Space-time intervals

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Now, let us set up a thought experiment.  We will assume we are stationary and at the origin of a space-time diagram. Then we will, at time t=0, send out numerous synchronised clocks at various speeds along the x-axis. Then we will note where each clock is as it reaches particular times.

Say we wait for a clock to reach the time t0.  In our frame of reference, the time will be

\displaystyle t=\gamma {{t}_{0}}

At that time, the clock has reached the distance x=vt.  The previous equation can be written as

\displaystyle t=\frac{{{{t}_{0}}}}{{\sqrt{{1-{{{\left( {\frac{v}{c}} \right)}}^{2}}}}}}

Now, let us multiply by the denominator, and then square the expression to get

\displaystyle {{t}^{2}}\left( {1-{{{\left( {\frac{v}{c}} \right)}}^{2}}} \right)=t_{0}^{2}

Next, we substitute in v=x/t to get

\displaystyle {{t}^{2}}\left( {1-\frac{{{{x}^{2}}}}{{{{c}^{2}}{{t}^{2}}}}} \right)=t_{0}^{2}

or

\displaystyle {{t}^{2}}-\frac{{{{x}^{2}}}}{{{{c}^{2}}}}=t_{0}^{2}

Multiplying by c2 will give us

\displaystyle {{\left( {ct} \right)}^{2}}-{{x}^{2}}={{\left( {c{{t}_{0}}} \right)}^{2}}

We will set s=ct0, and, if we can start at some other place than the origin, we will define the so-called space-time interval, Δs, as

\displaystyle {{\left( {\Delta s} \right)}^{2}}={{\left( {c\Delta t} \right)}^{2}}-{{\left( {\Delta x} \right)}^{2}}

Or, if we allow motion along all three spatial dimensions, as

\displaystyle {{\left( {\Delta s} \right)}^{2}}={{\left( {c\Delta t} \right)}^{2}}-{{\left( {\Delta x} \right)}^{2}}-{{\left( {\Delta y} \right)}^{2}}-{{\left( {\Delta z} \right)}^{2}}

This can be seen as a kind of distance between two points in spacetime. Every point with the same space-time interval from a point can be reached within the same time if you move with a constant velocity.

On the page linked below, you can explore this.  There are two points you can move, but I suggest you keep the one at the origin there and move the other one. Before you click the Invariant hyperbolas -on button, I suggest you try the exercises. Above the graph, you can see the coordinates of the points where the x-coordinate is shown as x/c.  Then you can see s2/c2and s/c.

  1. Move the blue circle (point)  to about (0,4). This means four time units into the future. What is the value of s/c?
  2. Move the point, trying to keep the same value of  s/c. What curve are you tracing?  Now you can click the Invariant hyperbolas -on button to see if you got the right curve.
  3. Move the point down to (0, –4). As you can see, you get the same space-time interval as from (0,0) to (0, 4).  The space-time distances are the same.
  4. Move the point to (4, 0). As you can see, we get a negative interval squared. That means that our space-time interval will be an imaginary number. That indicates that we cannot reach the point with any speed less than or equal to c.
  5.  Move the point to a point on the diagonals. As you approach the diagonals, the space-time interval goes towards 0.  The time will slow down to a halt, and you will reach infinity before any time has passed.
  6. A harder one – Try to figure out what the hyperbolas in the fields outside the light cones represent.
Up a level : The Theory of Relativity
Previous page : Space-Time diagrams - a first look
Next page : Invariance of Space-time intervalsLast modified: Jun 12, 2025 @ 17:11