Let's say a spaceship leaves Earth and is travelling at relativistic speeds. The spaceship's commander observes a supernova that took place somewhere ahead of the spaceship and then sends a signal to Earth to inform it of the supernova. On Earth, what will happen first, the reception of the signal or the observation of the supernova?

Common sense tells us that the light from the supernova will reach Earth first but doing the calculations tells us the opposite. How can this be explained?

Here is the full enunciation of the problem + solving:

The commander of a spaceship which is moving away from Earth on the x-axis with speed v=0.8c observes a supernova at the following spatio-temporal coordinates:

x'=6.2*10^8 ls (light-seconds)

y'=4*10^18 ls

z'=0

t'=6*10^8 s

measured from the spaceship's reference frame and reports the event to Earth.

What are the spatio-temporal coordinates of the supernova from the Earth's reference frame? Also, will Earth receive the information from the commander before or after the astronomers on Earth observe the supernova?

Applying the Lorentz transformations yields:

x=(x' - v*t')/sqrt(1-v^2/c^2)=2.3*10^8 ls

y=y'=4*10^8 ls

z=z'=0

t=(t' - v/c^2*x')/sqrt(1-v^2/c^2)=1.733*10^8 s

The results match those of my teacher's although I find the results pretty strange. x < x' although it should be vice-versa because the commander of the spaceship sees space contracted. Also t < t', but since time flows slower on Earth from the reference frame of the ship, it should be vice-versa. My guess is that my teacher got this wrong by misusing the Lorenz transformations.

closed as unclear what you're asking by ACuriousMind♦, John Rennie, Kyle Kanos, HDE 226868, gigacyanAug 21 '15 at 19:25

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• Can you explain how doing the calculations tells us the opposite i.e. update your question to give the working that shows this? – John Rennie Aug 19 '15 at 15:32
• Well because since the light from the supernova passed the spaceship it is already on its way to Earth and sending a signal afterwards makes it impossible for it to arrive before the light. That's what my common sense told me. – Adrian Dinu Aug 19 '15 at 15:39
• What calculations did you do that said otherwise? – Jim Aug 19 '15 at 15:40
• OK, but your question says: doing the calculations tells us the opposite. Can you explain how you arrived at this conclusion? – John Rennie Aug 19 '15 at 15:40
• I will explain it but can't right now, I have to leave and will return in 2hrs. Brb. – Adrian Dinu Aug 19 '15 at 15:42

1) It's the wrong Lorentz transformation. The Lorentz transformation takes the unprimed coordinates to the primed coordinates, but we need to obtain the unprimed coordinates from the primed coordinates. The correct "inverse" Lorentz transformation is: $$x = (x' + vt')/\gamma$$ $$y = y'$$ $$z = z'$$ $$t = (t' + vx'/c^2)/\gamma$$
2) Alternately, you can still use the "forward" Lorentz transformation, but you really should relabel your coordinate systems so that the ship coordinates are unprimed. In this case, your equations are correct, but you have the wrong frame velocity. If the ship is moving in the $+\hat x$ direction, then the ship sees the Earth moving in the $-\hat x$ direction. Therefore the relative velocity used in the "forward" Lorentz transformation should be $v=-0.8c$, not $v=+0.8c$.