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Is the Poynting Vector Frequency Dependent?

  1. May 10, 2010 #1
    This is not a homework question. Please do not delete it. I am 57 years old and trust me I am not in school. Doe the magnitude of the Poynting vector depend on the frequency of the wave hitting the surface?
  2. jcsd
  3. May 10, 2010 #2
    Nope. The poynting vector simply gives you the momentum density of the field at some point.
  4. May 10, 2010 #3

    Andy Resnick

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    Yes- the Poynting vector is defined as c/4*pi (E x B), where 'x' is the cross product. Since (for optics) this frequency is so fast, this specific time dependence is usually neglected in favor of paying attention to 'slowly varying' temporal changes.
  5. May 10, 2010 #4


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    The Poynting vector is NOT the momentum density of the field.
    It is the intensity and represents the energy per unit time per unit area transmitted by the fields. It depends on the frequency only to the extent that E and B do. As usually used, the Poynting vector is averaged over time, much like the power in an AC circuit, so any frequency dependence averages out.
  6. May 10, 2010 #5
    Yes, it is. I suggest you study up if thats what you think. It can also be interpreted as energy flux density of the field. Stating it is frequency dependent is ill-defined.
  7. May 10, 2010 #6
    How does E and B depend on the frequency? I've looked all over and haven't seen anything that shows a proportional relationship, although it would seem that is the case. For example, wouldn't a blue monochromatic light be transfering more energy per unit time than a red monochromatic light?
  8. May 10, 2010 #7
    Its not. The only thing that depends on frequency like that is energy per photon. Total rate is independant, as is magnitude of e and b.
  9. May 10, 2010 #8
    Anything that is time dependent is frequency dependent, and vice versa. If you just Laplace transform a time dependent variable you will get fields that are frequency dependent. It just depends on how you want to look at it. The poynting vector is most definitely frequency dependent. Not exactly sure about the magnitude. Need to look it up.
    Last edited: May 10, 2010
  10. May 10, 2010 #9
    You're confusing the matter. Fourier transform can be performed on any function. The field is independant of how you choose to represent it.
  11. May 10, 2010 #10


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    The Poynting vector can be frequency dependent. There are a few definitions for the Poynting vector. The simplest is that we choose ExH. Since E and H are going to be time dependent, then the Poynting vector is also time dependent. More often than not though, we generally take the time average which can make the Poynting vector independent of frequency. This is because taking the time average is equivalent to
    [tex]\left< \mathbf{E}(\mathbf{r},t) \times \mathbf{H}(\mathbf{r},t) \right> = \frac{1}{2} \Re \left\{ \mathbf{E}(\mathbf{r}) \times \mathbf{H}^*(\mathbf{r}) \right\} [/tex]
    When looking at the fields as a single time harmonic frequency, then the frequency dependence cancels out due to the complex conjugate. However, if we have a superposition of frequencies, then:
    [tex] \mathbf{S} = \frac{1}{2} \Re \left\{ (E_1e^{-i\omega_1 t} \hat{e}_1 + E_2e^{-i\omega_2 t} \hat{e}_2) \times (H_1^*e^{i\omega_1 t} \hat{h}_1 + H_2^*e^{i\omega_2 t} \hat{h}_2) \right\}[/tex]
    [tex] \approx \Re \left\{ E_1\hat{e}_1 \times H_1^*\hat{h}_1 + E_2\hat{e}_2 \times H_2^*\hat{h}_2 + E_1\hat{e}_1 \times H_2^*\hat{h}_2 e^{i(\omega_2-\omega_1)t} + E_2\hat{e}_2\times H_1^*\hat{h}_1 e^{i(\omega_1-\omega_2)t} \right\} [/tex]
    [tex] \approx E_1H_1 + E_2H_2 + (E_1H_2+E_2H_1)\cos(\omega_1-\omega_2) [/tex]

    Where I have taken the liberty of assuming that the E and H coefficients are purely real and are orthogonal for simplicity. Thus, we can see that when we have a signal that has multiple frequencies, that the frequency content of the signal will affect its Poynting vector.
    Last edited: May 10, 2010
  12. May 11, 2010 #11
    I don't think that is what the op was asking. You're talking about dependence in the sense that something depends on position or time, only you have transformed it to Fourier space. I am pretty sure the op was talking about a fundamental dependence like energy = h*f. Fields have no fundamental dependence on space and time. Invariance of the field equations under these transformations is what leads to the conservation theorems.
  13. May 11, 2010 #12
    The Poynting vector depends on E and B. If E and B are frequency dependent then the Poynting vector will be as well. One example i can think of where this is true is EM radiation penetrating a metal in which case the strength of the fields in the light is dependent on the frequency of the light and therefore the poynting vector is dependent on the frequency of the light.

    However the E and B fields don't have to be from radiation. there can be a Poynting vector when there are only static fields. in which case the term frequency doesn't have any bearing.
  14. May 12, 2010 #13
    [tex]E=E_0 \sin(\omega t - \phi) [/tex]
    In normal, linearly polarized plane waves E is proportional and perpendicular to B. [tex]\mathbf{B} = \frac{1}{c_0} \hat{\mathbf{k}} \times \mathbf{E}[/tex]

    No unless the number of photons is the same, but usually we are talking about amplitude, and then the power is simply proportional to the amplitude squared.
  15. May 13, 2010 #14
    Well, I'd say it is frequency dependent due to the relation to the radiation pressure. . .

    [tex] P=\frac{ \langle S \rangle }{c} [/tex]

    Pressure is just the momentum flux through a surface, and photons of course carry momentum and energy.

    The energy flux through an element dA from a source with a spectrum [tex] I_{\nu} [/tex] such as a black body, or the sun is

    [tex] F_{\nu} = \int I_{\nu} \cos(\theta) d\Omega [/tex]
    where [tex]\theta[/tex] is the angle of incidence and [tex] p_{\nu}=\frac{E}{c} [/tex]. The momentum transferred per photon of frequency [tex] \nu [/tex] is
    [tex] 2p_{\nu}cos(\theta) [/tex]

    So the radiation pressure is

    [tex] P=\frac{1}{c} \int I_{\nu} cos^2(\theta) d\Omega = \frac{\langle S \rangle }{c} [/tex]
  16. May 13, 2010 #15
    I appreciate everyone's response. It has been interesting to read. Here is a more concrete example that motivated the question in the first place. Suppose there is an inertial frame moving in the positive x-direction toward a stationary monochromatic light source with freq f with respect to a laboratory frame. A stationary light source of the same freq f is located "behind" the moving frame. We know relative to an observer in the moving frame, the light arriving from ahead will be blue-shifted and the light arriving from behind will be red-shifted. But I have also learned, according to Einstein's special theory, that the intensity of the light will be affected too. That is, the intensity of the blue-shifted light will increase and the intensity of the red-shifted light will decrease. Since the Poynting vector is an intensity entity, joules per second per meter squared, then the magnitude of the Poynting vector relative to this moving observer will be greater on the blue-shifted side and less on the red-shifted side. So, it appears that frequency and the Poynting vector(intensity) are linked, at least in this relativistic sense. Am I misunderstanding something here?
  17. May 13, 2010 #16
    That's an interesting point. Looking at the lorentz transformation of the fields boosted along the x direction, where Ey and Bz are the field, and Ex, Ez, By, Bx = 0. (poynting vector points in x direction)

    [tex]E'_y = \gamma (E_y - \beta B_z)[/tex]
    [tex]B'_z = \gamma(B_z - \beta E_y)[/tex]

    [tex]|E' \times B'| = \gamma^2 (E_y B_z - \beta E_y E_y - \beta B_z B_z +\beta^2 B_z E_y) = \gamma^2 ((1 + \beta^2) E_y B_z -\beta (E_y^2 + B_z^2))[/tex]
    Last edited: May 13, 2010
  18. May 13, 2010 #17
    Something must be changing if intensity increases with respect to the moving frame. Going back to the basic definition of intensity, we have energy density times velocity. Relative to the moving frame the intensity does increase with respect to the blue shifted light. Now of course the speed of light is still c in this frame, so something else must account for the increase in the energy density times velocity formula. There must be an increase in the energy density of the light wave with respect to the moving frame per unit of time. Blue shifted light photons in a given volume of space would give a greater energy density than red shifted light photons in the same given volume of space.
  19. May 13, 2010 #18
    I was afraid someone was going to reply before the edits went through. See brainfart amendment.
  20. May 15, 2010 #19
    Very interesting discussion, e2m2a.
    Concerning your original question:...
    and your follow up question....
    Of course, you are missing something and that is the experimental verification.
    The Nichols and Hull Experiment verified the Maxwell (Poynting) formula quantitatively. It is not commonly realized, however, that the experiment was done in three differeent wavelengths and the conclusion was that "the radiation pressure depends only upon the intensity of the radiation and IS INDEPENDENT OF THE WAVE-LENGTH". (Astrophysics Journal, Vol.17,pp. 351).

    The thing you apparently have forgetten in your analysis is that (in wave mechanics) "intensity" is NOT the Poyting Vector....it is the TIME AVERAGED Poynting Vector, <S> , and, as "Born2bwire' mentioned, the TIME AVERAGED Poynting vector formula removes the frequency dependence....meaning that the momentum delivered is frequency independent.
    Written in terms of the electric fiield only, the total Time Averaged Poynting Vector (Intensity) is simply.....
    <S> = (ec/2)(E^2).... (e is the vacuum permittivity.....and E is the MAXIMUM E field)

    Thus the radiation pressure, P = <S>/c , is not predicted by Maxwell (Poynting) to be frequency dependent, (at least not in the same inertial frame), and that is exactly what was empirically verified by Nichols and Hull.

    Now, not knowing the formula you are using for a relativistic "intensity", a comparison seems somewhat ambiguous.

    Having said all that, I can imagine how, in different (boosted) frames, the intensity (even based on time averaging) may change with doppler....(based on the more general Poynting Theorem involving time rate of change of energy density).

    Last edited: May 15, 2010
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