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The energy of an em radiation

  1. Aug 28, 2012 #1
    the energy of em radiation (or photon) is proportional to the frequency of the radiation.
    the em radiation (or photon) is composed of oscillating Electric and Magnetic fields.
    as such this energy must be stored in the oscillating electric and magnetic fields constituting the radiation.
    therefore the energy of the em radiation (or photon) must be related to the Magnitude of Electric and Magnetic fields associated with it.....(i.e. the Amplitude of the em field constituting it).....(because greater the magnitude of electromagnetic field , greater would be the energy stored in it and vice-versa).
    What is inappropriate here?
     
  2. jcsd
  3. Aug 28, 2012 #2

    berkeman

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    You are mixing concepts from a single photon in with general EM radiation (consisting of many photons).

    For a single photon, you are correct that E = hf (f = frequency). So given that energy, you can calculate the magnitude of the associated electric and magnetic fields for that single photon.
     
  4. Aug 28, 2012 #3
    I think you are confusing the energy of a single photon (which is only a function of frequency) to the energy contained in electromagnetic radiation, which is made of a bunch of photons. Also remember that photons don't have an electric and magnetic fields to contribute to their energies (they are electrically neutral) but virtual photons create electric and magnetic fields.
     
  5. Aug 28, 2012 #4
    Re: berkeman

    ok, for a single photon the energy E = hv .
    Given that energy I calculate the magnitude of the electric and magnetic fields associated with that single photon.
    I get a definite answer. Doesn't this mean that its energy is related to the magnitude of em fields associated with it (as knowledge of the former allows me to calculate the latter or vice-versa)?
    sorry if it appears stupid, but I am confused....
     
  6. Aug 28, 2012 #5
    Re: soothsayer

    ok..but isn't the energy associated with a single photon of electromagnetic nature? and if that is so why isn't it related to the amplitude of the em fields associated to it?
    please bear with my confusion..
     
    Last edited by a moderator: Aug 29, 2012
  7. Aug 28, 2012 #6

    jtbell

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    Re: berkeman

    Classically, the E and B fields at a given point are associated with an energy density (joules / m3) ##u## at that point:

    $$u = \frac{\epsilon_0}{2} E^2 + \frac{1}{2 \mu_0} B^2$$

    Therefore, in order to associate magnitudes of E and B fields to a photon, you need to associate a volume with the photon. But photons are quanta of energy, and volume does not enter into their fundamental description. In general, you can't think of a photon as a small localized object, as a sort of "fuzzy ball," as far as I know. The "spatial extent" of a photon depends on the method that you use to produce it, that is, it's basically a result of the production apparatus and not a fundamental statement about photons in general.
     
  8. Aug 28, 2012 #7
    Re: soothsayer

    No, the energy associated with a single photon is not of an electromagnetic nature. That is to say, photons are not made of tiny quantum electromagnetic fields, but rather, photons are the building blocks of electromagnetic fields.

    This is also incorrect thinking, for the same reason as above. It does not make sense to talk about the "magnitude of the electric and magnetic fields associated with a single photon." Photons are fundamental particles, they are neither made of E/M fields nor can a lone photon create an E/M field.
     
  9. Aug 28, 2012 #8
    Re: berkeman

    But you may certainly find the average density of photons (number of photons per unit volume) corresponding to a monochromatic em wave.
     
  10. Aug 29, 2012 #9
    Re: soothsayer

    But then as photon is what we may conceive as the fundamental building block of em energy, can't we consider it as the smallest bit/quanta/lump of the same em energy? Wouldn't it make some sense then (to talk about the corresponding em field)?
     
  11. Aug 29, 2012 #10
    Re: berkeman

    This is definitely true.

    Ehh, a single particle alone does not a field make. It's sort of a semantic argument; you're trying too hard to apply CLASSICAL E/M principles and equations to a totally QUANTUM situation. As Dickfore mentioned above, the amplitude of an E/M field is related to the density of photons in the field, and obviously, it makes no sense to talk about the density of photons in a photon. You cannot look at a single grain of sand and ask what the size of its beach is.
     
  12. Aug 29, 2012 #11
    Re: soothsayer

    I have to disagree with most of what soothsayer has posted above as being wrong. The above quote is correct.

    Photons are the quantization of the EM field -- end of story. Going back to the middle of the story, jtbell indicated above a way to calculate the energy density of the EM field. Then given a certain volume in space, one can directly calculate the (mean) number of photons in that volume. The assumption is that a continuous EM wave propagates throughout that volume. A more useful metric would be to determine the intensity (J/s/m^2 = W/m^2) of an EM wave that passes through a surface. Then you could determine the number of photons that pass through the surface per unit time.

    The important concept to note here is that as you increase the energy/power of the EM wave, it's the number of photons the increases. Why? Because of E=hf. More energy, more photons.
     
  13. Aug 29, 2012 #12
    Re: soothsayer

    How is that different from what I told him? :confused:
     
  14. Aug 29, 2012 #13
    Re: berkeman

    My last post seems to have been posted at the same time as the above quote. The quote clears up some of what I disagreed about with soothsayer.

    True that "it makes no sense to talk about the density of photons in a photon," however, the OP essentially asked about the density of photons in the (classical) field. As you said, this is a valid inquiry. But, going further, a single photon is (in the language of QFT) a excitation of the EM field. As such, you cannot say that it has nothing to do with electric and magnetic fields since it IS an EM field.
     
  15. Aug 29, 2012 #14
    Re: soothsayer

    I think he is complaining about this statement:

    I was going to question why you think that and what are your references. As far as I know, the only reason to believe a photon is not 100% EM energy is that there needs to be something to hold the energy focalized so that it doesn't dissipate or spread in space as waves tend to do.
     
  16. Aug 29, 2012 #15

    sophiecentaur

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    Re: berkeman

    This is a difficult area because that statement suggests that not all photons with the same energy are 'identical'. I don't see that this can be true. I would rather say that the 'effective extent' of a photon relates only at the period of interaction with the source or receiver and is more to do with what goes on at each end than with any idea of 'size' for the photon.
    If you think in terms of a resonant system absorbing or emitting a quantum of energy, the time it would take to change its energy state would relate, in a classical way, to the Q factor (God knows what that means in the context of QM) of the resonant system. I would tend to think that QM systems which could undergo the same energy change would not be likely to have this same 'Q factor' so one type of system could produce a photon and this same photon could expect to be absorbed by a totally different type of system.
     
  17. Aug 29, 2012 #16
    Re: soothsayer

    I was not trying to say that photons are completely unrelated from EM radiation, what I was trying to get across is that you cannot determine the energy of a photon via classical computation for energy in an electromagnetic field, as the OP wanted to do. You can't say: "Well, for a single photon, the amplitude of its electric field is this and the amplitude of its magnetic field is this, so it's energy should be this: derived from the strength of its E and B fields."

    It was inaccurate of me to say that the energy of a photon was not associated with electromagnetic energy, but what I was trying to say is that the energy of a photon is not determined by the strength of some quantum EM field that is contained in it, it's indeed determined purely by frequency of light.
     
  18. Aug 29, 2012 #17

    sophiecentaur

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    It would be a meaningful thing to relate the density of a stream of photons to the energy density of an EM 'beam' - which would give you the values of the fields. But that would really be putting it 'the other way round' and very different from assigning an actual set of fields to an individual photon. For a start, you couldn't give the photon an actual 'area' because that is undefinable so you couldn't, consequently, give it a value of energy flux or field values. It can be regarded as extending over the whole of the wave front and over a range of possible positions on any 'ray path' you could assume.
    It's just not a good idea to keep trying to impose 'old fashioned' values on photons. It may give an illusion of better understanding but it doesn't actually get you anywhere. They are 'photons' and photons are like photons - nothing else.
     
  19. Aug 29, 2012 #18
    Thank you, that was exactly my point but put a little bit more succinctly.
     
  20. Aug 30, 2012 #19
    Interesting point about the beam. That seems like a simple experiment to perform and control. Have there been experiments generating meaningful data where some sort of side-by-side arrangement and density of photons is involved? I suppose a wide beam laser experiment would qualify.
     
  21. Aug 30, 2012 #20

    sophiecentaur

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    I don't quite see what sort of "experiment" you think would tell you anything about this. It is surely well enough established what the photon energy is at a given frequency. You only need to divide the total energy flux by hf to find how many photons are involved. But that would tell you nothing (it would be meaningless) or imply anything about the 'fields' associated with each photon. You might as well ask "How many houses are there in a brick?" The simple Mathematical operation of taking the reciprocal of "how many bricks are there in a house?" would give you a numerical answer but what would it mean? - One brick is 1/2000 of a front wall plus 1/2000 of a back wall plus 1/1000 of some foundations etc. etc. But it isn't, is it? A brick is just a brick and it is just a contribution to the part of the house where you happen to find it.
    If you are trying to build a picture of a photon as a 'little squiggle' of fields, somewhere in space, you are onto a loser. That is nothing like any of the valid models of the photon that we use these days.
     
  22. Aug 31, 2012 #21
    I disagree with this. By quantizing the EM field, what you are left with is a quantum of the field, i.e. the field of one photon. Under canonical quantization, the electric and magnetic fields (if you want to be pedantic, the electric field and vector potential) are elevated to the status of Hermitian operators. If you were to quantize the field in a cubic box of side L, then the "magnitude" of the electric-field operator is

    [tex]e = \sqrt{\frac{\hbar\omega}{2\epsilon_0 L^3}} .[/tex]

    I was under the impression that the common interpretation of e is that it is the electric field of one photon. Indeed, this make sense in terms of basic intuition. For example, if a single photon were radiated into completely empty space, then L would be so large that at any arbitrary location we would most likely never be able to detect the photon since it's field would be too weak to interact with our instruments (intuitively, send out a single photon into the whole of the universe and chances are you'll never find it). On the other hand, an extremely confined photon would exhibit such a large-valued field that it would more than certainly interact with whatever matter surrounds it.
     
  23. Aug 31, 2012 #22

    sophiecentaur

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    You need to ask yourself, if it's all down to "common intuition", why was QM such a revolutionary idea and why has it taken so long to sort it out? It just can't be 'that obvious' or the great minds wouldn't have been struggling so long.
    I am not sure that I have ever read this "common interpretation" in anywhere of substance. ("e" usually stands for the electronic Charge and charge is not Field). If you consider the definition of what a Field is, then it is a set of values (scalar or vector) over a region of space; it is defined at a point. Which point would you choose to apply your definition of the field of a photon? In the case of the two slits experiment, would you apply it to each slit or just one. And what about a diffraction grating? Suddenly you would need to have a field at every slot in the grating. In fact, there is no reasonable way to describe the photon in terms of a field distribution in space because the same photon would hava a different field distribution, according to the situation it's in. We can have no knowledge about the 'position' of a photon in space, nor any notion of its 'size'.
     
  24. Aug 31, 2012 #23
    It took 1300 years from the fall of Rome for Newton to appear and formalize modern physics. It then took ~230 years to formalize particle-based quantum mechanics. Within 20-30 years from that, quantum field theory was formalized. In the grand scheme of things, quantum theory didn't really take that long.

    I didn't say that the result was obvious but merely that the results relate to some sort of basic intuition if you make the "correct" interpretation.

    Clearly I used the variable e to stand for the "magnitude" of the electric-field operator of quantum mechanics. It does not stand for charge; this is obvious. Give me better LaTeX commands and next time I'll make a script capital "E" for you.

    If you've never even quantized the EM field before, then Google "quantization of the electromagnetic field." The 2nd edition of Sakurai does it, most/all books on quantum optics do it, even some books on quantum field theory do it.

    As with any quantum-mechanical operator, the operator yields statistics/amplitudes related to measurements only after you make projections on quantum states (i.e. calculating matrix elements). True, the field at a specific point is not well characterized in this scheme, but the field associated with the photon is still defined. Also note that I was very careful to say that the result I quoted applies for a field quantized "in a cubic box of side L." This works well for cavity fields (cavity QED is a hot research topic) and for forming statistics of a photodetector; however, other quantization schemes may exist.
     
  25. Aug 31, 2012 #24

    sophiecentaur

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    I was being a bit dumb about the "e" thing!
    I also take your point about the 'bound photon' situation but this is only relevant when it is actually bound - which is not the case when you consider a photon on its way from A to B, which is the situation when you have a flux of energy.
    My argument against quantising a photon field still holds when you can't specify where and when you are measuring it, I think.
     
  26. Aug 31, 2012 #25
    Can someone clear something up for me? It seems like there are two types of photons to deal with here. We have electromagnetic radiation: alternating E and B fields, which are made up of force-carrying virtual photons, are they not? But these are not the photons that we see when the radiation reaches us, are they? What's the difference? Seems like there would be force carrying and non force carrying photons in EM radiation.
     
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