07 Jul 2025

Length Contraction and Time Dilation

Operational definitions of proper time and proper length, followed by complete Lorentz-transformation derivations.

bsc semester-i special-relativity time-dilation length-contraction

Relativistic measurements compare events. An event has coordinates $(t,x,y,z)$; a time interval needs two clock readings, while a length needs the two endpoint positions measured at the same time in the measuring frame. Throughout, $S’$ moves at speed $v$ along $+x$ relative to $S$, with $\lvert v\rvert<c$.

Time dilation

Let a clock be at rest in $S’$. Two ticks occur at the same position, so $\Delta x’=0$. Its reading

\[\Delta\tau\equiv\Delta t'\]

is the proper time. From the inverse Lorentz transformation,

\[\Delta t =\gamma\left(\Delta t'+\frac{v\Delta x'}{c^2}\right) =\gamma\Delta t'.\]

Hence

\[\boxed{\Delta t=\gamma\Delta\tau}.\]

Because $\gamma\ge1$, the coordinate-time interval in $S$ is at least as large as the proper time recorded by the single clock: a moving clock accumulates less elapsed time between the same two events.

The same result follows from a light clock. If the mirrors have rest separation $D$, one transverse half-tick lasts $\Delta\tau/2=D/c$. In $S$, the light crosses a diagonal while the clock moves $v\Delta t/2$:

\[\left(\frac{c\Delta t}{2}\right)^2 =D^2+\left(\frac{v\Delta t}{2}\right)^2.\]

Using $D=c\Delta\tau/2$,

\[c^2\Delta t^2=c^2\Delta\tau^2+v^2\Delta t^2,\] \[\Delta t^2(1-\beta^2)=\Delta\tau^2, \qquad \Delta t=\gamma\Delta\tau.\]

The ray leaves the lower mirror at its initial position and reaches the upper mirror after the clock has moved $v\Delta t/2$; the final lower and upper mirror positions show the proper transverse separation $c\Delta\tau/2$.

Lorentz contraction

Let a rod be at rest in $S’$ with endpoints $x_1’$ and $x_2’$. Its proper length is

\[L_0=x_2'-x_1'.\]

To measure its length in $S$, record both endpoint positions simultaneously: $\Delta t=t_2-t_1=0$. Applying the direct position transformation to the two endpoint events,

\[\begin{aligned} L_0=\Delta x' &=\gamma(\Delta x-v\Delta t)\\ &=\gamma\Delta x. \end{aligned}\]

Therefore

\[\boxed{L=\Delta x=\frac{L_0}{\gamma} =L_0\sqrt{1-\frac{v^2}{c^2}}}.\]

Only the dimension parallel to the relative motion contracts; $y’=y$ and $z’=z$ leave transverse lengths unchanged. The simultaneity condition $\Delta t=0$ is essential: endpoint positions recorded at different times do not define the rod’s length in $S$. At $v=0$, $\gamma=1$ and $L=L_0$; as $\lvert v\rvert$ increases, $L<L_0$.

The time-dilation, light-clock, and simultaneous-endpoint substitutions are verified in the Maxima worksheet; every printed residual is zero.