EM wave amplitude question

1. Nov 27, 2005

Simons

argh hello guys, I've been trying to find out a simple answer to this question but all the resources I've looked at seem to skim over it or contradict each other, but simply:

Do electromagnetic waves have a variable amplitude? "Intensity" is not necisarilly the same, if you're using the photon model, because "brighter" can just be "more". I understand EM waves are oscillations in 2 fields...surely there is some specific amplitude to a wave, but as the formula for the energy of a photon has only wavelength and a constant, how can there be variable amplitude? (I assume a greater amplitude vibration would have more energy)

If the amplitude can vary, what happens when the waves interfere, such as destructive interference, where does the energy go when the wave is canceled out?

2. Nov 27, 2005

Renge Ishyo

It is my understanding that the amplitude in terms of an electromagnetic wave refers only to the number of photons being delivered to an area at a given time, it does not increase the energy of a single photon. The energy of a single photon is dependant on the frequency of the em wave by E=hf. The photoelectric effect is what led to the discovery of this. If a coherant monochromatic light source emits light at a certain frequency that does not provide enough energy to eject an electron from a metal surface, you cannot simply amplify the wave to supply the necessary energy to eject the electron (the increased intensity will have no effect, no matter how bright you make it, you simply will be tossing more and more ineffective photons at the electron and none of them will have enough energy to dislodge it).

If you are talking about the amplitude of a single photon I have two questions for you: 1) Are you sure a single photon is a wave in and of itself? The wave properties of light that we have observed experimentally exist as a superposition of a very large number of photons (even the diffraction experiment where they sent one photon through a slit at a time didn't "reveal" the interference pattern until a great many photons went through the slit). It makes sense to me at least that the wave properties/amplitude for such a wave can be related to a large number of photons as well (but I am no quantum physicist so I don't claim to be an expert on this sort of thing). 2) Let's say a single photon did have an amplitude. Is it possible to measure it?

3. Nov 27, 2005

Staff: Mentor

You need to be careful when you're mixing classical concepts such as electromagnetic waves with quantum-mechanical concepts such as photons.

Classically, the irradiance $E_e$ of an electromagnetic wave (the power carried by the wave through a surface of unit area perpendicular to the direction the wave is traveling, i.e. joules per second per m^2 or watts per m^2) is proportional to the square of the amplitude (maximum value) $E_{max}$ of the electric field in the wave:

$$E_e = \frac{1}{2} \epsilon_0 c E_{max}^2$$

In terms of photons, the same irradiance can be described in terms of a number of photons per second per m^2, which I'm going to call $n$ (I don't remember the usual symbol for it, offhand):

$$n = \frac{E_e}{hf} = \frac{\epsilon_0 c E_{max}^2}{2hf}$$

$E_e$, $n$ and $E_{max}$ all vary simultaneously as the light spreads out from a source, or is focused into a small region.

Whenever you have destructive interference somewhere, you must have constructive interference somewhere else. The energy is simply redistributed so that the areas of constructive interference receive more energy than they otherwise would have.

Last edited: Aug 2, 2007
4. Nov 27, 2005

Simons

Alright so a photon is considered an indivisible 'unit' of EM radiation...Can more than one photon occupy the same location? If they are descrete particles, that seems impossible, but if simply small fractions of the overall waves in the EM field, do they "stack" and increase the total displacement at that point?

As for this:
say two waves come from opposite directions and intersect so that the electric field of one is at a max and the electric field of the other is at a max in the opposite direction right when they meet, what happens (and does this cause the destruction of individual photons)? If this energy is somehow accounted for elsewhere, is energy transfered from here?

5. Nov 27, 2005

Staff: Mentor

In this situation, you get a standing wave in which the points of minimum and maximum oscillation of the field (the nodes and antinodes respectively) are at fixed locations. At the nodes the amplitude of oscillation is zero. At the antinodes the amplitude is twice the amplitude of one of the original waves (assuming the two original waves have equal amplitudes). The energy density in a wave is proportional to the square of the amplitude, so the energy density at the antinodes is four times the energy density in each of the original waves. But when you average the antonodes together with the nodes, you get an overall average energy density equal to twice the energy density in one of the original waves (that is, the sum of the energy densities in the two original waves).

Last edited: Nov 27, 2005
6. Nov 27, 2005

Renge Ishyo

My understanding is that photons do "stack" depending on the intensity of the wave (the greater the intensity, the more photons "n" are present per area as described in the third post down from the top). It seems that the energy of a photon at a given frequency is a fixed amount (i.e. as far as I know you cannot combine two photons together at a single point and create a photon with "more energy"), and that is what led to the idea that photons themselves behave like discrete units (or particles).

Are they discrete particles? Here is the problem, the photon is described almost by metaphor. To my understanding we don't fully know what it is so much as we try to describe what it is like. There are many things that we cannot explain about a photon by considering it strictly as a particle (the fact that it seems to completely lack mass/matter for one thing), and as you have just discovered there are some things that you cannot describe about photons by applying the logic of a wave to it (increasing the intensity of the wave IS predicted to increase the energy of the wave using the ideas of a wave from classical physics, but experimental evidence via that photoelectric effect showed that not to be true).
Does this help or make sense? I hope so, because you are reaching the limit of my knowledge in physics

7. Nov 27, 2005

Simons

ayyyyyyyyyyyyyy

is it too late to defect to scalar fields or ether theory?

I get what you're saying about standing waves and nodes and so forth...thats classical mechanical wave properties...makes sense. But it seems pretty clear to me that the evidence for wavelike nature of light outweighs particle evidence, maybe the photoelectric effect is just misinterpreted?

8. Nov 27, 2005

Renge Ishyo

Actually, study a little further and you will see that there is plenty of evidence that light has particle like properties as well as wave like properties (the photoelectric effect was just the breakthrough discovery that light could not be described as a wave alone). In fact, go a bit further and you will find out that particles can behave like waves too. For example, it is possible to create the same interference patterns that you see with light using *electrons* via an electron beam.

9. Nov 27, 2005

Staff: Mentor

About eighty years too late, sorry.

I suggest that you scroll down to the Quantum Physics area here and study the past several months' worth of material. You'll probably find several threads in which someone tries to convince the rest of us that photons are a bunch of bullpucky, and the responses will contain much useful information.

This general topic pops up with some regularity, so I hope you'll forgive me and anyone else here who is too tired/lazy/whatever to dive into all this yet again. :zzz:

10. Aug 2, 2007

Gaucho

None of the replies I saw so far actually answers the question. I also like to know how the math wave energy totally overlooks amplitude, when we all know that E is proportional to A2- And to think that all EM has the same amplitude is creasy. Radio modulates amplitude to deliver power.

11. Aug 2, 2007

Claude Bile

Firstly, I don't think it's helpful to discuss this in terms of energy, I think it is easier intuitively to look at everything from the point of view of power and irradiance. One can then link the classical with the quantum as follows;

The classical amplitude of an EM wave is proportional to the irradiance which has units W/m^2. The irradiance is also proportional to the photon flux (no of photons/second) for any given wavelength (neglecting shot noise).

Claude.

12. Aug 2, 2007

Staff: Mentor

Correction: that should be "the square of the classical amplitude".

13. Aug 3, 2007

Gaucho

Well, that is the problem. We are so “entrenched” in the quantum mechanics which works so well in some cases, fail on other cases, and forbid us from solving the real problem. We keep adding bandages to an atomic model we all know is not correct. That is unbearable, and it is not scientific at all. I want the facts, not the theory. Anybody knows? A sure appreciate the help.

14. Aug 3, 2007

Staff: Mentor

Quantum electrodynamics has been tested in many cases with very high precision. I don't recall hearing of a case in which its predictions disagree with experiment, to date. What case(s) do you have in mind?

15. Aug 5, 2007

Claude Bile

Whoops . Well spotted.

Gaucho; I can't see this problem you are talking about, it all ties up rather neatly in my mind. Remember that the energy in terms of photon flux varies with frequency. A photon flux of 500 photons/sec/m^2 corresponds to a greater power in the UV part of the spectrum than it does in the infrared part of the spectrum since each photon has more energy as per E = hf.

Claude.

16. Aug 11, 2007

belliott4488

I think you cause us to wonder if you know what science is, when you say
Science is nothing more than theory, tested by experiment. I don't know what a scientific "fact" would be, other than a very well-confirmed experimental observation, although even that is always open to refinement by improvements in precision.

You seen to suggest that there is something obviously wrong with QM, even though the more fully developed theories, such as QED, have been successfully tested to a higher degree of precision than any other theory in history. What in the world are you talking about?

17. Aug 11, 2007

ZapperZ

Staff Emeritus
But see, your intention IS the central theme here, because you are basing your objection not based on emperical evidence, but based your PERSONAL TASTES! You categorize something as "ridiculous", not as "experimentally false". In physics, and in science, the latter is the only way in which any principle or theory can be falsified. Calling something "ridiculous" has zero value unless you are a prominent physicist and your opinion is respected. Do we have any reason to grant you that?

I also strongly suggest that you review the PF Guidelines before proceeding any further.

This thread should not be hijacked into a "QM is wrong" topic, which has been discussed ad nauseum. If anyone believes he/she has experimental evidence that contradicts QM, submit it for publication, or if you are unable to do that, go to our IR forum.

Zz.

Last edited: Aug 11, 2007
18. Aug 11, 2007

belliott4488

Thanks, Zz - that was a more useful response than mine, which probably served only to fan the flames ...