# Consequences of space-/time-/light-like separations

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1. Jul 23, 2015

### "Don't panic!"

I'm trying to prove the following statements relating to space-like, time-like and light-like space-time intervals:

1. There exists a reference frame in which two space-time events are simultaneous if and only if the two events are space-like separated.

2. There exists a reference frame in which two space-time events are coincident at a single spatial point if and only if the two events are time-like separated.

3. If two events are light-like separated, then there are no reference frames in which they are simultaneous or coincident at a single spatial point.

I can't seem to find any notes that a can verify my attempt with so I'm hoping that people won't mind taking a look at my workings on here to see if they're correct.

Consider the space-time interval $$\Delta S^{2}= (\Delta x^{0})^{2}-(\Delta\mathbf{x})^{2}$$ where we use the metric signature $(+,-,-,-)$.
We then perform a spatial rotation of our coordinate system such that one of the space-time events $x^{\mu}$ is located at the origin of our reference, $$x^{\mu}=(0,0,0,0)$$ The other space-time event that we consider $y^{\mu}$ is then spatially aligned along the $z$-axis in our coordinate system, $$y^{\mu}=(y^{0},0,0,y^{3})$$

Given this, we shall now consider each of the (numbered) cases above.

1. Space-like interval :

First, let the two events $x^{\mu}$ and $y^{\mu}$ be simultaneous in our inertial frame of reference $S$, i.e. $x^{0}=0=y^{0}$. It then follows that the space-time interval between them is given by $$\Delta S^{2}=(x^{\mu}-y^{\mu})^{2}=(0-0)^{2}-(0-y^{3})^{2}=-(y^{3})^{2}$$ Now, $(y^{3})^{2}>0$ and so clearly $\Delta S^{2}<0$. Therefore, if the two events are simultaneous in $S$, then they are space-like separated.

Next, let the two events $x^{\mu}$ and $y^{\mu}$ be space-like separated. We have then that $$\Delta S^{2}=(x^{\mu}-y^{\mu})^{2}=(y^{0})^{2}-(y^{3})^{2}<0\quad\Rightarrow\quad\vert y^{0}\vert < \vert y^{3}\vert$$ Thus, we can choose $\beta =v=\frac{y^{0}}{(y^{3})}$ (in units where $c=1$). From this, we see that $\beta <1$ as required. Performing a Lorentz boost along the $z$-axis we can relate the coordinates $x^{\mu}$ and $y^{\mu}$ in $S$ to their expressions in another inertial frame $S'$ $$x'^{0}=\gamma\left(0-\beta 0\right)=0\; , \qquad x'^{3}=\gamma\left(0-\beta 0\right)=0$$ and $$y'^{0}=\gamma\left(y^{0}-\beta y^{3}\right)=0 \; , \qquad y'^{3}=\gamma\left(y^{3}-\beta y^{0}\right)$$ where $\gamma =\frac{1}{\sqrt{1- \beta^{2}}}$.

Hence, in $S'$ we see that the two space-time events have the following coordinates $$x'^{\mu}=(0,0,0,0) , \qquad y'^{\mu}=(0,0,0,y'^{3})$$ and thus are simultaneous in this frame.

2. Time-like interval :

This follows a very similar approach to the space-like case.
First, let the two events $x^{\mu}$ and $y^{\mu}$ be coincident at a single spatial point in our inertial frame of reference $S$, i.e. $\mathbf{x}=\mathbf{0}=\mathbf{y}$. It then follows that the space-time interval between them is given by $$\Delta S^{2}=(x^{\mu}-y^{\mu})^{2}=(0-y^{0})^{2}-(0-0)^{2}=(y^{0})^{2}$$ Now, $(y^{0})^{2}>0$ and so clearly $\Delta S^{2}>0$. Therefore, if the two events are spatially coincident in $S$, then they are time-like separated.

Next, let the two events $x^{\mu}$ and $y^{\mu}$ be time-like separated. We have then that $$\Delta S^{2}=(x^{\mu}-y^{\mu})^{2}=(y^{0})^{2}-(y^{3})^{2}>0\quad\Rightarrow\quad\vert y^{0}\vert > \vert y^{3}\vert$$ Thus, we can choose $\beta =v=\frac{y^{3}}{(y^{0})}$ (in units where $c=1$). From this, we see that $\beta <1$ as required. Performing a Lorentz boost along the $z$-axis we can relate the coordinates $x^{\mu}$ and $y^{\mu}$ in $S$ to their expressions in another inertial frame $S'$ $$x'^{0}=\gamma\left(0-\beta 0\right)=0\; , \qquad x'^{3}=\gamma\left(0-\beta 0\right)=0$$ and $$y'^{0}=\gamma\left(y^{0}-\beta y^{3}\right) \; , \qquad y'^{3}=\gamma\left(y^{3}-\beta y^{0}\right)=0$$ where $\gamma =\frac{1}{\sqrt{1-\beta^{2}}}$.

Hence, in $S'$ we see that the two space-time events have the following coordinates $$x'^{\mu}=(0,0,0,0)\; , \qquad y'^{\mu}=(y'^{0},0,0,0)$$ and thus are spatially coincident in this frame.

3. Light-like interval :

In this last case it is trivial, as given two events $x^{\mu}$ and $y^{\mu}$, if they are light-like separated, then $$\Delta S^{2}=(x^{\mu}-y^{\mu})^{2}=(y^{0})^{2}-(y^{3})^{2}=0$$ and thus it is impossible to find a frame in which they are either simultaneous, or spatially coincident, as either one would change the interval into a space-like or a light-like interval. The interval is Lorentz invariant, so this clearly cannot be the case.

Last edited: Jul 23, 2015
2. Jul 23, 2015

### DEvens

What part are you having trouble with? It looks ok.

3. Jul 23, 2015

### "Don't panic!"

Nothing in particular, I just wasn't sure whether I'd constructed my argument correctly - particularly whether I've done the Lorentz transformation parents correctly?! Was seeking clarification more than anything really.

4. Jul 23, 2015

### Mentz114

If we have two points $P(t_1,x_1)$ and $Q(t_2,x_2)$ with $\Delta t = t_2-t_1,\ \Delta x=x_2-x_1$ the transformed interval $\Delta t' - \Delta x'$ is
$\gamma \Delta t +\beta\gamma \Delta x - (\gamma \Delta x +\beta\gamma \Delta t )= \gamma(1-\beta)(\Delta t - \Delta x)$

From this you can work out quickly the answers to your question. I can't see anything wrong with what you've done, I'm just proposing slightly more compact way.

5. Jul 23, 2015

### "Don't panic!"

Ok cool, thanks for the tip. Glad I've understand it correctly.