Infrared Divergence Question

The infrared light and visible light are both described by photons and the photon's energy is E=h.f which f is the frequency. The only difference between visible light and infrared light is the value of this frequency f.
The value of f depends on the frame from which we measure it. Thus the same beam of light can be visible light relatively to one frame and infrared light relatively to another frame. This is called Doppler effect.

 

It is not so. Take a permanently working laser that emits a coherent wave - the wave is absolutely analogous to a radio-wave. Or take a radio-antenna and rapidly switch the generator on and off. The radiowave will not surround the antenna.

A radio-wave in the MW or LW diapason is a coherent beam/stream of long photons. The energy is determined with the number of photons (=amplitude), OK?

Look, it is not so difficult. If you know a current in antenna J(t), you can calculate the emitted wave as a function of this current. There is no IR divergence in such a calculation.

This divergence arises in calculations when photon-current coupling is considered as a small perturbation, i.e. in the initial approximation there is no coupling as if there were no current (no current - no radiation). The perturbative approach works fine for high frequencies because the probability to radiate an energetic photon is small, but this approach fails for very low frequencies. The solution is not to consider the current as a perturbation but to calculate the radiated field as in the radio-science - with a given (known) current.

 

No, I did not agree. I said considering radiation by the perturbation theory was not correct since it is not weak coupling.

Radio waves are coherent photons, a flow of coherent photons. Their fields (amplitudes) are added indeed. In the low- and middle-wave radio diapason (LW, MW) the radio-stations use the amplitude modulation. That means they vary the field amplitude, i.e., the number of photons (which is very big actually). The theoretical calculation gives exactly the coherent photons in radiation if you consider the current J(t) as a known function of time (they go as one). And one photon is very large itself. It contains (consists of) many-many wavelengths.

The filed tension E determines the force (eE) in the charge equation motion in the receiver.

 

That's not correct. Light and radio behave the same way. It sounds like you are comparing the per photon case in one with the total case in their other.

That's a function of the typical sources one finds, not of the waves themselves. You also have to realize that the number of photons is much larger for radio - per unit energy, you have a billion times more, plus the fact that most radio transmitters are more energetic.

I have never heard of such a thing. Can you point me to a reference?

Infrared divergence as usually discussed means something else. Again, can you provide a reference?

 

I think there are some issues here of confusion between effects of the way we model interactions of the photon field with charge carriers and the effects of the field's quantum mechanical nature.

First, the only fundamental difference between visible light and radio waves is frequency. What seems to be causing confusion here, however, is that light, particularly at lower frequency, is only very rarely seen in single photon states. So, let me be a little pedantic (or, perhaps, pedagogical).

If we Fourier transform the photon field and then quantize it, we find that the natural basis for describing states of the field is the number basis. In other words, at each frequency we can have a state consisting of an arbitrary number of photons (we can call this state [itex]|n\rangle_k[/itex]. Quantum mechanics says that any superposition of these states is also a valid state. So, we can have monochromatic light that doesn't have a definite number of photons. (Generally [itex]|\psi\rangle_k = \sum_n a_n |n\rangle_k[/itex].) In fact, normal classical-looking light is usually in something akin to a coherent state, [itex]|\alpha\rangle_k = \exp\left(-\frac{|\alpha|^2}{2}\right)\sum_{n=0}^\infty\frac{\alpha^n}{\sqrt{n!}}|n\rangle_k[/itex].

At much higher energies, most light-producing processes produce single photon states, which is why it might appear that high energy light seems fundamentally different from low energy.

The IR divergence is a different issue entirely. It emerges from our perturbative description of interactions. In particular, in perturbative calculations it looks like there is infinite probability of emitting 0 energy photons. The solution to this problem is generally to recognize that there it's not possible to distinguish between emitting such photons and emitting and reabsorbing such photons. And, as it happens, when you add together such processes, the divergence disappears.

 

It is not exact. In fact, no re-absorption happens. The finite cross section is just the inclusive in all possible soft photons and it is calculated with a given current.

 

Including the soft loop photons, which are, at least naively, interpreted as emitting and reabsorbing a photon. In this case, the problem comes from the fact that very low energy loop photons are almost on-shell and are almost indistinguishable from extremely low energy emitted photons.