Here's the scenario I'm working off of:
Person A has an FTL radio, and Person B is exactly one light-minute away with the other FTL radio. For simplicity's sake, lets say that FTL is only 200% of c.
It's 8:00 pm when Person A calls Person B. The signal arrives at 8:00:30 (on watch A, but he doesn't know that), and Person B responds with a signal. Person A receives that signal at 8:01 by his own watch.
There's an understandable error here. You used a simple Newtonian coordinate transformation to get from Person A to Person B. This only holds within their shared reference frame.
Specifically, you assumed that a signal traveling at 2xC would cross a distance of one light minute in thirty seconds. It could then be returned at 2xC to reach the sender one minute after it sent.
And this is true - within the reference frame of Person A (Alice?) and Person B (Bob).
But you're using Newtonian transformations, which don't reflect the way the universe actually works. Instead, you need to use a Lorentz transformation between two reference frames.
Now, you're correct to say that this shouldn't matter assuming that Alice and Bob are sending signals to each other, stationary, in the same reference frame. Keep this in mind: Alice and Bob
share a common reference frame since there is no velocity difference between them.
But there's a problem.
What does an observer driving past at 3/5 lightspeed, moving away from Alice and towards Bob, see?
To this observer, Alice is receding at 3/5 lightspeed, and Bob is approaching at 3/5 lightspeed.
Let's say our observer, Charlie, sees Alice fire the 2xC signal at time 0 from coordinate 0 (we'll treat the line between Alice and Bob as a one-dimensional axis.)
In Alice's reference frame, the signal reaches Bob 30 seconds after she fires it. The coordinates of the reception, Event B, are, in Alice and Bob's shared reference frame, t = 1/2 minute, x = 1 lmin (since Bob is one light-minute away from Alice, and Alice is at x = 0 lmin.)
We can say that Event B, the signal reception, has coordinates 1/2, 1 in Alice and Bob's reference frame. Since the coordinates of Event A, the signal transmission, are 0,0, Alice and Bob agree that the reception occurred after the transmission. t = 1/2 is clearly after t = 1, right?
But what are the coordinates of the reception event to Charlie?
We need to use the Lorentz transformation to get the time of the event in Charlie's reference frame. Behold:
t' = γ(t - ux/c^2)
Gamma is going to be the Lorentz transformation constant. I'll just tell you that right here, it's 5/4.
We can treat the speed of light as '1' since that's what we're using as our velocity unit.
T = .5 minutes, the time coordinate of event B, the signal reception
U = 3/5, the speed differential between Charlie and Bob when Bob receives the signal (i.e. when event B occurs)
X = 1, since that's Bob's coordinates when event B occurs.
Work it out and we get t', the time of event B in Charlie's speeding reference frame, to be:
t' = -0.125
Remember that our time unit is in minutes, so we transform that to seconds and get...
t' = -7.5 seconds.
(The time coordinate of the original event at 0,0 transformed into Charlie's reference frame is still t = 0; I worked out the transformation to make sure.)
In your example, Alice fires a superluminal single to Bob at a speed of 2xC. It covers the distance of one light-minute in 30 seconds, and Bob and Alice both agree the signal reaches Bob 30 seconds after transmission.
But to Charlie, flying between them on a 3/5 C rocket...
...
the signal reaches Bob 7.5 seconds before it is ever sent by Alice.And remember,
all reference frames are equally valid. There is nothing to say that Alice and Bob's shared reference frame is the correct one, since there is no single correct reference frame.
The upshot of this is that, for any given superluminal signal, an observer in a different reference frame can interpret that superluminal motion as motion
back in time.As a little demonstration, by the way, just change the value of T in the above equation from .5 to 1, indicating that it took the signal a full light-minute to reach Bob and therefore was only traveling at C instead of 2xC.
In Charlie's speeding reference frame, the time coordinate of the reception becomes
5/4*(1-(3/5*1)/1^2) = .5, which is perfectly sensible since it occurs after the transmission event which happens at t = 0 for Charlie.