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Simple electron orbit question

  • #1

Homework Statement



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)?

Homework 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]

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!
 

Answers and Replies

  • #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.
 
  • #3
Doc Al
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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.
 
  • #4
Awesome! That definitely makes sense, and I wish I would have thought of it!

Thanks, Doc Al.
 
  • #5
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.
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?
 
  • #6
nrqed
Science Advisor
Homework Helper
Gold Member
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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?
Your speed "v" is an implicit function of time, and r is a function of time as well.
 
  • #7
Hootenanny
Staff Emeritus
Science Advisor
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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
 
  • #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:
  • #9
Doc Al
Mentor
44,892
1,144
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?
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.)


Is this the idea?

...
Exactly!
 
  • #10
Thanks again. I'm finished, at last! :)
 

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