Why do two observers measure the same order of events if we are inside the light cone?

(e.g. if $ds^2 > 0$ time-order is preserved according to the classical mechanics book I am reading, but it doesn't give any proof of this) I assume there is some simple geometrical argument that I am missing. And why do two observers measure possible different order of events if we are outside the light cone?


For a geometrical argument, you're looking for basically what Ron posted. But you can also argue this one mathematically: as you may know, the difference between two spacetime events is represented by a time difference $\Delta t$ and a spatial difference $\Delta x$. Under a Lorentz boost, these quantities transform like this:

$$\begin{align}c\Delta t' &= \gamma(c\Delta t - \beta\Delta x) \\ \Delta x' &= \gamma(\Delta x - \beta c\Delta t)\end{align}$$

Now, the spacetime interval is $\Delta s^2 = c^2\Delta t^2 - \Delta x^2$. For a timelike interval, $\Delta s^2 > 0$, this means $c\Delta t > \Delta x$, assuming that both differences are positive (and you can always arrange for that to be the case). Using the Lorentz boost equations, you can see that in this case, $c\Delta t'$ has to be positive. So for two events separated by a timelike interval, if one observer (in the unprimed reference frame) sees event 2 later than event 1, any other observer (in the primed reference frame) will also see event 2 later than event 1.

On the other hand, suppose you have a spacelike interval, $\Delta s^2 < 0$. In this case, $\Delta x > c\Delta t$, so it is possible to get $c\Delta t' < 0$ for a specific velocity (namely $\beta > \frac{c\Delta t}{\Delta x}$). So if one observer (in the unprimed reference frame) sees event 2 later than event 1, it's still possible for another observer (in the primed reference frame) to see them in the reverse order.

  • $\begingroup$ in this sense time-order is associated with the plus or minus sign of t. interesting thanks, david... ron's answer is a little over my head... I need to look at that some more... $\endgroup$ – Bozostein Oct 30 '11 at 13:31
  • $\begingroup$ My argument is no less mathematical because it doesn't use symbols $\endgroup$ – Ron Maimon Oct 30 '11 at 19:48

To get a feel for Lorentz 'rotations' in spacetime, you might want to have a look at this GIF: motion along world line

Notice the events outside the light cone to move up and down in response to the accelerations of the reference frame, and as a result, these can end up at both sides of the 'now' of the observer at the origin. This is not the case for events within the light cone. It is these latter events that can have an influence on the observer at the origin.

  • $\begingroup$ i don't understand this diagram at all... $\endgroup$ – Bozostein Oct 30 '11 at 19:50
  • $\begingroup$ The animation shows the changing views of spacetime along the world line of a rapidly accelerating observer. In this 1+1 dimensional animation, the light cone takes the shape of two diagonal lines. The dashed curve is the spacetime trajectory ("world line") of the observer. Note that the view of spacetime changes when the observer accelerates (the slope of the world line at the apex of the light cone denotes his instantaneous velocity). The order of events within the light cone, and in particular those along the world line of the observer do not change. $\endgroup$ – Johannes Oct 30 '11 at 22:22

The circles in geometry are the curves with

$$ x^2 + y^2 = C $$

In relativity, the analog of circles are hyperbolas:

$$ t^2 - x^2 - y^2 - z^2 = C $$

These curves, unlike circles, are disconnected hyperbolas. For any x,y,z, and positive C, there are two solutions for t, positive and negative, and they are never closer than 2C in t. The two branches of the hyperbola go up in time, and down, and define the forward and backward branch of the hyperbola.

Much as a rotation takes points around a circle, a lorentz transformation takes points along the hyperbola. Those Lorentz transformations which rotate the point continuously cannot move points from the upper hyperbola to the lower hyperbola.

Any timelike interval is either in the forward or backward hyperbola, and is either strictly to the future, or to the past. Null intervals too, by continuity.

  • $\begingroup$ ron i'm a little confused... I understand the concept of a circle and I believe that indeed that is the equation for a hyperbola... what I do not understand is what you mean by "rotate a point continuously cannot move points from the upper hyperbola to the lower hyperbola" . to rotate a point on the circle you just spin it... how can you rotate a point on a hyperbola? $\endgroup$ – Bozostein Oct 30 '11 at 13:41
  • $\begingroup$ perhaps I do not understand what is meant by "rotating a point". is the problem. about what axis? $\endgroup$ – Bozostein Oct 30 '11 at 13:42
  • $\begingroup$ @Bozostein: changing frames in relativity is just like a rotation of time and space into each other, except that the pythagorean theorem has a minus sign for time intervals. See this answer: physics.stackexchange.com/questions/12435/… $\endgroup$ – Ron Maimon Oct 30 '11 at 19:50
  • $\begingroup$ Elementary rotations in higher dimensions are not "about an axis". They are "in a plane". So you can ask "rotation in what plane?". The rotation in relativity is in the plane defined by the time axis and the velocity vector. The relativistic rotation tilts the time-path of a stationary observer, which is just parallel to the time axis, to be the tilted path for a moving observer. $\endgroup$ – Ron Maimon Oct 30 '11 at 19:52

I will restate the answers by Ron and DavidZ in slightly different words. The interval is $ds^2=dt^ 2-dx^2$ in natural units. Thus, $dt^ 2=ds^ 2+dx^ 2$. Thus, if $ds^ 2$ is positive (i.e., the interval is timelike) then no transformation that leaves $ds^ 2$ invariant can make $dt^ 2$ zero as $dx^2$ remains non-negative by definition. Thus, no transformation which is connected to the identity transformation can change its sign because in order to do so it would have to first make $dt$ zero which is forbidden as $dt^2$ cannot go to zero.


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