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Zero-Point Energy, Relativity

  1. Nov 12, 2008 #1
    1. The problem statement, all variables and given/known data
    Consider an electron confined in a region of nuclear dimensions (about 5fm). Find its minimum possible KE in MeV. Treat as one-dimensional. Use relativistic relation between E and p.

    2. Relevant equations
    KE = p2/(2a) = [tex]\hbar[/tex]2/(2ma2)
    p = h/[tex]\lambda[/tex]
    E = hf
    E2 = (mc2)2 + (pc)2
    E = mc2 + KE

    3. The attempt at a solution
    First attempt: [tex]KE = p^2/(2m) = (pc)^2/(2mc^2) = (E^2-(mc^2)^2)/(2mc^2)
    Thus, K2mc^2=E^2-(mc^2)^2 = K^2 + 2Kmc^2 + (mc^2)^2. [/tex]However, here everything cancels out and I get:
    K^2 = 0, which can't be right. (Answer in book is 40 MeV)

    Another approach: [tex]E^2=(pc)^2+(mc^2)^2=(mc^2)^2 + 2mc^2K + K^2 [/tex](setting the invariant energy squared equal to the energy squared where E = K + mc^2

    This leads to a quadratic equation: [tex]K = -2mc^2 \pm[/tex] [tex]\sqrt{(mc^2)^2+(pc)^2}[/tex]
    K = -2mc^2 [tex]\pm[/tex] [tex]\sqrt{(mc^2)^2+(hc/\lambda)^2}[/tex]

    When I plug in those values, I don't get 40 MeV. I feel like I'm missing something very simple and fundamental. This shouldn't be a hard problem. Any help would be appreciated, thanks.
  2. jcsd
  3. Nov 12, 2008 #2
    KE=p^2/2m is only valid for p<<mc

    You should only use E = sqrt[(mc2)2 + (pc)2] and therefore,
    KE = E - mc^2 = sqrt[(mc2)2 + (pc)2] - mc^2.
  4. Nov 12, 2008 #3
    That was indeed my guess on the second approach. If I take just that:
    KE = [tex]sqrt[(mc^2)^2 + (pc)^2] - mc^2 [/tex], I still don't get there. Using the formula [tex]p = h/\lambda[/tex]

    If I do as I did on the "second approach" then setting energies equal, I also have trouble. I'm a bit puzzled because I am sure I am missing something simple, and this is just straightforward.
  5. Nov 12, 2008 #4
    You should transform the statement that the electron is confined to small region in space to a statement about the uncertainty in position.

    Then, you use the uncertainty relation to find a minimum for the expectation value of p^2.

    That's the value you substitute in your relativistic relation between E and p, and you extract K from that, as you already correctly noted.
  6. Nov 12, 2008 #5
    You get 40Mev if you use p=hbar/lambda.
  7. Nov 12, 2008 #6
    Yes...and the correct explanation for that is the one I posted.
  8. Nov 12, 2008 #7
    Mmkay, I think I've tried that.

    I have [tex]\delta p\delta x = \hbar/2.[/tex]

    Thus, when I use the KE form from before:[tex]

    KE = sqrt[(mc^2)^2 + (\hbar c/\delta x*2)^2] - mc^2 [/tex]

    I must be doing incorrect math... sigh.
    Last edited: Nov 12, 2008
  9. Nov 12, 2008 #8

    Where's the math mistake?

    [tex]-.511MeV + sqrt[.511^2 + (6.58E-22*3E8/(2*5E-15))^2)
  10. Nov 12, 2008 #9
    What you wrote down is the minimum uncertainty condition, which is only satisfied for special cases such as the gaussian wave packet.

    Normally people use [tex]\delta p\delta x \sim \hbar[/tex].

    Actually, since it is an order of magnitude estimation, you can't really say it's wrong even if you use values like 2*hbar or hbar/2.

    In my opinion, any value between 10MeV and 100Mev is OK for the answer.
  11. Nov 12, 2008 #10
    \delta x is not equal to 5fm, the better estimate is 2.5fm: it's the uncertainty in the position, and we can assume the expectation value of the position of the electron corresponds to the center of the region.

    Apart from that, I agree with weejee: an order of magnitude estimate will have to do: the actual physics is too complicated to get the "real" answer.
  12. Nov 12, 2008 #11
    I see. It all makes so much more sense now, lol. Just take away that darn factor...

    New question on this. I figured I should use minimum uncertainty for [tex]\delta x[/tex] since we were confining it to a box of size 5E-15m. Do you suppose my problem here was assuming the box was [tex]\delta x[/tex]? After all, [tex]\delta x[/tex] is usually only one side of the wave packet (akin to radius, not diameter). Thus, if I have a region of some size, then [tex]\delta x[/tex] is half the width of the region. Then when I plug in for minimum uncertainty, I get 2*0.5*width...

    Thanks for the help.
  13. Nov 12, 2008 #12
    Right, borgwal, I think that was the problem. Thanks again!
  14. Nov 12, 2008 #13
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