Physical interpretation of j in E&M field equations

  1. Hi all. I've been trying to study microwave and electromagnetic engineering . I'm not sure how I should interpret j in some of the field equations. For example, for the field equations for a rectangular waveguide resonant cavity are:

    E[itex]_{y}[/itex] = E[itex]_{0}[/itex] sin[itex]\frac{\pi x }{a}[/itex] sin [itex]\frac{l \pi z}{a}[/itex]
    H[itex]_{x}[/itex] = [itex]\frac{-j E_{0}}{Z_{TE}}[/itex] sin[itex]\frac{\pi x}{a}[/itex] cos [itex]\frac{l \pi z}{d}[/itex]
    H[itex]_{z}[/itex] = [itex]\frac{j \pi E_{0}}{k \eta a}[/itex] cos[itex]\frac{\pi x}{a}[/itex] sin [itex]\frac{l \pi z}{d}[/itex]

    What is the physical interpretation of j and -j in the H field in the x z direction? Does that indication that they are 90° out of phase of the E field? Does it indicate phase in the sense of time or space? Or should i think of them as derivatives of phasors? I know that the fields are derived from the more general phasor form of Maxwell's equations:
    ∇ × E = - jωμH
    ∇ × H = jωεE
    for which jω = [itex]\frac{\partial E}{ \partial t }[/itex] and E is E[itex]_{0}[/itex] e[itex]^{j \omega t}[/itex]
    which makes sense to me as I believe you can interpret jω as the sinusoidal frequency. But once the E and H fields have been derived as above it's no longer jω just j, so I've lost the sense of there meaning in the H fields. Could someone please explain how I should interpret them. Or anything else I seem to have screwed up in my thinking. Thanks a lot.
     
  2. jcsd
  3. dlgoff

    dlgoff 3,149
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    2014 Award

    [​IMG]

    I got this from http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html



    Regards
     
  4. Thanks, but I don't think that's correct in this instance. In my book current and current density are usually denoted at J0 or I0. I'm fairly certain that in this instance j is the same as the imaginary unit, i, indicating that it's coefficient is the imaginary part of a complex number. Also these field equations were originally derived from;
    ∇×E = -jωμH
    ∇×H = jωεE
    so everything should be in terms of the E and H fields.
     
  5. dlgoff

    dlgoff 3,149
    Science Advisor
    Gold Member
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    Okay. j is imaginary. Let's try again.

    [​IMG]

    http://www-ece.rice.edu/~daniel/262/pdf/lecture02.pdf
     
  6. In complex number [itex] j=e^{j\frac{\pi}{2}}=1∠90^o\;\hbox { and } \; -j=e^{-j\frac{\pi}{2}}=1∠-90^o[/itex]

    But I think you should write the formulas in vector form, I am only guessing the direction of the vectors by the subscripts like Ex is in x direction. What you gave are only the scalar value of the vectors.

    Regarding to the disappearance of the jωt, your equation is in phasor form where

    [tex]\vec E= Re[\tilde E e^{j\omega t}] \;\hbox{ where } \;\tilde E=\hat y E_0e^{-j\beta R}[/tex]

    BTW,

    [tex]\frac {\partial E}{\partial t}= j\omega E_0 e^{j\omega t} \;\hbox { not }\; j\omega[/tex]

    Where your Eo has to contain the phasor function.
     
    Last edited: Nov 12, 2012
  7. Thanks guys.
    OK Yungman, so you think the j represents a 90° phase shift in the orientation of the coordinates of the H fields. That's kind of something I was thinking, but I can't figure out how a 90° shift like j puts you on the x axis, while a -90° shift from -j puts you on the z axis. Also i was thinking because the field were derived from;
    ∇×E = -jωμH
    ∇×H = jωεE
    that they must be related to the frequency jω, although ω does not seem to be factored into the fields in this case.
    Looking at what dlgoff posted seems kind of close, but that would seem to indicate that the H field are purely imaginary, and can not be measured. I don't think this is the case though.

    I could post the full derivation from the textbook I'm referencing if that would help??
     
  8. I change my original post, I don't think the j has anything to do with the direction of H. If you look at the phasor equation, [itex]j=e^{j\frac{\pi}{2}}[/itex] is a phase shift between the E and H ALONG the direction of propagation. This has nothing to do with the direction of E in y, H in x and z.

    Remember if

    [tex] E= jE_0 e^{-j\beta R}= E_0e^{-j(\beta R - \frac {\pi}{2})}[/tex]
    [tex] E= -jE_0 e^{-j\beta R} = E_0e^{-j(\beta R + \frac {\pi}{2})}[/tex]

    Where R is at the direction of propagation. This only mean the peak of the E and H is off by 90 degree along the direction of propagation, NOT the angle between E and H.

    Edited: equations has been modified.
     
    Last edited: Nov 12, 2012
  9. I draw out a TEM wave propagates in z direction. j is just a simple phase shift of the H. It has nothing to do with H in y direction and E and H are in perpendicular direction to each other.
     

    Attached Files:

  10. Yeah, that interpretation seems to make more sense. What is also confusing to me though is I thought that in general the E and H fields were in phase unless in a medium with a complex impedance. These seem to show that in a vacuum that the E and H fields are 90° out of phase. I'll have to dig further in the book and do a little more research to see if i can figure out what's going on. Thanks a lot for the hard work on helping me with this!
     
  11. Only when TEM wave travel in lossless medium, both are in phase. But yours is in cavity. I just go by straight interpretation of your equations.

    In fact, you should leave the equation in exponential form to take into account of the j before translating to sin and cos function as j represents a phase shift of ∏/2 and is together with the βR. so you should get something like [itex]sin(\frac{πR}{a}+π/2)[/itex]

    From your equation:

    E[itex]_{y}[/itex] = E[itex]_{0}[/itex] sin[itex]\frac{\pi x }{a}[/itex] sin [itex]\frac{l \pi z}{a}[/itex]
    H[itex]_{x}[/itex] = [itex]\frac{-j E_{0}}{Z_{TE}}[/itex] sin[itex]\frac{\pi x}{a}[/itex] cos [itex]\frac{l \pi z}{d}[/itex]
    H[itex]_{z}[/itex] = [itex]\frac{j \pi E_{0}}{k \eta a}[/itex] cos[itex]\frac{\pi x}{a}[/itex] sin [itex]\frac{l \pi z}{d}[/itex]

    The direction of propagation is in xz plane as your propagation constant has [itex] \sin\frac{\pi x }{a} \sin \frac{l \pi z}{a}[/itex]. Obviously it is TE wave as E is normal to the direction of propagation and not H.
     
    Last edited: Nov 12, 2012
  12. Yes. The equations are for fields in a cavity but do not specifiy if they are in a lossless medium or not. That would depend on β and Z0 I think. Also these are TE waves, not TE waves.
     
  13. I just add the last part of into the last post. Yes, it is a TE wave from your formulas as the direction of propagation is in xz plane. As I said, I don't know TE wave, I have no comment on this. But medium being lossless is only one part of the big picture, there might be other factor that add a phase shift between the two waves. Case in point, yours is TE in cavity.
     
    Last edited: Nov 12, 2012
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