I was thinking if I built a device with 7 clocks, synchronized to each other, one in the middle, one up, down, left, right, behind and in front of me, say 1 meter away, and I fired a laser from the middle and split it into six pieces would the light always arrive at all six "corners" at the same time?

If we accelerated (say to the right) up to half the speed of light and fired the laser again, wouldn't the laser have to travel farther to reach the right sensor (as it's moving at half the speed of light) and travel less distance (as it is approaching the light at half the speed of light)? So the time the light arrives at the right sensor should be greater than the time it takes to arrive at the left sensor.

I'm assuming you will tell me that this is not true, the light will still arrive at all the sensors at the same time. Why would this be true?

Isn't the whole apparatus moving at half the speed of light in one direction? Why would that make no difference? In that case, why couldn't I just accelerate to half the speed of light again, to the right? (Then I will be going at the speed of light - half the speed of light, twice, but in each instance I was considered to be standing still.)

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    $\begingroup$ how much relativity have you learned? $\endgroup$
    – Jim
    Commented Sep 7, 2014 at 22:09
  • $\begingroup$ Can you feel the Earth moving at 30km/s relative to the sun? Can you feel the sun moving at 230km/s relative to the center of the galaxy? Why not? $\endgroup$
    – CuriousOne
    Commented Sep 7, 2014 at 22:37
  • $\begingroup$ "how much relativity have you learned?" Obviously, not a lot. I've read the train and two lightning bold thought experiment. I would assume our frame of reference would always be the Earth, or where ever we are starting our journey from. If a space ship accelerated away from the Earth up to 99.9C, then to us they are travelling very fast, but to them they are standing still. That makes sense to me. However, can the people on the spaceship then assume they are standing still and accelerate from zero to 99.9C again? $\endgroup$ Commented Sep 8, 2014 at 23:47
  • $\begingroup$ Yes the people in the spaceship can do that, no problem with that. $\endgroup$
    – Héctor
    Commented Sep 8, 2014 at 23:50
  • $\begingroup$ If, throughout the universe, one observer sends information to another, about their relative speed, and so on, can't we get an overall picture of who is going very fast and who is going slow. Won't the people at the edge of the universe be going near the speed of light and the people closer to the center be going a lot slower, relative to each other? It's not a matter of observing each other. Each observer sends relative information down the line. (That's why I had synchronized clocks, so people can send information to each other, regardless of their frame of reference.) $\endgroup$ Commented Sep 8, 2014 at 23:52

1 Answer 1


In relativity the notion of simultaneity is relative to the obeserver. While one observer (the one "standing") see all the rays of light arrive at the same time, another observer (the one going near the speed of light) will see one ray arriving before another. The paradox here is you think simultaneity can be defined in an absolute way independent of the observer, the true is that notion is frame dependent.


Suppose there are three observers, observer 2 go at a half of the speed of light relative to observer 1 ($ v_{21} = \frac{c}{2}$), and observer 3 go half of speed of light relative to observer 2 ($ v_{32} = \frac{c}{2}$),the observer 3 simulate what you said about accelerate to half of the speed of light again. Then, at what speed observer 3 is moving respect observer 1? the speed of light? No, because the Galilean law for adding velocities is not true in special relativity, instead you have $$ v_{31} = \frac{v_{21}+v_{32}}{1+\frac{v_{21}v_{32}}{c^2}} = \frac{\frac{c}{2}+\frac{c}{2}}{1+\frac{1}{4}} = \frac{4}{5}c<c $$ So is not problem, you can still accelerate to near of speed of light again, and still be under the speed of light for all inertial observers . All inertial observers are on equal footing.

  • $\begingroup$ did you mean the photon's frame of reference on your last sentence? $\endgroup$
    – Héctor
    Commented Sep 8, 2014 at 23:04
  • $\begingroup$ I made a edit with the answer. $\endgroup$
    – Héctor
    Commented Sep 8, 2014 at 23:19
  • $\begingroup$ I'm not worried about simultaneity. I'm pondering if we can judge our speed relative to a photon. From what you said, the one going near the speed of light will see one ray arrive before another. That's exactly the information I want to deduce. How fast we are going relative to the photons. $\endgroup$ Commented Sep 8, 2014 at 23:22
  • $\begingroup$ It's not possible to be in the frame of reference of the photon, they are not inertial observers, so it's not a meaningful question ask "what a photon see?", even the notion of proper time it's not defined in a null frame. $\endgroup$
    – Héctor
    Commented Sep 8, 2014 at 23:25
  • $\begingroup$ So, it's not possible to know if you are getting close to the speed of light in any particular direction? $\endgroup$ Commented Sep 8, 2014 at 23:27

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