I am trying to show that if the Lorentz Invariant Interval is positive $$c^2 \Delta t^2 - \Delta x^2>0$$ then there exists a reference frame $S^{'}$ where $\Delta x^{'}=0$.


consider a pair of reference frames $S$ and $S^{'}$.

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$S^{'}$ is sliding away from $S$ at constant velocity $v$ along a common $x$, $x^{'}$ axis.

Consider two events $E1$ and $E2$.

From the perspective of an observer in $S$, $E1$ happens at $(x_{1},t_{1})$ and $E2$ happens at $(x_{2},t_{2})$.

From the perspective of an observer in $S^{'}$, $E1$ happens at $(x^{'}_{1},t^{'}_{1})$ and $E2$ happens at $(x^{'}_{2},t^{'}_{2})$.

We can relate the events in the two reference frames using the Lorentz transformations.

$$x^{'}=\gamma(x- v t)$$ $$t^{'}=\gamma(t- \frac{v}{c^2} t)$$

where as usual: $\gamma = (1- \frac{v^2}{c^2})^{-\frac{1}{2}}$

The differences between the events can be denoted.

$$\Delta x = x_2 -x_1$$ $$\Delta t = t_2 -t_1$$ $$\Delta x^{'} = x^{'}_2 -x^{'}_1$$ $$\Delta t^{'} = t^{'}_2 -t^{'}_1$$

Using the Lorentz transformations we are able to express the space and time separations between the events in terms of the coordinates in the first reference frame and $v$ the speed of $S^{'}$ relative to $S$. $$\Delta x^{'} = \gamma (\Delta x - v \Delta t)$$ $$\Delta t^{'} = \gamma (\Delta t - \frac{v}{c^2}\Delta x)$$

Attempt 1.

So my task is prove that a reference frame exists where $\Delta x^{'}=0$.

I guess I can do this by defining such a reference frame. How do I define a reference frame?

I guess I can define a reference frame $S^{'}$ by choosing its velocity $v$. (This is the velocity relative to the other reference frame).

I see that I could choose $v=\frac{\Delta x}{\Delta t} $. If I plug this $v$ into $\Delta x^{'} = \gamma (\Delta x - v \Delta t)$. Then I get: $$\Delta x^{'} = \gamma (\Delta x - v \Delta t)$$ $$\Delta x^{'} = \gamma (\Delta x - \frac{\Delta x}{\Delta t} \Delta t)$$ $$\Delta x^{'} = 0$$

But this is nothing to do with the Lorentz Invariant Interval being positive??

Help appreciated!

How do I show that the positive-ness of the Lorentz Invariant Interval implies that there exists a reference frame $S^{'}$ with $\Delta x^{'}=0$?


1 Answer 1


Lorentz transformation works only if $|v|<c$. Otherwise, $\gamma$ becomes complex.

However, when $c^2\Delta t^2 -\Delta x^2 \le 0$, then $v^2=\frac{\Delta x^2}{\Delta t^2}\ge c^2$ or $|v|\ge c$.


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