# Show that a reference frame exists where the spatial separation is zero.

## question

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 $$\Delta x^{'}=0$$.

## context

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

$$S^{'}$$ 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 $$\Delta x^{'}=0$$?

Lorentz transformation works only if $$|v|. Otherwise, $$\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$$.