I'm reading about special relativity and looking at the Lorentz transformations.
I'm reading that:
If $v > c$, we can find a frame in which $t_2' < t_1'$, i.e. a signal arrived before being sent.
I am trying to satisfy myself that that is true. I am beginning with the invariant spacetime interval:
$c\Delta t^{2} - \Delta x^{2} = c\Delta t'^{2} - \Delta x'^{2}$
I assume that I'm trying to show that $\Delta t' < 0$, but I'm not seeing how to do that.
What you want to do is draw a light cone (LC) based at the first event $t_1$ and draw the invariant sets of the Lorentz transformations, which are hyperboloids (H) - "outside" the light cone - and paraboloids (P) - "inside" the light cone.
If $v<c$ the second event is "inside" the light cone (say $t_2 > t_1$, i.e. the future light cone) and Lorentz transforming you'll be moving it along a paraboloid, it stays "inside" the light cone and we always have $t_2 > t_1$.
If $v>c$ the second event is "outside" the light cone and a Lorentz transform is moving it along a hyperboloid, meaning you can use a Lorentz transform to cross the $x$-axis/the $t_1$ mark and thereby reverse the order of the two events in the transformed system.
I presume that $v$ is the speed of some message that's FTL, so let's assign some spacetime coordinates to the events where the message is sent and received. For simplicity let's assume $c=1$ and that all coordinate frames are centered at the event where the message is sent, i.e. $(t_1, x_1) = (0,0)$ are the coordinates of event S (message sent) in all relevant frames. Event R (message is received) has in the Earth frame coordinates $(t_2, v t_2)$ (the message travelled for time $t_2$ at speed $v$ in the positive $x$ direction).
Now let's consider this in a reference frame boosted by $u$ relative to Earth, where $-1 < u < 1$ so we can use an ordinary Lorentz transformation, and with the origin arbitrarily stipulated to also be the event S (to simplify the math). Then the coordinates of event R in the new frame are found with a straightforward Lorentz transformation using $u$:
$(t_2', x_2') = \gamma (t_2 - ux_2, x_2 - u t_2) = \gamma (t_2 - uvt_2, vt_2 - ut_2)$, where $\gamma = \frac{1}{\sqrt{1 - u^2}}$.
We wish to solve for $t_2' < 0$, i.e. $\gamma(t_2 - uvt_2) < 0$. We know $\gamma > 0$ so we can ignore that factor, and just require that $t_2 < uvt_2$, i.e. $1 < uv$. But $v > 1$, so this is satisfied whenever $u > \frac{1}{v}$. So for example if the original FTL message was sent at $10c$ in the Earth frame, it will be traveling backwards in time in the frame of an STL spaceship moving at greater than 0.1c relative to the Earth and in the same direction as the FTL message.
