EM Wave Phase - Fewer Photons at 0 Degrees?

[Drakkith]

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)

 

[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.

 

[Drakkith]

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.

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?

 

[berkeman]

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?)...

 

[Drakkith]

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.

 

[berkeman]

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?

https://www.physicsforums.com/showthread.php?t=631949

.

 

[Drakkith]

Not really, thanks for the link though.

 

[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.

 

[nsaspook]

Not really, thanks for the link though.

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

 

[Drakkith]

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

Honestly, neither am I. Thanks though.

 

[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.

 

[Drakkith]

Recall, for example, that when the electric field is at zero, the magnetic field is maximized and vice-versa.

Is it? I was unaware of that.

there is no difference in the rate of photons.

Got it, thanks!

 

[sophiecentaur]

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.

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.

 

[Drakkith]

Shot noise is due to the discrete charges on electrons in amplifiers in electronic circuits and not the energy in individual photons.

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.

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.

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

 

[Born2bwire]

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.

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.

 

[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.

 

[sophiecentaur]

Recall, for example, that when the electric field is at zero, the magnetic field is maximized and vice-versa.

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.

 

[Drakkith]

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?

 

[Studiot]

So the energy imparted into an antenna by a radio wave is always constant if we have a steady signal?

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 anyone carrier frequency (and sensibly constant for FM), for a steady modulating signal.

 

[Drakkith]

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 anyone carrier frequency (and sensibly constant for FM), for a steady modulating signal.

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

 

[Studiot]

How can it be if the fraction is a small region around zero amplitude?

That is why we need to count over complete cycles.

 

[Drakkith]

How can it be if the fraction is a small region around zero amplitude?

That is why we need to count over complete cycles.

Hmm. Now I am confused. This is exactly what my original question is about. (Or so I thought)

 

[Studiot]

You have to either work in terms of proper quantum photons and forget waves or in terms of waves and forget photons. You cannot work in terms of impinging balls in either case.

I advised the wave approach as being much easier.

Phase only has meaning in terms of the wave model. It has no meaning in terms of the photon model.

 

[Drakkith]

Phase only has meaning in terms of the wave model. It has no meaning in terms of the photon model.

Sorry, I'm coming into this problem after thinking about radio telescopes. Since I use a CCD camera in my own telescope, I naturally think of photons. So, if the energy transferred to the antenna is less when the phase is near 0, how does the photon model explain that?

 

[sophiecentaur]

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?

An absorber in a particular position will see the wave passing it and will 'see' all possible phases in sequence as it goes past. As it happen, with EM, the energy arrives in sinusoidal bursts but that is not relevant to the photons at all - different aspect of the whole thing or to the current in the antenna because any 'pulsing' effect is smoothed out by the reactive energy in the antenna and its fields. This is because any energy transfer will take more than one cycle to take place (basing this on classical ideas of a wave coupling into an absorbing circuit) The electrons in the absorber / antenna (if you really insist on not using the notion of current) are just moving along the line of the E field in a phase 'just behind' the variations of the passing field.

Unfortunately, this thread is turning into a hotchpotch mixture of classical and quantum ideas and seems to be getting nowhere, as a consequence. There is so much 'good stuff' written about these matters (even on the Net) that it really would be better to read through wikki and elsewhere rather than to try to manufacture a dodgy personal picture of all this.

 

[sophiecentaur]

Sorry, I'm coming into this problem after thinking about radio telescopes. Since I use a CCD camera in my own telescope, I naturally think of photons. So, if the energy transferred to the antenna is less when the phase is near 0, how does the photon model explain that?

If you are looking at light from a distant source, you will not be looking at a coherent source. Individual photons will each be associated with 'their own' wave function. This could be analogous to what you would receive if you tuned a 10MHz wide band receiver to the FM bands and listened to the signal from all the VHF transmitters on Earth when you were a very long way away - just getting one or two photons from each transmission.
That doesn't mean anything as the photon is not a particle that exists at a particular part of a wave so how can there be a "phase near 0"? It exists in an undefined region of space until it actually interacts with the CCD in your camera - at which time the wave function collapses and the charges in the CCD acquire the energy. Whether or not it does that, hits another bit of the CCD or just gets absorbed and wasted, is down to statistics. (see the problem of this hybrid approach?)

There is a very significant difference between RF signals from 'regular' transmitters and what you get from normal light sources. You can really treat narrow band, CW RF signals as a set of photons with the same wave function (this is like laser physicists describe the degenerate states of the photons of coherent light from a laser). For light from a normal incoherent source, the photons each have their own state but the optics works classically in much the same way in most respects (apart from the speckles that you get with laser light).

 

[Drakkith]

Alright, thanks.

 

[Studiot]

Don't forget that the EM wave in the air is a traveling wave.

This induces a standing wave in the antenna. So one part of the antenna is always at zero and another part is always at a maximum.

This is what I meant by the 'organ pipe' model.


Have you seen this? - It is the simplest model I know of antenna action.

 

[Drakkith]

Have you seen this? - It is the simplest model I know of antenna action.

Nope. I have not studied EM theory or antennas before.

 

[Studiot]

Here is a very quick rundown before I knock off.

If you blow across the top of a partly filled bottle you can generate a resonant sound or tone within the bottle. The tone or frequency depends upon the length from the water surface (don't waste beer on this) to the mouth as shown.

This is equivalent to an organ pipe, closed at one end as shown with a standing wave inside.

The air at the closed end cannot move freely so forms a permanent node.

If we place a conductor rod or wire in an EM field and ground one end we find the same effect in the conductor.

The EM field has a voltage between points in the field. These induce a voltage in the conductor between the ends. Since one end is grounded its voltage cannot change, but the other end chages in sympathy with the field changes.

The current waveform is of the opposite phase, shown dashed.

If we have an organ pipe open at both ends or ground our rod in the middle, both ends are free to vary as shown. This is called a half wavelength dipole as an antenna.

 

[Drakkith]

Alright, thanks Studiot. I didn't think about induced waves in the antenna.