Apologies, this thought has been bugging me and I am... Simple

Consider two observers.

One on Earth (stationery) One travelling away from Earth at some appreciable % of c

If the spaceship sent a message back saying "a year ago we saw an event amd then all these things took place"

What can those on Earth say about when that reported event took place? Did it happen a year ago "for Earth"? Can we calculate when it happened? Is that meaningful? If they listed everything that happened in their year would it seem empty to us because for us nearly two years had passed (plus travel time of message)?

EDIT to add a concrete example:

  • A ship flies past Earth and two observers share the event.
  • On the ship they get some decorations and food together and have a party.
  • 1 year later, ship time, they send a message with all the data about the party and every event that has happened since.
  • It arrives at Earth but Earth experienced say 3 years, plus the message travel time.
  • To Earth it would seem like the crew waited ages to have the "we saw Earth" party? Right?

My question, stated more clearly, is: if we have shared just one event, and all the ship can do is send ship times of events that happened subsequently, is it meaningful to say "this event in the log coincided with this event on Earth"?


4 Answers 4


There's a straightforward formula for converting times and distances between observers moving relative to one another; it is called the Lorentz transformation and the people on Earth may use it to convert from "spaceship years" and "spaceship meters" to "Earth years" and "Earth meters". The transformation is continuous, so there are no gaps or empty spaces, although typically durations will stretch or shrink as you convert between reference frames.

If you look at the Lorentz transformation you'll see that (a) something that happened one spaceship year ago probably didn't happen one Earth year ago, and (b) exactly how things change depends not only on the time between events but also the distance, which is why what is "simultaneous" depends on the observer (if they're moving relative to one another).

  • $\begingroup$ I edited my main question to have a concrete example. But i think your answer confirms what I was thinking?! "Simultaneously" doesn't really make sense in these situations? If we passed each other and shared Christmas cards and agreed to open them in exactly a year, we cant say that we both opened at the same time, nor can we calculate the same time, unless we share another event $\endgroup$
    – Avisian
    Dec 15, 2022 at 4:14
  • $\begingroup$ Not sure I agree about "nor can we calculate the same time"; each can calculate what the other's clock will show at any given instant of their own time, it's just that the planes of simultaneity ("same time") are different for the two observers. $\endgroup$
    – Eric Smith
    Dec 15, 2022 at 11:54
  • $\begingroup$ In other words, they could open their presents "at the same time" relative to the spaceship, or "at the same time" relative to the Earth, but these are different. Google "relativity of simultaneity" for details. $\endgroup$
    – Eric Smith
    Dec 15, 2022 at 12:18
  • $\begingroup$ Only with perfect knowledge though? $\endgroup$
    – Avisian
    Dec 15, 2022 at 17:20
  • $\begingroup$ No, perfect knowledge is not required, you just have to pick one or the other definition of "simultaneous", and recognize that it won't mean things happened at the same time coordinate for the other person. Just as what happened "in the same place" is relative, so is what happened "at the same time". $\endgroup$
    – Eric Smith
    Dec 15, 2022 at 18:32

You should be able to compute this by doing two calculations:

  1. Firstly, convert the time of 1 year in the moving frame to proper time in the (apparent) stationary frame.

  2. Next, calculate the time passed in the stationary frame for transporting the message. Add them and you have the total time in the stationary frame.

For the first calculation you can use the time dilation formulate. For the second calculation you will need the distance between Earth and the moving observer at the time of the message being sent.

  • $\begingroup$ See this is where my head went. But then if 14 years happened on earth, and 10 on the ship. And the ship reports 10 years of events, wouldnt the data also include the time dilation? IE the amount/density of data would be different $\endgroup$
    – Avisian
    Dec 15, 2022 at 3:58

Let's try some explicit numbers.

  1. Right before it sends its message, the spaceship takes 3 different angle measurement of its position relative to 3 reference stars.

  2. Spaceship sends a message towards Earth using light, including this 3 reference angles to 3 reference stars.

  3. This signal is received on Earth, and the spaceship's location at the time of message sending is determined using triangulation from the 3 reference angles sent within the message. This distance is determined to be 1 light year away from Earth.

  4. In this way, we know for sure that this message must have been sent exactly 1 year ago from the rocket (because speed of light is c as measured from any frame).

  5. The message says that, 1 year before the spaceship sends this message, an important event A happened.

  6. The question is whether we can ascertain how many years ago on Earth this important event A occurs?

  7. We can't be perfectly sure, because the spaceship would have been accelerating with respect to Earth in complicated ways in the past 1 year.

  8. Thus its onboard clock has ticked differently from Earth. The clock must have ticked some smaller amount from Earth's clock due to its accelerations with respect to Earth's inertial clock, but the exact difference is ticks depends on the exact history of the spaceship's motion in the past 1 year.

  9. Thus, an event A that takes place exactly 1 year ago on the spaceship would have taken place more than 1 year ago in Earth's clock.

  10. Add this to the exactly 1 year it took for the message to reach Earth.

  11. Thus the same important event A must have taken place > 2 years ago on Earth's clock. (?)

  • $\begingroup$ 1) 1 light year in what frame of reference, they are lorentzcontracted and don't use the same ruler lengths, 4) only if we know the distance in advance $\endgroup$
    – Yukterez
    Dec 15, 2022 at 1:42
  • $\begingroup$ @Yukterez edited with method to determine distance of spaceship from earth, does it solve the distance problem? $\endgroup$
    – James
    Dec 15, 2022 at 1:49
  • 1
    $\begingroup$ Do you know what geolocation is? That finds a position on the surface of the earth, you wouldn't use that for the position of a spaceship, I think you mean ordinary triangulation $\endgroup$
    – Yukterez
    Dec 15, 2022 at 2:06
  • $\begingroup$ @Yukterez corrected. Does triangulation solve the distance issue between Earth and our spaceship? $\endgroup$
    – James
    Dec 15, 2022 at 2:17
  • $\begingroup$ So, if i modify this to remove acceleration: ship flies past Earth and they share an event. On the ship they have a party. 1 year after they send a message with all the data about the party and every event that has happened since. It arrives at Earth but Earth experienced say 3 years, plus the message travel time. To Earth it would seem like the crew waited ages to have the "we saw Earth" party? Right? $\endgroup$
    – Avisian
    Dec 15, 2022 at 4:06

It would not seem empty. Let's say instead that the ship transmitted a message for every second that passed on its mission clock – just the reading on its clock $\tau$:

12:00:00 = $\tau_1$

12:00:01 = $\tau_2$

12:00:02 = $\tau_3$


The observer on Earth would receive those signals by his clock $t$ at equally spaced intervals greater than 1 sec. The interval measured by the Earth clock between each signal would be

$$\Delta t=\left(1-\frac {v^2}{c^2} \right)^{-1/2} \Delta \tau \approx \left( 1+\frac{1}{2}\frac{v^2}{c^2} \right)\Delta \tau$$

$$\Delta t=\left(1-\frac {v^2}{c^2} \right)^{-1/2}(\text{1 sec} ) \approx \left( 1+\frac{1}{2}\frac{v^2}{c^2} \right)(\text{1 sec} )$$

depending on the speed $v$.

This video might help you visualize it.

UPDATE: Try to couch your original question in terms of the example I've given above, by assigning numbers to events. One additional thing you need is for the two observers to agree on a single event as a time reference. For example, the moment the rocket flies past the Earth is agreed to be $t=\tau=0.$

Then your ship-based observer can say more clearly:

At this moment the mission elapsed clock is $\tau_B = \text{4 years, 5 days, 12:00:00}$. At $\tau_A = \text{3 years, 5 days, 12:00:00}$, we observed an alien outside our window.

The Earth-based observer can calculate the corresponding Earth timeline $t = t_0 + \Delta t$ from the knowledge that $t_0=\tau_0=0$, and the formula above.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Buzz
    Dec 15, 2022 at 22:32

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