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Pi, momentum, and kinetic energy

  1. Feb 22, 2010 #1
    There is a geometric way one can show the relation between kinetic energy and momentum which is a mathematical curiosity in my opinion. Let the mass of an object be equal to 2 PI. Then:

    P = 2 pi v

    KE = 1/2 (2 pi ) v sq
    or
    KE = pi v sq

    Hence, graphically, if we set v = r, where r is the radius of a circle, we have a circle whose circumference is momentum, 2 pi v, and whose area is the kinetic energy, pi v sq. Thus, for the case where the mass is 2 pi, we see the kinetic energy is geometrically bounded by the momentum of the object. Interesting.
    Wonder what would happen to this circle at relativistic velocities?
     
    Last edited: Feb 22, 2010
  2. jcsd
  3. May 26, 2015 #2
    ive been thinking about pi and energy lately.

    to simplify things here, ur just describing cosx/2. ie graph cosx...d/dt cosx is -sinx. pi rad of -sinx is (-1, +1 or "total ke" until it goes back down to -1 ie pe). relativistically speaking, the extreme case is analogous to an infinite amp, (staying with cos waves.) ie lim v---> c dt----->0, hence large distance(high amp) and no oscillation.
     
  4. May 26, 2015 #3

    Drakkith

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    Staff: Mentor

    Well, momentum in SR can take the form: d2dec44ba56c41a31b4d334b144b51d6.png where m0 is the invariant mass and gamma is the lorentz factor:
    daa6c3acdd12b843a2f3bf4d9d757a28.png

    The relativistic relation between kinetic energy and momentum is given by:

    7dbc3a29632b88dcf0645840b1bf5a53.png

    I haven't tried turning this relationship into a circle, but you're welcome to try.
     
  5. May 26, 2015 #4
    Radius ---> infinity.
     
  6. May 26, 2015 #5
    Also just to add: the classic example of this in relativity is in the expansion of the universe and why we see "light years away" not a distance away at far distances.
     
  7. May 26, 2015 #6

    Drakkith

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    Staff: Mentor

    This has nothing to do with the thread topic.
     
  8. May 27, 2015 #7
    I personally don't think there is any significance here. For the cases where v is not equal to r, the properties of a circle disappear. There is no fundamental property here that can be generalized to all cases of v, and is nothing more than a result of simply choosing the variables to get the desired outcome.
     
  9. May 27, 2015 #8
    I'm a little agree with it, for the special case won't exist if the special amount ##\pi## is replaced with any others.
     
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