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Radioactive decay in relativistic frames

  1. Dec 4, 2003 #1
    If you take two identical radioactive samples, place one on Earth and another on a near-lightspeed spaceship, and compare them some time later, will the one left on Earth have undergone comparatively more radioactive decay than the one on the spaceship?

    If the experiment is repeated by leaving one on Earth and another on a planet with the same radius as Earth but made entirely of lead, what, if any, will the difference be?

    Finally, how would the radioactive decay of a test source on the surface of Earth and its centre compare?
     
  2. jcsd
  3. Dec 4, 2003 #2

    Janus

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    Yes. (assuming that the second sample is brought back to a state where it is motioness wrt the first when the comparision is made. )

    The sample sitting on Earth will decay faster.
    The sample on the surface will decay faster.
     
  4. Dec 4, 2003 #3

    mathman

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    This relativistic effect on decay time has been obseved. Muons produced from cosmic rays come to earth at speeds close to c. As a result they live longer than expected.
     
  5. Dec 5, 2003 #4
    Does that mean that time dilation slows time down to a near standstill as you near the centre of a black hole, but the second (excuse the pun) you hit the centre, time goes discontinuously back to normal?

    Or is it generally accepted that some kind of bridging theory must be made between physics as we know it and the physics at a singularity, thus making the above interpretation speculative?
     
  6. Dec 5, 2003 #5

    mathman

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    An outside observer sees something falling into a black hole slow down and never fall in. In fact, to the outside observer, the black hole never formed.

    On the other hand, an object falling in would be in a different reference frame and would simply pass the event horizon and quickly be destroyed.
     
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