Suppose in a inertial reference frame $S$, an event $A$ occurs at $(ct_A, x_A, y_A, z_A)$ and event $B$ occurs at $(ct_B, x_B, y_B, z_B)$.

Now the invariant interval of these two events is,

$$I = -c^2 (t_A - t_B)^2 + (x_A - x_B)^2 + (y_A - y_B)^2 + (z_A - z_B)^2 = -c^2 \Delta t^2 + \Delta \bar x^2,$$

where I'm using the $(-, +, +, +)$ metric.

Now there can be $3$ particular cases of interest corresponding to time-like, space-like and light-like events.

For $I = 0 \implies c^2 \Delta t^2 = \Delta \bar x^2$, events are light-like.

For $I < 0 \implies c^2 \Delta t^2 > \Delta \bar x ^2$, events are time-like and a reference-frame $\bar S$ exists(accessible by appropriate Lorentz Transformation) for which these two events occur at the same location. The velocity(magnitude and direction) can be computed.

For $I > 0 \implies c^2 \Delta t^2 < \Delta \bar x^2$, events are space-like and a a reference frame $\bar S$ exists(again accessible by appropriate Lorentz Transformation) for which these two events are simultaneous.

I know how to calculate the velocity(direction and magnitude) of the $\bar S$ frame relative to the $S$ frame in case of a time-like event. I also know how to calculate the magnitude of velocity of the $\bar S$ frame relative to the $S$ frame for a space-like event.

How to find the direction of the $\bar S$ frame relative to $S$ for a space-like event?

  • $\begingroup$ The direction of motion for the barred frame in this case would be on a line connecting the two events spatially, as that is when the length contraction yields proper results. $\endgroup$
    – R. Rankin
    Nov 3, 2017 at 8:34
  • $\begingroup$ @R.Rankin: Didnt get your point! $\endgroup$
    – sbp
    Nov 3, 2017 at 8:46
  • $\begingroup$ In other words, in the (1+1)-case, $\frac{\Delta x}{\Delta t}$ is a slope. If that slope has magnitude less than $c$ (the timelike case), then that slope is your velocity. If that slope has magnitude greater than $c$ (the spacelike case), then (c/slope) is your velocity. In the (3+1)-case, you have to express the "reciprocal slope" as @Frobenius suggests. $\endgroup$
    – robphy
    Nov 3, 2017 at 14:08
  • $\begingroup$ Yup. But my question was to find the direction of velocity and not the magnitude. The latter is trivial using LT. $\endgroup$
    – sbp
    Nov 3, 2017 at 17:51

5 Answers 5


This problem would have given me fits when I first learned relativity, but since I've started using the geometric point of view as the tool I reach for first it is almost trivial.

The direction is easy: boost in the direction from the earlier to the later event as measured in your current frame. Why? Because you want the space-like axis to tilt upward in that direction.

Getting the speed is also surprisingly easy. You need the new space-like axis to have a slope (in your current coordinate system) equal to $(\Delta x)/(c \,\Delta t)$, which is exactly the $\beta$ of the boost you need.


Here is an image which should correspond to the graphical solution of dmckee:

enter image description here

You have to proceed as follows:

  • Put event A into the coordinate origin (by coordinate displacement).

  • Then connect both events by a line x' which is the space axis of the researched new reference frame

  • Then draw accordingly the time axis ct' which is the new space axis x' mirrored on the 45° axis (α=β).

  • In your coordinates, this new time axis ct' is the worldline of the researched reference frame, and the direction of its relative velocity.


Taken another way, just define the $x$-axis as the one that spatially separates them, so that they are both located along the line $y=0,z=0$. You believe that one happened at $x=0, w=0$ (where $w=ct$) and the other happened at $x=\xi,w=\tau$ for $\tau < \xi.$ If you boost with $\beta = \tau/\xi$ then you will find that the Lorentz transformation looks like:$$w' = \gamma~(w - \beta~x) = \gamma~(\tau - \frac\tau\xi~\xi) = \gamma~0 = 0,$$and both events will be simultaneous. The proper distance between them is then of course $$x' = \gamma~(x - \beta~w) = \frac{\xi~(1 - \beta^2)}{\sqrt{1-\beta^2}} = \xi\sqrt{1 - \beta^2}=\sqrt{\xi^2-\tau^2},$$as it must be to leave the interval invariant.

The direction is therefore basically just "from the one in the past towards the one in the future."

A nice way to think about relativity is that as you accelerate, the clocks ahead of you tick faster in proportion to both your acceleration and their distance from you, and the clocks behind you tick slower. This makes the "direction" pretty clear; there are many directions to choose from but the easiest is to go in the direction from the earlier event to the later one. Imagine two clocks that are in-sync in the reference frame you want to be in at these two distant places, they both count down to "0" at their respective events: in the reference frame where they are not in-sync then the earlier-event clock might be at "200 s" while the later-event clock might be at "300 s," we need to make that later-event clock tick faster while the earlier-event one ticks slower.


enter image description here

The answer concerns the case $\:I > 0$, so let in a frame $\:\mathrm{S'}\:$ two events taking place by a space interval $\:\Delta\mathbf{x}^{\boldsymbol{\prime}}\:$ and time interval $\:\Delta t^{\boldsymbol{\prime}}\:$ apart with \begin{equation} \left\Vert\dfrac{\Delta\mathbf{x}^{\boldsymbol{\prime}}}{\Delta t^{\boldsymbol{\prime}}} \right\Vert^{2} >c^{2} \nonumber \end{equation} These two events are causally independent.

Now, we seek for frames $\:\mathrm{S}\:$ moving with respect to $\:\mathrm{S'}\:$ wherein these two events happen simulta- neously. Without loss of generality let such a system $\:\mathrm{S}\:$ be in Standard Configuration to $\:\mathrm{S'}\:$ and moving with velocity $\:\mathbf{v}\:$ with respect to it. Then for the Lorentz Transformation between them we have (see Figure) \begin{align} \Delta \mathbf{x} & = \Delta\mathbf{x}^{\boldsymbol{\prime}}+(\gamma-1)(\mathbf{n}\boldsymbol{\cdot} \Delta\mathbf{x}^{\boldsymbol{\prime}})\mathbf{n}-\gamma \mathbf{v} \Delta t^{\boldsymbol{\prime}} \tag{01a}\\ \Delta t & = \gamma\left( \Delta t^{\boldsymbol{\prime}}-\dfrac{\mathbf{v}\boldsymbol{\cdot} \Delta \mathbf{x}^{\boldsymbol{\prime}}}{c^{2}}\right) \tag{01b}\\ \mathbf{n} &=\dfrac{\mathbf{v}}{\Vert\mathbf{v}\Vert} \tag{01c} \end{align} For these two events to be simultaneous in the frame $\:\mathrm{S}\:$, equation (01b) yields \begin{equation} \Delta t =0 \quad \Longleftrightarrow \quad \Delta t^{\boldsymbol{\prime}}-\dfrac{\mathbf{v}\boldsymbol{\cdot} \Delta \mathbf{x}^{\boldsymbol{\prime}}}{c^{2}}=0 \tag{02} \end{equation} This means that the frame $\mathrm{S}$ must move with velocity $\mathbf{v}$ such that its projection $\mathbf{v}_{\Vert}$ on the space interval $\Delta\mathbf{x}^{\boldsymbol{\prime}}$ satisfies

\begin{equation} \Vert \mathbf{v}_{\Vert}\Vert =\dfrac{c^{2}}{\left\Vert\dfrac{\Delta\mathbf{x}^{\boldsymbol{\prime}}}{\Delta t^{\boldsymbol{\prime}}} \right\Vert} (<c) \tag{03} \end{equation}

So there exists an infinite number of velocities.

A choice parallel to $\Delta\mathbf{x}^{\boldsymbol{\prime}}$ is

\begin{equation} I > 0 :\quad \mathbf{v}=\dfrac{c^{2}}{\left\Vert\dfrac{\Delta\mathbf{x}^{\boldsymbol{\prime}}}{‌​\Delta t^{\boldsymbol{\prime}}} \right\Vert^{2}}\dfrac{\Delta\mathbf{x}^{\boldsymbol{\prime}}}{\Delta t^{\boldsymbol{\prime}}}\,, \quad \Vert \mathbf{v}\Vert =\dfrac{c^{2}}{\left\Vert\dfrac{\Delta\mathbf{x}^{\boldsymbol{\prime}}}{\Delta t^{\boldsymbol{\prime}}} \right\Vert} < c \tag{04} \end{equation}

enter image description here

For a 3D version of above Figure click here : Figure 3D


Given a pair of spacelike related events there are infinitely many reference frames where they appear to be simultaneous (with different velocities with respect to a given initial reference frame). I will pick out the simplest one in the following.

(I use $c=1$.) Let ${\bf e}_x$ and ${\bf e}_t$ be the unit vectors tangent to the $x$ axis and the $t$ axis of the refernce frame $S$.

Let us suppose that the two considered events $A$ and $B$ are along $x$ at different times and spaces (with a spatial rotation we always can obtain this configuration).

$\bar{S}$ henceforth denote the reference frame where $A$ and $B$ are simultaneous.

Consider the spacelike vector $${\bf V}:=(t_A-t_B) {\bf e}_t + (x_A-x_B) {\bf e}_x\:.$$ This vector is in the rest 3-space of $\bar{S}$ ant thus it is normal to the time axis ${\bf e}'_t$ of the reference frame $\bar{S}$.

I other words ${\bf e}'_t$ may be either parallel or anti-parallel to the timelike vector $${\bf E} = (t_A-t_B) {\bf e}_x + (x_A-x_B) {\bf e}_t$$ since $$\eta({\bf E},{\bf V}) = -(t_A-t_B) (x_A-x_B) + (x_A-x_B) (t_A-t_B)=0\:.$$

Actually ${\bf e}'_t$ can also have components along ${\bf e}_y$ and ${\bf e}_z$ and there is no way to fix them. The simplest choice is to assume these components vanishing.

${\bf e}'_t$ is future directed as due if and only the ${\bf e}_t$ component of ${\bf e}'_t$ is positive. Summing up, with the choice made for ${\bf e}'_t$, $${\bf e}'_t = \frac{(t_A-t_B) {\bf e}_x + (x_A-x_B) {\bf e}_t}{\sqrt{(x_A-x_B)^2 - (t_A-t_B)^2}}\quad \mbox{if $x_A-x_B>0$}$$ $${\bf e}'_t = \frac{-(t_A-t_B) {\bf e}_x -(x_A-x_B) {\bf e}_t}{\sqrt{(x_A-x_B)^2 - (t_A-t_B)^2}}\quad \mbox{if $x_A-x_B<0$}$$ The spatial component of ${\bf e}'_t$ is the direction of the velocity of $\bar{S}$ with respect to $S$:

If $x_A-x_B>0$ then the velocity is parallel to ${\bf e}_x$ if $t_A>t_B$ otherwise it is anti-parallel.

If $x_A-x_B<0$ then the velocity is parallel to ${\bf e}_x$ if $t_A<t_B$ otherwise it is anti-parallel.


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