Blew me away that chatGPT could solve this one
The Puzzle
There are three people, A, B and C. C is a girl and A and B are her suitors who have departed on interplanetary voyages and are soon to return. B has never been intimate with C.
At the instant t=0, A and B are both at a distance R from C. That is, assume C is a point and draw two radii out from C. Assume, just for simplicity, that the radii are at right angles.
A is currently stationary w.r.t. C and B is moving toward C at a velocity v.
A and B are not friends. If B finds out that C has gotten pregnant, he will assume that A is the culprit and take hostile action (fire a missile) against A which could have lethal consequences if it is early enough.
It so happens that C is pregnant, and, at time t=0, "news" of this event is broadcast (from C's position) by emitting photons in all directions such that, in C's frame of reference, stationary observers at A's and B's positions would receive them at the same time. This "news" is not broadcast by C but by another entity at the same location as C, or very near to C.
From A's frame of reference, the time taken for the news to arrive is R/c because A is stationary w.r.t the source that emitted the photon. From B's frame of reference, C is moving towards the source, so the time taken to intercept the photon will be less than R/c (that is, R/(c+v)).
From B's frame of reference, the fact that C is moving towards B at velocity v is irrelevant in regards to how long the photon takes to arrive because it was emitted at t=0, when the separation was R and the fact that the source has moved since then is irrelevant. The photon has to travel R/c and so it will take R/c before B receives the "news".
From C's frame of reference, the situation is similar to what A observes. B is travelling toward the photon and will therefore intercept it before A does and will therefore take hostile action towards A.
A and B do not know what the news is, but C does, though she is not responsible for the transmission. Given that she believes B will receive the unwelcome news before A does, is she right to worry that A is in danger?
 |
| A simple puzzle with a relatively common mistaken assumption |
The Answer
Set the stage (C’s rest frame = A’s rest frame):
Put C at the origin. A sits at x=R (at rest). B sits at y=R and moves straight toward C with speed v. At t=0 (C’s clock) a flash goes out from C in all directions.
Light travel to each:
To A: tA = R/c.
To B: solve c·t = R − v·t ⇒ tB = R/(c+v) < R/c.
So in C’s (and A’s) frame, B receives the news first.
“But in B’s frame shouldn’t it still be R/c?”
No. The R you quoted is the C-frame separation at the C-frame instant t=0. In B’s own frame, simultaneity is different: at the moment the flash is emitted (t'=0 at C), B is farther than R away—specifically γ·R awayγ = 1 / √(1 − v²/c²)—because of relativity of simultaneity. Transforming the emission and reception events shows B measures a travel time t'B = γ·R/c, not R/c. This is consistent with light still going at c in every frame.
Can B’s reaction endanger A before A gets the news?
This is the crux. Let:
- Eemit: flash at C,
t=0. - EB: B receives, at
tB = R/(c+v). - EA: A receives, at
tA = R/c.
c) from EB reach A before EA?
Compute in C’s frame:
Δt = R/c − R/(c+v) = (R·v)/(c(c+v)).
Spatial gap: |Δr| = R√[1 + (c/(c+v))²].
Minkowski interval: c²Δt² − |Δr|² = −(2c/(c+v))·R² < 0.
Conclusion
- C is right that B hears first in her frame.
- B is wrong to think the travel time is
R/c—mixing distances from one frame with times from another. - Most importantly: Although B hears first, B cannot act on A before A hears. Causality protects A.
Bottom line: C can worry about B being angry sooner, but she need not worry about immediate danger. Gossip can travel fast, but not faster than light.
