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I Explaining transverse theory of light using EM theory

  1. Jun 6, 2017 #1
    Selection_002.png

    just below eqn 3.25, it is said:
    At any given time, Ex is constant
    for all values of x, but of course, this possibility cannot
    therefore correspond to a traveling wave advancing in the
    positive x-direction.

    Why can't Ex be a constant?
     
  2. jcsd
  3. Jun 6, 2017 #2

    Dale

    Staff: Mentor

    Because then it wouldn't be a wave. It would be a straight line, not a wave.
     
  4. Jun 6, 2017 #3
    Wave is propagating in x- direc.
    em wave is transverse. This means that the varying component has to be perpendicular to the direction of wave propagation.
    So, it's the y or z component of the E which has to vary, not the x- component of the E.
    So, the unvarying component i.e.x- component of the E could be constant.
    What is wrong with this argument?
     
  5. Jun 6, 2017 #4

    Dale

    Staff: Mentor

    No, this is exactly wrong. A plane wave propagating in the x direction varies along x and t, not along y and z. You have this completely backwards. Just look at the equation for a plane wave to see.
     
  6. Jun 6, 2017 #5
    I didn't get what you meant by this.
    What I want to say is : for an em plane wave propagating in the x direction , it is y or z component of the E which varies along x and t,not the x component of E.
    That is Ey(x,t) or Ez (x ,t) could vary, while Ex(x,t) is constant.
    Now the book says that Ex(x,t) has to be zero.
    So, my question is why can't it be any nonzero constant?


    Selection_004.png Selection_005.png


    In the first paragraph, what is meant by the phrase "em waves in equilibrium"?
    Does it mean that energy density of the em wave remain constant?
     
  7. Jun 6, 2017 #6

    Dale

    Staff: Mentor

    Because then it wouldn't be a wave.

    If f is a wave in the x direction then ##\partial f/\partial x \ne 0##. So since ##\partial E_x/\partial x =0## it is not a wave. There may be a field in the x direction, but it doesn't make a wave.
     
  8. Jun 7, 2017 #7

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    A free em. wave obeys (in HL units with ##c=1##)
    $$\vec{\nabla} \cdot \vec{E}=\vec{\nabla} \cdot \vec{B}=0, \quad \vec{\nabla} \times \vec{E}+\partial_t \vec{B}=0, \quad \vec{\nabla} \times \vec{B}-\partial_t vec{E}=0.$$
    Now make a plane-wave ansatz (to be seen as finding the properties of the Fourier modes to be used in a Fourier-transform representation of the true fields, which are always "wave packets" of finite energy and momentum),
    $$\vec{E}=\vec{E}_0 \exp(-\mathrm{i} \omega t+\mathrm{i} \vec{k} \cdot \vec{x})+\text{c.c.},$$
    $$\vec{B}=\vec{B}_0 \exp(-\mathrm{i} \omega t+\mathrm{i} \vec{k} \cdot \vec{x})+\text{c.c.}.$$
    Here ##\vec{E}_0## and ##\vec{B}_0## are functions of ##\omega## and ##\vec{k}##. Now using the first two equations you get
    $$\vec{k} \cdot \vec{E}_0=\vec{k} \cdot \vec{B}_0=0.$$
    This already implies that the em. field is transverse.

    Now take the curl of the third equation. For the here used Cartesian components you get
    $$\vec{\nabla} \times (\vec{\nabla} \times \vec{E})=-\Delta \vec{E}=-\partial_t \vec{\nabla} \times{\vec{B}}=-\partial_t^2 \vec{E},$$
    i.e.,
    $$(\partial_t^2-\Delta) \vec{E}=0.$$
    Plugging in the plane-wave ansatz for ##\vec{E}## gives the dispersion relation
    $$\omega=\pm |\vec{k}|.$$
    The same holds true for ##\vec{B}##, which equation you find by taking the curl of the 4th Maxwell equation and use the third to eliminate ##\vec{E}##, leading to the wave equation
    $$(\partial_t^2-\Delta) \vec{B}=0.$$
    Now given ##\vec{E}## we can evaluate ##\vec{B}## from the 3rd equation,
    $$\partial_t \vec{B}=-\vec{\nabla} \times \vec{E}=-\mathrm{i} \omega \vec{B}=-\mathrm{i} \vec{k} \times \vec{E}_0 \exp(\ldots)=-\mathrm{\omega} \vec{B}_0 \exp(\ldots).$$
    This gives
    $$\vec{B}_0=\hat{k} \times \vec{E}_0,$$
    i.e., given ##\vec{E}_0## with ##\vec{k} \cdot \vec{E}_0##=0 gives uniquely ##\vec{B}_0##, i.e., ##\vec{E}## and ##\vec{B}## are perpendicular to each other and perpendicular to ##\vec{k}##.
     
  9. Jun 7, 2017 #8
    2017-06-07-215315.jpg
    Thank you, Vanhees 71.

    What is meant by c.c. and HL units?

    Now, I got Dale's argument, too.
    If Ex is not 0, then k has to be zero and k being 0 means there is no wave.

    Thank you, Dale,too.


    This is true for plane wave.
    In general ,is it always true that component of E in the direction of wave propagation is zero?

    Any wave could be written as a combination of plane waves .
    For each plane wave, component of E in the direction of wave propagation is zero.
    So, using superposition principle ,for any wave, component of E in the direction of wave propagation is zero .

    Is this o.k.?
     
    Last edited: Jun 7, 2017
  10. Jun 7, 2017 #9

    vanhees71

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    Science Advisor
    2016 Award

    HL units = Heaviside-Lorentz units; c.c.=complex conjugate. Since the Maxwell equations are linear differential equations with real coefficients you can just evaluate everything for the exp function, which makes the calculations a lot more easy than using the trigonometric functions sin and cos.
     
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