 figure 1: A sketch of the Fizeau apparatus

# Introduction:

In the same way that a pure mathematician depends heavily upon mathematical rigour, a competent applied mathematician depends heavily upon scientific rigour which would be impossible without a carefully defined system of measurement units. For concreteness, let’s consider the modern definition of the meter.

Unlike sound, which travels through a medium, light does not require an aether. For this reason, the speed of light is constant in any inertial frame and this makes it very convenient for defining the meter. More precisely, the definition of the metre by the International Bureau of Weights and Measures(BIPM) is given by:

The metre is the length of the path travelled by light in vacuum during a time interval of 1/299 792 458 of a second.

Today, NIST and the BIPM have rather sophisticated methods for measuring the speed of light precisely which makes the quantitative measurements of European scientists more than one hundred years ago all the more impressive. In this article I would like to focus specifically on the Fizeau apparatus, developed in 1848-1849, for measuring the speed of light.

# Fizeau’s experimental setup:

In a modern adaptation of Fizeau’s experimental setup, we have a laser, a spinning wheel with regularly-spaced teeth on its perimeter, and a mirror some distance away from the wheel. This system is setup so the light wave may travel back to its source along the path of least action, a straight line, via the mirror.

There are a few key variables in this experimental setup:

$$\omega$$: The wheel’s number of rotations per second.

$$N$$: The number of regularly-spaced teeth on the perimeter of the wheel.

$$d$$: The Euclidean distance between the centre of the wheel and the centre of the mirror.

$$R$$: The radius of the wheel.

$$l$$: The width of the wheel.

In order to define realistic quantitative relationships between these variables it is important to consider important physical constraints on the entire system.

## Physical constraints on Fizeau’s apparatus:

### Constraints due to the speed of sound:

To avoid shock waves, the radius and number of revolutions per second must satisfy:

\begin{equation} 2 \pi R \cdot \omega < v_{\text{sound}} \end{equation}

where $$v_{\text{sound}}\approx 343 \; m \cdot s^{-1}$$, so if we suppose that $$R \approx 10^{-1} \; m$$:

\begin{equation} \omega \lesssim \frac{343 \; m \cdot s^{-1}}{,628} = 546 \; \text{rev} \cdot s^{-1} \approx 3430 \; \text{rad} \cdot s^{-1} \end{equation}

### Material constraints:

There is another constraint on $$\omega$$ due to the compressive strength of the material the wheel is made of. If we use glass which has a high compressive strength of $$10^9 \text{Pa}$$:

\begin{equation} \frac{\lVert F_{\text{rot}} \rVert}{A_{\text{disk}}}= \frac{\lVert F_{\text{rot}} \rVert}{\pi \cdot (l/2)^2} \lesssim 10^9 \; kg \cdot m^{-1} \cdot s^{-2} \end{equation}

and using the symmetry of the wheel, we find that:

\begin{equation} \lVert F_{\text{rot}} \rVert = M \cdot \omega^2 \int_{0}^R r \cdot {dr} = \frac{M \cdot \omega^2 \cdot R^2}{2}= \frac{M\cdot v^2}{2} \end{equation}

and if we assume that $$R \approx ,10 \; m$$ and that glass has a density of $$2500 \; kg \cdot m^{-3}$$:

\begin{equation} M = \pi \cdot R^2 \cdot l \cdot 2500 \; kg \cdot m^{-3} \end{equation}

which implies that we have:

\begin{equation} \lVert F_{\text{rot}} \rVert = \frac{M \cdot \omega^2 \cdot R^2}{2}= \frac{\omega^2 \cdot (\pi \cdot l) \cdot R^4 \cdot 2500 \; kg \cdot m^{-3}}{2} \end{equation}

and the ratio of the magnitude of force over the cross-sectional area is given by:

\begin{equation} \frac{\lVert F_{\text{rot}} \rVert}{\pi \cdot (l/2)^2} = \frac{2 \cdot \omega^2 \cdot R^4 \cdot 2500 \; kg \cdot m^{-3}}{l}= \frac{5000 \cdot \omega^2 \cdot R^4}{l} \end{equation}

and if we make the reasonable assumption that $$\frac{R}{l}=10$$:

\begin{equation} \frac{\lVert F_{\text{rot}} \rVert}{\pi \cdot (l/2)^2} \approx \frac{5000 \cdot 10^{-4} \cdot \omega^2}{10^{-2}}= 50 \cdot \omega^2 \end{equation}

which implies that we have:

\begin{equation} \omega^2 \leq \frac{10^9}{50} = 2 \cdot 10^7 \; \text{rad}^2 \cdot s^{-2} \implies \omega \leq 4.4 \cdot 10^3 \; \text{rad} \cdot s^{-1} \approx 700 \; \text{rev} \cdot s^{-1} \end{equation}

From our last two calculations on the constraints on $$\omega$$, we may conclude that the speed of sound is the dominant constraint. Naturally, this constraint vanishes if the experiment is operated in a vacuum but this would be prohibitively expensive for an apparatus with large dimensions.

### Energetic considerations:

As no scientific institution has an inexhaustible energy budget, it’s also important to consider the energy requirements of this rotational system:

\begin{equation} \text{Power} = \frac{\Delta W}{\Delta t} \end{equation}

where the formula for work is given by:

\begin{equation} \Delta W = \frac{1}{2} \cdot I_{\text{rot}} \cdot \omega^2 \end{equation}

\begin{equation} I_{\text{rot}} = \frac{1}{2} \cdot M \cdot R^2 \end{equation}

so we have:

\begin{equation} \Delta W = \frac{1}{4} \cdot (\pi R^4 \cdot l \cdot 2500 kg \cdot m^{-3}) \cdot \omega^2 \approx 2,3 \cdot 10^4 \; \text{Joules} \end{equation}

and if this is the work done every second, we have:

\begin{equation} \text{Power} = \frac{\Delta W}{\Delta t} = \frac{2,3 \cdot 10^4 \; \text{Joules}}{1 \; s} = 2,3 \cdot 10^4 \; J \cdot s^{-1} \end{equation}

so we may use a 24 kWH battery.

Now, if we assume a rate of ,17 euros per kWH the energy bill for running this experiment will probably exceed 2 euros and we have not yet considered the energy lost to friction, nor the cost of operating the laser.

## Equations that must be satisfied in order to determine the speed of light:

If we assume that gaps and ridges on the perimeter of the wheel are regularly spaced and of equal length, we may determine the speed of light $$c$$ when the product:

\begin{equation} 2 \cdot N \cdot \omega \end{equation}

satisfies:

\begin{equation} \frac{2 \cdot d}{c} = \frac{1}{2 \cdot N \cdot \omega} \end{equation}

Now, if we assume a reasonable value for $$N$$ and choose $$\omega$$ to be approximately equal to the number of revolutions per minute of a Formula 1 engine:

\begin{equation} c = 3 \cdot 10^8 \; m \cdot s^{-1} \end{equation}

\begin{equation} N = 10^4 \end{equation}

\begin{equation} \omega = 250 \; rev \cdot s^{-1} < 700 \; rev \cdot s^{-1} \end{equation}

then the distance $$d$$ between our mirror and the rotating wheel must satisfy:

\begin{equation} d = \frac{c}{4 \cdot N \cdot \omega} = \frac{3 \cdot 10^8 \; m \cdot s^{-1}}{4 \cdot 10^4 \; \text{ridges per rev} \cdot 250 \; rev \cdot s^{-1}} \end{equation}

which simplifies to:

\begin{equation} d = \frac{3 \cdot 10^8 \; m \cdot s^{-1}}{10^7} = 30 \; m \end{equation}

so this Fizeau apparatus would fit inside a basketball court.

## Discussion:

The apparatus I propose is much smaller than the original by Fizeau which required a distance of 8 km between the mirror and the light source. However…the space, time, and energy and material considerations for the apparatus I describe may still seem impractical to a university physics lab. But, practicality is just one consideration of course.

I personally think that it would be really cool to show up at a basketball court to witness the re-enactment of a brilliant experiment to measure the speed of light.

1. Mise en pratique for the definition of the metre in the SI. BIPM. 2019.

2. Catchers of the Light: The Forgotten Lives of the Men and Women Who First Photographed the Heavens. Hughes, Stephan ArtDeCiel Publishing. 2012.

3. Sur une expérience relative à la vitesse de propagation de la lumière. H. Fizeau. 1849.