Will moving observer see time dilation?

Question

Please look at this animation.

The green dots and red dots in the animation represent spaceships. The ships of the green fleet have no velocity relative to each other, so for the clocks onboard of the individual ships, the same amount of time elapses relative to each other, and they can set up a procedure to maintain a synchronized standard fleet time. The ships of the "red fleet" are moving with a velocity of 0.866 of the speed of light with respect to the green fleet. The blue dots represent pulses of light. One cycle of light-pulses between two green ships takes two seconds of "green time", one second for each leg.

As seen from the perspective of the reds, the transit time of the light pulses they exchange among each other is one second of "red time" for each leg. As seen from the perspective of the greens, the red ships' cycle of exchanging light pulses travels a diagonal path that is two light-seconds long. (As seen from the green perspective the reds travel 1.73 $( {\displaystyle {\sqrt {3}}} {\displaystyle {\sqrt {3}}})$ light-seconds of distance for every two seconds of green time.)

The animation cycles between the green perspective and the red perspective, to emphasize the symmetry.

Image and Extract Source Time Dilation Wikipedia

I have calculated, how many ticks moving clock makes during time of travel and how many ticks any clock at rest.

While moving makes 3, any clock at rest makes 6. Well, moving clock dilates. But observer at moving RED clock will think like that: when I started, my clock showed 12 and GREEN clock at rest showed 12. Then, some later, my RED clock shows 3 and GREEN clock at rest shows 6. That means, clocks at rest (time in reference frame) runs faster.

Then RED clock forgets, that it was in motion and becomes one at rest. The GREEN clock now moves in RED frame and dilates. But, I am afraid to say, now the GREEN clock measures that RED clocks run faster.

Well, let’s imagine a sand clock with two flasks A and B. The sand is in the bottom in flask B, flask A is empty. Then we turn the clock upside down. Sand falls into flask A. Now flask B is empty.

Flask A is empty and flask B is empty. There is less sand in the flask B from the point of view of A and vice versa. Paradox. Reciprocity of observations. Symmetry.

If two observers move relatively to each other with velocity 0.9 c, how each of them can be “at rest”? Who moves with velocity 0.9 c? They both? Nobody? GREEN first, RED then?

Isn’t it the same sort of symmetry as in the sand clock?

What is physical reality? One mutual space or different frames that describe it?

Is it necessary to replace once chosen reference frame which describes mutual for the both observers space with a new one?

Finally, can observer think that he is in a state of motion, but not at rest?

Answer 1

If two observers move relatively to each other with velocity 0.9 c, how each of them can be “at rest”? Who moves with velocity 0.9 c? They both? Nobody? GREEN first, RED then?

Let me call the Red observer "Alice" and the Green observer "Bob." Alice sees herself at rest, and relative to her, Bob moves with velocity $+0.9c.$ Bob sees himself at rest, and relative to him, Alice moves with velocity $-0.9c.$ They both agree on the relative speed between them.

What is physical reality? One mutual space or different frames that describe it?

The best way to describe the underlying reality is as a 4-dimensional space $(w, x, y, z)$ where we interpret $w$ as $ct,$ a time coordinate multiplied by the speed of light. Let me write the symbol $\Delta X$ as a shorthand for $X_1 - X_0,$ the difference in the symbol $X$ from some initial state $X_0$ to some final state $X_1.$

In normal 3D space, the distance between two points is related to these coordinates by the 3D Pythagorean theorem, $s^2 = \Delta x^2 + \Delta y^2 + \Delta z^2,$ and the transforms which preserve this are translations and rotations. Every coordinate system which is a translation or rotation of another agrees on these distances. The reason why we say it's all one space, and do not focus on this coordinate or that coordinate, is because these translations and rotations are allowed.

In this 4D space, the distance between two events (points in spacetime, positions plus a moment in time) is best understood as the "spacetime interval" $I = \Delta w^2 - \Delta x^2 - \Delta y^2 - \Delta z^2.$ The transforms which preserve this are the rotations, translations, time-translations, and the so-called Lorentz boosts, which mix a space coordinate with the time coordinate. Preserving this interval means, among other things, that everyone agrees that a bubble expanding with speed $c$ maps to another bubble expanding at speed $c$, in other words, everyone agrees on the speed of things moving at the speed of light. Similar to the 3D space example, the fact that these transforms are possible means that this is one unified space, not just separate coordinates.

In terms of the hyperbolic sine and cosine these boosts are (mixing $z$ with $w$ for example) $$w' = w \cosh\phi - z\sinh\phi,\\z'=z\cosh\phi - w \sinh\phi.$$ For very small $\phi$, $\cosh\phi \approx 1$ while $\sinh\phi\approx\phi.$ The low-velocity limit where Alice moves past Bob with velocity $v$ in the $z$-direction has $z' = z - v t$ and so for small velocities this corresponds to $$t' = t - \frac{v}{c^2} z,\\z' = z - v~ t.$$ This means that there is a feature of our world that you have never noticed before, which is that when you start moving forward with speed $c$, clocks that you thought were in sync are no longer in-sync. If one of them is "in front" of the other (in terms of the direction that you accelerated), then the one "in front" now shows an earlier time than the one "in back". This "de-synchronization" is not just important, it is at the core of relativity.

This is important so I will say it outright: all of the rest of relativity, all of the time dilation and length contraction stuff, is just built out of a lot of little de-synchronizations one on top of another on top of another, as the $z - v t$ terms shift through the $t - vz/c^2$ terms and vice versa. You need to know how to do matrix diagonalization and exponentiation to prove this, but even if you can't prove this, you can appreciate it. The only mystery in special relativity is that clocks de-synchronize.

In turn this is probably the core of your misunderstanding of this graphic. You are probably thinking implicitly that the clocks of one side see the clocks of the other side as synchronized, but they don't.

Answer 2

Hopefully, I have understood you properly, and even better if you can get an answer from a specialist in the field, but here is my attempt at a response to your questions until someone better informed comes along.

If two observers move relatively to each other with velocity 0.9 c, how each of them can be “at rest”? Who moves with velocity 0.9 c? They both? Nobody? GREEN first, RED then?

This is why it's called relatively. If there is no absolute space and time, then I don't see the problem of who is at rest arising, neither of them are, they simply disagree on their judgements of velocity and time.

One mutual space or different frames that describe it?

I might be misunderstanding you here, apologies for that, but different frames are vital, connected by appropriate transformations to ensure that physical laws are universal, ( as far as we know). If you want to call that a mutual space, that's what you can call it, but if by mutual you are bringing in absolute time and space, no then it's not mutual.

Is it necessary to replace once chosen reference frame which describes mutual for the both observers space with a new one?

I apologise, I don't follow you, but my wording above might answer you, I am not sure how you define mutual.

Finally, can observer think that he is in a state of motion, but not at rest?

I think you have a typo here, but anybody with a knowledge of GR, or simply an awareness that absolutes don't exist will know he is not at rest, ever, in any situation.