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A question about energy carried by a electromagnetic wave

  1. Jun 20, 2011 #1
    My teacher told us energy at a point of a progressive wave is proportional to square of amplitude, it is true for both mechanical and electromagnetic wave.

    But in the other lesson, he also tols us the energy of the EM wave is equal to hv, where h = Planck constant, v = frequency of the wave. So gamma ray has a higher energy than radiowave, but if the frequency is the same, such as X-ray and gamma ray, they have same energy.

    But the EM wave still have large or small amplitude, so the energy of a EM wave depends on th frequency or amplitude or both?
     
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  3. Jun 20, 2011 #2

    Bill_K

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    The energy of a classical EM wave depends on the amplitude. The formula E = hν applies to the energy of a single photon.
     
  4. Jun 20, 2011 #3

    vanhees71

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    You have to distinguish between classical em. waves (which are quantum theoretically socalled coherent states, which are the superposition of states of any photon number corresponding to a high average photon number). For a classical electromagnetic wave in free space the energy density is given by (in Heaviside-Lorentz units)

    [tex]\epsilon(t,\vec{x})=\frac{1}{2} [\vec{E}^2(t,\vec{x})+\vec{B}^2(t ,\vec{x})].[/tex]

    For a single-photon state, which is not a state that can be visualized as a classical em. wave, the total energy is [itex]E=\hbar \omega=h f[/itex].
     
  5. Jun 20, 2011 #4

    Born2bwire

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    Both pictures remain consistent though. If you were to take an electromagnetic wave and, while keeping the amplitude constant, change the frequency you would observe a change in the rate of photons measured per unit time. In fact, at very low frequencies and/or very low intensities you can get granularity noise due to the low rate of photons in what is called shot noise.
     
  6. Jun 20, 2011 #5
    we have just studied a little of the wave in the syllabus, so how to distinduish the classical EM wave and the single-photon state? And so for the example of gamma ray, which of the two equations can be applied? Since we have confused E = hv with E α A^2 in the example of gamma ray.
     
  7. Jun 20, 2011 #6
    You need to differentiate between a single photon and a group of photons. Photons are bosons, which means they do not obey the Pauli exclusion principle. So several photons can and do occupy the exact same quantum state. The energy E = hv is the energy of a single photon, whereas the energy that is proportional to the square of the electric field is the energy of all the photons in the same energy state. When you measure the amplitude of the electric field of a wave, you are measuring the combined effect of all the photons in that same energy state. So it also matters how many photons presently. I think the http://en.wikipedia.org/wiki/Photoelectric_effect" [Broken] demonstrates these concepts nicely.
     
    Last edited by a moderator: May 5, 2017
  8. Jun 20, 2011 #7

    vanhees71

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    There is a profound difference between a photon, which more precisely should be called "single-photon state". These are states with the precise photon number 1. Coherent states are superposistions of all photon-number states from 0 to [itex]\infty[/tex]. The number of photons is thus not fixed but the probability distribution for a number of photons is given by a Poisson distribution. As an intro see

    http://en.wikipedia.org/wiki/Coherent_states
     
  9. Jun 22, 2011 #8
    So the energy of a EM wave is depends on the energy of a single proton, and the amplitude of the electric field is depends on how many proton in the same energy state?
     
  10. Jun 22, 2011 #9
    It's photons that constitute light, not protons. Yes, this is basically correct. The photoelectric effect is a good example of this: You shine a beam of light on a material and electrons are knocked off. An individual electron only interacts with an individual photon, so the kinetic energy of the electrons depends on the individual photon energy and thus the frequency according to E=hv. Increasing the beam intensity applies more photons and therefore knocks out more electrons, but does not speed up each individual photon. To do that, you have to raise the frequency.
     
  11. Jun 23, 2011 #10
    Thanks all of you. Now i have understanded the difference.
     
  12. Jun 23, 2011 #11
    This should read: "...but does not speed up each individual electron. To do that, you have to raise the frequency.
     
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