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Group Velocity and Phase Velocity

  1. Oct 14, 2006 #1
    Obtain the group velocity [tex]v_g[/tex] and phase velocity [tex]v_p[/tex] of 7.0 MeV protons/electrons. Write each answer as a multiple of the speed of light [tex]c[/tex].

    My work:

    1. Finding the group velocity:

    [tex]v_g = \frac{\partial \omega}{\partial k} = \frac{\partial \left( E / \hbar \right)}{\partial \left( p / \hbar \right)} = \frac{\partial E}{\partial p}[/tex]

    [tex]v_g = \frac{\partial E}{\partial p} = \frac{\partial}{\partial p} \left( \sqrt{p^2c^2 + m^2c^4} - mc^2 \right) = \frac{pc^2}{\sqrt{p^2c^2 + m^2c^4}}[/tex]

    [tex]v_g = \frac{pc}{\sqrt{\left( pc \right) ^2 + \left( mc^2 \right) ^2}} \times c[/tex]

    If [tex]E = pc = 7.0 \times 10 ^6 \mbox{ eV}[/tex] and the mass of protons and electrons are known, it is possible to obtain [tex]v_g[/tex].

    Assuming

    [tex]m = 938.27 \mbox{ MeV}/c^2[/tex] for a proton
    [tex]m = 0.51100 \mbox{ MeV}/c^2[/tex] for an electron

    the values [tex]v_g \approx 0.007460 \times c[/tex] (for protons) and [tex]v_g \approx 0.9973 \times c[/tex] (for electrons) are obtained.

    2. The phase velocity is simply

    [tex]v_p = \frac{c^2}{v_g} = \frac{c}{v_g} \times c[/tex].

    I believe there is a mistake in my approach; those numbers don't look right. Any help is highly appreciated.
     
    Last edited: Oct 14, 2006
  2. jcsd
  3. Sep 27, 2009 #2
    Why the -mc2?
     
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