# Classical model of light

1. May 23, 2005

### Cheman

Classical model of light....

NB - let us consider light as a purely classical electromagnetic phenominum - no mention of quantum. ( The reason for this is that is feel that it is important i have an indepth undertanding of the classical theory of light, which will therefore allow me to see more clearly where the probablems which led to the development of quantum theory arose. )

In classical wave theory, we treat light (and the rest of the EM spectrum) as if it results from an electric (and therefore magnetic) field, which imposes a force on other charged objects. However, in order to account for the formation of a wave, we say that this field takes time to “travel”; this speed being 3*10^8 m/s.

Eg – Suppose we take an electron which is oscillating up and down, and a proton which is responding to that. ( NB : for the sake of simplification we shall say that the proton can ONLY move up and down; it can not accelerate towards the elctron.)
This would therefore produe a situation like the one found through the link below, with the proton on a slight time delay to the motion of the elctron, as it must wait for the change in the field to reach it.

Now, effectively what we have here is a light ray travelling between the 2 particles. Ie – an oscillating electric field wave. The energy possessed by this wave is dependant upon its amplitude. ( ie – its intensity.) However, this amplitude becomes smaller and therefore weaker as the wave permeates through space, since the strength of an electric field is inversely proportional to the distance from the source squared. However, any charged particle in the path of the light will still be pushed up and down in pperiodic motion; just to a lesser extent.

My questions arise from what people apparently thought the results of the photoelectric effect should have been, before they knew anything about quantum. According to the AS text book, if we used the wave model then we would suppose that “waves gradually transfer energy to all the elctrons near metal saurface. Eventually a large number of electrons gain enough energy to leave the metal.” Now, obviously this is not what is observed, but my question is in fact on how this should ties in with classical wave theory.

How can the waves gradually transfer energy to the electrons? All a light ray is is an electric (and therefore magnetic) field which will cause an electron to periodically oscillate “in time” with the wave – how does this electron gain energy from this? Does the light’s amplitude in fact decrease as it interferes with the electron, which absorbs energy as it passes through it meaning the light amplitude is not just dependant on distance, or at least something along those lines?

Also, if a charged particle continues to gain more and more energy as light is shown on it, does it oscillate with an increasing amplitude compared to the set amplitude of the light? WHy do charged particles not moved faster and faster and more wildely therefore when exposed to othermoving charged particles around them? (because surely this effectively the same thing? )

Last edited by a moderator: May 2, 2017
2. May 25, 2005

### Cheman

Erm... would anyone like me to explain anything i've said in my question,since no one seems to have answered yet? :uhh:

3. May 25, 2005

### Crosson

You are focusing on a simplified model that does not tell the whole story. It is better to think of the photoelectric effect in terms of scattering (collisions between light rays and electrons, that involve an exchange in energy just like a billiard ball collision).

4. May 25, 2005

### krab

Think of a metal surface. The electrons do exactly what you describe; they move in response to the electric field, making it very small, actually "shorting out" the electric field, so that the light wave has no choice but to reflect off the metal surface.

This makes no sense; you are comparing an electric field amplitude to a position amplitude.
They do. Charged particles scatter off each other, sometimes gaining energy, sometimes losing. Eventually, they have a distribution of energies, whose average can be thought of as their temperature.

5. May 25, 2005

### Staff: Mentor

Think of pushing someone on a swing, with a small periodic force. Do it the right way and you can make the amplitude of the swing's motion increase with time, even though each push is small, and has the same strength.

6. May 26, 2005

### Cheman

Jtbell, you are implying that the light is effectively making the electrons resonate - but why would this be the case? Yes, it is indeed true swings will resonate at their natural frequency - but light has different frequencies and therefore cant b making the elctrons resonate surely?

7. May 27, 2005

### Staff: Mentor

Well, we know that classical models don't work. But for the moment, we have to look at this from the point of view of a 19th-century physicist who doesn't know anything about the detailed structure of matter yet. All he knows is that electrons (cathode rays) sometimes get kicked out of a metal when light shines on it.

Lots of binding forces can be at least approximated by a simple harmonic oscillator potential, so this would seem to be a reasonable first guess at the behavior of whatever force normally keeps the electrons in the metal. This leads to the conclusion that some frequencies of electromagnetic radiation might be better than others at jiggling the electrons out, because of resonance. Off-resonance, the energy transfer wouldn't be as efficient, but you might still get some photoelectric effect, although it would take a brighter light or longer time.

At leat that seems like a plausible train of thought to me, although I haven't actually seen any contemporary references that describe this.

8. May 27, 2005

### Cheman

Ah, ofcourse, i forgot to consider the fact that the nature of matter was not yet properly esablished. Thanks.

As you have said, according to the classical model, some frequencies of light would cause electrons to be ejected due to resonance in the photoelectric effect - however you also said "Off-resonance, the energy transfer wouldn't be as efficient, but you might still get some photoelectric effect, although it would take a brighter light or longer time" - how would a forced, and therefore relatively small, oscillation cause electrons to be ejected, and why owuld this depends on intensity and time length?

9. May 27, 2005

### Pas De Trois

all EM is an electric + - dumbell tumbling over and under each other at their frequency rate and emitting a magnetic wave at right angles to their plane. their motion and velocity is controlled by my UGE (universal gravitational ether), which is responsable for all gravitational forces as well as the nuclear forces, it causes an ab extra force; therefor, eleminating the need for a medium of exchange. It is geven of by all matter, so, it comes with equal intensity from all directions, so the universe is infinite. EM ages and turns into 3K, which comes to us with equal intensity from all directions, and verifies the infinity of the universe, and supports my theory and turns the BB universe into a bs (bad science) universe. if you care to hear more, let me know.

10. May 27, 2005

### Staff: Mentor

The simple-harmonic-oscillator treatment is only an approximation, for small oscillations. Consider a mass on a spring: eventually a real spring will break, or deform, or the mass flies off because it wasn't fastened very tightly to the spring in the first place. Or a person on a swing: if the amplitude gets big enough, something interesting (and probably painful) is likely to happen.

Or consider a fanciful example: drill a hole through the earth and let a mass oscillate back and forth in it, under gravity. Give it a little push every time it goes past a certain point. Of course, we ignore air resistance and the effects of the earth's rotation. If I recall correctly, as long as the motion stays "within" the earth (and assuming the earth has uniform density), the motion is in fact simple-harmonic (although this isn't crucial here; the important thing is that the mass is "bound" by the earth's gravitational field).

Eventually the maximum-distance points will extend beyond the earth's surface (the mass rises out of the hole and falls back into it). Beyond this point, the motion is no longer simple-harmonic, but still periodic. Going still further, the mass will eventually reach escape velocity at the earth's surface, and never fall back.

I won't try to speculate further, because I don't know what kind of models people actually tried in the late 1800s to explain the photoelectric effect. But any model would have to have the electrons bound to the metal or atoms at short range or small energies, but unbound further out or with higher energies.

11. May 28, 2005

### krab

Wow. Are you God?
How deflating. God can't spell.

12. May 28, 2005

### Cheman

Cool. By saying "Off-resonance, the energy transfer wouldn't be as efficient, but you might still get some photoelectric effect, although it would take a brighter light or longer time" though Jtbell you are suggesting that ocillations can build up even if the forced frequency is not equal to the natural frequency - but how can this be? The building up would be resonance and surely this only occurs when the forced oscillations is equal to the natural frequency?

13. May 28, 2005

### Staff: Mentor

I'm at home and all my books are at the office, so I'm going from memory here. Consider a harmonic oscillator driven by a periodic force. Usually, books discuss the steady-state solution for the motion, that is, after the force has operated for a long enough time that the system has settled down into a steady state. The amplitude is maximum if the driving frequency equals the natural frequency of the oscillator, but it is not zero if the driving frequency is off-resonance. Instead, as the driving frequency moves away from the natural frequency, the amplitude decreases smoothly. How fast the amplitude decreases off-resonance depends on the amount of damping in the system. The formula for the steady-state amplitude is:

$$A = \frac {a_0}{\sqrt {(\omega_o^2 - \omega^2)^2 + 4 \gamma^2 \omega^2}}$$

where $\omega_0$ is the natural frequency, $\omega$ is the driving frequency, and $\gamma$ is a constant that contains the effects of damping.

Try a Google search on "driven harmonic oscillator" and you might be able to get a more complete picture.

Last edited: May 28, 2005