A travels away from B at 0.9c. But B does so acc. to A. Now how come A alone returns younger?

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  • 1 decade ago
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    A will say that the clock of B is slow, since according to A, B is moving.

    B will say that the clock of A is slow, since according to B, A is moving.

    To compare the time in both clocks, AT LEAST one of the observers must perform a complete circuit.

    Upon returning to the point of departure, the clocks may be compared.

    The one who executes circular motion cannot apply the formula for time-slow during the accelerated motion. The formula is valid for inertial frames only.

    Thus the observer, who executes the circuit and returns, finds that his clock is slow on comparison with the other clock.

    At first reading you will not be convinced with this answer.

    But remenber, to compare the time, atleat one of the two must execute an accelerated motion.

    Think over this again and again till you catch up the point.

  • 1 decade ago

    Acceleration. Motion is relative, acceleration is not. If A and B are together at the same time and A speeds up to .9c he is accelerating, so the situation is no longer symetrical.

    If they each head away from the original spot at .45c their ages will remain the same, relatively.

  • Because B was always older. A never traveled from B, B is constantly moving away from A. So, since A was before B it will always B younger

  • Anonymous
    1 decade ago

    Relative velocity

    A is moving but B isn't

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