Refraction
//Tampan//, Paminggir people. Sumatra 19th century, 58 x 61 cm. From the library of Darwin Sjamsudin, Jakarta. Photograph by D Dunlop.
Tampan, Paminggir people. Sumatra 19th century, 58 x 61 cm. From the library of Darwin Sjamsudin, Jakarta. Photograph by D Dunlop.

Consider some particle P characterized by its wavevector $\overline{ \kappa }$ and the total number of quarks it contains $N$. Report on any changes relative to a frame of reference F which is characterized using $\tilde{ \kappa }$ the average wavevector of the quarks in F. The momentum of P in the F-frame is defined as

$\begin{align} \overline{p} \equiv \frac{ h }{ 2 \pi } \left( \overline{ \kappa }^{ \sf{P}} \! - N^{ \sf{P}} \, \tilde{ \kappa }^{ \sf{F}} \right) \end{align}$

Let the frame of reference be formed from a gravitational component $\mathbb{G}$ and a smaller part $\mathbb{S}$ that surrounds P so that

$\overline{ \kappa }^{ \sf{F}} = \overline{ \kappa }^{ \mathbb{G}} + \overline{ \kappa }^{ \mathbb{S}}$

When gravitational effects are completely negligible

$\overline{\kappa} ^ { \mathbb{G} } = (0, 0, 0)$

and then P's momentum is

$\begin{align} \overline{p} = \frac{ h }{ 2 \pi } \overline{ \kappa }^{ \sf{P}} + \frac{h N^{ \sf{P}} }{ 2 \pi N^{\sf{F}} } \overline{ \kappa }^{ \mathbb{S}} \end{align}$

Now by definition, the mechanical energy of P is $E \equiv \sqrt{ c^{2}p^{2} + m^{2}c^{4} \vphantom{\sum^{2}} \ }$. So if P is a photon with no mass

$\begin{align} E ^{2}= c^{2}p^{2} = \left( \frac{ h c \kappa^{\sf{P}} }{ 2 \pi } \right)^{2} + \frac{ h^{2} c^{2} N^{ \sf{P}}}{2 \pi ^{2} N^{\sf{F}} } \ \overline{\kappa}^{ \sf{P}} \! \! \bullet \overline{\kappa}^{ \mathbb{S}} + \left( \frac{ h c N^{ \sf{P}} \kappa^{\mathbb{S}} }{2 \pi N^{\sf{F}} } \right)^{2} \end{align}$

Also by definition, the wavelength of P is $\lambda \equiv 2 \pi / \kappa ^{ \sf{P}}$. Combining these relationships gives a messy expression for the wavelength

$\begin{align} \lambda =\frac{ 1 }{ \sqrt{ \; \left( \frac{ E }{h c } \right)^{2} - \frac{ N^{ \sf{P}}}{2 \pi ^{2} N^{\sf{F}} } \ \overline{\kappa}^{ \sf{P}} \! \! \bullet \overline{\kappa}^{ \mathbb{S}} - \left( \frac{ N^{ \sf{P}} \kappa^{\mathbb{S}} }{ 2 \pi N^{\sf{F}} } \right)^{2} \ \ \vphantom{{\overline{\kappa}^{ \sf{P}}}^{2}} } } \end{align}$

For clarity, set $\overline{\kappa} ^ { \mathbb{S} } = (0, 0, 0)$ above to obtain $\lambda_{\sf{o}}$ the wavelength for a photon in a non-dispersive medium

$\begin{align} \lambda_{ \sf{o}} \equiv \frac { hc }{ E } \end{align}$

These wavelengths are complicated functions of P and F. But many different combinations of photons and media are usefully characterized by the ratio

$\begin{align} \eta \equiv \frac { \, \lambda_{ \sf{o}} }{ \lambda } \end{align}$

Then in environments where there is no dispersion, $\lambda = \lambda_{ \sf{o}}$ and motion is described by $\eta =1$. Definition: the number $\eta$ is called the index of refraction.

Right.png Next step: counting bundles.
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