= Fizeau experiment path-length convention
{c}
{title2=$L=2\ell$}
Use the schematic below to count the <water> traversals explicitly. Each stationary tube has laboratory length $\ell$. In the displayed setting, upper-tube <water> flows right and lower-tube <water> flows left, each at <speed> $u>0$. Beam A goes right through the upper tube, exchanges tubes at the far optics, and returns left through the lower tube: it follows the flow twice. Beam B goes right through the lower tube and returns left through the upper tube: it opposes the flow twice. Each complete beam path therefore contains total <water> length $L=2\ell$, not $\ell$ or $4\ell$. The connecting air paths and fixed optics are assumed to contribute no flow-dependent <phase difference>.
\Image[/fizeau-counterflow-paths.png]
{title=Counterflow path accounting: beam A follows the water in both tubes, beam B opposes it, and each beam traverses total water length 2ℓ}
{id=image-fizeau-counterflow-paths}
{description=Original schematic of the tube exchange and recombination, not a scaled optical ray trace. Parallel coloured lines separate overlapping counterpropagating beams for readability. The grey arrows show <water> flow; reversing both grey arrows exchanges the roles of A and B.}
Let $v_+$ and $v_-$ be positive <phase velocity> magnitudes along and against the <water> flow:
$$
v_+=\frac{c/n+u}{1+u/(nc)},\qquad
v_-=\frac{c/n-u}{1-u/(nc)},\qquad
\Delta t=t_B-t_A=L\left(\frac1{v_-}-\frac1{v_+}\right).
$$
For a laboratory source of <frequency> $\nu=c/\lambda_0$, where $\lambda_0$ is its vacuum <wavelength>, define the signed <interference fringe> offset $N$ by the propagation <phase difference> $\Delta\phi=2\pi\nu\Delta t=2\pi N$. This is an equivalent phase delay between steady beams; no stopwatch resolves their transit-time difference. Subtract the zero-flow baseline and expand to first order:
$$
\Delta t\simeq\frac{2Lu}{c^2}(n^2-1)=\frac{4\ell u}{c^2}(n^2-1),\qquad
N(u)\simeq\frac{2Lu}{c\lambda_0}(n^2-1)=\frac{4\ell u}{c\lambda_0}(n^2-1).
$$
Thus one shifted fringe corresponds to one cycle of <phase difference>. If the stationary pattern has spatial <interference fringe spacing> $d$, a simple readout model is $I(x)=I_0[1+\mathcal V\cos(2\pi x/d+2\pi N)]$, where $\mathcal V$ is the <interference visibility>. The maxima shift by $-dN$ with this choice of screen coordinate. A sign assigned to “rightward” motion depends on the image orientation; the magnitude in fringe units does not.
Reversing both flows gives $N(-u)=-N(u)$. The displacement between equal and opposite settings therefore has magnitude
$$
\left|N(u)-N(-u)\right|=2|N(u)|\simeq\frac{8\ell |u|}{c\lambda_0}(n^2-1).
$$
The factor $2$ from comparing opposite flows is additional to the factor $2$ already included in $L=2\ell$. Repeating a round trip in just one flowing tube would instead put a beam with the flow once and against it once; that geometry does not produce the two same-sign contributions drawn here.
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