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Inertia and momentum, the modern view.

  1. Jul 8, 2003 #1
    The classical quantity momentum is proportional to the spacelike rate of change of phase of a quantum mechanical amplitude. The classical quantity energy is proportional to the timelike rate of change of the phase. The factor of proportionality is Planck's Constant h. We represent a moving object as a superpostion of waves, a wave train or wavegroup. The probability that we will find the object represented by the wavegroup, e.g. an electron, is the absolute square of its amplitude, so the object classically moves at the group velocity. The group velocity is:


    where p=momentum, E=rest energy, k=wavenumber and ω=frequency and C the speed of light,

    so that inertia is m=E/C2 classically or

    i= ω/C2 quantum mechanically

    As long as we don't change the rest energy or total momentum (wavenumber or total frequency) of a wavegroup it will continue to propagate in a straight, unaccelerated line. The property of inertia arises from the combination of quantum mechanics and special relativity, no deep mystery there.

    And that's the modern view.
  2. jcsd
  3. Jul 8, 2003 #2
    sorry this doesn't make sense to me. How can a wavegroup have just one value of &omega and just one value of k?
  4. Jul 8, 2003 #3
    A wavegroup is a superpostion of several waves with the same value of rest energy (rest frequency) but slightly different values of momentum (wavenumber). The kinetic energy is p2C2/2E, so it differs slightly for each wave component, this causes the "bump" in the interference pattern to move with time, and that is why the object moves. The formula for the group velocity is an average of the different values.

    You can find a more complete description in just about any book on quantum mechanics.
  5. Jul 8, 2003 #4
    But Tyger, you give a formula which links energy to momentum:
    So, how can we have different values of momentum p while energy E is constant?
  6. Jul 8, 2003 #5
    Re: Re: Inertia and momentum, the modern view.


    That formula gives the group velocity for a wave group with an average momentum p and rest energy E. The individual momenta and kinetic energy for each wave in the group will be different.

    I was trying to present the basic idea involved without writing a book on the subject, so some of the details are missing.
  7. Jul 9, 2003 #6
    Re: Re: Re: Inertia and momentum, the modern view.

    Yes, OK. Now I think I understand. Up to now, I only knew the formula
    vg = d(omega)/dk.
    Now, since
    omega(k) = hbar * E(k)
    = hbar * c * ((m0 c)^2 + (hbar k)^2)^(1/2),
    we get
    vg = hbar * c * 1/2 * ((m0 c)^2 + (hbar k)^2)^(-1/2) * 2 hbar k * hbar
    = hbar * c * k ((m0 c)^2 + (hbar k)^2)^(-1/2)
    = pc^2/E

    So the wavegroup will travel with this velocity if the expectation value (average) of momentum is p. Like you say. And that's indeed the classical particle velocity. OK.

    But, since different omegas are involved, the wavegroup will of course disperse very quickly, and the spatial distribution will soon be 'smeared out' over a large area. Well, you can state that it represents a 'moving object'. But not a 'moving particle'. There are no trajectories of particles in QM.
  8. Jul 9, 2003 #7
    err, sorry if thats stupid question, but is there in this explanation anything that does not depend on axiom of inertia? Like C is const, h is const, frequency doesn't change on its own, uniformity of time, group velocity...?
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