# I Explaining Light intensity with EM Field Oscillations?

1. Jan 2, 2017

### JohnnyGui

I have understood that the frequency of an EM wave is caused by the frequency by which a charged particle oscillates, which causes its electrical field to periodically change its strength with respect to a fixed location point at a distance from that particle.

The more energy (heat) you add to that charged particle, the faster that particle oscillates and thus the higher the frequency of the EM wave.

If adding energy merely increases the frequency of the oscillation of a charged particle and thus the frequency of the EM wave, then what causes an increase in intensity of an EM wave? How can one explain an increase in intensity in terms of electrical field oscillation without (if possible) mentioning the increase in number of photons?

2. Jan 2, 2017

### Staff: Mentor

Depending on how you add the energy, you can increase either the amplitude or the frequency or both.

The intensity is (very loosely) determined by the number of wave crests arriving per unit time and the amount of energy delivered by each individual wave crest, so all else being the same, increasing either one will increase the intensity.
Photons have nothing to do with anything here. This is a classical situation described by classical electrodynamics with no quantum effects involved.

3. Jan 2, 2017

### JohnnyGui

Could you please explain which way of adding energy would increase the intensity and which way it would increase the frequency?

Two questions regarding this.
1. Can the amount of energy delivered by each individual wave crest be translated into the amount of displacement a charged particle oscillates in?
2. If intensity is also determined by the number of wave crests arriving per unit time, doesn't this mean that frequency is a bit tied to intensity?

4. Jan 2, 2017

### Staff: Mentor

Take a charge $2q$ and oscillate it at one gigacycle per second and you will get radiation with the same frequency and greater amplitude than from a charge of $q$ oscillating at that frequency.
One way or another, that delivered energy depends on the work done, which you find by integrating the force (which is proportional to the strength and amplitude) along the displacement distance. Depending on the exact conditions, you can more energy from a greater displacement, or more force along the same displacement, or a combination of both.
Mmmmm..... yes..... that's why I said "all else being the same, increasing either [frequency or amplitude] will increase the intensity".

5. Jan 2, 2017

### JohnnyGui

Ah ok, so it seems that it depends on how strong a particle is charged if everything is held the same?

I was surprised since many readings say that intensity is independent from frequency.

6. Jan 5, 2017

### JohnnyGui

@Nugatory : I have one question that's a bit outside of this subject.

From what I understand, electrons of certain elements always send out a particular set of specific wavelengths when they're being excited like when given energy to them

If atoms of an element only give out a specific set of wavelengths when energy is added to them, why does an element then change color when energy in the form of heat is gradually given to them? From what I know, increasing the temperature of an element would make it look red and increasing the temperature further would eventually make it look blue.

Is it because the electrons of the element's atoms can't absorb the specific energy packets from that total given heat energy so that the atoms as a whole would be oscillating instead, giving the observed colors based on the atoms' oscillating frequency instead of the electrons' alone?

7. Jan 8, 2017

### tech99

With respect, I don't think increasing frequency will increase the intensity. If we raise the frequency of a transmitter, which is itself driving the motion of the electrons in the antenna, the Power Flux Density and Amplitude of the radiated fields remain unaltered. I think there may be confusion here from the concept of photons, where at higher frequencies the photons contain the energy of the wave in bigger packets.

8. Jan 9, 2017

### Staff: Mentor

Yes, the time-averaged intensity of a classical electromagnetic wave (W/m2 arriving on a surface perpendicular to the direction of propagation, averaged over a whole number of cycles) is $I = \frac 1 2 \varepsilon_0 c E_0^2$, where $E_0$ is the amplitude (maximum value) of the electric field, regardless of the frequency. This is the time-averaged square of the magnitude of the Poynting vector, $\langle S \rangle$. See for example Griffiths 3rd edition, section 9.2.3.