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Velocity of Klein-Gordon particles of mass m

  1. Mar 29, 2013 #1
    1. The problem statement, all variables and given/known data

    Show that [tex]ψ(x,t)=Ae^{i(kt-ωt)}[/tex] is a solution to the Klein-Gordon equation [tex]\frac{∂
    ^2ψ(x,t)}{∂x^2}-\frac{1}{c^2}\frac{∂^2ψ(x,t)}{∂t^2}-\frac{m^2c^2}{\hbar^2}ψ(x,t)=0[/tex] if [tex]ω=\sqrt{k^2c^2+(m^2c^4/\hbar^2)}[/tex] Determine the group velocity of a wave packet made up of waves satisfying the Klein-Gordon equation. Show that [itex]E=\sqrt{p^2c^2+m^2c^4}[/itex] for these particles and show that speed v is equal to the group velocity.

    3. The attempt at a solution

    The last proof, show that the speed = group velocity, is where I'm having trouble. I've done the rest. The solution I got for group velocity is
    [tex]V_g=\frac{∂ω}{∂k}=\frac{kc^2}{\sqrt{k^2c^2+(m^2c^4/\hbar^2)}},[/tex] so now the way I've been trying to solve this is taking the expression for E and manipulating it trying to achieve Vg.

    Using equations like [itex]E=\hbar\omega,E=\frac{p^2}{2m},p=mv,E=\frac{mv^2}{2},p=\hbar k[/itex] among others, I cant get it to equate.

    I can get down to [itex]V_g=\sqrt{c^2+v^2}[/itex] but that's the closest I've got.

    Just a subtle hint at a starting point would be great. Thanks.
     
    Last edited by a moderator: Mar 29, 2013
  2. jcsd
  3. Mar 29, 2013 #2

    vela

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    You need to use relativistic formulas. In particular, the velocity of a particle is given by p/E.
     
  4. Mar 30, 2013 #3
    Using v=p/E i can solve it only if vph= v

    From what i understand the phase velocity is the velocity of a particular point on a single wave of the packet. that's analogous to speed of the particle yes? so v is indeed equal to vph
     
  5. Mar 30, 2013 #4

    vela

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    I'm not following your reasoning. Can you elaborate a bit more?
     
  6. Mar 30, 2013 #5
    [tex]V_g=\frac{kc^2}{\sqrt{k^2c^2+m^2c^4/\hbar^2}}[/tex]

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

    [tex]p=\hbar k[/tex]

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

    [tex]=\frac{k}{\sqrt{k^2c^2+(m^2c^4)/\hbar^2}}[/tex]

    [tex]V_{ph} V_g = c^2 ∴ V_{ph}=c^2/V_g[/tex]

    but i need [tex]V_{ph}=V_g/c^2[/tex]...
     
  7. Mar 30, 2013 #6

    vela

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    You're not going to get that because your expression for the group velocity is
    $$V_g = \frac{kc^2}{\sqrt{k^2c^2+m^2c^4/\hbar^2}} = \frac{kc^2}{\omega} = \frac{c^2}{\omega/k} = \frac{c^2}{V_\text{ph}}$$
     
  8. Mar 30, 2013 #7
    exactly.

    So then how do i show that speed v = Vg if not by direct substitution.

    i dont suppose it would be enough to sub in c=w/k

    solves it in like 2 lines, end up with w/k=v

    but that assumes v=c which isnt the case for particles with mass m....
     
  9. Mar 30, 2013 #8

    vela

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    I'm not sure why you took ##p/E## to be the phase velocity. It's the group velocity. (If you include the factors of c, it's actually ##\frac{V_g}{c} = \frac{pc}{E}##.)
     
  10. Mar 30, 2013 #9
    okay now im not following.

    in your first post you said the velocity of a particle is given by p/E

    i know that v=Vg but how do i show that.

    if i take Vg=p/E and sub in E and play around i get nowhere.

    how do i reduce p/E down to v..
     
  11. Mar 30, 2013 #10

    vela

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    It's an established result from relativity. Try dividing ##p=\gamma mv## by ##E=\gamma mc^2##.
     
  12. Mar 30, 2013 #11
    thats works. i knew i needed a formula from outside of the ones i used to derive it all cause it goes back to just crap. but the v/c^2 gives me the missing c^2 ive needed.

    Thanks a lot! bit of a dicky proof
     
  13. Mar 30, 2013 #12
    In fact, if you assume that the phase velocity is the velocity of the particle you will see that the particle travels faster than the speed of light! On the other hand, the group velocity is always smaller than c. So if relativity is right, physical states must be a superposition of plane waves. However, you'll have an uncertainty in the position of the particle as well on the momentum.
     
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