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Counting electromagnetic modes in a rectangular cavity and boundary conditions

  1. Jul 23, 2012 #1
    The electric field in a cubical cavity of side length L with perfectly conducting walls

    E_x = E_1 cos(n_1 x \pi/L) sin(n_2 y \pi/L) sin(n_3 z \pi/L) sin(\omega t)
    E_y = E_2 sin(n_1 x \pi/L) cos(n_2 y \pi/L) sin(n_3 z \pi/L) sin(\omega t)
    E_z = E_3 sin(n_1 x \pi/L) sin(n_2 y \pi/L) cos(n_3 z \pi/L) sin(\omega t)

    with E_1 n_1 + E_2 n_2 + E_3 n_3 = 0.

    In counting the number of modes, the counting is restricted to non-negative
    values of n_1, n_2 and n_3. Is there a simple way to show that

    a) any mode in which one or more of the n_1, n_2 and n_3 are negative, can
    be written as a linear combination of the modes that are included in the
    counting and

    b) the modes that are included in the counting are all independent?

    Also, is the perpendicular component of the magnetic field on the surface
    of a plane conductor required to be zero? The vanishing electric field in the
    conductor only implies that the the time derivative of the perpendicular
    component of the magnetic field vanishes.

  2. jcsd
  3. Jul 24, 2012 #2


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    Staff: Mentor

    If n1<0, a transformation
    n1 -> -n1
    E2 -> -E2
    E3 -> -E3
    gives the same fields, as you swap two signs in the second and third equation (and none in the first). In the 4th equation, all signs are swapped, therefore the relation holds.
    In a similar way, you can restrict n2 and n3 to be positive.

    Consider the partial derivatives at the origin (x=y=z=0): [itex]\del_x E_x = E_1 n_1 \frac{\pi}{L}sin(\omega t).
    Each n1 gives a unique value here. The same can be done with the other components.

    I would expect that this is not required, but I do not know it. As you have a time-dependent electric field, Maxwell's equations should give a field somewhere. If the component orthogonal to the surface is 0, it has to be parallel to the conductor surface.
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