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Time Domain Expression

  1. May 24, 2007 #1
    1. Problem Statement
    Write time-domain expression for electric field e(r,t) of unifrom plane wave with amplitude Eo and frequency f= 1 GHz propagating in free space. The plane wave is propagating in yz plane, 30 degrees from the +y-axis, and 60 degress from -z axis. It is linearly polarized in the x-direction. Find the corresponding h(r,t) field as well.

    2. The attempt at a solution
    If e(r,t) = Re{E(r)exp(iwt)},
    where r = x ax + y ay + z az
    and w = 2 * pi * f
    and k = k [cos(theta)*az + sin(theta)*ay]

    Since this vector in the yz-plane is 30,60,90.
    Shouldn't r be : r = sqrt(3)*ay + az

    Then do dot product of r and k to get the exponential of e(r,t)

    Is this way off base, i think i'm getting confused with k and r, because of the propagation direction versus the polarization direction.

    If someone could help me understand this a bit better, i would greatly appreciate it. Thanks.
  2. jcsd
  3. May 27, 2007 #2


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    Staff Emeritus
    Science Advisor
    Gold Member

    Yeah, vector waves can be a bit tricky at first.

    The direction of the wavevector k is the direction of propagation. This is given to you (in the yz plane at such and such angle...) A unit vector in this direction would be:

    [tex]\frac{\sqrt{3}}{2} \mathbf{a}_y - \frac{1}{2}\mathbf{a}_z[/tex]


    [tex]\mathbf{k} = k(\frac{\sqrt{3}}{2} \mathbf{a}_y - \frac{1}{2}\mathbf{a}_z)[/tex]

    where [itex] k = |\mathbf{k}| [/itex]

    Since you know f, you know [itex] \omega [/itex]. Furthermore, since you know that the wave propagates in free space, you can determine k (the wavenumber) from [itex] \omega [/itex]

    The direction of E is the direction of polarization (i.e. the polarization is just the direction in which the electric field vector points). Note that the magnitude of the electric field vector actually osciallates in time and in space, so that when we say that E has a given polarization direction, what we mean is that it oscillates from a maximum in that direction, down to zero, and then to a maximum in the negative of that direction (i.e. in the opposite direction or antiparallel) and then back again. This is what is meant by linear polarization. In this case, the direction is given (it's the x-direction):

    [tex] \mathbf{E}(\mathbf{r}, t) = \mathbf{a}_x E_0 e^{i \mathbf{k} \cdot \mathbf{r} - \omega t} [/tex]

    One final note: r is a variable: r = xax + yay + zaz in Cartesian coordinates (it's just the position vector). E is a function of r and t, i.e. the magnitude of the electric field vector depends on where and when you look.
    Last edited: May 27, 2007
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