Distant events and simultaneity I've been fiddling on this problem for a couple hours now and I'm getting more confused as time moves on, so I have a couple questions. Here it goes:

An event $A$ takes places at the origin of the $S$ frame at $t = 0$. Another event $B$ takes place 10 seconds later, located $2.5\times 10^9\,\text{m}$ away from $A$. Find the velocity of $S'$ relative to $S$ for which the events $A$ and $B$:
  
  
*
  
*(a) take place at the same point
  
*(b) take place simultaneously
  
*(c) In the case of (a), what is the $t'$ delay between $A$ and $B$?
  


What I have so far:
\begin{align*}
    V &= \frac{\Delta x}{c\Delta t}\,c \\
    &= \frac{2.5\times 10^9}{\left(3\times 10^8\right)\cdot 10}\,c \\
    &\approx 0.833\,c
\end{align*}

\begin{align*}
    \gamma &= \left(1-V^2\right)^{-1/2} \\
    &\approx \left(1-0.833^2\right)^{-1/2} \\
    &\approx 1.809
\end{align*}

\begin{align*}
    t' &= \gamma\left(t-Vx\right) \\
    t'_B &\approx 1.809\left(10-0.833\left(\frac{2.5\times 10^9}{3\times 10^8}\right)\right) \\
    &\approx 5.528\,\text{s}
\end{align*}

My issues:


*

*What is the right way to find $t'_A$? Using kinematics I found that $t_A \approx 9.167\,\text{s}$, thus $x_A \approx 9.167\,c$, therefore $t'_A\approx 2.764\,\text{s}$. By $t_A$, I mean the time that event $A$ takes to reach the (moving) location of $B$'s source. Am I completely off-track, or is it the right thing?

*For (a), what would be your interpretation of "at the same point"? I'm not sure about what a "point" represents here, besides "location", and having $A$ and $B$ at the same location would not make much sense for that question. Maybe it means "with the above parameters"? I dunno.

*For (b), would it be in the context of $S$ or $S'$? As in, $t_A = t_B$, or $t'_A = t'_B$?


Thank you very much.
 A: First let's deal with "What does it mean 'at the same point'?" question. Actually this question has little to do with relativity theory and can be discussed in terms of usual mechanics.
Imagine a train, one time per second a man on a train turns on and then turns off a lamp which is staying on his table. So we have a series of events "lamp is turned on". Because the train is moving these events have different coordinates. But this is only in the frame of reference of a person who stands on ground! In the frame of reference of a person who is on the train all these events are happening at the same point. Right here, in front of him, on a table.
Now back to question (a). If a person starts traveling with a speed $V=5/6 c=0.833c$ at the point ant at the moment of event A, he would arrive to location of event B at right time. He would observe that events A and B happened right in front of him. At the same point (in his frame of reference).
So, your answer to (a) looks correct.
Now to part (c). Because the traveling guy is moving fast, his time is going slower. Again, your answer is correct. According to his watches the travel took about 5.528 seconds: $\Delta t'=\Delta t / \sqrt{1-v^2/c^2)}$
(This is easier to calculate than to understand. I suggest to look for Twin Paradox discussions here on stackexchange)
Part (b) now.
It is possible that for some two events $X$ and $Y$ in some frame of reference event $X$ happened earlier and in some other frame of reference event $Y$ happened earlier and yet in some other frame of reference both events happened simultaneously. But this is not for all pairs of events! If you can actually travel from $X$ to $Y$ than in ALL frames of reference event $X$ happened earlier than event $Y$.
We have exactly such a case: it is possible to travel from A to B. So in all frames of reference A happened earlier than B.
More rigorously: spacetime interval defined as $s^2 = \Delta r^2 - c^2 \Delta t^2$ between two events is always the same (in all frames of reference). Let's write it down for two frames of reference: first one is the one where both events happened in the same time, second one is where both events happened at the same place:
$$ \Delta r_1^2 - c^2 \Delta t_1^2 = s^2 = \Delta r_2^2 - c^2 \Delta t_2^2$$
$$ \Delta r_1^2 = - c^2 \Delta t_2^2$$
This can not hold unless we have some complex periods of time or distances.
