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Homework Help: Simple electron orbit question

  1. Apr 6, 2007 #1
    1. The problem statement, all variables and given/known data

    Since an electron on a circular orbit around a proton has a centripetal acceleration, it should radiate energy according to the Larmor relation

    [tex] \frac{dE}{dt} = -2/3(q^2/4\pi\epsilon_o)(a^2/c^3) [/tex]

    where [itex]q, a, \epsilon_o[/itex] and [itex]c[/itex] are respectively the electron charge, its acceleration, the vacuum permittivity and the velocity of light in a vacuum. Therefore, in classical, mechanics, it should spiral and crash on the nucleus. How long would this decay take, supposing that the size of the initial orbit is 10[itex]^{-10}[/itex]m and the nucleus is a point charge (radius=0)?

    2. Relevant equations

    [tex] a = \frac{F_{coulombic}}{m} = \frac{v^2}{r_n} = \frac{1}{m} \frac{q^2}{4 \pi \epsilon_o r^2_n} [/tex]

    3. The attempt at a solution

    I can easily do this if I calculate the centripital acceleration, [itex]a[/itex], out at 10[itex]^{-10}[/itex]m, and treat it as constant in the Larmor relation.

    My question is, do you think this is what the problem wants me to do? Or do you think I have to set up an integral somewhere to vary the acceleration with the distance from the nucleus?

    I tried getting a function [itex]a(r)[/itex] by integrating [itex]a[/itex] with respect to [itex]r[/itex] from [itex]r=0[/itex] to [itex]r=10^{-10}[/itex]m, but then I realized that the integral would be infinite, since [itex]r[/itex] is in the denominator. Which makes me think they want me to treat [itex]a[/itex] as a constant.

    Any ideas or thoughts?

    Thanks!
     
  2. jcsd
  3. Apr 6, 2007 #2
    My last paragraph was really stupid. a is already a function of r. I did try to find the average acceleration by integrating from 0 to r and dividing by r. But of course the integral is infinite, and that's the problem.
     
  4. Apr 6, 2007 #3

    Doc Al

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    Staff: Mentor

    Try this: First express E as a function of radius. Then take the derivative with respect to time and set it equal to the Larmor formula; integrate to find time.
     
  5. Apr 6, 2007 #4
    Awesome! That definitely makes sense, and I wish I would have thought of it!

    Thanks, Doc Al.
     
  6. Apr 7, 2007 #5
    Doc Al,

    But if you have E as a function of radius, and you differentiate with respect to time, you get 0, since there is no t in that function. Is this what you intended?
     
  7. Apr 7, 2007 #6

    nrqed

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    Gold Member

    Your speed "v" is an implicit function of time, and r is a function of time as well.
     
  8. Apr 7, 2007 #7

    Hootenanny

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    Staff Emeritus
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    Perhaps it would be helpful to note that;

    [tex]v = \frac{ds}{dt} = 2\pi\cdot\frac{dr}{dt}[/tex]

    Edit: Dammit patrick
     
  9. Apr 7, 2007 #8
    Is this the idea?

    [tex]

    \textnormal{Let A, B, and C be the product of the constants in the functions E, a, and dE/dt respectively. Then,} \\
    [/tex]

    [tex]
    E = \frac{-A}{r} \ \textnormal{(1)}\\
    [/tex]

    [tex]
    a = \frac{B}{r^2} \ \textnormal{(2)}\\
    [/tex]

    [tex]
    \frac{dE}{dt} = C a^2 = \frac{C B^2}{r^4}\ \textnormal{(3)} \ \textnormal{(Lamar Relation)} \\
    [/tex]

    [tex]
    \textnormal{Then, using (1)}, \\
    [/tex]

    [tex]
    \frac{dE}{dt} = \frac{A}{r^2} \frac{dr}{dt} \textnormal{(4)} \\
    [/tex]

    [tex]
    \textnormal{Setting (3) and (4) equal, we get:} \\
    [/tex]

    [tex]
    \frac{A}{r^2} \frac{dr}{dt} = \frac{C B^2}{r^4} \\
    [/tex]

    [tex]
    A r^2 dr = C B^2 dt \\
    [/tex]

    [tex]
    \int{A r^2}dr = \int{C B^2}dt \\
    [/tex]

    [tex]
    \frac{A}{3} r^3 = C B^2 t \\
    [/tex]

    [tex]
    t = \frac{A r^3}{3C B^2} \\
    [/tex]


    So is this a valid expression for time t needed for the electron to crash into the point nucleus, given an initial distance, r, from the point nucleus??

    Thanks.
     
    Last edited: Apr 7, 2007
  10. Apr 8, 2007 #9

    Doc Al

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    Staff: Mentor

    Since the electron is crashing into the nucleus, r must be a function of time and thus E is a function of time (as noted by nrqed and Hoot). If
    [tex]E = E(r)[/tex]

    Then
    [tex]dE/dt = E'(r) dr/dt[/tex]

    Start by finding E(r) explicitly, where E is the total energy. (Use the Bohr model.)


    Exactly!
     
  11. Apr 8, 2007 #10
    Thanks again. I'm finished, at last! :)
     
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