If a person moves in a rocket with a constant acceleration of 1 G, then sooner or later he will reach about the speed of light. At the same time, time will flow slower for him than for the surrounding universe. But I heard that from the point of view of theory of relativity, the surface of the Earth accelerates relative to the surrounding space with a constant acceleration of 1G. That is, in fact, moving in a rocket with an acceleration of 1G and being on the surface of the Earth are the same. However, in the rocket there will be a time change effect, but on the Earth we do not feel this. Why is that?

  • $\begingroup$ Well, for starts, you don't actually accelerate while standing on the ground, so you speed does not increase as you stand there. $\endgroup$
    – Jon Custer
    Jan 29 at 14:32
  • $\begingroup$ But. It should be like this. As far as I know, we experience the G-force due to the acceleration of the Earth's surface relative to the surrounding space. Which seems to fall into the Earth, doesn't it? $\endgroup$ Jan 29 at 15:42
  • $\begingroup$ I got it from this video: youtu.be/XRr1kaXKBsU $\endgroup$ Jan 29 at 15:45
  • $\begingroup$ You experience the g force because you are not accelerating relative to the earth. In free fall you don't feel the force since you are responding to it. $\endgroup$
    – Jon Custer
    Jan 29 at 16:16

3 Answers 3


but on the Earth we do not feel this

In the accelerated ship they don't feel it either. Time dilation is not a matter of feeling, it is a comparison between different clocks.

But it is true that a rocket passing by the earth, say at 99% of the light speed, starting then to accelerate to reverse its relative velocity, sooner or later will be back for a second meeting.

When it happens, comparing the clocks readings, the time recorded by the ship will be far less than that recorded by a clock at the earth. The fact that both are all the time under the same acceleration doesn't matter. It is a misconception to think that acceleration is the cause of time dilation.

Our clocks here in Earth ticks a little bit slower than another clock far from the gravitational field of our planet and static with respect to us. But the $\Delta t$ of the comparison with the clock in the rocket is far greater than that tiny difference at such very high relative velocities.


Not really, an accelerated observer in a flat space time (in fact, any observer in any space time) can't reach the speed of light. What you are describing right now are Rindler coordinates.

The reason you don't observe time dilation is because, time dilation always refer to one frame of reference with respect to another frame of reference. In the case of being on the surface of Earth, with respect to whom? With respect to an inertial observer (i.e. an observer free falling through space with an acceleration of 1G), then you would get exactly the same time dilation as with the case of the rocket. In fact, they are the same scenario, but with reversed roles.

Indeed, the behavior of a stationary observer on a gravitational field and an accelerated observer in flat space time are indistinguishable. That is (more or less) the reason why general relativity predicts curved trajectory of light when passing near massive objects, or black holes.

In fact, in these Rindler coordinates, one can speak of a horizon in the past of the non-inertial observer, very similar to how a stationary observer can talk of a horizon when nearby a black hole. The effect is just very small.

  • 1
    $\begingroup$ In the case of being on the surface of Earth, with respect to whom? - for example with respect to someone on the Moon. Let's say the rocket accelerated for 2 months. That is, 60 days passed on the rocket. During this time, the rocket accelerated to 0.17C. Then the rocket turned around and began to slow down with the same 1G. When the rocket returned to Earth, 4 months had passed on the rocket. How much time has passed on Earth during this same time? Why has so much more time passed on Earth than the man on the rocket relative to an observer from the Moon? $\endgroup$ Jan 29 at 15:27
  • $\begingroup$ After all, if we take into account that the Earth surface moves upward in space-time (since space-time falls into the Earth), then the distance traveled by a man on the Earth surface and a man in a rocket will be the same $\endgroup$ Jan 29 at 15:32
  • $\begingroup$ With respect to someone that is falling towards the gravitational field (Assuming a constant gravitational field). How much time has it passed? I don't know, you'd have to integrate. In fact it is actually the acceleration to change the direction what makes the twin paradox not really a paradox, but an oversimplification. $\endgroup$
    – dolefeast
    Jan 29 at 17:40

then sooner or later he will reach about the speed of light

The "about" is important. Just wanted to stress nothing ever reaches exactly the speed of light no matter how hard or how long it accelerates.

To make it a fair comparison, imagine we have infinitely many copies of the Earth spread around the universe and all at rest with respect to each other. We also have infinitely many copies of the rocket and they also spread around the universe and also initially at rest with each other and the many Earths. When the rockets all start accelerating with constant proper acceleration of 1G the Earth observer will see each copy of the rocket that happens to pass near the Earth to pass faster than the previous one. The rocket observer will similarly see each copy of the Earth that happens to pass near his window as passing faster than the previous one. When looked at like this you see there is a symmetry.

In your scenario, you are basically asking why the Earth observer sees everything (except the rocket) as at rest with respect to her, while while the rocket observer sees everything racing progressively faster past him. Let's modify things a bit. Let's remove everything in the universe except The Earth and the multitude of rockets. Now the rocket observer experiencing artificial acceleration is asking why he sees everything (except The Earth) as stationary (where 'everything' is all the co-accelerating rockets), while the Earth observer sees 'everything' as racing progressively faster past her.

Let's remove acceleration from the equation. Lets imagine a rocket cruising along a constant 0.9c. This rocket pilot sees the whole universe as moving at 0.9c in the opposite direction. We conclude the Earth must be stationary because it basically at rest with the rest of the universe and the rocket must be moving, but in relativity, the rocket pilot is entitled to consider himself as stationary and the whole universe is going at 0.9c for random reason.

The rocket observer will observe that if he lets go of an apple it will accelerate to the the back of the rocket and hit the rear wall in just the same time it would take an apple to fall from the same 'height' on the Earth.

In General relativity the equivalence principle only really holds locally. In curved spacetime the speed of light at distant locations can appear to be different to the local speed of light and the equivalence principle does not hold over great distances.

  • $\begingroup$ Thank you for your answer, but I didn't catch the essence of the example with asymmetry, could you rephrase? $\endgroup$ Jan 29 at 15:36
  • $\begingroup$ @nilecrocodile OK, had a go at rephrasing. $\endgroup$
    – KDP
    Jan 29 at 15:56
  • $\begingroup$ Thank you! More or less clear. The only question, isn't true that space-time falling into the Earth? I was inspired by this video youtu.be/wrwgIjBUYVc?t=517 (the link with timestamp). And it says that space "falls" into massive objects with acceleration. Essentially, this means that the entire universe is moving towards the Earth, just like the rocket in your example... essentially the difference between the rocket and the Earth is only that space disappears inside the Earth, while for the Rocket it stays behind it... Or is this video also an allegory for clarity? $\endgroup$ Jan 29 at 16:11
  • $\begingroup$ I think that is the waterfall analogy and it has its problems. If you think of photons as little motorboats in this flowing river, then the photons near the edge of the waterfall that are going outwards are going slow against the flow while those going towards the waterfall are going progressively faster. An observer standing near the waterfall (hovering) would see photons going at different velocities, but general relativity says an observer always see the speed of light locally as being constant in all directions. As far as I am aware, the analogy is not part of mainstream science. $\endgroup$
    – KDP
    Jan 29 at 16:22
  • $\begingroup$ A second problem with the analogy is this. Imagine you are on the bank holding an apple in the water flow. When you release the apple, the flowing water exerts a drag on the apple accelerating it away from you, to to able to drag the apple there must be friction between the water and the apple. Now way upstream where the water is almost stationary you shove the apple away from you. It comes to a stop because of the drag of the water, but we know in flat space when we push an object it continues forever. There is no drag in flat space. $\endgroup$
    – KDP
    Jan 29 at 16:31

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