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Many formulas in Special relativity have v - velocity. And theory says there is no absolute velocity only relative and it cannot be greater than speed of light.

Consider this thought experiment: Empty space, two asteroids A and B, are stationary relative to each other at very close distance. There is an astronaut on asteroid A, he kicks asteroid B and distance between asteroids begin to increase. Then astronaut takes a rock from his asteroid A and throw it in the direction of B, this gives some acceleration (momentum), he notice the distance between asteroids increase more rapidly now. So he continue to repeat this process...

Does astronauts (A) velocity relative to B = change of distance between A and B per tick in special relativity?

Will astronauts (A) velocity relative to B exceed speed of light?

Special relativity states that objects with mass cannot have velocity greater that speed of light, so does it mean there are no such 2 reference frames in universe which move faster than C ? I'm pretty sure there are lots of rocks in universe whose distance change much faster than C.

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    $\begingroup$ Relative velocities, as viewed by a third observer, are allowed to go past $c$. It's the velocity of B as measured by A that cannot exceed $c$. $\endgroup$ – Chris Feb 4 '18 at 1:45
  • $\begingroup$ @Chris Does this mean astronaut will not be able to throw more rocks? or he will throw them but they will not give acceleration? $\endgroup$ – Alex Burtsev Feb 4 '18 at 1:52
  • $\begingroup$ No. You can keep throwing rocks as long as you like. You just can't accelerate anything past $c$. You can keep getting closer and closer without limit. $\endgroup$ – Chris Feb 4 '18 at 1:56
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    $\begingroup$ No. From his point of view, the rock he throws will just fly away, nothing special about it. He feels the full accleration, but his time is dilated, so that acceleration is smaller in his original reference frame. $\endgroup$ – Chris Feb 4 '18 at 2:04
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    $\begingroup$ You should probably just crack open a SR textbook. Those are all pretty elementary questions in relativity. $\endgroup$ – Chris Feb 4 '18 at 6:46
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There is no absolute velocity only relative.

This is not something new that comes out of Special Relativity. It has existed since Galileo. Inertial frames are relatively moving with respect to each other at constant velocity.

Does astronaut A's velocity relative to B = change of distance between A and B per tick in special relativity?

Yes, Special Relativity does not change the definition of velocity as change in distance per tick. But, "distance" and "tick" is observer-dependent. So, I guess the better way to phrase this is "Astronaut A's velocity relative to B = change of distance between A and B (as seen from B) per tick (as measured by B).

Will astronaut A's velocity relative to B exceed speed of light?

Never. That is a result from Relativity.

I'm pretty sure there are lots of rocks in universe whose distance change much faster than C.

It's not about how many rocks you have. The inertia of astronaut A increases as he approaches the speed of light. As astronaut A's speed gets closer to the speed of light, the same force causes less and less acceleration, so that you never reach the speed of light.

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  • $\begingroup$ "The inertia of astronaut A increases" so he can measure his inertia and determine his absolute speed? "the same force causes less and less acceleration" so he can measure how much acceleration he gets per force and determine his absolute speed? $\endgroup$ – Alex Burtsev Feb 4 '18 at 10:18
  • $\begingroup$ The inertia of an object depends on the velocity of the object as seen from a frame. You see, even inertia of an object is dependent on which frame you are watching the object from, so you can't. $\endgroup$ – PhyEnthusiast Feb 4 '18 at 11:00
  • $\begingroup$ Forces and accelerations are always frame-dependent in relativity. $\endgroup$ – PhyEnthusiast Feb 4 '18 at 11:01
  • $\begingroup$ en.wikipedia.org/wiki/Acceleration_(special_relativity) is helpful in understanding this I think. $\endgroup$ – PhyEnthusiast Feb 4 '18 at 11:03
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    $\begingroup$ Just learn about it first and then ask about something you want clarification about here at StackExchange $\endgroup$ – PhyEnthusiast Feb 4 '18 at 11:13

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