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How to show group velocity = particle velocity?

  1. Sep 7, 2010 #1
    1. The problem statement, all variables and given/known data

    In an electron wave, show that the group velocity = particle velocity of the electron

    2. Relevant equations

    E = [tex] p^{2}c^{2} + m^{2}c^{4}[/tex]

    3. The attempt at a solution

    [tex]

    V_{g} = \frac{d\omega}{dk} = \frac{dE}{dp} = \frac{d}{dp} \sqrt{(p^{2}c^{2} + m^{2}c^{4})}
    [/tex]

    [tex]
    =\frac{1}{2} \frac{(2pc^{2})}{\sqrt{(p^{2}c^{2} + m^{2}c^{4})}}
    [/tex]

    [tex]
    =\frac{pc^{2}}{\sqrt{p^{2}c^{2} + m^{2}c^{4}}}
    [/tex]


    after this point I'm stuck, I tried factoring out [tex] m^{2}c^{4} [/tex] and ended up with
    [tex]

    \frac{p}{m\sqrt{(\frac{p}{mc})^2 +1}}

    [/tex]

    but I'm still not sure what to do from here. Any help would be greatly appreciated! Thanks
     
    Last edited: Sep 7, 2010
  2. jcsd
  3. Sep 7, 2010 #2
    Multiplying both sides by [tex]\sqrt{p^{2}c^{2} + m^{2}c^{4}}[/tex] and then squaring both sides, will have you end with something nicer
     
  4. Sep 9, 2010 #3
    Thanks for the hint, however I'm still really stumped on what to do next. I divided and squared both sides and I get this:

    [tex]

    V_{g}^{2}(p^{2}c^{2}+m^{2}c^{4})=pc^{2}

    [/tex]

    I've tried subbing in [tex]p=mv[/tex] and I get:

    [tex]

    V_{g}^{2}(m^{2}v^{2}c^{2}+m^{2}c^{4})=mvc^{2}

    [/tex]

    Pull out an [tex]m[/tex] and [tex]c^{2}[/tex]

    [tex]
    V_{g}^{2}(mv^{2}+mc^{2})=v
    [/tex]

    and now I'm stuck again =(
     
  5. Sep 9, 2010 #4
    You have made a mistake:

    [tex]p=\gamma m v\neq mv[/tex]
     
  6. Sep 9, 2010 #5
    Sorry, what's [tex]\gamma[/tex]? I've always learned that Momentum = Mass*Velocity?
     
  7. Sep 9, 2010 #6
    Relativistic Momentum is redefined as [tex]p=\gamma m v[/tex] where [tex]\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]

    (An alternative way to look at it is that mass scales with velocity, that is, that relativistic mass is [tex]\gamma m_0[/tex] where [tex]m_0[/tex] is the object's rest mass, as measured from its rest frame.
    Personally I don't like this take on the subject, and greatly prefer the redefinition of momentum instead, since redefining mass as relativistic mass makes for some inconsistent applications of the concept in formulas. Or at least, it is far more confusing than redefining momentum.)

    On a related note, you can find that the expression for kinetic energy is [tex]K=(\gamma-1)mc^2[/tex], this is something that can be derived (What is the work you apply to bring a mass from 0 velocity to a velocity v)

    A useful algebraic identity for this exercise is [tex]\gamma ^2 = 1+\gamma ^2 \frac{v^2}{c^2}[/tex]
     
  8. Sep 13, 2010 #7
    Would this work also?

    [tex]


    V_{g} = \frac{d\omega}{dk} = \frac{dE}{dp} = \frac{d}{dp} \frac{p^{2}}{2m}

    =\frac{1}{2m} \frac{d}{dp} p^{2}

    =\frac{1}{2m} 2p

    =\frac{p}{m}

    =\frac{mv}{m}

    =v
    [/tex]
     
  9. Sep 14, 2010 #8
    You did not square the right side. It should be:

    [tex]V_{g}^{2}(p^{2}c^{2}+m^{2}c^{4})=p^2c^{4}[/tex]
     
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