A.5.8 (HL)—Space-time interval

Define the interval

For two events separated by $\Delta t$ and $\Delta x$, the space–time interval is $(\Delta s)^2=(c\Delta t)^2-(\Delta x)^2$. Although observers can measure different $\Delta t$ and $\Delta x$, the value of $(\Delta s)^2$ is invariant between inertial frames.

Calculate carefully

Read the time and position differences between the same two events. Convert $\Delta t$ into the distance $c\Delta t$, square both terms, and subtract the spatial term: $(c\Delta t)^2-(\Delta x)^2$. Keep the sign; a negative result is physically meaningful.

Compare frames

Calculate the interval from either frame’s coordinates. Matching values demonstrate invariance and provide a check on transformed coordinates. For a light signal, $(\Delta s)^2=0$; this null interval is consistent with $\Delta x=c\Delta t$.

Worked example from local Question Bank row 32720

For two events with $c\Delta t=100\,\mathrm{ly}$ and $\Delta x=20\,\mathrm{ly}$,

$$(\Delta s)^2=(100)^2-(20)^2=10{,}000-400=9{,}600\,\mathrm{ly^2}$$

Any inertial frame must calculate the same $9{,}600\,\mathrm{ly^2}$ from its own coordinate differences.

Common trap

Do not replace the subtraction with addition, and do not take an absolute value before reporting. The sign distinguishes the interval type and is part of the answer.