The Tachyonic Antitelephone
Two "instant" messages, one reply from the future visualized with a Minkowski diagram
Earth sends a signal that travels instantaneously in Earth's rest frame to a ship
8 light-years away, receding at 0.8c. The ship replies instantaneously
in its own rest frame the only frame it has direct access to. The reply reaches
Earth 6.4 years before the original message was sent. No clock ever runs
backward; the paradox lives in the geometry of simultaneity.
v = 0.00c · γ = 1.00 (Earth frame)
Earth worldlineShip worldlineHop 1 · instant in Earth frameHop 2 · instant in ship frameLight cone
How to read it
Time runs upward, space runs right, and units are chosen so light travels at 45°.
The dashed lines are the light cone of event A. Drag the slider and notice they
never move. The light cone is the invariant structure of spacetime.
The faint grid is Earth's coordinate system. As you boost toward the ship's frame,
it shears into parallelograms: lines of constant Earth-time tilt. That tilt is
relativity of simultaneity, drawn directly.
At v = 0 (Earth's frame): hop 1 is flat because "instant in Earth's frame" means
"along a line of constant Earth-time." Hop 2 runs along the ship's simultaneity
slice, which has slope v/c² = 0.8 in Earth's coordinates.
At v = 0.8c (the ship's frame): the roles swap. Hop 2 is now flat, the ship's
reply really is "instant" from its own honest point of view and hop 1 is the
tilted one. Neither frame is wrong.
In every frame, event C sits below event A on Earth's worldline. A and C both
lie on that worldline, so their order is frame-independent: every observer in the
universe agrees Earth received the answer before asking the question.
The calculation
γ = 1/√(1−0.8²) = 5/3
A = (t, x) = (0, 0) · B = (0, 8) → ship frame: t′ = γ(t − vx) = −10.67 yr
Ship replies at constant t′; Earth is at x′ = −0.8t′ = 8.53 ly at that moment
Transform back: t = γ(t′ + vx′) = −6.4 yr · x = γ(x′ + vt′) = 0 ✓
The mechanism generalizes: for any signal at speed u > c in one frame, a subluminal
partner frame moving at v > c²/u exists in which the signal goes backward in time
and since u > c implies c²/u < c, such a frame always exists, no matter how mild
the FTL. Two hops chained across a relative velocity close the loop. FTL, relativity,
and causality: pick two.