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Maxwell's equation in reciprocal space

  1. Jan 7, 2015 #1

    naima

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    I found this http://people.sissa.it/~benassi/capitolo1/node2.html [Broken]
    in the vacuum the equations would be
    q B = q E = 0
    ##q \times E = (\omega / c) B##
    ##q \times B = -(\omega / c) E##

    Is there a typo? there is no t derivative.
    If E = 0 B would have to be null.
    Has B to be allways orthogonal to E (elliptic polarisation?)

    thanks.
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Jan 7, 2015 #2
    It's just Fourier analysis: By writing ##\mathbf{E}(\mathbf{x},t)=\intop_{\mathbb{R}}\intop_{\mathbb{R}^3}\mathrm{d}^3\mathbf{q}\mathrm{d}\omega\mathbf{E}(\mathbf{q},\omega)e^{i\mathbf{q}\cdot\mathbf{x}-it\omega}## and similarly for the magnetic field, one has, for example, MIII as (with usual assumptions re: sufficient well-behavedness of the physical fields)

    ##0=\nabla\times\mathbf{E}(\mathbf{x},t)+\frac{\partial}{\partial\,t}\mathbf{B}(\mathbf{x},t)=\intop_{\mathbb{R}}\intop_{\mathbb{R}^3}\mathrm{d}^3\mathbf{q}\mathrm{d}\omega(i\mathbf{q}\times\mathbf{E}(\mathbf{q},\omega)-i\omega\mathbf{B}(\mathbf{q},\omega))e^{i\mathbf{q}\cdot\mathbf{x}-it\omega}##

    and for this to hold for arbitrary momentum-space fields one has (technically, almost everywhere) one of the equations you quote, ##\mathbf{q}\times\mathbf{E}=\omega\mathbf{B}##. The rest are obtained in similar way.

    Alternatively, these can be obtained by considering a plane wave ansatz ##\mathbf{E}=\mathbf{E}_0e^{i\mathbf{q}\cdot\mathbf{x}-it\omega}## – it's essentially the same thing, but slightly more informal.

    The time derivative disappears because in what the site calls "reciprocal space" one considers a frequency variable ##\omega## instead of a time variable ##t##. And yes, in this form in vacuum the equations imply that ##\mathbf{q},\mathbf{E},\mathbf{B}## are all mutually perpendicular.​
     
    Last edited: Jan 7, 2015
  4. Jan 7, 2015 #3

    naima

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    With linear polarisation when E is null B is maximal.
    How can we hav ##q \times E = \omega B##? B would not be maximum but null too?
     
  5. Jan 7, 2015 #4
    I'm honestly not sure what you mean by that? In the standard simple linearly polarised plane wave, both electric and magnetic fields reach their minima (and maxima) simultaneously.

    EDIT: Are you talking about something like this? Note that the amplitudes of both electric and magnetic fields become zero at the same point, as per MIII.
     
  6. Jan 7, 2015 #5

    vanhees71

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    These are the fields in Fourier space as explained in #2. For a plane wave you have ##\vec{E}(\omega,\vec{q})=\text{const}## and ##\vec{B}(\omega,\vec{q})=\text{const}##. The free Maxwell equations in #1 are correct, i.e., we have
    $$\vec{q} \cdot \vec{B}=\vec{q} \cdot \vec{E}=0, \quad \vec{q} \times \vec{E}=\omega/c \vec{B}, \quad \vec{q} \times \vec{B}=-\omega/c \vec{E}.$$
    Solutions to the 3rd and 4th equation automatically satisfy also the 1st and 2nd. So we need to deal only with the latter two equations.

    Now take the 3rd equation and vector-multiply it with ##\vec{q}##. This gives
    $$\vec{q} \times (\vec{q} \times \vec{E})=-\omega^2/c^2 \vec{E}.$$
    The left-hand side can be expanded to
    $$\vec{q} \times (\vec{q} \times \vec{E})=\vec{q} (\vec{q} \cdot \vec{E})-\vec{q}^2 \vec{E}=-\vec{q}^2 \vec{E}.$$
    Thus, if you don't want the trivial solution ##\vec{E}=0##, you necessarily must have
    $$\omega^2=c \vec{q}^2,$$
    which is the dispersion relation for free electromagnetic waves.

    Now you can choose ##\vec{E}## as an arbitrary (generally complex) vector. Then from the 3rd equation it follows
    $$\vec{B}=\frac{c}{\omega} \vec{q} \times \vec{E}$$
    and
    $$\vec{E} \times \vec{B}=\frac{c}{\omega} \vec{E} \times (\vec{q} \times \vec{E})=\frac{1}{|\vec{q}|} \vec{q} \vec{E}^2.$$
    This tells you that ##\vec{E}##, ##\vec{B}##, and ##\vec{q}## build a orthogonal set of vectors (if ##\vec{E} \in \mathbb{R}^3##, i.e., for a linearly polarized wave, it's a right-handed basis).
     
  7. Jan 7, 2015 #6

    naima

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    Thank you.

    I had a misconception in mind. As the hamiltonian is a quadratic function of B and H, i thought that when E = 0, B had to be maximal for energy conservation. Why is it false?
     
  8. Jan 7, 2015 #7

    naima

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    Thank you Vanhees

    I see that you find in the energy momentum reciprocal space solutions with E and B orthonormal.
    You will have to sum them to get E and B in the real space time. E1.B1 = 0 and E2.B2 = 0 do not give (E1 + E2). (B1+B2) = 0.
    Maxwell equations in the vacuum have solutions With E and B not orthogonal.
    Take two constant vectors in space time. their curl divergence and time derivative are null.
    They obey to these equations. (we also have E(z=0,t=0) = E(ct,t)!
    So what have we to add to prove that E and B are orthogonal?
    Have you a textbook with these details?
    I read in Ballentines the paragraph on normal modes but did not find what i am looking for.
     
  9. Jan 8, 2015 #8

    vanhees71

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    But you got it right. Why do you think that the electric and magnetic components should be orthogonal? There are two Lorentz invariants you can built from the field-strength tensor, namely ##F_{\mu \nu} F^{\mu \nu}## and ##^{\dagger}F_{\mu \nu} F^{\mu \nu}##. In 3D notation these are ##\vec{E}^2-\vec{B}^2## and ##\vec{E} \cdot \vec{B}##. Why should either of them necessarily vanish?

    What I derived is a momentum-polarization eigenbasis of the electromagnetic field. You can build each field as "superposition", i.e., a Fourier transform wrt. ##\vec{q}## and a sum over the two polarization states (e.g., the helicity of the modes, i.e., in left- and right-handed circular polarization, which is most natural from the point of view of the representations of the Poincare group).
     
  10. Jan 8, 2015 #9

    naima

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    I thought you were talking about orthogonality in real space.

    This thread appears as with no answer on my PC on the middle of the screen. Do you understand why?
     
  11. Jan 8, 2015 #10

    vanhees71

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    Then I don't understand your question. I've given you the general solution of the Maxwell equations in energy-momentum representation. So in principle all your questions should be answered. So, what's still not answered?
     
  12. Jan 10, 2015 #11

    naima

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    I think that i found what bothered me.
    We saw that Maxwell's equations with no source terms have solutions with ##E_{Tr}## and B not orthogonal.
    But in the real vacuum they are so we have to add something.

    When we are very far from the charges which created the fields, we have plane waves.
    And plane waves are such that ##E_{Tr}## and B are orthonormal. So we have to study only plane waves when we consider physical c propagating waves in the vacuum.
    I do not know if they have to be monochromatic.
    Please tell me if this is correct.
     
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