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EM waves - wave equation derivation

  1. Jun 16, 2008 #1

    Something has been bothering me about deriving the wave equation for a plane EM wave. We were showed this derivation in class and had to reproduce it but something is not making sense to me...

    The derivation is as follows:

    Suppose you have a plane EM wave (in a vaccuum) travelling in the x-direction. The E-field is in the y-direction and the B-field in the z-direction.

    Take a rectangular loop in the xy-plane with length [itex]dx[/itex] and height [itex]l[/itex].
    Using the maxwell equation:
    [tex]\oint \vec{E} \cdot \vec{dl} = - \frac{ d \phi_B}{dt}[/tex]
    [tex]E(x+dx,t)l - E(x,t)l = -\frac{d}{dt} B(x,t) l dx[/tex]
    [tex]\frac{E(x+dx,t) - E(x,t)}{dx} = - \frac{d}{dt} B(x,t)[/tex]
    [tex]\frac{ \partial E(x,t)}{\partial x} = - \frac{ \partial B(x,t)}{\partial t}[/tex]

    And similar for B.

    Now what bothers me is the calculation of [itex]\phi_B[/itex]... I am told that we are assuming B to be constant over the infinitesemal distance dx. However, we are not assuming the same for E? We are using the change in E-field explicitly to arrive at the partial derivative for E, but we ignore it for B, isn't that strange?

    Also, similarly, since we are advancing x with dx, shouldn't we advance t with dt also?
    I guess ignoring this is a safe 'approximation' to make since dt = dx / c which is even more small...

    What is going on here? Why can we do this?
    Last edited: Jun 16, 2008
  2. jcsd
  3. Jun 17, 2008 #2
    About [tex]\phi_B[/tex] you don't have to compute a spatial variation of [tex]B[/tex], so you can take an average (on space) value of [tex]B[/tex]; about dt, you can take it so small that the fields don't vary appreciably during that time; even if a time derivative of fields compare in the equation, its derivation don't use the fact that time varies:
    you compute [tex] \oint \vec{E} \cdot \vec{dl} [/tex] and [tex]\phi_B[/tex] at a fixed instant of time, then you write that the first equals the -time variation of the second.
    Last edited: Jun 17, 2008
  4. Jun 17, 2008 #3
    I still don't understand why we can ignore the change in B over the distance dx, but we cannot ignore the change in E over the same distance dx..?
  5. Jun 17, 2008 #4
    If you wouldn't ignore the spatial variation of B, what would you write instead of B(x)? To compute the flux [tex] \phi_B[/tex] you should in that case integrate B(x) over the rectangular region; you know from the average value theorem of math analysis that such an integral equals the area of the region multiplied by the function you want to integrate, computed in an opportune average point P inside that region; call it B(P). This represent an average value of B(x) inside the region. When dx goes to zero, that average point P becomes a point with coordinate x.
    Last edited: Jun 17, 2008
  6. Jun 17, 2008 #5
    Yeah that makes sense, but in other words, we are ignoring the change of B over this distance, since otherwise B(x) would not be an average value, it would be something like B(x + 1/2 dx), but since dx is infinitesemal this doesn't make sense...

    All I'm trying to say is, why can we safely ignore the change in B when we cannot ignore (we even explicitely use it) the same change in E?
  7. Jun 17, 2008 #6
    Because the equation explicitely depends on the spatial variation of E in the little rectangle, but not on the spatial variation of B. It's quite te same as if you want to compute the force on an object and its kinetic energy in a small interval of time dt: for the force you need to know acceleration, so you need to know how the velocity v varies in that interval of time; for kinetic energy you are not interested in how velocity varies in dt, so you can take an average value of it in the interval dt.
    Hope to have clarified, because I don't know how to explain it better than this way. :smile:
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