# What does the line of constant space time (interval) represent?

In this diagram, what do the hyperbola represent. I understand that for a given hyperbola t^2-(x/c)^2 is constant, but in an answer given to this question: What spacelike, timelike and lightlike spacetime interval really mean?

The person writing the answer says "Events on a given hyperbola must, under a given frame boost, remain on that hyperbola". Can someone clarify on this line? How can events be moving positions on space time diagrams? Surely they have a fixed position, and are just read with different numbers based on the coordinate system of the observer moving through it?

• In that answer, above the diagram, it says "the effect of a frame shift is to slide events around on hyperbolae of constant $\Delta s^2$." So events stay on the same hyperbola when you do a boost. – PM 2Ring May 19 at 15:49
• I'm not sure I am reading the diagram correctly. If you take an event that occurs on the orange observers x = 0 axis. It cant also occur at a moving observers x' = 0 axis. Which is what keeping the events on the same hyperbola looks like it does? – Vishal Jain May 19 at 16:01

All points on a given hyperbola are separated from the origin by the same spacetime interval $$\Delta s$$. We can say that not only is $$\Delta s$$ frame invariant, so are the hyperbolas.

Let's suppose we have two events, A and B, with A at the origin, $$x_A = 0$$, $$t_A = 0$$; and B is at $$x_B = 3.0$$, $$t_B = 5.0$$. (Same units for $$x$$ and $$t$$. I'm using an example from Thomas A. Moore's book A Traveler's Guide to Spacetime.)

In the $$x$$-$$t$$ frame,

$$\left(\Delta s\right)^2 = \left(\Delta t\right)^2 - \left(\Delta x\right)^2 = \left(5.0\right)^2 - \left(3.0\right)^2 = \left(4.0\right)^2.$$

Now consider the $$x'$$-$$t'$$ frame, moving at some speed $$v$$ relative to the $$x$$-$$t$$ frame. We assume the $$x$$-$$x'$$ axes are parallel, and that the origins of the two frame overlap at event A. We have two options on how to represent the $$x'$$-$$t'$$ frame. The first, as shown in your figure, is to draw the $$x'$$-$$t'$$ axes on the same diagram with the $$x$$-$$t$$ axes. Obviously, nothing already drawn on the figure will change when the $$x'$$-$$t'$$ axes are added to the figure, including the hyperbolas and the positions on the figure of the spacetime points A and B.

We can use the hyperbolas to calibrate the scales on the $$x'$$ and $$t'$$ axes against the scales on the $$x$$-$$t$$ axes, and then we can identify the coordinates of A and B in $$x'$$-$$t'$$ system. Suppose that $$v$$ is such that we get $$\Delta x' - 5.0$$ and $$\Delta t' = 6.4$$. (As an exercise, you can work out the value of $$v$$ from these differences.)

Then, for the $$x'$$-$$t'$$ frame, we get

$$\left(\Delta s'\right)^2 = \left(\Delta t'\right)^2 - \left(\Delta x'\right)^2 = \left(6.4\right)^2 - \left(-5.0\right)^2 = \left(4.0\right)^2.$$

The second option is to redraw the figure with the $$x'$$-$$t'$$ axes in the normal orientation. The hyperbolas will not change at all because, like $$\Delta s$$, they are invariants. Neither will the position of event A since it is at the origin in both frames. But the position of event B must change because the $$x'$$-$$t'$$ axes are now drawn differently. Since $$\Delta x' = -5$$ event B will now 'hop' to the left side of the $$t'$$ axis on the new figure. But it will nevertheless fall onto the same hyperbola as on the original figure.

Why does B fall on the same hyperbola? Because a given hyperbola always represents all points at a common spacetime interval from the origin. Event B is separated from event A by the same $$\Delta s$$ in all inertial reference frames. It doesn't matter how the axes are drawn. So event B must fall onto precisely the same hyperbola when the figure is redrawn. In addition to hopping to the left of the $$t'$$ axis, event B has to move toward the top of the figure.

For any value of $$v$$, we will get $$\left(\Delta s\right) = (4)^2.$$ If we take event A to define the origin of all possible inertial frames moving with speed $$v$$ relative to the original $$x$$-$$t$$ frame, then event B must fall onto the same hyperbola in all inertial frames. And because the hyperbolas are themselves invariant, we can say B is falling onto literally the same hyperbola in all frames. Adjusting the value of $$v$$ will do one of two things to the figure, depending on which version of the figure we use.

For the original version, in your post, changing the value of $$v$$ does not shift the positions of the A and B events on the figure. It causes the $$x'$$-$$t'$$ axes to either close in on or open up from the $$v = c$$ axis. For the alternative version, where the $$x'$$-$$t'$$ axes are drawn in the normal orientation (so really we have a new alternative figure for each value of $$v$$), the point B will slide along $$\left(\Delta s\right)^2 = \left(4\right)^2$$ hyperbola. So it is in this sense that events can be 'moving' points on a spacetime diagram.

I think you're correct that saying the events 'move' is a little confusing. The events are not moving in a physical sense.

Hope this helps.