What is the absolute stationary frame of reference?

Does there have to be one?

Suppose there are two people. Each person is on a different planet. The two planets however are moving apart from each other with great speed. Say 90% the speed of light. Is there any difference in measured time between the two people? If a third person traveled from one planet to the other at 95% the speed of light relative to the planet that he left than he'd approach the other planet at a speed of 5% the speed of light. This is relatively a much slower speed. So how would each person on the different planets and in the rocket ship measure time?

Update:

Would you people actually answer the question? Ok I added the velocities wrong. Guess what? It's completely irrelevant to the point I'm asking about. Suppose istead that the 3rd person tries to head towards the other planet at .9c. In other words coming to a rest with respect to the other planet. Now how does time get recorded by each observer? For the person on each of the planets why should one's time reference change over the other's?

Update 2:

Actually to add velocities like some of you mention is to assume that one person is at a stationary frame of reference. So why is that frame of reference stationary for the person on one planet and not the other one.

4 Answers

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  • Anonymous
    1 decade ago
    Favorite Answer

    According to the accepted physics by now, there is none. The experimental proof is the Michelson-Morley experiment. There used to be an absolute stationary frame called the luminouferous aether, that the speed of light derived from the Maxwell equations apart from any reference frame considered to be true, but the experiment I mentioned proved that if there exist an aether it must be stationary relative to the earth which is revolving around the son, and thus accelerating, either the earth is the center of the universe or the stationary frame is accelerating which means its not stationary. (Michelson used his interferometer to detect the change in the speed of light as the speed of the earth changed ,and the result was null.)

    One more thing, if two observers are moving relative to each other with the velocity v=9c/10 ( c = 299792458 m/s ) and an object is moving from one to another with the velocity u= 95 c/100 relative to one of the observers the relative velocity relative to the other is not (according to the consequences of the Special theory of relativity) simply the sum of the velocities ,it is

    u1=(u+v)/(1+uv/c^2)

  • 1 decade ago

    According to the relativity principle, there an absolute frame of reference doesn't have to exists. Yet, some cosmologist believe in the existiance of an absolute frame of reference.

    It's not critical for your example, though. You should take into account that the distance the object has to travel depends on the frame of reference - if one car tries to catch another car which is orginally driving a few meters in head, the distance it needs to run is only a few meters from the car's own frame of reference, yet from a stationary frame of reference tit may take several kilometers, yet the time needed is the same.

  • 1 decade ago

    I'm afraid that there is a misconception in your question. At speeds that represent a significant fraction of the speed of light, you cannot get away with simply adding/subtracting velocities in the classical (Galilean) way that you have (where you have ended up with 5% the speed of light).

    Rather, you have to include the effects of special relativity, and use Lorentzian addition of velocities. See http://en.wikipedia.org/wiki/Velocity-addition_for... and go from there.

  • 1 decade ago

    i heard my teacher teaching this.. but i was doing something else.. but if i heard right, each one will calculate and get diff answers

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