# Oscillating electron has drift velocity too?

1. Oct 17, 2008

### loonychune

Just hoping you might explain the physics here for me.

An electron is placed in an oscillating electric field, say,

$$E = E_0\sin(\omega t)$$

and as a result accelerates in x at,

$$a = a_0\sin(\omega t)$$

(grouped some constants in a_0). Solving for the resulting motion yields,

$$x = \frac{a_0}{\omega}t - \frac{a_0}{{\omega}^2}\sin(\omega t)$$

In the second term we see the expected oscillatory motion. However, in the first term we have something corresponding to uniform velocity! The electron drifts as well as oscillates!

Why?

Thanks

Damian

2. Oct 17, 2008

### loonychune

I should probably be a little clearer as to the velocity of the electron:

$$v(t) = \frac{a_0}{\omega} -\frac{a_0}{\omega}\cos\omega t$$

EDIT: Something to ponder:

If the electric field we apply is:

$$E = E_0\cos\omega t$$

then the resulting velocity is:

$$v(t) = -\frac{a_0}{\omega}\cos\omega t$$

Is this physically possible and why, physically, the difference?

Last edited: Oct 17, 2008
3. Oct 17, 2008

### physics girl phd

I suspect there are errors in the ways you are applying boundary values / initial conditions if you're looking for simple oscillation. If you try to simultaneously impose conditions of v=0 at x=0 at t=0, that doesn't fit the general oscillating case (where speed is greatest at zero displacement and force is greatest at the place of maximum displacement). Just let
$$v (t=0) = - \frac{q E_0}{m \omega}$$

Part of the problem is that you never really "drop an electron in, or conversely "start" or "stop" that incoming planer E-field -- its supposedly uniform in time and space and the electron is in steady state. Check out this simulation of an electron on a wire: http://phet.colorado.edu/simulations/sims.php?sim=Radio_Waves_and_Electromagnetic_Fields" [Broken] In the simulation, you do get some initial slosh in an antenna if you suddenly start or stop the oscillation, and the response of the slosh is determined by the resistance of the wire and the capacitance of the system (since the wire is essentially a capacitor with a positive plate and one end and a negative plate at the other end)... eventually with the resistance dissipating the energy, and the capacitance pushing the charge back to an even distribution if you stop the signal and let the wire "discharge."

Also think about adding a mass to a vertical spring... the oscillation isn't about the unstretched length of the spring... there's offset because of the mass. Often I like to think of electrical systems as gravity systems (just with charge relating to mass).

By the way, typically you try to solve the differential equation ( -- not integrating step-by-step imposing boundary conditions... which might be wrong, as in this case). In this case the problem is a second order HOMOGENEOUS ordinary differential equation (i.e. there is no constant in the force... what you would expect if there were a constant offset to the field, which would cause uniform acceleration. This site has some info about diffeq's: http://mathworld.wolfram.com/OrdinaryDifferentialEquation.html" [Broken] and you can link to solving second order homogeneous diffeq's). You really should have a textbook on the matter, and preferably take a course as well. The generic version to this would have sine and cosine forms... that you'd then fit with best boundary conditions.

Last edited by a moderator: May 3, 2017
4. Oct 17, 2008

### loonychune

I'm still none the wiser however.

If you would, pick out the stages in the following and explain where we couldn't in fact do that.

An electron with mass m is initially at rest in the ionosphere. It is suddenly subjected to an electric field,

$$E = E_0\sin\omega t$$

See, can you not do this? You have an electron that's at rest and you apply some field that goes as sin(wt).

Thus, it accelerates as,

$$a = \frac{-eE_0}{m}\sin\omega t$$

Then,

$$v(t) = \int^{t}_{0}a_0\sin\omega t$$

Therefore,

$$v(t) = \frac{a_0}{\omega} - \frac{a_0}{\omega}\cos\omega t$$

5. Oct 17, 2008

### loonychune

I don't mean to be an arse and force the issue, but I have the idea that, since the particle IS at rest at x = 0 and should in fact be at a maximum here but isn't until the sine function 'catches up' then some of the kinetic energy goes into just moving the electron and not oscillating (which it would if E = E_0coswt)