I'm new to thinking about relativity, as well as phenomena related to the speed of light, time dilation, length contraction, simultaneity, etc. I am pondering a paradox.


  • Motion is relative.
  • Nothing can travel faster than the speed of light relative to anything else.

Imagine 3 points in space, A, B, and C, in alignment along the same line. A is 5 light years away from B and B is 15 light years away from C, which means that A is 20 light years away from C.


It takes an observer at A 20 years to receive a radio signal sent from C. It takes an observer at B 15 years to receive a radio signal sent from C.

Imagine a radio signal is sent from C at t=0. Also at t=0, an observer at A begins a trip toward B at half the speed of light relative to both B and C, going from 0 to 0.5c instantaneously (no acceleration). Let's call this observer "Superman."

Because Superman starts at A and is traveling at the half the speed of light relative to B, and A and B are 5 light years apart, it will take Superman 10 years to reach B, at time=10 years.

Also at time=10 years, the radio signal sent from C has traversed 10 light years of space between C and B and has 5 light years more to travel before it reaches B. It will take the radio signal, which is traveling at the speed of light, 5 years to travel the remaining 5 light years.

My question is: Will Superman be able to receive the radio signal at B at time=15 years? If so, how does this not violate the principle that nothing can travel faster than the speed of light relative to anything else?

If Superman had stayed at point A, the radio signal would have moved towards him at the speed of light, eventually reaching him 20 years after it had left C. But the signal reaches B after only 15 years. It seems that Superman should be able to fly to B and receive the signal in less than 20 years. But then Superman and the radio signal would have moved faster than the speed of light relative to each other.

Finally, what bearing does the answer to this question have on our ability to observe objects in space, such as stars and planets? Are we unable to do anything to affect the amount of time it takes for us to see things (i.e. light) in space, even if we got up and moved to another place in the universe light years away from Earth? Does this all come down to "simultaneity?"

  • $\begingroup$ You should investigate something called the Lorentz transformation. This says that coordinate distances and times are different for differing reference frames. When Superman starts moving, he no longer measures things as someone who remains on A (or B or C). $\endgroup$
    – Bill N
    Commented Mar 16, 2016 at 21:25

1 Answer 1


Special relativity doesn't forbid saving time by meeting in the middle. It also doesn't forbid you from time-dilating yourself by an arbitrarily large amount to get somewhere without personally waiting a long time. For example, see this web page about relativistic flight times:

Here are some of the times you will age when journeying [with constant 1g acceleration] to a few well known space marks, arriving at low speed:

d               Stopping at:               T [your age on arrival]      
4.3 ly          Nearest star              3.6 years
27 ly           Vega                      6.6 years
30,000 ly       Centre of our galaxy     20 years
2,000,000 ly    Andromeda Galaxy         28 years

When superman changes speeds, he switches inertial frames. In his new frame, where he's moving at 0.5c away from A towards C, C is closer (due to length contraction) and already sent the signal years ago (due to the relativity of simultaneity). Those two effects will account for the saved years from his perspective.

To get a better idea of how changing frames looks, I recommend reading this website which has nice diagrams like this one:


Also this video shows the difference between Galilean (classical) and Lorentz (relativistic) transforms quite clearly.

And you can get some interactive experience by playing velocity raptor.

  • $\begingroup$ Thank you very much for your response and the links. I will check them out. $\endgroup$ Commented Mar 17, 2016 at 3:26

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