Why don't people on Earth experience a same time effect as when accelerating in a rocket with the same G-value?

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.