Simple electron orbit question

[WolfOfTheSteps]

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

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

where $q, a, \epsilon_o$ and $c$ 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$^{-10}$m and the nucleus is a point charge (radius=0)?

Homework Equations

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

The Attempt at a Solution

I can easily do this if I calculate the centripital acceleration, $a$, out at 10$^{-10}$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 $a(r)$ by integrating $a$ with respect to $r$ from $r=0$ to $r=10^{-10}$m, but then I realized that the integral would be infinite, since $r$ is in the denominator. Which makes me think they want me to treat $a$ as a constant.

Any ideas or thoughts?

Thanks!

 

[WolfOfTheSteps]

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.

 

[Doc Al]

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.

 

[WolfOfTheSteps]

Awesome! That definitely makes sense, and I wish I would have thought of it!

Thanks, Doc Al.

 

[WolfOfTheSteps]

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?

 

[nrqed]

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.

 

[Hootenanny]

Perhaps it would be helpful to note that;

$$v = \frac{ds}{dt} = 2\pi\cdot\frac{dr}{dt}$$

Edit: Dammit patrick

 

[WolfOfTheSteps]

Is this the idea?

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

$$E = \frac{-A}{r} \ \textnormal{(1)}\\$$

$$a = \frac{B}{r^2} \ \textnormal{(2)}\\$$

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

$$\textnormal{Then, using (1)}, \\$$

$$\frac{dE}{dt} = \frac{A}{r^2} \frac{dr}{dt} \textnormal{(4)} \\$$

$$\textnormal{Setting (3) and (4) equal, we get:} \\$$

$$\frac{A}{r^2} \frac{dr}{dt} = \frac{C B^2}{r^4} \\$$

$$A r^2 dr = C B^2 dt \\$$

$$\int{A r^2}dr = \int{C B^2}dt \\$$

$$\frac{A}{3} r^3 = C B^2 t \\$$

$$t = \frac{A r^3}{3C B^2} \\$$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.

 

[Doc Al]

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
$$E = E(r)$$

Then
$$dE/dt = E'(r) dr/dt$$

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

Is this the idea?

...

Exactly!

 

[WolfOfTheSteps]

Thanks again. I'm finished, at last! :)