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Setting c=1, how does that work?

  1. Feb 25, 2007 #1

    quasar987

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    1. The problem statement, all variables and given/known data
    In the usual units where c=3x10^8 m/s and g=10m/s², what is g in the units where c=1?

    2. Relevant equations

    3. The attempt at a solution

    I said, let's redefine the meter like so: [itex]\tilde{m}:=3\times 10^8 m[/itex] so that now [itex]c=1\tilde{m}/s[/itex]. And then I calculated g by substituting m for [itex](3\times 10^8)^{-1}\tilde{m}[/itex]... but is this what the question is asking? Because I might as well have chosen to derefine the seconds and since it is s² that pops in in the units of g, the answer would have been different. And the question seems to talk like there is only one answer: "what is g in the units where c=1?"
     
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  3. Feb 25, 2007 #2

    arildno

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    Dearly Missed

    The question is, indeed, ill-posed.
    Perhaps you are meant to use a standard meter as your new length scale as well, but who knows?
    Perhaps you are meant to use the second as your new time scale..
     
  4. Feb 25, 2007 #3

    George Jones

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    In the context of interstellar travel with g = 1, it is often convenient to use years and lightyears, and g then also has a fairly nice value.
     
  5. Feb 25, 2007 #4

    arildno

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    Eeh, you meant c=1, George?
    Of course using both c=1 and g=1 make those two values very nice, indeed..
     
  6. Feb 25, 2007 #5

    George Jones

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    Yikes!

    Instead of "with g = 1", I meant "with proper acceleration a = g".

    Thanks arildno.
     
  7. Feb 25, 2007 #6

    Meir Achuz

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    c=1 LY/Y, as any reader of the New York Times knows.
    Just calculate g in LY/Y^2. That proves that God exists,
    and God is 1.
     
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