= Solution
The <Michelson-Morley experiment> splits coherent <light> into two perpendicular arms, reflects the beams and recombines them to observe <interference fringes>. Rotating the apparatus interchanges each arm's orientation relative to a proposed stationary <luminiferous aether>. The historical experiment used multiple reflections to increase the effective optical path and a rotatable support to compare orientations.<On the Relative Motion of the Earth and the Luminiferous Ether by Albert Abraham Michelson and Edward W. Morley>[1887, printed pp. 333–341: optical arrangement, rotation and fringe-shift comparison]{ref}
\Image[/past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2016/paper-1-interferometer.png]
{title=Original schematic of perpendicular round-trip optical paths in a Michelson interferometer}
For equal one-way effective arm lengths $\ell$, a classical stationary-aether model gives
$$
t_\parallel=\frac\ell{c-v}+\frac\ell{c+v}\simeq\frac{2\ell}{c}\left(1+\frac{v^2}{c^2}\right),\qquad t_\perp=\frac{2\ell}{\sqrt{c^2-v^2}}\simeq\frac{2\ell}{c}\left(1+\frac{v^2}{2c^2}\right).
$$
A right-angle rotation should reverse their difference, producing a <fringe shift> of approximately $2\ell v^2/(\lambda c^2)$. The observed orientation-dependent shift was much smaller than the stationary-aether prediction. In <special relativity>, the apparatus rest frame has the same vacuum <speed of light> along both arms, so no such orientation-dependent shift is expected. The result supports <special relativity> but does not, by itself, uniquely establish every postulate or exclude every modified aether theory.
For the rocket, take its tail at $x'=0$ and nose at $x'=L_0$, with coincident frame origins when $t=t'=0$. Denote emission, reflection and return by $A,B,C$. In the rocket's <inertial frame>, their coordinates $(t',x')$ are
$$
\boxed{A:(0,L_0),\quad B:(L_0/c,0),\quad C:(2L_0/c,L_0).}
$$
The inverse <Lorentz transformation> is $x=\Gamma(x'+ut')$, $t=\Gamma(t'+ux'/c^2)$, where $\Gamma=(1-u^2/c^2)^{-1/2}$. Thus the same <spacetime events>, in coordinates $(t,x)$, are
$$
\boxed{\begin{aligned}
A&:\left(\Gamma uL_0/c^2,\ \Gamma L_0\right),\\
B&:\left(\Gamma L_0/c,\ \Gamma uL_0/c\right),\\
C&:\left(\Gamma(2L_0/c+uL_0/c^2),\ \Gamma(L_0+2uL_0/c)\right).
\end{aligned}}
$$
There is no requirement that emission have $t=0$ in this chosen laboratory coordinate system.
The nose-to-tail travel time is $\Delta t_{AB}=\Gamma L_0(1-u/c)/c$. If $L$ is the laboratory length measured at one laboratory time, a left-moving pulse and the right-moving tail close their initial separation at speed $c+u$. Therefore
$$
L=(c+u)\Delta t_{AB}=\Gamma L_0(1-u^2/c^2),\qquad \boxed{L=L_0/\Gamma.}
$$
This derives <length contraction> from the pulse's travel. Comparing the positions of the two endpoints at different laboratory times would not measure the rocket's laboratory length.
Back to article page
