master formula used in the primary solution:
$$ r = \frac{\mu}{\left|\frac{d\mu}{dy}\right|} \implies r = \frac{\mu_0 – kh}{k} $$