Consider the following two points, or events as they are more commonly called, in SpaceTime:
Event 1: $(x,t) = (0,0)$
Event 2: $(x,t) = (a,0)$
As you can see they are merely two separate locations, Event 2 could be any point in space. The word 'event' here is a misnomer, because it carries with it the connotation of time, and the two events are simultaneous in the unprimed frame. In other words the events are simply separate points in space.
For two coordinate systems in standard configuration, the general expression from the Lorentz transformations relating the t' coordinate to the t coordinate with $c=1$ is:
$t' = \gamma(t - vx)$
So
$t'_0 = \gamma(t_0 - vx_0)$
$t'_0 + ∆t' = \gamma(t_0 + ∆t -v(x_0+∆x) )$
Taking the difference yields
$∆t' = \gamma(∆t - v∆x )$
So for the two events at the start of my question, that are simultaneous in the unprimed frame, we have
$∆t' = \gamma(0- va )$
According to the relativity of simultaneity, what is one moment in time in the unprimed frame, corresponds to two moments in time in the primed frame.
This is all because $∆t'$ isn't zero, and that is the case because $\Delta t'$ is a function of spatial coordinates.
Now suppose there is a particle at location $(a,0)$ whose $instantaneous$ speed is greater than zero. Let its instantaneous speed be the constant $v_0$. Define $a=v_0 dt >0$. Therefore $∆t'=\gamma(-vv_0dt)$. Therefore $∆t' <0$, but amounts of time are strictly non-negative.
Suppose $\Delta t′<0$ simply means that event 2 occurred before event 1 in the primed frame, and $\Delta t′>0$ means that in the primed frame event 2 occurred after event 1.
Let the particle start out at $a=-dx$, amount of time $2dt$ before event 1, and be moving to the right in the unprimed frame with speed $v_0$. Call that moment Event 0. Clearly $dx>0$ and $dt>0$.
By that supposition Event 0 occurred after Event 1, and Event 1 occurred after Event 2 in the primed frame. But the particle is moving rightwards faster than v. Therefore in the primed frame event 0 is before event 1, and event 1 is before event 2. Therefore event 1 is before and after event 2 in the primed frame.
To rigorously prove this requires the use of the Einstein velocity addition rule.
$u' =\frac {u-v}{1-uv}$
The Lorentz transformations give $t=\gamma (t'+vx')$
So $dt=\gamma(dt'+vdx')$
So $dt=\gamma dt'(1+v\frac {dx'}{dt'})$
Therefore Event 0 is $ (-v_0\gamma dt'(1+v\frac{dx'}{dt'}),-\gamma 2dt'(1+v\frac{dx'}{dt'})$
Now $\frac {dx'}{dt'}= \frac {u-v}{1-uv}$
And $u=\frac {dx}{dt}=v_0$.
So my question is,"do the Lorentz transformations lead to negative amounts of time?"