# I Fizeau's experiment and TRANSFORMATION OF VELOCITIES

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1. May 17, 2018

In 1851 Fizeau made a famous experiment which corroborated de Fresnel's drag coefficient of the luminiferous ether. In the experiment two light beams traveled through a tube of moving water (at 7cm per second), one moving against the water flow (let's called it beam A), and one for the water flow (beam B). Then, the two beams were detected at the same detector, making an interference patern. The results were in agreement with Fresnel's theorerical prediction. According to it, the velocities of each light beam must be:
[V][/B] = c/n + αv,
[V][/A] = c/n - αv,
where n is refraction index of the water, and α is the Fresnel's drag coefficient, equals to 1 - 1/n², v is the velocity of the water, and c is the velocity of light in vacuum.

Nowadays we can reinterpretate this experimental result with Einstein's transformation of velocities.
Let's consider two reference frames: the laboratory frame, and the water frame, moving in respect to the first. In the water frame the velocity of light is c/n. In the lab frame the velocity of light must be:
[c][/lab] = (c/n + v)/(1 - v/nc)
If we expand this expression and neglect terms of the order of (v/c)2 and higher, we obtain exactly the same results as predicted by Fresnel's theory.

Ok, so far so good.

But one may ask: "The principle of relativity teaches us that light moves with the same speed, no matter the frame of reference (lab frame, water frame, whatever). So, how can we explain the Fizeau's experiment?"

2. May 17, 2018

### Orodruin

Staff Emeritus
This statement is missing the often implicit but ultimately important "in vacuum". If you have a surrounding medium, such as water, its presence in itself breaks Lorentz invariance by singling out a particular frame (its rest frame) where the speed of light is isotropic.

Edit: Also, that is not the principle of relativity. It is the assumption that the speed of light in vacuum is an invariant quantity.

3. May 17, 2018

### pervect

Staff Emeritus
In a vacuum, the speed of light is a constant for all observers, a constant we call c.

In a non-vacuum, such as water, the analysis is more complex. In the rest frame of the water, we can say that v = c/n, where v is the speed of light in the water, c is (as before) the speed of light in a vacuum, and n is the refractive index of the water. This equation is only valid in the rest frame of the water, however, it's not valid in general. We could in principle determine the correct description of the speed in light-in-water in moving water by applying the Lorentz transform, but it'd take a fair bit of math.

4. May 17, 2018

Let me see if I'm getting it straight, guys.
If an observer at the lab mesures the refractive index of the moving water he/she will get a different value compared with the value that an observer in the water (which sees lab moving) measures? The contraction of the lengths make the density of the water and the refractive index change, so as the speed of light in the moving medium?

5. May 17, 2018

### Orodruin

Staff Emeritus
The index of refraction is a material property that tells you what the speed of the light in the medium is in the rest frame of the medium. It makes no sense to measure it in a moving medium.

6. May 17, 2018

### sweet springs

Maximum speed of interaction is called light speed because the light speed in vacuum is a familiar example. Light speed in media is out of it.

7. May 18, 2018

### pervect

Staff Emeritus
We can apply the formulae for relativistic velocity addition to find the velocity-of-light-in-water in a moving frame knowing the velocity-of-light-in-wanter in the water frame.

If we limit ourselves to one dimension of space (plus of course, time), we can use the simple form of the relativistic velocity addition law, https://en.wikipedia.org/wiki/Velocity-addition_formula.

If we have a frame moving with respect to the water at velocity u, then we can write the velocity-of-light-in-water in the moving frame, which we will denote by v' by using the formula

$$v \ominus u = \frac{v-u}{1 - \frac{u\,v}{c^2}}$$

with v = c/n to get

$$v' = \frac{\frac{c}{n}-u}{1- \frac{u}{n\,c} } = \frac{c-u\,n}{n - \frac{u}{c}}$$

We use a "relativisticx velocity subtraction" so that if u = c/n, the velocity-of-light in water, v', is zero.