The description of the passage from wave optics to geometrical optics claims that light rays are the integral curves of a certain vector field (the Poynting vector direction, normalized to 1). Here are the details, could you fill the blanks:

The wavelength $\lambda$ is much smaller than all other characteristic lengths.

  1. The setting is a "nice" medium (with spatially varying refraction index $n(x)$) through which an "almost plane" wave is propagating. The wave (if linearly polarized) turns out to be representable by $\vec{E}{(x,t)} = \vec{E}_0\exp(i(\chi{(x)}-\omega t))$ and $\vec{B}{(x,t)} = \vec{B}_0\exp(i(\chi{(x)}-\omega t))$ with constant $\vec{E}_0$ and $\vec{B}_0$.

  2. Maxwell's equations imply $(\vec{\nabla}\chi)^2=\frac{n^2\omega^2}{c^2}$ and a time-averaged Poynting vector $\vec{S}=\frac{c}{n}\vec{s}$, where $\vec{s}$ is the unit vector $\vec{s}=\frac{\vec{\nabla}\chi}{n\omega/c}$.

  3. The integral curves of the field of unit vectors $\vec{s}$ are the light rays.

  4. Working through the equations results for this for a ray trajectory $X(\tau)$ (where $\tau$ is just a parameter): $\frac{d}{d\tau}(n\frac{d\vec{X}}{d\tau})=\vec{\nabla} n$

How does one prove the jump from 2 to 3. Why follow the unit vectors and not the Poynting vectors themselves? Or even, why should the light rays be tangential to the Poynting vectors at all (besides intuition like "light rays should transport energy")?

Could someone give me the proof of 3. or point me to a reference?

  • 2
    $\begingroup$ The distinction between $\vec{s}$ and $\vec{S}$ is not terribly significant; the point of the unit vectors is that they point in the same direction as the Poynting vector, with unit magnitude. The field of unit vectors deforms continuously across phase boundaries, as does the Poynting vector. You could just as easily follow the Poynting vector, but it would be a little more confusing since the "size" of the vector would change shape as the index of the medium changed. This isn't necessary to see that the path of light propagation at any point in space is along the Poynting unit vector. $\endgroup$
    – KDN
    Feb 27, 2013 at 0:05
  • $\begingroup$ @KDN, intuitively your comment makes a lot of sense, thanks, however I hope to get a more detailed proof at some point. $\endgroup$
    – Krastanov
    Feb 27, 2013 at 0:07
  • 1
    $\begingroup$ Also, I don't know if there is any kind of "proof" for 3. I believe it is more of a definition. The geometric "light ray" is taken to be the path traced out by the Poynting vector of the electromagnetic field. $\endgroup$
    – KDN
    Feb 27, 2013 at 0:12
  • $\begingroup$ Related: Derivation of the ray equation. $\endgroup$ Feb 27, 2013 at 1:37

1 Answer 1


There's a number of interesting points to this.

  • The passage from 1. to 2. is not trivial. If you do the calculation, you will see that the laplacian $\nabla^2\vec{E}$ from the wave equation gives rise to the term in $\left(\nabla\chi\right)^2$ you mention as well as a term in $\nabla^2\chi$. This second term only goes away in the small $\lambda$ limit and it is the essence of the eikonal approximation. It is not a calculation you should wave away: work it out in full and implement the approximation, noticing that locally $\chi(\vec{x})=\vec{k}\cdot\vec{x}+\text{slow factors}$, where $\vec{k}$ is large. (You will of course need to quantify "slow".)
  • (The calculation that $\vec{S}=\frac{c^2}{n^2\omega}\nabla\chi$, on the other hand, is trivial.)
  • As KDN mentioned, the integral curves of $\vec{S}$ and its unit vector $\vec{s}$ are the same. This follows from the definition of integral curves: they are curves such that the vector field is tangent to them throughout. This is independent of the length of the vector. (In terms of the curve it corresponds to a reparametrization of the "time": it changes the speed but not the direction of the velocity.) Using a unit vector means that light rays will be parametrised by path length.
  • One can simply define light rays to be the integral curves of $\vec{s}$ and be happy about it, though of course that is simply missing the physics. The key fact about the light rays, so defined, is that they are everywhere normal to the surfaces of constant $\chi$, i.e. the surfaces of constant phase, i.e. the wavefronts. Plane waves propagate in straight lines normally to the wavefronts in free space, and so do light rays (so defined). It is the normal to the wavefronts that matters when working out Fresnel equations, and therefore the (so defined) light rays will obey Snell's law. Ultimately, proving 3. is a matter of definition: what are light rays? Write down any defining property and you'll be able to prove the integral curves of $\vec{s}$ obey it.
  • It is important to note that in isotropic media $\vec{s}$ is not only the local unit Poynting vector, but it is also the local unit wave vector. (Essentially, this is the same point as above.) Intuitively, light rays ought to follow wave vectors because it is wave vectors that tell light waves where to go. In a birefringent (not isotropic) medium the phase propagation direction (wave vector) and the energy propagation direction (Poynting vector) are not necessarily the same (and the snell law does not apply).
  • Proving 4. is an interesting exercise (i.e. do it!) but it is essentially trivial. It relies on the identity $\frac{d\vec{X}}{d\tau}=\vec{s}$, which defines light ray curves $\vec{X}(\tau)$, on judicious use of the total derivative $\frac{d}{d\tau}=(\frac{d\vec{X}}{d\tau}\cdot\nabla)$, and some interesting vector calculus manipulations. (Hint: prove $(\nabla\chi\cdot\nabla)\nabla\chi=\frac12\nabla\left(\nabla\chi\right)^2$.) Presumably you know by now that what you get is called the ray equation, what it means, and how to use it, or you would not have stopped there ;).

This looks like enough to get you going but if you have more questions, do ask.


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