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Relations between kinetic energy, momentum and velocity

  1. Sep 16, 2011 #1
    Dear Sirs,
    I have discovered these two formulas:

    p = (1-v^2/c^2) * dKE / dv
    v = dKE / dp

    where
    p – momentum;
    v – velocity;
    KE – kinetic energy.

    Everywhere are used relations with full energy instead of kinetic.
    Therefore would be nice to know why these two are not used?
    Are they correct?
     
  2. jcsd
  3. Sep 16, 2011 #2

    clem

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    E=KE+M. Since M is constant, E and KE have the same derivatives.
     
  4. Sep 16, 2011 #3

    PAllen

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    The first is correct. I haven't checked the second.
     
  5. Sep 16, 2011 #4

    PAllen

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    The second equation is also correct.
     
  6. Sep 16, 2011 #5

    robphy

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    The second equation is related to the definition of "work" (as in the "work-[kinetic]energy theorem").
    It has that form in both classical and special-relativistic dynamics: dK= v dot dp .
    Check out this page from Maxwell:
    http://books.google.com/books?id=lcdAAAAAIAAJ&pg=PA206
     
  7. Sep 17, 2011 #6
    Thank you very much.
    The second formula I derived by using classical Doppler effect and
    and two formulas for a quantum:
    E = h*ν and p = h*ν / c
    in a simple though experiment to accelerate a mirror with quanta.

    Now if we have some guess that mass may arise we may write like so
    dp = p2 – p1 = (m+dm)*(v+dv) – m*v = m*dv + v*dm

    and by using equation v = dKE / dp
    we may write:

    v = dKE / (m*dv + v*dm)

    when v->c dv -> 0 (this is experimental fact)

    so we may write:

    c = dKE / (0 + c*dm)

    or dKE / dm = c^2

    So I have got relation between mass and kinetic energy without relativity.

    What weaknesses do you see in this way of thinking?

    Thank you.
     
    Last edited: Sep 17, 2011
  8. Sep 17, 2011 #7

    PAllen

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    This is nonsense. In all these formulas, m is rest mass so dm is identically zero (relativity or otherwise).
     
  9. Sep 17, 2011 #8
    Thank you for your note.
    I was thinking about “full” mass in these formulas.
    But I am not saying I am right.
    Just I am looking for my mistakes.

    By thinking about all these things one funny question come to my minds.
    Maybe it do not worth to open a new tread.
    Just it is interesting how to solve it.

    Let say we have such situation:

    {star A}..........(body1)-->>................<<--(body2)...........{star B}

    Body1 moves to the right and body2 moves to the left.
    Without collision body1 would reach starB
    and body2 would reach starA.

    But what if bodies collide.
    Lets say we have big speed.
    Now body1 will think that body2 are very very massive and therefore will take him to the starA.
    He will feel himself like a ball against a train.
    For the same reason body2 will think that body1 is very massive and will take him to the starB.
    Something here is not good.
     
  10. Sep 17, 2011 #9

    PAllen

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    Relativistic mass is a confusing, unnecessary concept. Just deal in rest mass, energy and momentum.
     
  11. Sep 18, 2011 #10
    Yes, likely you are right.
    But lets see what will happen if we still would think in terms of
    current mass and resulting mass ( after it gets some small additional impulse dp ).
    By using m = E/c^2 and v = dE/dp
    for dp = m*dv + v*dm
    we may show that such view also is possible:

    Lets we have a body with momentum
    p = m * v
    and will support it with additional impulse dp
    We may imagine that momentum dp divides into dp1 and dp2 so that

    dp = dp1 + dp2

    where dp1 = dp * ( v^2 / c^2 )
    and dp2 = dp * (1 - v^2 / c^2 )

    And we may imagine that p1 is used to arise mass by equation
    dm = dp1/v

    and p2 is used to arise velocity by equation
    dv = dp2/m

    So we will have resulting momentum
    (p + dp) = (m + dm) * (v + dv)

    It gives the same results,
    but maybe it looks easier for brain to visualize the “mechanic” of the process.
    What do you think?
     
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