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If the spaceship, travelling at c/2 were to turn on only its forward facing headlights, then after one second (relative to us, the stationary external observers) we would see that the ship has moved half a light second, whereas the first photon to be emitted from the headlight one entire light second. Relative to the crew, however, the photon would have travelled only half a light second. Therefore, Special Relativity jumps in to explain that the crew would have only experienced half a second of time since the headlights were turned on. This means that only after another second (for us) passes, will the photon have travelled one light second relative of the ship, and the crew would measure one second of elapsed time. So, for every second that passes for the crew, two have passed for us. Here's where this breaks apart for me:

What if the crew we're to turn on their headlights in the opposite direction of the ship's motion? What I imagine would happen is that, since the speed of light is always the same, then after one second (for us), the first photon to come off of the back of the ship will have travelled one light second "backwards", whereas the ship half a light second forwards, so that the total distance between them is 1.5 light seconds. The issue is that during our one second, as explained above, the crew will have measured only half a second of elapsed time. Yet, if them look back, they will see how during that half second those photons will have travelled 1.5 light seconds! This would mean that after one second, the crew will see that the photon has travelled 3 light seconds.

If my question is not understandable, I can throw in a sketch.

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    $\begingroup$ "Special Relativity jumps in to explain that the crew would have only experienced half a second of time since the headlights were turned on" - That's incorrect. For $\frac{v}{c}=\frac{1}{2}$, the Lorentz factor is $\frac{1}{\sqrt{1-(1/2)^2}}=\frac{2}{\sqrt{3}}$. We would see time on the ship run a factor of $\frac{2}{\sqrt{3}}$ slower and we would see the ship's length contract by a factor of $\frac{2}{\sqrt{3}}$. The crew on the ship would see time in the rest of the universe run slower and would see lengths in the rest of the universe contracted by the same factor. $\endgroup$ – probably_someone Dec 8 '18 at 16:55
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/79331/2451 and links therein. $\endgroup$ – Qmechanic Dec 8 '18 at 18:50
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    $\begingroup$ Time dilation and length contraction are not the right frame for understanding this: the relativity of simultaneity is the right frame. And this arrangement is very similar to one on Einstein's train and platform gedanken experiments. $\endgroup$ – dmckee --- ex-moderator kitten Dec 8 '18 at 22:18
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If you'd thrown in the right sketch, you could have simply read off the answer:

enter image description here

According to the earth observer, there are stationary mileposts every light-second. The earth is at milepost $-1$, and the ship is at milepost $0$ when it emits its light beams going forward and backward (A). At time $1$, the backward beam reaches earth (B), the ship reaches milepost $1/2$ (C), and the forward light beam reaches milepost $1$ (D).

According to the ship observer, the same mileposts are located .866 light-seconds apart, and are traveling "backward" (i.e. toward where the earth is right now) at speed $1/2$. When I release the lightbeams (A), milepost $1$ is .866 light-seconds away, and traveling toward me at speed $1/2$, while the forward light beam travels toward it at speed $1$. They meet up (D) at time $.577$ and location $.577$. Meanwhile, milepost $1/2$ has been moving toward me at speed $1/2$, and it catches up with me at time $.866$ (C). Finally, the earth has been moving away from me at speed $1/2$ while the backward lightbeam has been chasing it at speed $1$; the lightbeam catches up with it (B) at time $1.73$ and location $-1.73$.

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    $\begingroup$ You put in guidelines for constant $x$, constant $t$ and constant $t'$, but not for constant $x'$ (nor do you exhibit the $x'$ axis). Not sure if it would help (there is a risk of overwhelming beginning readers, but the whole notion of Minkowski diagrams tends to do that in the first place. $\endgroup$ – dmckee --- ex-moderator kitten Dec 8 '18 at 22:21
  • $\begingroup$ @dmckee: But I did exhibit the $x'$ axis! It's labeled $t=0$, in blue. (I used blue instead of "prime"). (For that matter, I also exhibited a line for constant $x'$, though admittedly not several --- I still don't see the value or more.) $\endgroup$ – WillO Dec 8 '18 at 22:53
  • $\begingroup$ I'm having a little trouble understanding the sketch. Is A moving horizontally or diagonally? Why are the light beams facing the left and right corners? The scenario I proposed would have the light beams travel on one common line (like the front/rear headlights of a car). Thanks for taking your time! $\endgroup$ – Izak the coder Dec 9 '18 at 14:33
  • $\begingroup$ @Izakthecoder : The light beams are traveling due east and due west. If you ignore the blue frame and just look at the black frame (the frame of the earthbound observer, it shows that at time $0$, both beams are at location $0$, and at time $1$, the westbound beam has reached location $-1$, while the rightbound beam has reached location $+1$. (Time is measured vertically; distance is measured horizontally.) The blue frame is the ship observer's frame. The thick blue line is the time axis. Other lines parallel to it are omitted to avoid clutter.....(CONTINUED) $\endgroup$ – WillO Dec 9 '18 at 15:36
  • $\begingroup$ (CONTINUED) In the blue frame, the lightbeams are still traveling due east and due west. The mileposts (whose paths are shown by the black vertical lines) are all traveling due west. At time $0$, the lightbeams are location $0$. At time $.577$, the eastbound lightbeam and the first milepost cross paths. At time $.866$, the traveler (represented by the thick blue line) and the $1/2$ milepost cross paths. At time $1.73$, the westbound beam and the earth (milepost $-1$) cross paths. I hope this helps. $\endgroup$ – WillO Dec 9 '18 at 15:41

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