# Infrared Divergence Question

1. Sep 25, 2009

### Jedi_Sawyer

To my mind radio waves are different than light because they are described by fields whose energy is not a function of frequency but just the amplitudes. That they always have to surround their source, and are not uniquely associated with individual particles. Light on the other hand is described by photons and the photon's energy are dependant on frequency. A photon does not surround the source of it's emission, and it is associated with individual particles. Between the two descriptions is something called the infrared divergance. Physicist get around this difficulty with state equations where they either let the photon have effective mass by letting the action constant be one for light or zero for radio waves.

It is really hard to find understandable information on the infrared divergance problem. The infrared divergance came into existance when Einstein explained the photo electric effect with photons, thereby solving one problem while creating another problem.

2. Sep 25, 2009

### ghc

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.

3. Sep 25, 2009

### Bob_for_short

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.

Last edited: Sep 25, 2009
4. Sep 25, 2009

### Jedi_Sawyer

This is another example of how it is hard to get an understandable answer about the infrared divergence problem. You seem to agree that you have to give up on the photon model at low frequencies but so why did you use photons to explain radio waves above. Actually the whole photon-current coupling thing you mentioned above is what I don't understand as the fields around a photon are closed on themselves in one model but everything seems to be model dependent when it comes to the world of electrons/photons. I have real troubles with electrons as I don't know if the model is a spinning shell of charge to create the magnetic field or is it something else.

5. Sep 25, 2009

### Bob_for_short

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.

Last edited: Sep 26, 2009
6. Sep 26, 2009

Staff Emeritus
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?

7. Sep 26, 2009

### Parlyne

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 $|n\rangle_k$. 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 $|\psi\rangle_k = \sum_n a_n |n\rangle_k$.) In fact, normal classical-looking light is usually in something akin to a coherent state, $|\alpha\rangle_k = \exp\left(-\frac{|\alpha|^2}{2}\right)\sum_{n=0}^\infty\frac{\alpha^n}{\sqrt{n!}}|n\rangle_k$.

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.

8. Sep 26, 2009

### Bob_for_short

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.

9. Sep 26, 2009

### Parlyne

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.

10. Sep 29, 2009

### Jedi_Sawyer

Well thanks for the ideas. I am actually doing the calculations for 150 MHZ or 2 meters wave lengths and am working closely with some math and physics textbooks at this point.

Roger Penrose wrote about QFT in his book "The Road to Reality" I will quote a few sentences from chapter 26. p670 "do we get good physical answers that are in agreement with experiment?
I can only give very mixed answers to these questions. The issue of mathematical is particuarly troublesome, and the fairest answer to give, would be: 'No; not as things stand today,' "

p676 "There are also infrared divergences that we can regard as coming about from infinitely large distances, (i.e. from indefinitely small momenta). These are usually regarded as 'curable' by various means, often by restricting the type of question that is regarded as being physically sensible to ask of a system."

p677 "One may take the view that according to some future theory the divergent integrals should be replaced by something finite,"

Incidentally there is a good description of 'mass shell' momenta with figures for massive and massless cases on page 673 and how it relates to Feynman graphs and scattering matrix.