1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Complex orthogonality of electric and magnetic fields

  1. Jan 9, 2012 #1

    I have a hard time finding a justification that electric and magnetic fields are still orthogonal when presented in complex form. As far as I know the notion of orthogonality for complex vectors is not as intuitive as the one for real vectors. Notably, [itex]\vec{x}\cdot\vec{y}=0[/itex] does not imply that [itex]\Re(\vec{x})\cdot\Re(\vec{y})=0[/itex] (the former dot product is a complex one, the latter is a real one).

    Similarly, does the cross product relation between the electric field, the magnetic field, and the wave vector still hold?


    PS: I'm a novice in electromagnetism and a novice on this forum, please tell me if I did not respect any rule that I would be unaware of.
  2. jcsd
  3. Jan 9, 2012 #2
    Am I right by observing that all complex vectors that we meet in electromagnetism can be written as [itex]z\vec{x}[/itex] where [itex]z\in ℂ[/itex] but [itex]\vec{x}\in ℝ^{n}[/itex]?

    In this case the complex cross and dot products can be easily reduced to their real counterparts:
    [itex]z_1\vec{x_1}\odot z_2\vec{x_2}=(z_1z_2)(\vec{x_1}\cdot\vec{x_2})[/itex]
    [itex]z_1\vec{x_1}\otimes z_2\vec{x_2}=(z_1z_2)(\vec{x_1}\times\vec{x_2})[/itex]
    where [itex]\cdot[/itex] and [itex]\times[/itex] (resp. [itex]\odot[/itex] and [itex]\otimes[/itex]) are the real (resp. complex) dot and cross products.
    At least this somewhat allows to retrieve the geometrical interpretation... ?

  4. Jan 10, 2012 #3
    The orthogonality holds. The dot operation takes place on the real basis vectors. The complex coefficients just go along for the ride.
  5. Jan 10, 2012 #4


    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    This problem occurs usually when treating electromagnetic waves in terms of plane-wave modes. Then it is in most cases easier to simply write

    [tex]\vec{E}=\vec{E}_0 \exp[-\mathrm{i} (\omega_{\vec{k}} t-\vec{k} \cdot \vec{x})][/tex]

    with [itex]\vec{E}_0 \in \mathbb{C}^3[/itex], [itex]\omega_{\vec{k}}=c |\vec{k}|[/itex], instead of taking the real part, which is of course the physical field meant to be described here. An analogous formula also holds for the magnetic components of the field or for the four-potential of the field.

    Of course, this works only as long as you perform only linear operations like the standard differentialoperators, div and curl, or look at superpositions of fields, or cross and dot products with real vectors (like [itex]\vec{k} \cdot \vec{E}=0[/itex] for a free field in the vacuum).

    As soon as you calculate something beyond linear operations or multiplications with real scalars or vectors like the energy density [itex]\mathcal{E}=(\vec{E}^2+\vec{B}^2)/[/itex] (in the vacuum with Heaviside-Lorentz units) or the Poynting vector [itex]\vec{S}=\vec{E} \times \vec{B}[/itex], you have to take the real parts of the fields first and do then the products.
  6. Jan 10, 2012 #5
    Do you mean that we are indeed in the situation that I described in my second post?

    So you mean that [itex]\vec E\cdot\vec B=0[/itex] holds only for the real parts of both vectors? It seems contradictory to what Antiphon just wrote, isn't it?
  7. Jan 10, 2012 #6
    I think I've seen in some contexts some complex wavevectors [itex]\vec k[/itex]. Then, from what you say, it means that [itex]\vec{k} \cdot \vec{E}=0[/itex] does not hold in such a context?
  8. Jan 10, 2012 #7


    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    No, that's not what I said. Complex wave numbers appear, e.g., in the solution of the Maxwell equations in wave guides. The imaginary part in the wave number describes just the damping of the wave.
  9. Jan 10, 2012 #8
    But you wrote that the operation was valid only if one complex vector only was involved, right?

    Ok, so what does this tell me about the dot product?
  10. Jan 10, 2012 #9
    Actually the Poynting expression you have for S remains in the complex domain. The real part of S is energy that traverses the surface. The imaginary part of S represents reactive power (power which crosses back and forth across the surface on each cycle.)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook