I have an observer on Earth with an atomic clock, let's call this the unprimed frame with space coordinate x=0 and t. Universe is one dimensional.

I have a rocket ship sitting stationary in empty space to the far "left" of the earth on the x-axis. Because it is stationary wrt the Earth, rocket ship and earth have identical spacetime basis (up to a shift in space). The ship begins to accelerate to the right at t=0 with immense force such that relativistic effects can be felt quickly. Special relativity would say that the travelling space ship plane of simultaneity begins to rotate counter-clockwise in a standard Minkowski diagram. To me, this seems to mean that "things" to the left of the spaceship would advance in time more quickly (the further left, the further the time advances), while things to the right of the spaceship must go backwards in time. Since the rocket ship is to the left of the earth, won't it mean that the rocketship crew sees events on Earth going backwards in time?

  • $\begingroup$ "won't it mean that the rocketship crew sees events on Earth going backwards in time" - please note that seeing and observing are distinctly different concepts in SR. Would you please clarify which of the two you're asking about? $\endgroup$ Jul 29 '19 at 3:06
  • $\begingroup$ @AlfredCentauri What is the difference between seeing and observing? $\endgroup$ Jul 29 '19 at 5:00
  • $\begingroup$ I meant observe. I just assume that the light travel time can be skipped. Or we can use the replay trick of having an army of observers at every location in space. $\endgroup$ Jul 29 '19 at 5:51
  • $\begingroup$ You are completely right that time will go backwards. Actually i asked this question before (physics.stackexchange.com/questions/487179/…) and to put it in the simple way, if you don't assume that time goes backward you will have to deal with some paradoxes. See "My problem" it's exactly like yours. $\endgroup$
    – Paradoxy
    Jul 29 '19 at 10:13

If I turn my body 90 degrees, can I see the city of Moscow relocate itself by thousands of miles? After all, a minute ago it was 1000 miles in front of me, and now it's 1000 miles to my left.

When you change your velocity, you change frames, and therefore assign new time coordinates to existing events --- just as I assigned new space coordinates to Moscow when I turned my body. That doesn't mean that anything about the events, or about their relationships with each other, is changing, any more than the city of Moscow is moving through space.

  • $\begingroup$ But in the case of time, in the rocket ship, t'=0, t'=1, t'=2 is the history that the rocketship experiences. Before acceleration, there are certain events that have happened that, after acceleration, are now "behind" the rocketship's plane of simultaneity. So they would happen again. That means the rocketship crew would witness those events twice. $\endgroup$ Jul 29 '19 at 1:09
  • $\begingroup$ Relativity only allows us to understand that other frames of reference may have time advancing forward at different rates, never backwards. $\endgroup$ Jul 29 '19 at 1:34
  • $\begingroup$ @ShuhengZheng : When I turn 90 degrees, do I see Moscow in two different places? No, I see it in the same place, and I change my description of that place. Likewise, you see an event and assign a time to it; then you change your velocity and now you assign a new time to it. That doesn't mean you "see it twice"; it means you describe it two different ways. Which is perfectly fine. $\endgroup$
    – WillO
    Jul 29 '19 at 2:15

The Minkowski $t$ coordinate tells us what time an inertial observer would infer for an event based on a signalling procedure such as Einstein synchronization. It doesn't have any deeper philosophical significance than that. You don't have to do special relativity using Minkowski coordinates. You can choose any coordinates you like, or you can do special relativity using coordinate-free methods. Coordinates are just names that we give to points.

For an observer who is accelerating rather than inertial, we can, if we wish, define a notion of simultaneity based on the observer's instantaneous state of motion. Based on this, there can be distant points for which simultaneity is either nonexistent or not unique. If you try to cover spacetime with a curvilinear set of coordinates that agrees with this notion of simultaneity, then the coordinates will become badly behaved at the boundary where existence and uniqueness fail. You can't extend those coordinates past that boundary, because changes of coordinates need to be one-to-one.

I have a discussion of this kind of thing in section 3.9.3 of my SR book, which is free online: http://www.lightandmatter.com/sr/ .


No, it does not mean that the crew will see events on Earth going back in time. What they would 'see' is time on the Earth speeded-up.

In SR it is important to maintain a clear distinction between what one can see happening at a distant point and what is actually happening there.

You can only see an event when light from it reaches you, and once the light from the event has passed you, you can no longer see it.

Consider the following thought experiment. Suppose you are in the Earth's frame of reference at noon on Christmas Day in 2021 but exactly three light years to the left of the Earth (to use the terminology in your question). You look towards the Earth with a telescope that is sufficiently powerful to make out a large digital clock there. What you will see is the clock showing noon on Christmas Day in 2018, three years earlier. As you watch, you will see the clock moving forward second by second, the ticking of the clock in the telescope matching the ticking of the watch on your wrist. If you suddenly start moving towards Earth, what you will see through the telescope is the digital clock ticking more rapidly- time on Earth will seem to be speeding up! The reason for that is simply the relativistic Doppler effect- the light from each successive tick of the Earth clock has less distance to travel to meet you, so the ticks seem to happen faster.

The closer you get to Earth, the smaller will be the time gap between Earth's proper time and its time relative to your slanted line of simultaneity, so eventually you will arrive at the Earth in what is the present moment in both reference frames.

So, contrary to the expectation in your question, what you will see is time on Earth at some earlier date appearing to speed up.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.