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Thread: Kinetic Energy and the De Broglie Wavelength

  1. #1 Kinetic Energy and the De Broglie Wavelength 
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    Just something I noticed a little while ago, and want to know why I keep getting this contradictory result.

    From the de Broglie Equation,

    lambda (here denoted as l) =h/mv, where m =mass, v =velocity and h = Planck's constant.

    But frequency ( here denoted as f) = v/l

    Therefore,

    f = v/h/mv = mv^2/h

    The energy of a wave is given by E= hf.

    Therefore, E = hmv^h/h

    Therefore, E =mv^2

    Yet the energy should be equal to the kinetic energy of the particle, which is mv^2/2.

    Why is it that I keep getting this answer?


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  3. #2  
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    I'm not sure but I think it goes something like this:

    -First you omitted the Lorentz factor g=(1 - v^2/c^2)^(-1/2). With it the de Broglie equation is l = h/gmv. It's tempting to just assume that v<<c and therefore g=1, but let's keep it for now. This gives you the result E=gmv^2.

    -Second, E=hf gives you the total energy of the particle both kinetic and rest energy. In other words E = KE + RE, where RE=mc^2 with m being the rest mass of the particle. Also we know that E=gmc^2 where g is the Lorentz factor.

    Thus we get that for a particle with mass m E=gmv^2=gmc^2. Clearly that cannot be right. The only flaw in your reasoning that I can find is letting f=v/l. It makes sense in classical physics but the de Broglie hypothesis takes relativity into account and things get weird with relativity.


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  4. #3  
    Moderator Moderator Janus's Avatar
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    Some posts have been split off into a new thread here:

    http://thescienceforum.com/De-Brogli...gth-17809t.php

    as they pertained to a individual poster's personal theory
    "Men are apt to mistake the strength of their feelings for the strength of their argument.
    The heated mind resents the chill touch & relentless scrutiny of logic"-W.E. Gladstone


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  5. #4  
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    Well, my discussion with martillo got moved, but I have an answer for you. The velocity of the particle is not the phase velocity of its waveform. Rather the phase velocity of the wave is vp = c^2/v. If you use that to calculate the frequency of the wave and then apply E=hf you will get E=gmc^2 as we should. Again remember that this is the total energy of the particle E = KE + RE.

    Thus the kinetic energy is KE = E - RE, where E=gmc^2 and RE=mc^2. Therefore:

    KE = gmc^2 - mc^2
    = mc^2*(g-1)

    If you do a Taylor expansion on g the c^2 will drop out and you will get KE = 1/2*mv^2. The Newtonian kinetic energy is just a first order approximation to the true relativistic kinetic energy.
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