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The 100 Meter Dash

  1. Aug 28, 2004 #1
    An ancient greek philosopher, an ancient greek hero, and Shrodenger's cat walk into a bar. Ok, forget the cat. The philosopher gets so drunk that he decides to prove to the hero that he cant reach the finish line of a 100 meter dash because he would have to go half the remaining distance an infinite number of times.

    Now, we all know that the philosopher's reasoning was purposefully flawed, but DO WE KNOW FOR CERTAIN that movement is possible in a universe where distance is not quantized? After all, we haven't got such a universe to experiment on. Do we know for certain that a complex universe where distance and time are NOT quantized could even have self consistent rules? If the answer is yes, do we know that such a universe could support life?

    As a side note: an exercise left for the reader's calculator, how many times would the hero have to go half the distance before he got a planck length from the finish line? Bonus points for the person who has to press the fewest buttons to find the answer.
  2. jcsd
  3. Aug 28, 2004 #2
    If the universe is not quantized, then 100 meters has no meaning.
  4. Aug 28, 2004 #3

    Huh? Why? Did I use the wrong word? Should I have said "quantumized". The spell checker says that is not a word. How could the universe not being chopped up in Planck length sized units preclude "100 Meters" from having meaning? Its true that you couldn't assign it an integer number based on the underlying granularity of space, but who ever said that a length cant be based on something else?
  5. Aug 29, 2004 #4
    Words only have demonstrable meaning according to their function in a given context.

    In this case, quantized refers to the observation that nature comes in bits and pieces, specific quantities. However, we also observe that these quantities act like waves, which are not quantized. If another universe existed where nothing was quantized, then literally 100 meters would have no meaning.
  6. Aug 29, 2004 #5


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    My calculator expressed appreciation and no small surprise at having a problem left for it, and it says that N = 123 ought to work.

    it gets this number with almost no buttons pressed because it is

  7. Aug 29, 2004 #6


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    I suspected you might wonder why the reader's calculator thinks that is the answer, so I inquired and got this response.

    the idea is to find N

    [tex]\frac{100}{2^N}<1.616\times 10^{-35}[/tex]

    [tex]\frac{100}{1.616\times 10^{-35}}<2^N[/tex]

    [tex]\log_2 2^N = N[/tex]
    and for any number X
    [tex]\log_2 X = \frac{log_{10}X}{log_{10}2}[/tex]

    Taking log base two of both sides
    [tex]\frac{37 - \log_{10}1.616}{\log_{10}2}<N[/tex]

    and [tex]\inline{ \log_{10}1.616}}[/tex] is negligible so one looks for


    which takes few strokes
  8. Sep 1, 2004 #7
    Damn Marcus you are everywhere-- I will have to read this when its not one in the morning

    i have to get up soon
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