# Eikonal equation and trajectory of a ray of light

1. Mar 23, 2014

### PhilDSP

On page 122 of Born and Wolf's "Principles of Optics" the following equation for the trajectory of a ray of light is glibly derived in association with the eikonal equation.

$$\frac{d}{d \bf s} (n \frac{d \bf r}{d \bf s}) = \nabla n$$
where n is the index of refraction and r is the displacement vector

This equation is extremely interesting because much earlier J. J. Thomson developed it into an equation of motion for the electron. But details in its derivation are sadly missing in both places.

What are the limitations? Does the equation degenerate as the wavelength approaches zero? Does anyone have references to a more detailed derivation?

Last edited: Mar 23, 2014
2. Mar 26, 2014

### Khashishi

It is only approximately correct if the wavelength is much smaller than the typical variation length scale. In other words,
$|k^{-2}||\nabla k| \ll 1$
where k is the wavenumber.

If k goes to zero somewhere, then the eikonal approximation breaks down.

3. Mar 27, 2014

### UltrafastPED

You can derive the eikonal equation by using Fermat's principle and the calculus of variations.

4. Mar 27, 2014

### PhilDSP

Thanks Khashishi and UltrafastPED. That equation stands a bit apart from the eikonal equation and so doesn't necessarily inherit all of the eikonal equation's limitations. So it seems that the only way to know the details is to do the work of deriving that equation from scratch.

5. Mar 27, 2014

### UltrafastPED

I give an informal derivation of both forms of the eikonal equation, along with some discussion in lecture 10 of Notes on Analytical Mechanics: "Connection to Optics".

6. Mar 27, 2014

### PhilDSP

Excellent! Thanks again

7. Mar 28, 2014

### PhilDSP

J. J. Thomson specified, by the way, that ${\bf v}$ will represent the group velocity of a particle from which we can get the relationships

$d {\bf s} = {\bf v} dt$ and ${\bf v} = nc$

Hence
$$\frac{d}{d \bf s} (\frac{\bf v}{c} \frac{d \bf r}{d \bf s}) = \nabla n$$
and
$$\frac{1}{\bf v} \frac{d}{dt} (\frac{1}{\bf v} \frac{d \bf r}{dt}) = \frac{c}{\bf v} \nabla n$$
or
$$\frac{d^2 \bf r}{dt^2} = nc^2 \nabla n$$
Equating $\frac{d^2 \bf r}{dt^2}$ with the acceleration of the particle and allowing the particle to have mass we get
$${\bf F} = m{\bf a} = mnc^2 \nabla n$$
This applies to a [STRIKE]charged[/STRIKE] elementary particle only of course which is not close to any boundary

P.S. I've striked through the word charged as evidently the equation before inserting mass applies to a photon.

Last edited: Mar 28, 2014