01 Jul 2025

Frames, the Michelson–Morley Experiment, and Einstein's Postulates

Inertial and non-inertial frames, the ether-drift test, its null result, and the two postulates of special relativity.

bsc semester-i special-relativity inertial-frames michelson-morley

A reference frame is a coordinate system together with synchronized clocks. If two frames $S$ and $S’$ have parallel axes and $S’$ moves at constant velocity $v\hat{\mathbf x}$ relative to $S$, Newtonian kinematics gives the Galilean transformation

\[x'=x-vt,\qquad y'=y,\qquad z'=z,\qquad t'=t.\]

Differentiating twice,

\[u_x'=u_x-v,\qquad a_x'=a_x,\]

so Newton’s equation has the same form in every frame moving uniformly relative to another. Such a frame is inertial: a free particle has constant velocity, or equivalently $\mathbf a=0$.

If the frame origin has acceleration $\mathbf A(t)$, then $\mathbf r’=\mathbf r-\mathbf R(t)$ gives

\[\mathbf a'=\mathbf a-\mathbf A, \qquad m\mathbf a'=\mathbf F-m\mathbf A.\]

The extra term $-m\mathbf A$ is an inertial (fictitious) force. A frame whose origin accelerates, or whose axes rotate, is therefore non-inertial; rotating frames similarly require centrifugal and Coriolis terms. Special relativity relates inertial frames.

The ether-drift prediction

Nineteenth-century wave theory suggested that light propagated through a stationary ether. Consider equal interferometer arms of length $L$ in the pre-relativistic ether model, with the apparatus moving at speed $v$ through the ether. Light of ether-frame speed $c$ would then have different round-trip times along arms parallel and perpendicular to $\mathbf v$.

For a parallel arm of length $L$, the outward and return times would be

\[t_+=\frac{L}{c-v},\qquad t_-=\frac{L}{c+v},\]

hence

\[t_{\parallel} =\frac{L}{c-v}+\frac{L}{c+v} =\frac{2Lc}{c^2-v^2} =\frac{2L}{c}\frac{1}{1-\beta^2}, \qquad \beta=\frac vc.\]

For one transverse crossing of duration $t_\perp/2$, the mirror advances by $vt_\perp/2$ while the light covers $ct_\perp/2$. The right triangle therefore gives

\[\left(\frac{ct_\perp}{2}\right)^2 =L^2+\left(\frac{vt_\perp}{2}\right)^2,\]

and hence

\[t_\perp=\frac{2L}{\sqrt{c^2-v^2}} =\frac{2L}{c}\frac{1}{\sqrt{1-\beta^2}}.\]

In the ether model, the longitudinal closing speeds are $c-v$ and $c+v$; the transverse arm has the effective along-arm speed $\sqrt{c^2-v^2}$ because the actual ether-frame light path is diagonal.

Their predicted difference is

\[\Delta t=t_\parallel-t_\perp =\frac{2L}{c} \left[ \frac{1}{1-\beta^2}-\frac{1}{\sqrt{1-\beta^2}} \right].\]

For $v\ll c$, $(1-\beta^2)^{-1}\simeq1+\beta^2$ and $(1-\beta^2)^{-1/2}\simeq1+\beta^2/2$, so

\[\boxed{\Delta t\simeq\frac{Lv^2}{c^3}}.\]

Rotating the apparatus through $90^\circ$ exchanges the arms, changing the difference by $2\Delta t$. The corresponding predicted fringe displacement is

\[N=\frac{c(2\Delta t)}{\lambda} \simeq\boxed{\frac{2Lv^2}{\lambda c^2}}.\]

Michelson and Morley observed no displacement of the predicted size within experimental sensitivity: the outcome was null. No preferred ether rest frame was detected.

The two postulates

Einstein replaced the ether hypothesis by two statements:

1. Relativity principle: the laws of physics have the same form in every inertial frame.
2. Light-speed invariance: light in vacuum has the same speed $c$ in every inertial frame, independent of the motion of its source or observer.

The Galilean rule would give $u_x’=c-v$ for a light pulse and therefore contradict the second postulate. Space and time must transform together. The next lecture derives that transformation.

The longitudinal and transverse times, the time-difference expansion, and the rotated fringe shift are verified in the Maxima worksheet; every printed residual is zero.