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Answer to high-school question

  1. Sep 16, 2010 #1

    When I was studying physics in high-school, I often wondered what would be the meaning of sequence below in physics...

    m + mv + mv2 + mv3 + .....

    This is equivalent to

    [tex]\sum[/tex]mvn for n=0 to infinity

    where m is mass, and v is velocity.

    The though process was, that if mass (m) is conserved, momentum (mv) is conserved, and a form of energy (1/2 mv2) is conserved, then their sum probably should be conserved as well - resulting in above rudimentary sequence.

    I never studied physics further, but I always wanted to see what this sequence - if any, would mean in physics. Can some one point me to it? Since velocity (v) is directional, I could never figure out how it can be summed in a sequence like this...
  2. jcsd
  3. Sep 16, 2010 #2


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    Staff: Mentor

    Units don't allow summation. Adding kg to kg*m/s to J and so on doesn't make sense.
  4. Sep 16, 2010 #3
    Thank you Borek. It is so obvious, and I did not see it....this sequence has no meaning in physics.
  5. Sep 17, 2010 #4


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    As a purely mathematical sum, it does have sense:
    [tex]\sum mv^n= m \sum v^n[/tex]
    is a "geometric series". If -1< v< 1, its sum is
    [tex]\frac{m}{1- v}[/tex]

    But yes, Borek is correct. If v has units of "distance/time" then the series, or even just "m+ mv+ mv^2" has impossible units.
    Last edited by a moderator: Sep 19, 2010
  6. Sep 17, 2010 #5
    I remember I re-wrote it into this form:

    m + mv + 1/2 mv2 + 1/6 mv3 + ...
    m + 1/1! mv + 1/2! mv2 + 1/3! mv3 + ...
    m ( 1 + v/1! + v2/2! + v3/3! + ...)

    which is equal to

    mev (using exponential series)

    But then I could never think of any meaning of ev..what could possibly be e to the power of "velocity"?
  7. Sep 17, 2010 #6
    By taking it further...

    m*ev = some constant (k)
    ev = k/m
    v = log(k/m)
    where v is velocity and m is mass
  8. Sep 17, 2010 #7


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    Staff Emeritus
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    There is an issue with units again. Exponents must be unitless.
  9. Sep 20, 2010 #8
    HallsofIvy, Redbelly98 - thank you for taking time to answer my question. Yes, I realize there is issue with units. My curiosity is satisfied. Thanks.
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