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Homework Help: Relativistic electron in static electric field

  1. Jan 28, 2009 #1
    1. The problem statement, all variables and given/known data
    Electron placed in static electric field [tex]\vec{E}[/tex] = -[tex]\Psi[/tex][tex]\hat{x}[/tex] , its initial velocity is 0. Calculate V(t).

    2. Relevant equations


    p(t) = [tex]\gamma[/tex](t)mv(t)

    Gamma is 1/sqrt(1-v^2/c^2) of course

    3. The attempt at a solution

    This is how I go about it and want to know if I'm on the right track.

    i) First you multiply the electric field by the charge of an electron to get:

    [tex]F_{e}[/tex] = [tex]\frac{d}{dt}[/tex]p(t) = e[tex]\Psi[/tex]

    ii) Then you integrate wrt time to get:

    p(t) = e[tex]\Psi[/tex]t

    iii) Then you relate momentum to velocity by:

    p(t) = [tex]\gamma[/tex]mv(t)

    iv) Finally you solve for V(t) from the above equation, expressing gamma explicitly I get the following formula:

    v(t) = [tex]\frac{e{\Psi}t}{m}[/tex]*[tex]\frac{1}{ {\sqrt{1+ {\frac{ e^{2}{\Psi}^{2}t^{2} }{ m^{2}c^{2} }} }} }[/tex]

    Does this seem to be the right method? I have to integrate this eventually to get x(t)..
  2. jcsd
  3. Jan 28, 2009 #2
    This doesn't seem right because V(t) is unbounded as t goes to infinity and thus we would pass the speed of light correct?
  4. Jan 28, 2009 #3


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    Homework Helper

    No. This expression

    approaches c as t goes to infinity.
    Last edited: Jan 28, 2009
  5. Jan 28, 2009 #4
    So maybe I am on the right track. Just curious, how do you get that limit? Do you have to binomial expand the square root?
  6. Jan 28, 2009 #5


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    Homework Helper

    I don't know what you mean by that, but probably not. After you do about 100 million of these kinds of limits, you just start to smell the approach. In this case, I just know that the other stuff under the radical will be much larger than 1, so I ignore the 1. The rest is straightforward. I think the more mathy way to do it is to divide top and bottom by t or something ...
    Last edited: Jan 28, 2009
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