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Phase relationship is electromagnetic wave.

  1. Aug 31, 2006 #1
    What is the phase relationship between the magnetic component and the electric component of an electromagnetic field. Is it in phase, or 90 degrees out of phase. I see it both ways in a Google search ???
     
  2. jcsd
  3. Aug 31, 2006 #2

    quasar987

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    The electric and magnetic field oscillate in phase (i.e. when E is at its maximum, so is B, etc.), but their direction of oscillation is perpendicular.
     
  4. Aug 31, 2006 #3
    Thanks; that's what I always thought; but I'm finding lots of resistance to that notion in another forum.
     
  5. Aug 31, 2006 #4

    rbj

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    we probably should qualify that a little. for sure, in a plane wave, the E field and B field are in-phase. for a spherical expanding wavefront (the EM source is a point), at a distance that is many wavelengths away from the center point, the spherical wavefront is virtually identical to a plane wave so the E and B fields are in-phase. however much closer to the source, the E and B fields are not in-phase and, in the limit as you close in on the source, the E and B fields are 90 degrees out of phase.

    the same is true for the instantaneous pressure (difference from atmospheric) and particle velocity of a spherically expanding sound wave.
     
  6. Aug 31, 2006 #5

    Claude Bile

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    Ask them to prove using Maxwell's equations why E ought to be a maximum when dB/dt is a maximum. They will quickly see the error in their ways.

    Further to rbj's post I would also like to point out that waves in media (including hollow waveguides) and resonators also do not have their E and B fields in phase.

    Claude.
     
    Last edited: Aug 31, 2006
  7. Aug 31, 2006 #6

    quasar987

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    I don't know about spherical waves, but for monochromatic plane waves in a media, the equations

    [tex] \vec{E}\times \vec{k} = \vec{B}[/tex]

    [tex] \vec{B}\times \vec{k} = \vec{E}[/tex]

    still hold. Besides pointing to the perpendicularity of E,B and k, don't they also indicate that E and B are in phase?
     
  8. Sep 1, 2006 #7
    Thanks guys; a physicist finally showed the solutions to Maxwell's equations that required an in-phase relationship and the oponent conceeded. We were having trouble confusing a radiating EM wave with an LC circuit.

    This is the thread

    Vern
     
    Last edited: Sep 1, 2006
  9. Sep 1, 2006 #8

    quasar987

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    Really Claude, you're worrying me. Did I read you right when you said that in matter, E and B (D and H) are not in phase?
     
  10. Sep 3, 2006 #9
    Well maximum voltage occurs when the conductor is perpendicular to the field
     
  11. Sep 3, 2006 #10

    Claude Bile

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    Sorry, that was worded badly. I should have said not necessarily in phase. Materials with an imaginary component in their refractive index (typically conductors) will have E and B not in phase.

    If you have Griffith's intro to EM, in section 9.4 (of my 3rd edition), it walks through this case in detail.

    Apologies for the confusion (and the relatively late reply - I don't usually visit PF on weekends).

    Claude.
     
  12. Sep 4, 2006 #11

    quasar987

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    I appreciate the reply!
     
  13. Sep 16, 2010 #12
    Electric and magnetic field vectors are 90 degrees out of phase in electromagnetic wave propagation. Many text books and resources fail to make note of this. Worse even, some explanations for polarization confuse a second electric wave with the magnetic component of the first wave. Both feilds are able to propagate through space without a physical medium by cyclically inducing eachother. The magnitude of one vector results from the differential of the other, meaning one will be changing fastest as it aproaches zero magnitude, while the other aproaches maximum or minimum, hence, 90 degrees out of phase. By this, the vector sum of both components is constant, allowing a photon to consist of a single quantum of energy at all points in its propagation. If the fields were in phase, both would reach zero magnitude at the same point. At that point, the photon would consist of no energy at all and thus there would be no propagation. Some optical media lag one field from another to produce a phase shift. In these cases, the wave corrects it's phase back to 90 degrees by rotating it's axis of polarization, but things get more complicated at that level.
     
  14. Sep 16, 2010 #13

    Dale

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    Hi Psyrick, welcome to PF!
    This is not correct, the fields are in-phase. This is a fairly common question, here is my explanation a couple of days ago.
     
  15. Sep 20, 2010 #14
    If you look at Faraday's law of induction, you will find it supports my argument. Furthermore, if you are to dismantle my argument, you will need to explain how a photon can hold a uniform quantity of energy with ever-changing charge and both fields reaching zero magnitude at the same points.
     
  16. Sep 20, 2010 #15

    Dale

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    No, it doesn't. See my explanation above.

    Would you care to state Faraday's law of induction and show how it supports your position?
     
    Last edited: Sep 20, 2010
  17. Sep 20, 2010 #16
    Indeed, I can.


    This helps to describe a phase relationship where one field induces the other as it changes. A 90 degree phase relationship explains for how both fields can pass their energy back and forth between one another as they propagate through space. One field holds all the energy while the other aproaches zero and then cyclic repetition of induction as the field magnitudes continuously change while the total energy is conserved.
    If both fields fluctuate in matched phase, where then does the energy stored within the fields go to and come from as the field magnitudes change?


    References:

    Wilson, Jerry D., Anthony J. Buffa, and Bo Lou. College Physics. 6th ed. Upper Saddle River, NJ 07458, USA: PEARSON Prentice Hall, 2007. 657-659. Print.

    Ulaby, Fawwaz T. Fundamentals of Applied Electromagnetics. 5th ed. Upper Saddle River, NJ 07458, USA: PEARSON Prentice Hall, 2007. 255. Print.
     
    Last edited: Sep 20, 2010
  18. Sep 20, 2010 #17

    Dale

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    Tell me Psyrick, for a EM plane wave propagating in free space what is the number of turns in the loop, N? This expression is obviously not the general one (it is for EMF in a loop of circuit), and it is not applicable for a wave propagating in free space.

    The correct general expression is:
    [tex]\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}} {\partial t}[/tex]

    So a time-varying B field induces a spatially varying E field. For a sinusoidal plane wave the spatial variation is highest at the zero crossing and the time variation is also highest at the zero crossing. Therefore, they are in phase.

    The energy goes in the direction of the Poynting vector. Remember, the fields are varying in space and time, not just time.
     
    Last edited: Sep 20, 2010
  19. Sep 20, 2010 #18
    I have considered free space to act as a conductor in the plane as it also exibits a sub-infinite impedance.

    Can you describe the phase relationship considering time as a variable of constant rate? I would like to hear more explanation about the Poynting vector. Mathematically, it is simple to understand there being a cross product of two vectors, and how it results in propagation on a third axis. But do you mean to say that this vector carries the energy of both fields at the zero-cross and delivers it from one point in space to another? If so, what form of energy does it exist in at that point? How can an energy differential be energy at a point where it does not have any electrical or magnetic potential?
     
  20. Sep 21, 2010 #19

    Dale

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    Sure. Let's say that we are working with a monochromatic plane wave propagating in the z direction, and let's say that the phase between the E and B field is unknown and see if we can solve for it.

    So in units where c=1 we have:
    [tex]E=\left(\sin (k z-t \omega ),0,0\right)[/tex]
    [tex]B=\left(0, \sin (k z-t \omega+\phi ),0\right)[/tex]

    Now, let us plug these into Faraday's law:
    [tex]\nabla \times \mathbf{E} = \left( 0,k \cos (k z-t \omega ),0\right)[/tex]
    [tex]-\frac{\partial \mathbf{B}} {\partial t} = \left( 0,\omega \cos (k z-t \omega +\phi),0\right)[/tex]

    These two expressions are only equal for:
    [tex]\frac{\omega}{k}=c=1[/tex]
    and
    [tex]\phi=0[/tex]
     
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