Apologies if this has been asked and answered before, as I am still having doubts reconciling the symmetrical effects of time dilation (i.e each frame sees clocks of other frame slowing down), and I am not sure where I am making a conceptual mistake.

Assume that we have Earth and a distant planet 8 light years away (as per Earth's frame), and assume that the planet is at rest with Earth - and so clocks on both are somehow synchronized.

A space ship leaves the space station at 0.8c towards Earth, when the space-station clock reads 0. Assume it is able to accelerate to 0.8c instantaneously (or more realistically let us ignore effects of acceleration). The spaceship emits light signals every 24 hours (or one spaceship day) as per clocks on the spaceship. When the Earth clock also reads 0, a laboratory on earth also starts emitting signal every 24 hours (or one Earth day) as per Earth's clock.

At 0.8c, the Lorentz factor is 0.6. Also, as per relativistic Doppler effect both the laboratory on earth and space ship would receive signals at thrice (*3) the rate.

Thus as per Earth’s frame, by the time the space ship arrives at Earth, it’s clock would read 10 Earth-years (8/0.8 = 10).

Now I also can understand, that 8 light-years in Earth's frame is 4.8 light years in Space-ship frame (Length contraction: 8 * .6 = 4.8), and thus from the Spaceship’s point of view, it is stationary and Earth is approaching it at 0.8c from 4.8 Spaceship-light-years away. Thus in Space-ship's time it would take 6 Spaceship-years for Earth to reach it, 10 * 0.6 = 6 and so that seems to make sense.

It seems to me that here we took what we know in Earth's point of view (8 light years), and applied Lorentz factor to arrive at what Spaceship time/length would be, and then saw how the calculation from the Spaceship’s frame’s perspective would match.

Of course from Spaceship frame, Earth’s lengths should be contracted and time should be dilated. For it, Earth is coming towards it at 0.8c, from 4.8 Spaceship-light-years away, and it takes 6 Spaceship-years to reach it. Then for Earth's frame, that length would 4.8 * 0.6 = 2.88 Earth-light-years. And to travel that would take 2.88/0.8 = 3.6 Earth-light-years. Of course, since 6 * 0.6 = 3.6 and so it still seems to add up in the sense that Earth’s lengths are contracted and time is dilated.

But 3.6 Earth-years arrived here obviously doesn’t match 10 Earth-years in the other calculation. So I am not sure if this is right (even though 3.6 Earth-years and 6 Spaceship-years are related by the correct Lorentz Factor 0.6 for 0.8c speed).

I was then thinking even if both Earth and Spaceship disagree on lengths and times, what they must agree on (besides speed of light), must be how many signals earth sent out and how many earth-signals the spaceship got.

From the Spaceship’s point of view, the first signal was sent from Earth when it was 4.8 Spaceship light years away. Thus it would take 4.8 Spaceship years to reach the Spaceship. By that time, Earth would have have travelled (as per Spaceship) 4.8*0.8 = 3.84 Spaceship light years, and thus would be 4.8 - 3.84 = 0.96 Spaceship light years away. It takes 0.96/0.8 = 1.2 Spaceship years to cover the remaining distance. Thus the Spaceship receives Earth signals for 1.2 Spaceship years (at thrice the rate and thus 1.2 * 3= 3.6 light year) until Earth and Spaceship are coincident in the Spaceship frame.

But from Earth’s frame, it would have sent 10 Earth-years worth of signals before Spaceship arrives. Thus as per Earth, it would have sent sent 10*365 = 3650 signals (one signal per Earth-day), to spaceship, and so the space-ship ought to have received so many. But doing the calculation for Earth’s frame from Spaceship and applying the time dilation, I seem to arrive that Spaceship would have received Earth signals for 1.2 Spaceship years albeit 3.6 Earth years worth of signals i.e. 3.6 * 365 = 1314.

How can I account for all Earth signals received at the Spaceship. What am I missing (I know I went off the beam somewhere :-) - just can’t figure out where)

  • 1
    $\begingroup$ If you draw the spacetime diagram, the answers to all your questions will be clear. $\endgroup$
    – WillO
    Oct 5, 2015 at 15:13
  • $\begingroup$ "or more realistically let us ignore effects of acceleration". Unfortunately, you can't. Since the spaceship accelerates when leaving the station and accelerates again (in reverse) to slow down to land on Earth, it experiences less time than the station or Earth. Any object in a non-inertial frame (ie, a frame nonzero acceleration) ages slower than objects in an inertial frame. $\endgroup$
    – user854
    Mar 17, 2016 at 2:41
  • 1
    $\begingroup$ @barrycarter A large acceleration can have a huge effect on simultaneity of the two inertial frames of before and after acceleration, but it can happen in a very small region of proper time so could easily be ignored when computing the proper time of the path. The whole definition of proper time as an integral along a curve actually approximates the path with a series of small straight paths with impulsive accelerations. Those are the things you add up to get the Riemann sum. $\endgroup$
    – Timaeus
    Mar 19, 2016 at 10:01
  • $\begingroup$ @barrycarter And as for aging less, an inertial observer going from an event A to an event B ages more than any other observer ... that goes from the same event A and to the same event B. But that isn't happening in this question. $\endgroup$
    – Timaeus
    Mar 19, 2016 at 10:03

2 Answers 2


You write: "From the Spaceship’s point of view, the first signal was sent from Earth when it was 4.8 Spaceship light years away. Thus it would take 4.8 Spaceship years to reach the Spaceship."

Be careful. I, on earth, send out a light signal at the same time (according to me) that the space ship takes off. According to you (on the now-moving spaceship), I sent that signal 10.67 years ago (at time -10.67 by your clock). Also, according to you, I've spent that 10.67 years traveling towards you at speed .8. Therefore, when the light signal was sent, I was not just 4.8 light years away; I was 4.8 plus another (10.67 x .8) light years away --- a total of about 13.33 light years. So the light signal should take a total of 13.33 years to reach you. As it was sent at time -10.67 it should arrive at time 2.67. (I hope I got the arithmetic right!).

The right way to do this, of course, is not to grind through the arithmetic, but to draw the spacetime diagram, which makes everything clear from the start. The black line is the earth, the blue line is the distant planet, the red line is the traveler, and the gold lines are light rays, sent every two years from earth and every two years from the ship. The black numbers are earth times and the red numbers are traveler-times.

enter image description here

  • $\begingroup$ Thanks. It will take a bit to digest fully but I see that I made the mistake of assuming what is simultaneous for the Earth and simultaneous for the spaceship that is 8 earth-light years away. I am still a beginner in learning relativity, which is also why at this point, while I do see that space-time diagrams are the way to go, they seem to "abstract" the problem a bit too much for me :-), i.e. when I am still trying grasp an understanding of the underlying phenomenon. This is why I am grinding through the arithmetic (and yes exposing myself to the danger of conceptual mistakes). $\endgroup$
    – Ark
    Oct 5, 2015 at 20:29
  • $\begingroup$ @Ark : You are probably right that grinding through the arithmetic will pay off at your stage. But I suggest always trying to draw the picture too, and making sure the picture and the arithmetic are telling you the same thing. Eventually, you'll stop needing the arithmetic. $\endgroup$
    – WillO
    Oct 5, 2015 at 20:36
  • $\begingroup$ Yep. Thanks - although I still want to understand relativity of simultaneity in "lay man terms" (I do get the basic concept), just applying the math to the equations at en.wikipedia.org/wiki/Relativity_of_simultaneity helps me arrive that t' (spaceship) is -10.67 years and x is -13.33 light years (origins coincide at t,t' = 0 and x,x' = 0 at the planet of course). Thanks again! $\endgroup$
    – Ark
    Oct 5, 2015 at 23:31
  • $\begingroup$ @Ark: The picture I drew does not include the lines of simultaneity (because I didn't want to overclutter). But for the traveler, there's a line running from 6.4 on the black vertical axis to 0 on the blue axis, and the others are all parallel to that. From that, it's instantly clear that the simultaneity line through the black 0 hits the red line way below 0. Arithmetic shows it to be -10.67, but a little experience with the geometry will let you eyeball the figure and make a good ballpark guess without calculating. $\endgroup$
    – WillO
    Oct 5, 2015 at 23:53
  • $\begingroup$ This is incorrect. If you start out 8 light years away and get closer, you will never be "a total of about 13.33 light years" in any frame of reference. $\endgroup$
    – user854
    Mar 17, 2016 at 2:46

This may or may not be helpful.


The diagram above shows the following in the planet frame:

  • The solid blue line is Earth, which remains 8 light years distant from the planet at all times. I've drawn dots for t=-8 to t=10.

  • The solid green line is the planet, which remains at x=0 at all times. I've drawn dots for t=0 to t=18.

  • The solid red line is the ship, which travels from the planet to Earth in 10 years, covering 8 light years, at least in the planet's frame. Note that the dots are drawn every year of planet time, not every year of ship time.

  • The dashed blue lines are light-speed signals traveling from Earth to the planet, intersecting the ship as they pass.

  • The dashed red lines are light-speed signals traveling from the ship to Earth.


  • The x and t scale are not equal on this graph. Because we're dealing with a distance of 8 light years and a timeframe of 26 years, I couldn't find a good way to create a graph with x and t scaled equally.

  • I chose x as horizontal and t as vertical. Some people do it the other way.

Things To Notice:

  • When the ship leaves the planet, it sees Earth's signal from t = -8. In other words, it sees the Earth as it looked at t = -8. This isn't the time it assigns Earth, it's just what Earth light it's currently receiving.

  • By the time the ship arrives, it sees Earth at t = 10. Again, this isn't the time it assigns Earth, just what it actually sees.

  • Thus, considering only the light actually reaching the ship, 18 years passes on Earth during the ship's 10 year journey. Notice that this has nothing to do with dilation or contraction, it's pretty much just the Doppler Effect.

  • In your example, Earth only starts sending signals at t=0. In this case, the ship receives the first signal at about 4.44 years (still in the planet's frame), a little before it's halfway through it's journey.

  • Since the ship still receives the last signal at t=10, we note the ship receives 11 signals during its 10 year journey, but the first signal doesn't arrive until t = 4.4 (Earth time, not ship time), so it really receives the 11 signals in 5.56 years (one signal every 0.505 years or so).

  • The first signal the ship sends doesn't arrive until Earth time t = 8. This makes sense, since, in this frame, the Earth is 8 light years away.

  • The last signal the ship sends arrives at t = 10 (just as the ship is landing). Thus, the Earth sees 10 signals arriving in the space of two years, one signal every 0.2 years.

Now, let's do a Minkowski transform into the ship's frame.

Just for fun, we'll animate the change as we move from the planet's frame to the ship's frame (for the purposes of the animation, I've removed the labels giving Earth's time).


Focusing on the final image:


While this diagram is important, let's zoom in a bit:


Notice what we see here:

  • The ship now remains at x=0 throughout its journey. In other words, we are now looking at things from the ship's point of view.

  • When the ship leaves, it still sees light from Earth at t = -8 Earth time.

  • When the ship arrives, it still sees light from Earth at t = 10 Earth time.

  • During its journey, the ship still sees 11 signals (if we ignore those where Earth t < 0).

  • The first signal the ship sends still arrives at Earth time t = 8, and the last signal still arrives at Earth time t = 10.

In other words, in terms of what the ship and Earth actually see, nothing has changed.

The only difference is that we can now talk about ship time and ship distance, instead of planet time and planet distance.

Note that if we look at t = 0, we see the Earth at x = 4.8 (the Lorentz contraction) and at Earth time t = 6.4.

In some sense, however, this is artificial. The light from Earth at t = 6.4 will still reach the ship at the same point it is journey as it did before.

Further, the Earth that's 4.8 light years away is not Earth at t = 0, but Earth at t = 6.4.

Ultimately, regardless of the coordinate system, an observer in a given frame will see the same event at the same time and same distance.

The Lorentz contraction and simultaneity are artificial constructs that insures this happens.


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