Wave Theory of Light: Energy Independent of Frequency?

[cesiumfrog]

Isn't the SHO done by the mass-spring system A CLASSICAL WAVE description?

No. Simple harmonic motion is not a propagating wave.

If you'd used a chain of coupled mass-springs, that would be different.

If you are telling me that the frequency isn't a part of the total energy, then it means that there is no time dependence at all in the instantaneous energy.

It is incorrect to claim that the total energy increases with frequency. Nonetheless, there is a time dependence in the instantaneous energy: it oscillates from one femtosecond to the next, which is why the conventional measure of intensity is technically the average per unit time.

For example, if I have a point (or a surface) in a particular location, and I have a plane monochromatic EM wave passing through that point and I look at the number of oscillation that crosses through that point, I see the time dependence in the magnitude of the E-field. If I have a charge at that point, that oscillating field will be doing work onto the charge. If I have the same EM field, with the same amplitude, but with a larger frequency, I see MORE oscillation of that charge, meaning more work was done onto the charge by the field in the same time period.

This is where I think you are becoming confused; the energy transmitted in an electromagnetic wave is a different concept to the energy transferred to an electron (indeed, you are aware that if the light wave is incident on a metal, much of the energy is merely reflected and so forth).

Now, a driven mass-spring oscillator is analogous to an antenna in the path of an EM wave. The electrons behave like the mass whilst the field they produce in the antenna acts as the restoring spring. For a constant driving amplitude, the energy transferred to the electrons will increase as the frequency approaches a resonance (although it will be a maximum, since higher frequency does not give the current enough time to gather).

[..]There IS a time dependence! In fact, I make use of this effect everytime I run our accelerator! I shoot a 10 fs laser pulse at our photocathode at different RF phase[..]

We both agree the the higher frequency wave rises faster (and then falls faster, while the lower frequency is sustaining it's maximum longer). For a very short pulse envelope, I've no doubt controlling the phase will affect the total energy of the pulse itself.

What I'm disagreeing with is the content of your initial post:

The higher the frequency, the more cycles in that unit time that you measure, the more energy you will get.

Your assumption that each cycle has equal energy is incorrect; the wattage is constant with amplitude and divided equally among whatever number of cycles occur in the second. Unless you are switching a pulse so quickly for it to matter which part of the cycle you cut, increasing the number of cycles per second does not increase the energy.

Note that the topic of this thread is the energy of classical light waves, not the photelectric effect. Classically, you would not expect ten minutes of blue light to have a different effect to ten minutes of red light (of equal intensity), because both transmit the same amount of energy to your apparatus. Hence it was a surprise that a different effect is seen, more so once shown (by then going to UV) it isn't just a result of one colour happening to be on a resonance.

 

[lightarrow]

If you are telling me that the frequency isn't a part of the total energy, then it means that there is no time dependence at all in the instantaneous energy. For example, if I have a point (or a surface) in a particular location, and I have a plane monochromatic EM wave passing through that point and I look at the number of oscillation that crosses through that point, I see the time dependence in the magnitude of the E-field. If I have a charge at that point, that oscillating field will be doing work onto the charge. If I have the same EM field, with the same amplitude, but with a larger frequency, I see MORE oscillation of that charge, meaning more work was done onto the charge by the field in the same time period.

Sorry if I get into your discussion. What you say is infact very interesting. It seems to me you have found a classical reason to introduce the quantum of electromagnetic field!
As cesiumfrog says, from Poynting theorem it comes that energy per unit time is only given by |E |^2 and not by frequency. Now we know that, with a higher frequency of a field with given amplitude, the number of photons per unit time is lower and this, together with the fact the energy of every photon is higher, mantain the equality of energy.

But you, rightly, says that a *continuous* EM field of the same amplitude but higher frequency makes an higher work on an electron, e.g., and this gives a paradox, in my opinion.

So, the solution is that the field cannot be continuous!

I'm probably making some mistake.

 

[ZapperZ]

Sorry if I get into your discussion. What you say is infact very interesting. It seems to me you have found a classical reason to introduce the quantum of electromagnetic field!
As cesiumfrog says, from Poynting theorem it comes that energy per unit time is only given by |E |^2 and not by frequency. Now we know that, with a higher frequency of a field with given amplitude, the number of photons per unit time is lower and this, together with the fact the energy of every photon is higher, mantain the equality of energy.

But you, rightly, says that a *continuous* EM field of the same amplitude but higher frequency makes an higher work on an electron, e.g., and this gives a paradox, in my opinion.

So, the solution is that the field must not be continuous!

I'm probably making some mistake.

No, cesiumfrog is correct. In my case, I WAS applying it to the excitation of of a charged particle, and the example would be the line antenna case or even the photoelectric effect. When we do such energy transfer, then yes, the frequency does come in for such a process, even in classical terms. It is only when we deal with the different intensity versus a fixed frequency is there now an issue with classical wave.

Zz.