# EM Wave Phase Question

1. Oct 26, 2012

### Drakkith

Staff Emeritus
Let's say we have a radio wave coming into a receiver. When the phase is near 0, does this mean that fewer photons are being absorbed by the receiver compared to when the phase is near 90? (Hope I'm using the right terms here)

2. Oct 26, 2012

### Studiot

Not sure I follow the question. What you you mean by phase coming into receiver?

The signal excites an antenna before entering the receiver circuitry.

3. Oct 26, 2012

### Drakkith

Staff Emeritus
When the signal excites the antenna, the elecrons oscillate back and forth correct? So when the phase is near 0, are there fewer photons absorbed during that period of time? (Since less energy is available to move the elctrons) Or have I misunderstood an EM wave?

4. Oct 26, 2012

### Staff: Mentor

I think he's asking how a radio frequency EM wave is composed of photons. Presumably it is, but I'd be interested in hearing the explanation as well. Seems kind of strange to think of a 40 meter EM wave composed of a bunch of in-phase 40 meter wavelength photons (but maybe it is?)...

5. Oct 26, 2012

### Drakkith

Staff Emeritus
Don't all EM waves interact through photons? (Even if you pretty much can't detect individual ones at radio wavelengths)
Edit: I'm just using a radio wave as my example since you can easily measure the phase.

6. Oct 26, 2012

### Staff: Mentor

Here is a thread where the discussion of photons and EM radiation came up. I was surprised by some of the answers given by our sharper physicists, and I meant to go back and study the thread more when I had time (which hasn't happened yet). Does it help?

.

7. Oct 26, 2012

### Drakkith

Staff Emeritus
Not really, thanks for the link though.

8. Oct 27, 2012

### Studiot

I'm still not quite sure what you are asking, Drakkith.

I prefer the organ pipe explanation or model for the action of an antenna. Quantum explanations via photons are much much more difficult and I don't find they add anything to understanding.

Remember that the antenna responds to the carrier wave, we really want the receiver to respond only to the signal or modulating wave.
This is actually true of the simplest receiver ( a crystal set).

If this is what you are looking for I will post more.

9. Oct 27, 2012

### nsaspook

I think your question is related to the visualization of a linear polarized (in a circular polarized wave the amplitude is constant and the field vector rotates) 2D EM wave on paper as a set of changing magnitudes and polarities with a zero reference point of the electric field. In free space the E and B fields are in phase and EM energy is real but near field (at the antenna, in a transmission line) this is not true so the relationship between them changes. The phase of the electric field polarity and resultant EM energy levels become complex as the E and B fields are out of phase when the space they travel is reactive.

When we transmit an EM wave with information the energy of that information channel can be encoded in ways that might not translate directly to photon density levels at the receiver antenna(s). Examples: MIMO/Radio OAM

http://www.intel.com/support/wireless/sb/CS-025345.htm
http://en.wikipedia.org/wiki/MIMO

A very nice EM visual tool: http://www.enzim.hu/~szia/emanim/emanim.htm [Broken]

Last edited by a moderator: May 6, 2017
10. Oct 27, 2012

### Drakkith

Staff Emeritus
Honestly, neither am I. Thanks though.

11. Oct 27, 2012

### Born2bwire

It sounds like you are asking whether or not the rate of photons varies when an electromagnetic wave's amplitude goes from maximum to zero. For that, there is no difference in the rate of photons. Recall, for example, that when the electric field is at zero, the magnetic field is maximized and vice-versa. Phase has no bearing, it's a way of specifying the relative position in time that we are looking at the wave. The rate of photons can become sporadic when we have very weak signals. This is due to the granularity that arises and is called the shot noise. But once the signal strength grows beyond a very modest value, the photon rate is fairly consistent since we move out of the quantum behavior of the light into the classical regime.

12. Oct 27, 2012

### Drakkith

Staff Emeritus
Is it? I was unaware of that.

Got it, thanks!

13. Oct 28, 2012

### sophiecentaur

Shot noise is due to the discrete charges on electrons in amplifiers in electronic circuits and not the energy in individual photons. The energy of a photon for RF frequency EM waves is just soooo low that I don't think they are identifiable at all. Photons of visible light can be detected with a photomultiplier but that's about it - and they have more than a million times the energy of even mm microwaves.

This thread is again showing the problems that arise when you try to look at photons as being little bullets (or even very big bullets - in the case of Long Wave RF). You really have to think differently to get any sense out of photons.

14. Oct 28, 2012

### Drakkith

Staff Emeritus
I think he means if you reduce the intensity of the radiation enough, not the individual photon energy. I encounter this regularly in my astrophotography. Most targets are so dim that you might get about 10-20 photons over the course of a several minute exposure in some areas, leading to very low SNR. This forces you to spend multiple hours just getting exposures in order to have a quality picture.

Nah, I just didn't understand how an EM wave works and that the energy is transferred irregardless of the phase.

15. Oct 29, 2012

### Born2bwire

Yeah, shot noise is commonly discussed in reference to electronic circuits as a result of the discrete nature of electrical charge. However, it can also be applied to photons when you have such a weak source that the behavior of the electromagnetic wave falls into a quantum or perhaps quasi-quantum regime. This then can be applied to photon detectors. Say you have an electromagnetic source of constant power but variable frequency. At very low power and low frequencies, you could still detect a steady rate of photons with your detector. But as you increase the frequency, each photon carries a higher energy which means that the rate of the photons decreases correspondingly for the same power. At very high frequencies, the rate becomes low enough that the detected rate of photons becomes sporadic over short time scales as we start detecting the variation in the number of the photon count due to quantum statistics. But I think that this effect is only evident over short time scales. As one draws out the time scale, the sample size becomes large enough that you start losing the quantum variations (i.e. the photon rate detected over large time periods becomes stable).

Drakkith has it spot on from what I am thinking about. The shot noise in photons can be seen when you build up a picture, for example, using a CCD. With a very weak source, the picture is very noisy, fuzzy, and dark after a very short exposure time. That is, the rate is so low that not enough photons have been absorbed over a given area to allow one to discern what regions are light and which are dark. As you increase the exposure time, the picture resolves and the SNR decreases as you increase the effective number of photons that have been absorbed.

16. Oct 29, 2012

### sophiecentaur

But, to return to the OP question. A photon is not a little bullet and the model of a 'wave' consisting of 'bunches of' photons, corresponding to peaks and empty bits corresponding to troughs of the wave is nonsense. A photon is just the amount of energy involved when an EM wave is emitted or absorbed. True, there is also momentum transferred so the photon can be treated (with care) as having properties a bit like a particle but its properties don't include a 'size' or extent. It doesn't exist at peaks or troughs or at any other particular point on a wave so the 'phase' is not a meaningful concept.

17. Oct 29, 2012

### sophiecentaur

Drakkith is right to question this. Unlike physical waves (e.g. sound or water surface) in which the sum of powers is constant and shared Kinetic Energy and Potential Energy, so making the phases in quadrature, the phase between E and H fields is zero for progressive EM waves in space. However, waves on a wire ( such as the waves along an antenna, are almost exactly in quadrature. 'Something funny' happens in the near field of a radiating antenna as the quadrature phase changes to zero phase difference at a great distance.

18. Oct 29, 2012

### Drakkith

Staff Emeritus
So the energy imparted into an antenna by a radio wave is always constant if we have a steady signal? What is the oscillation of the electrons in antenna then? What are they doing when the phase is near 0 and near +90 and -90?

19. Oct 29, 2012

### Studiot

Remember that the energy of an oscillator depends on its frequency, so the energy in the wave will be constant over one complete cycle at any one carrier frequency (and sensibly constant for FM), for a steady modulating signal.

20. Oct 29, 2012

### Drakkith

Staff Emeritus
But is it steady for fractions of a cycle? I think that's what was said earlier.