View Full Version : Electron in an Electric Field


oxymoron
03-09-04, 07:01 AM
Imagine an electron cloud in 'empty' space. You can picture it roughly as a spherical cloud around a positive nucleus. When you apply an Electric field, the electron cloud distorts. Further, the electric field is harmonic.

I will assume there is some force F which is trying to restore the system to equilibrium. By applying the Electric field momentarily, the electron cloud will oscillate about its equilibrium point with a 'resonant frequency' given by w_0. From the wave equation w_0 = sqrt(k/m) where k and m are related to the properties of the electron. Obviously this is an undriven scenario.

I want to analyze this better. I have plotted a graph of angular frequency as a function of amplitude (So I have amplitude up the y-axis and angular frequency along the x-axis).

Solving the DE m[x" - w_0 x] = Fcoswt we obtain a solution in the form

x(t) = F/(w_0 ^2 - w^2).

I take the natural frequency (w_0) of an electorn to be 3x10^-15Hz and by letting F = q/m = 1 I developed the following plot.

As I start at w = 0 the amplitude of the oscillating electron cloud is 0. As I increase w the amplitude slowly grows. As w approaches the natural frequency (or resonant frequency) the amplitude grows extremely quickly until at w = w_0 the amplitude is 'off the scale'. At this point I guess the electron cloud is 'in phase' with the electric field and will proceed to oscillate without bound.

As I increase w further, the amplitude quickly decreases at the same rate that it increased until it reaches 0 again.

The biggest question is...
At w = w_0 where the infinity occurred the is a PHASE CHANGE of pi radians. So If I plot frequency as a function of phase difference the plot will look like a step. Where for

0 < w < w_0 : Phase = 0.

w >= w_0 : Phase = pi

Does this make sense to anyone. Sorry, it is a bit of a read but I need a good understanding of this.

Helloween
03-10-04, 05:35 PM
i believe the scenario you have set up is actually a driven harmonic oscillator. the Fcoswt in your equation would actually be the driving force. for an undriven system, your equation looks like:

(m*x")+(b*x')+(k*x)=0

i.e. the sum of your forces are equal to zero. b*x' being the damping force, (which i think is zero in your case) and k*x being the restoring force.

oxymoron
03-10-04, 11:40 PM
That is correct Helloween. It is a driven harmonic oscillator with no damping.

But the trouble I am having is that when w < w_0 the does the phase lag by 90degrees, 180degrees, ?? I do not know. And what exactly happens when w = w_0 in terms of phase changing?

Helloween
03-10-04, 11:58 PM
well, first, i'm not really sure i'd agree with this equation : DE m[x" - w_0 x] = Fcoswt
m*w_0*x does not have units of force.

second, your solution: x(t) = F/(w_0 ^2 - w^2). does not have units of length, and it does not have time dependence. i'd check the solution if i were you, there should be some time dependence and a cosine term.

1100f
03-11-04, 02:35 AM
well, first, i'm not really sure i'd agree with this equation : DE m[x" - w_0 x] = Fcoswt
m*w_0*x does not have units of force.

The DE of the driven HO is m[x" - w_0^2 x]


second, your solution: x(t) = F/(w_0 ^2 - w^2). does not have units of length, and it does not have time dependence. i'd check the solution if i were you, there should be some time dependence and a cosine term.

The solution of the DE is x(t) = A(w)cos(wt) where A(w) = F/(w_0 ^2 - w^2).

1100f
03-11-04, 03:29 AM
Imagine an electron cloud in 'empty' space. You can picture it roughly as a spherical cloud around a positive nucleus. When you apply an Electric field, the electron cloud distorts. Further, the electric field is harmonic.

I will assume there is some force F which is trying to restore the system to equilibrium. By applying the Electric field momentarily, the electron cloud will oscillate about its equilibrium point with a 'resonant frequency' given by w_0. From the wave equation w_0 = sqrt(k/m) where k and m are related to the properties of the electron. Obviously this is an undriven scenario.

I want to analyze this better. I have plotted a graph of angular frequency as a function of amplitude (So I have amplitude up the y-axis and angular frequency along the x-axis).

Solving the DE m[x" - w_0 x] = Fcoswt we obtain a solution in the form

x(t) = F/(w_0 ^2 - w^2).

I take the natural frequency (w_0) of an electorn to be 3x10^-15Hz and by letting F = q/m = 1 I developed the following plot.

As I start at w = 0 the amplitude of the oscillating electron cloud is 0. As I increase w the amplitude slowly grows. As w approaches the natural frequency (or resonant frequency) the amplitude grows extremely quickly until at w = w_0 the amplitude is 'off the scale'. At this point I guess the electron cloud is 'in phase' with the electric field and will proceed to oscillate without bound.

As I increase w further, the amplitude quickly decreases at the same rate that it increased until it reaches 0 again.

The biggest question is...
At w = w_0 where the infinity occurred the is a PHASE CHANGE of pi radians. So If I plot frequency as a function of phase difference the plot will look like a step. Where for

0 < w < w_0 : Phase = 0.

w >= w_0 : Phase = pi

Does this make sense to anyone. Sorry, it is a bit of a read but I need a good understanding of this.

One of the mistakes that you are doing is to treat the atom classically. This cannot be done. In order to see what is happening in the case of an atom in an oscillating electric field, is to take the solutions of the Schrodinger equation and to treat the interraction of the electric field as a perturbation in the framework of the time dependent perturbation theory. You can find here a developement for the case of a sinusoidal perturbation and also for the case of an harmonic electric field.
http://electron6.phys.utk.edu/qm2/modules/m10/time.htm

oxymoron
03-11-04, 05:10 AM
One of the mistakes that you are doing is to treat the atom classically. This cannot be done. In order to see what is happening in the case of an atom in an oscillating electric field, is to take the solutions of the Schrodinger equation and to treat the interraction of the electric field as a perturbation in the framework of the time dependent perturbation theory. You can find here a developement for the case of a sinusoidal perturbation and also for the case of an harmonic electric field.

Thankyou for the response 1100f but I am pretty sure I am not meant to go into it this much. I only need some sort of explanation of the phase difference when you increase the frequency from less than the natural frequency to more than it. i.e. from w < w_0 to w > w_0. Why is there a phase change??

That link has a lot of 3rd year Quantum Physics stuff and I am in 2nd year. 1100f, is that your website??

1100f
03-11-04, 05:41 AM
Thankyou for the response 1100f but I am pretty sure I am not meant to go into it this much. I only need some sort of explanation of the phase difference when you increase the frequency from less than the natural frequency to more than it. i.e. from w < w_0 to w > w_0. Why is there a phase change??

The differential equation that you gave is in fact the differntial equation of the classical harmonic oscillator.
There are two answers to your question. The first is that the mathematical solution gives this change of phase.
The second is by intuition, or experiment.
Supose that you hold in your hand a pendulum. Try to do the following:
let it swing by itself. you see that it has a natural frequency. Now if you begin to move your hand, let's say from right to left and left to right, in a much slower rate than the natural frequency of the pendulum, you will see that the pendulum follows the motion of your hand. However if you move your hand much faster than the natural frequency of your pendulum, you'll see that the mass will almost not move, it doesn't succeed in folowing your hand's motion. When your hand is on the right side, since the mass didn't move (almost) the pendulum will make an angle to the left and vice-versa. So you see that the pendulum swings to the opposite of your hand.
So intuitively, you can say that bellow the natural frequency, the system just follows the driving force. At higher rates, it is not able to do this and this is why you have this phase difference



That link has a lot of 3rd year Quantum Physics stuff and I am in 2nd year.
Well, good luck with your studies, and have fun studying physics.


1100f, is that your website??
No, I just found it by searching:"time dependent perturation theory" on altavista, and I got his one.

BTW, I found that it is a page of a 2-parts course in quantum mechanics:
the first part is well suited for second year.
Anyway these are the links for the two parts:
http://electron6.phys.utk.edu/qm1/ and http://electron6.phys.utk.edu/qm2/

oxymoron
03-11-04, 06:50 AM
Great! That is exactly what I thought! Thanks for that 1100f.

Helloween
03-11-04, 09:11 AM
ok, this solution has some interesting features. x(t) = A(w)cos(wt +/- phi) where A(w) = (F/m)/(w_0 ^2 - w^2).

well, first, this is the solution after all of the transcient oscillations have died out. if you notice, when you drive the system at w=w_0, the amplitude blows up.that is, you are driving at the natural frequency of the system, so your force is always adding energy by working with the natural frequency.

second, all you have to do to solve for phi, the phase constant, is to put the solution back into the original equation and solve. you'll be able get out of that a system of two equations using some trig id's and separating the sine and cosine term.