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Coplanar Electromagnetic Fields

  1. Sep 27, 2013 #1
    1. The problem statement, all variables and given/known data
    The permittivity of a given medium is given by the equation

    [tex]ε =
    \begin{pmatrix}
    ε_1 & 0 & 0\\
    0 & ε_1 & 0\\
    0 & 0 & ε_2
    \end{pmatrix}
    [/tex]

    A wave is traveling in the [itex]\hat{k}[/itex] direction in this medium where the unit vector [itex]\hat{k}[/itex] is defined as [itex]\hat{k} = \hat{x}sin(θ)+\hat{z}cos(θ)[/itex]. All fields are proportional to [itex]e^{-i*(\overline{k}\bullet\overline{r})}[/itex] and [itex]\overline{E} = \overline{E_0}e^{-i*(\overline{k}\bullet\overline{r})}[/itex] where [itex]\overline{E_0}[/itex] is a constant vector.

    Show that:
    a.[itex]\overline{E_0}, \overline{D_0}[/itex] and [itex]\hat{k}[/itex] are coplanar.
    b.[itex]\overline{H_0}[/itex] is perpendicular to this plane.
    c.[itex]\overline{D_0}[/itex] is perpendicular to [itex]\hat{k}[/itex].

    2. Relevant equations
    Determinant of 3x3 matrix
    Maxwell–Faraday equation
    Cross Product

    3. The attempt at a solution
    I'm seeing this as more of a math problem than a physics problem.

    a.) This is what I am thinking. If I can show that [itex]\overline{E_0}, \overline{D_0}[/itex] and [itex]\hat{k}[/itex] vectors are linearly dependent then this proves that they are coplanar. I can do this by setting up a 3x3 matrix where each column corresponds to one of these vectors and then take the determinant of said matrix. If that determinant is equal to zero then it proves that they are coplanar. In this case, I need to know all of the vectors.

    [itex]\hat{k} = \hat{x}sin(θ)+\hat{z}cos(θ)[/itex]

    [itex] \overline{E_0} = E_x\hat{x} + E_y\hat{y} + E_z\hat{z}[/itex]

    [itex]\overline{D} = ε\overline{E} = (ε_1E_x\hat{x} + ε_1E_y\hat{y} + ε_2E_z\hat{z})e^{-i*(\overline{k}\bullet\overline{r})} = \overline{D_0}e^{-i*(\overline{k}\bullet\overline{r})}[/itex]

    so

    [itex]\overline{D_0} = ε_1E_x\hat{x} + ε_1E_y\hat{y} + ε_2E_z\hat{z}[/itex]

    Now we set up the equation and take the determinant like so
    [tex]det
    \begin{pmatrix}
    E_x & ε_1E_x & sinθ\\
    E_y & ε_1E_y & 0\\
    E_z & ε_2E_z & cosθ
    \end{pmatrix}
    [/tex]

    [itex]= E_x(ε_1E_ycosθ)-ε_1E_x(E_ycosθ)+sinθ(E_yε_2E_z-E_zε_1E_y)[/itex]
    [itex]=E_yE_zsinθ(ε_2-ε_1)[/itex]

    This last line here should equal zero to show that the these three vectors are coplanar. This is also where I run into a problem. If the vectors are coplanar then at least one of the variables above(excluding epsilons) must be equal to zero. Now I could just assert that E_y or E_z is equal to zero, its not like they can't be but beside asserting it, I don't know how to show that one of these variables is zero. I'm stuck here. There is obviously some piece of information I am lacking.

    b.) For this part, I was thinking of using the cross product to figure out if vectors are perpendicular. The magnitude of the cross product is as follows:

    [itex]|a \times b| = |a||b|sinθ[/itex]

    In the case that vectors are perpendicular to each other, θ = 90 deg so

    [itex]|a \times b| = |a||b|[/itex]

    If I take the cross product of [itex]\overline{H_0}[/itex] and [itex]\overline{D_0}[/itex] and compute its magnitude, and then compute the magnitude of [itex]\overline{H_0}[/itex] and [itex]\overline{D_0}[/itex] separately and multiply, they should be equal. In other words:

    [itex]|\overline{H_0} \times \overline{D_0}| = |\overline{H_0}||\overline{D_0}|[/itex]

    To get [itex]\overline{H}[/itex] and ultimately [itex]\overline{H_0}[/itex], I think I just need to use a form of the Maxwell–Faraday equation and crunch the variables.

    c.) This is a repeat of part b but using vectors [itex]\overline{D_0}[/itex] and [itex]\hat{k}[/itex]

    My biggest concern is in part A. I feel that there is a piece of info that I am missing and it is impeding me from completing the other parts.

    Thanks for the help!
     
  2. jcsd
  3. Sep 28, 2013 #2

    TSny

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    Use the Faraday-Maxwell equation to see how the orientations of ##\hat{k}, \overline{E}_0##, and ##\overline{H}_0## are related.
     
  4. Sep 28, 2013 #3
    I'm not exactly sure what you mean. I know that the Maxwell–Faraday is the following:

    [itex]\nabla \times \overline{E} = -iω\overline{B}[/itex]

    [itex]\nabla \times \overline{E} = -iωμ\overline{H}[/itex]

    This can solve for [itex]\overline{H}[/itex]

    [itex]\frac{\nabla \times \overline{E}}{-iωμ} = \overline{H}[/itex]

    Since [itex]\overline{E_0} = \hat{x}E_x+\hat{y}E_y+\hat{z}E_z[/itex]
    and [itex]\overline{k} = k_0\hat{k} = k_0(\hat{x}sin(θ)+\hat{z}cos(θ))[/itex]
    and [itex]\overline{r} = \hat{x}x+\hat{y}y+\hat{z}z[/itex]
    and [itex]\overline{k}\bullet \overline{r} = \hat{x}xsin(θ)+\hat{z}zcos(θ)[/itex]
    then the Maxwell–Faraday equation should pop out an H field that looks like

    [itex]\overline{H} = [\hat{-x}k_0\frac{E_ycos(θ)}{ωμ} +\hat{y}k_0\frac{E_xcos(θ)-E_zsin(θ)}{ωμ}+\hat{z}k_0\frac{E_ysin(θ)}{ωμ}]e^{-i(\overline{k}\bullet \overline{r})}[/itex]

    where
    [itex]\overline{H_0} = \hat{-x}k_0\frac{E_ycos(θ)}{ωμ} +\hat{y}k_0\frac{E_xcos(θ)-E_zsin(θ)}{ωμ}+\hat{z}k_0\frac{E_ysin(θ)}{ωμ}[/itex]

    After all that, I don't see how that helps me with part A although I guess that helps me for part B.
     
  5. Sep 28, 2013 #4

    TSny

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    For any vector function ##\vec{V}(\vec{r})## of the form ##\vec{V}_0e^{-i \vec{k} \cdot \vec{r} }## where ##\vec{V}_0## is a constant vector, see if you can show that ##\vec{\nabla} \times \vec{V}(\vec{r}) = \vec{k} \times \vec{V}_0 \;(- i e^{-i \vec{k} \cdot \vec{r} })##
     
  6. Sep 28, 2013 #5
    Ok. Doing the left side,

    [itex]\nabla \times \overline{V} = \begin{pmatrix}
    \hat{x} & \hat{y} & \hat{z}\\
    ∂_x & ∂_y & ∂_z\\
    V_xe^{-i(\overline{k}\bullet\overline{r})} &V_ye^{-i(\overline{k}\bullet\overline{r})} &V_ze^{-i(\overline{k}\bullet\overline{r})}
    \end{pmatrix} [/itex]

    [itex] = \hat{x}(∂_yV_ze^{-i(\overline{k}\bullet\overline{r})}-∂_zV_ye^{-i(\overline{k}\bullet\overline{r})}) - \hat{y}(∂_xV_ze^{-i(\overline{k}\bullet\overline{r})}-∂_zV_xe^{-i(\overline{k}\bullet\overline{r})}) + \hat{z}(∂_xV_ye^{-i(\overline{k}\bullet\overline{r})}-∂_yV_xe^{-i(\overline{k}\bullet\overline{r})})[/itex]

    [itex]=(-ie^{-i(\overline{k}\bullet\overline{r})})[\hat{x}(yV_ze^{-i(\overline{k}\bullet\overline{r})}-zV_ye^{-i(\overline{k}\bullet\overline{r})}) - \hat{y}(xV_ze^{-i(\overline{k}\bullet\overline{r})}-zV_xe^{-i(\overline{k}\bullet\overline{r})}) + \hat{z}(xV_ye^{-i(\overline{k}\bullet\overline{r})}-yV_xe^{-i(\overline{k}\bullet\overline{r})})][/itex]

    Now going to the right side

    [itex]\overline{k}\times \overline{V_0} = \begin{pmatrix}
    \hat{x} & \hat{y} & \hat{z}\\
    k_x & k_y & k_z\\
    V_xe^{-i(\overline{k}\bullet\overline{r})} &V_ye^{-i(\overline{k}\bullet\overline{r})} &V_ze^{-i(\overline{k}\bullet\overline{r})}
    \end{pmatrix}[/itex]

    [itex]= \hat{x}(k_yV_z-k_zV_y) - \hat{y}(k_xV_z-k_zV_x) + \hat{z}(k_xV_y-k_yV_x)[/itex]

    which means k_x = x and k_y = y and k_z = z. I wouldn't have to have said that if I generalized 'k' and 'r' but I think this shows, more or less, that the two equations are equal.

    Are you trying to say to use this formula inside of the Maxwell-Faraday law like so?

    [itex]\nabla \times \overline{E} = iω\overline{B}[/itex]

    [itex](\overline{k} \times \overline{E_0})(-i*e^{-i(\overline{k}\bullet \overline{r})}) = iωμ\overline{H}[/itex]

    [itex](\overline{k} \times \overline{E_0}) \frac{(-e^{-i(\overline{k}\bullet \overline{r})})}{ωμ} = \overline{H}[/itex]

    Since I believe [itex]\overline{H} = \overline{H_0} e^{-i(\overline{k}\bullet \overline{r})}[/itex] the last equation in the line above can be further simplified.

    [itex] \frac{-(\overline{k} \times \overline{E_0})}{ωμ} = \overline{H_0}[/itex]

    Doesn't this final equation show that H_0 is perpendicular to the vector k and vector E since that is the definition of the cross product? I guess that solves part B. That equation you mentioned was very useful TSny. What exactly clued you into getting that equation? It isn't obvious to me. There may be similar equations which help out with parts A and C .
     
  7. Sep 28, 2013 #6

    TSny

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    Yes. Now, there is another Maxwell equation that involves the curl of H and the time derivative of D. It will tell you some information about the orientation of D.

    Well, I first learned it from a textbook and you'll find it in many E&M texts. It's used quite a lot. Another nice relation is ##\vec{\nabla} \cdot \vec{V} = \vec{k} \cdot \vec{V}_0 (-i e^{-i\vec{k} \cdot \vec{r}})## which might be useful for part (c).
     
  8. Sep 28, 2013 #7
    I understand that if I use the previous equation you showed me in the Maxwell-Ampere Law then I will get some result that shows D is perpendicular to the k and H vector. If that is true then doesn't that imply that the D_0 vector, E_0 vector and k hat are coplanar by definition?

    You take the cross product of k and E to get H. This vector H is perpendicular to the plane created by k and E. If you take the cross product of k and H then you will get a perpendicular vector D. Doesn't this perpendicular vector D HAVE to be in the same plane as k and E or am I thinking naively?
     
  9. Sep 28, 2013 #8

    TSny

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    Yes.

    Well, the Maxwell-Ampere implies that D is perpendicular to k and H. The Maxwell-Faraday law implies that B is perpendicular to k and E. Thus, you can deduce that k, D, and E are all perpendicular to H. Hence, k, D, and E must be coplanar since that all lie in the plane perpendicular to H. I'm not quite sure what you mean when you say that this follows "by definition".

    Yes, that's right.
     
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