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Poynting vector / time averaged energy?

  1. Nov 6, 2011 #1
    Something has always bothered me about finding the energy of an EM wave -

    What justification is there for taking the time average of the wave?
    I know we do this to find the energy of an EM wave, but I haven't seen this before in any of my courses.

    Isn't this a violation of conservation of energy?
    If I remember correctly, If E=Acos(w(r,t)) and the energy density is 1/2ε|E|^2 , then the energy density = 1/2ε|Acos(w(r,t))|^2. We can do the same for B - they are in phase and so then the net energy density must vary as cosine^2. Which can't make sense if it is to agree with cons. of energy. As far as I know from mechanics, the total energy of a closed system is always constant and independent of time.

    Yet for these waves we take the time average. Why?

    I'm assuming I'm wrong here - I just want to know where the time averaging idea came from, so i can learn how it is derived/proven.
     
  2. jcsd
  3. Nov 6, 2011 #2
    As you say the total energy - which is the volume integral of the energy density - is conserved in EM.

    For a plane wave the integral over the whole space is not finite, it only make sense for a wave packet which goes to zero sufficiently fast (in physical situations it is the field generated by a smooth current charge limited in space).
    In your example the energy density is "moving" in space as time varies, we use the time average because for a sin wave it is equal to the spacial average
     
  4. Nov 6, 2011 #3
    I'm not sure I understand.

    After thinking about it for a good while, energy density itself seems strange to me - I don't think we used it in mechanics either. How would you define energy density for in say, the motion of a pendulum?

    I feel like energy density is exclusively applied to light, but I'm not certain. I don't remember using it anywhere else. do you?
     
  5. Nov 6, 2011 #4

    Bill_K

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    elegysix, We use energy density whenever energy is spread out over a continuous region. For example a compressed spring is one-dimensional and we use energy per unit length. A stretched rubber sheet is two-dimensional and we use energy per unit area. And when we squeeze a three-dimensional object like a rubber ball we talk about energy per unit volume.
     
  6. Nov 6, 2011 #5

    Dale

    Staff: Mentor

    You don't need any justification. If you take the time average then you get the average, if you want an instantaneous then you don't take the average.

    No.

    Here is my favorite page on the conservation of energy in EM:
    http://farside.ph.utexas.edu/teaching/em/lectures/node89.html

    If you draw a closed surface, then the EM energy flux that enters the surface is equal to the change in the field energy inside the surface plus any work done on matter inside (eq 1034).
     
  7. Nov 7, 2011 #6
    You take the time average for practical reasons (many sensors cannot respond to the fast fluctuations of EM fields and thus output the average.) I am not aware of any fundamental physical laws that depend on taking the time average. Any derivation that uses the time average energy could just as well use the instantaneous energy, if careful with the math, and get the expected answer.
     
  8. Nov 8, 2011 #7
    ok, I've read about it some more on the link provided above. I understand it somewhat better but I still have a question.

    suppose we pulsed a laser so fast that only a single photon escaped (in the particle sense). I assume the field of the photon would still oscillate like we said before, but how can the energy of a single particle fluctuate on its own?
     
  9. Nov 8, 2011 #8

    Dale

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    The energy of a single photon is the expectation of the Hamiltonian of that photon. The expectation involves the integral over all space. So just because the Hamiltonian is zero at some specific location at some point in time does not imply that the expectation, integrated over all space, is zero.
     
  10. Nov 8, 2011 #9
    In trying to respond, I have realized a fundamental concept that is unclear to me.

    When we write E=Acos(w(r,t))

    Is this the electric field of a discrete point as it travels through space? such as a photon traveling along r(t)?
    Does it say anything about the field in regions surrounding the photon, or is it just the field at the point where the photon exists?

    (I'm trying to interpret from a particle perspective so I can use it with my earlier example)

    As with any other E field source I would expect it to drop off in space as 1/r^2, where r is the distance from the photon or region. Does that happen? why not?

    As I understand it, this is the electric field of a photon as it travels along r(t).
    Not the field in space, just the field at the point in which the photon exists.
    Is this wrong? why?
     
  11. Nov 8, 2011 #10

    Dale

    Staff: Mentor

    No, this equation is the equation of a plane wave which fills all of space and time.

    It doesn't represent a particle, it is a classical EM field.

    You are talking about a spherical wave or a dipole, not a plane wave. Plane waves don't drop off.
     
  12. Nov 8, 2011 #11
    so where do photons come into the picture?

    how are photons and the E&M fields related?
     
  13. Nov 8, 2011 #12

    Dale

    Staff: Mentor

    I wouldn't recommend looking into photons at this point. You should learn classical EM well before trying to learn quantum mechanics. It is an unnecessary complication for understanding your earlier questions.
     
  14. Nov 9, 2011 #13
    Ok. well I'll just write where I'm going with these questions and maybe you could answer them for me.

    If the field from a single photon fluctuates on its own (assuming there are E&M fields of a photon), then I have to ask why isn't that a violation of cons. of energy?

    If it doesn't fluctuate, the I have to ask why the field in a plane wave does?

    If there are no fields associated with a photon, then I have to ask why not?
    photons are the particles of light, and light has these fields, right?
     
  15. Nov 9, 2011 #14
    The Poynting vector is not Lorentz invariant, it is one part in four of another tensor. It contains terms of divergence in both E and B (electric and magnetic charge). It's amazing it's useful at all.
     
  16. Nov 9, 2011 #15

    vanhees71

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    The Poynting vector is the energy-current density and part of the symmetric energy-momentum tensor of the electromagnetic field. It's built by the time-space compononents of this tensor, the time-time component is the energy density.

    For a free field (i.e., if there are no currents and charges around), the integral over these four components give the energy-momentum fourvector of the electromagnetic field.

    For more details on Lorentz invariance and the physical meaning of the energy-momentum tensor, see any textbook on electromagnetism, e.g., Jackson.
     
  17. Nov 9, 2011 #16

    Dale

    Staff: Mentor

    I already answered this question in post 9. The Hamiltonian of a photon involves an integral over space. This integral is constant over time, so energy is conserved.

    Do you understand how energy is conserved in classical EM even though the fields regularly go to zero at any given point?
     
  18. Nov 9, 2011 #17
    yes, I am trying to ask if a single photon's "charge" varies - being the cause of the changing E field.
    An electron's charge doesn't vary, but the field in space around it can.

    I say "charge" for lack of a better word...
     
  19. Nov 9, 2011 #18
    Photons don't have charge and photons don't create fields. They are the field.
     
  20. Nov 9, 2011 #19

    Dale

    Staff: Mentor

    chrisbaird is correct.

    Again, I recommend you learn classical EM first. I don't think that QM is helpful at all for understanding this topic.
     
  21. Nov 9, 2011 #20

    jtbell

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    Staff: Mentor

    And when you are ready to study photons seriously, look up "quantum optics" and then ask questions in the Quantum Physics forum. Some posters there know a lot about quantum optics, e.g. Ctugha.
     
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