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Electron under gravity

  1. Mar 9, 2008 #1
    1. The problem statement, all variables and given/known data
    An electron at rest is released from rest and falls under the influence of gravity. In the first centimeter, what fraction of potential energy lost is radiated away?

    2. Relevant equations
    Lienard-Wichert potential for the electron of charge e is given by:
    [tex]\phi = \frac{e}{R(1 - \beta \cdot \hat R)}[/tex]

    In this case the charge is accelerated by a gravity (a = g).

    3. The attempt at a solution

    Reference: Electrodynamic Radiation by Marion and Heald

    The problem hasn't mentioned whether the speed of the electron is relativistic.

    If the speed of the electron is less than c ([itex]\beta << 1[/itex]) then [itex] R(1 - \beta \cdot \hat R) \rightarrow 0[/itex] and the potential can be written as:
    [tex]\phi = \frac{e}{R}[/tex]
    The accelerated field can be given as:
    [tex]\vec E = \frac{e}{c^2 R^3}\left((\vec R \cdot \vec g)\vec R - R^2\vec g\right)[/tex]

    I don't know how the potential energy can be calculated here. If the direction of g and R are the same, shouldn't E = 0?

    I assumed a non-relativistic case here. Am I going wrong here?
  2. jcsd
  3. Mar 9, 2008 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Do you really think that something can reach relativistic speeds after falling for 1 centimeter?

    Hint: Look up the Larmor formula for energy radiated by an accelerating charge.
  4. Mar 11, 2008 #3
    Thanks, I got it! Here goes:
    Considering the motion of the electron along the y-axis.
    The dipole moment is [itex]\vec p = -ey\hat y[/itex] and [itex]y^2 = {1\over 2}gt^2[/itex]
    [tex]\vec p = -{1\over p}get^2\hat y[/tex]

    [tex]\ddot{\vec p} = -{1\over p}ge\hat y[/tex]

    By Larmor's formula, the radiated power is given by,

    [tex]P = {2\over 3c^3}(ge)^2[/tex]

    Now, the time t it takes to fall a distance 'h' is given by: [itex]h = {1\over 2}gt^2[/itex]

    [itex]t = \sqrt{{2h\over g}}[/itex]

    So the energy released in falling a distance 'h' is: E = Power x time

    [tex]E_{rad.} = Pt = {2g^2e^2\over 3c^3}(\sqrt{{2h\over g}})[/tex]

    Meanwhile the potential energy lost is PE = mgh.

    So the fraction radiated is:
    [tex]f = {E_{rad}\over E_{pot}} = {2g^2e^2\over 3c^3}(\sqrt{{2h\over g}} \times {1\over mgh} = {2e^2\over 3mc^3}(\sqrt{{2g\over h}})[/tex]

    Just another question, why does the dipole moment appear when there is only a single charge involved?
    Last edited: Mar 11, 2008
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