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Prove perpendicularity of E field and direction of EM wave

  1. Nov 26, 2013 #1
    1. The problem statement, all variables and given/known data
    Show that the E and B field of a plane EM wave are both perpendicular to its direction of propagation in a non-dispersive and transparent medium.

    2. Relevant equations
    Maxwell's equatons.


    3. The attempt at a solution

    Well, I know how to easily show that E and B fields are perpendicular to each-other (Faraday's/Ampere's law). However, how am I supposed to show one of them is perpendicular to the direction of propagation of the EM wave?

    And is it normal to ask such questions on an exam for 2nd year university students? This question was only worth 5% in the 2011 exam...
     
    Last edited: Nov 26, 2013
  2. jcsd
  3. Nov 26, 2013 #2

    Simon Bridge

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    That would be a pretty normal question, yeah. You'd normally do it by remembering the exercise in class.
    You should have a relationship between the E B and v vectors off Maxwell's equations ... how would you normally show the perpendicularity of two vectors?
     
  4. Nov 26, 2013 #3
    I would dot them, or equate the cross product including one of the vectors, with the other.However, to do any of this I need to find the direction of v. How do I do that?

    All I know is that in the solutions manual, the prof just wrote divE = 0 => E is perpendicular to k. this guy is amazing
     
  5. Nov 26, 2013 #4

    BruceW

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    you can either remember "divE = 0 => E is perpendicular to k" by heart, or remember the form of a plane wave in a vacuum, so then you can derive "divE = 0 => E is perpendicular to k". I guess it depends on how much time you have to do each question.
     
  6. Nov 26, 2013 #5

    Simon Bridge

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    You don't need the direction of v, since it is arbitrary, you need the relationship.

    The relationship comes from Maxwell's equations ... go look.
    Do you know how to derive the result for EM radiation?

    You can pick a direction for v, then show that EM waves propagating in that direction must have E and B vectors perpendicular to v, or, from the equations for E and B find ExB=cv (c=proportionality)

    You'd buy k.E=0 means E is perpendicular to k right?
    How does v relate to k?
    How does k relate to div?
     
  7. Nov 27, 2013 #6
    How? All you get from maxwells equation is the wave equation for the B and E fields, and that they move with the velocity c.

    How am I too get k? At best i can insert the general form of the wave function for E into the wave equation for E, but then k would be inside the cosine term...

    Could someone derive the relationship for me? Please?
     
  8. Nov 27, 2013 #7
    I don't understand at all the prof's logic, so it would do me no good remembering it.

    Only link between divE and k I can think of, is that k.E should equal to divE, but I can't see how that will get me anywhere.
     
    Last edited: Nov 27, 2013
  9. Nov 27, 2013 #8

    BruceW

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    well, that is the tricky bit done already. Showing that divE = k.E is the part that you could derive, instead of just remembering it. So I'm guessing that you don't understand why the prof says divE=0 ?
     
  10. Nov 27, 2013 #9
    No, the tricky bit is why k.E = 0 (or why k.E = divE). This is what I fail to understand. Can you help me?

    Also, it's expected that I'm able to derive these results purely by logical use of the given equations on the go, not remember them from lectures.
     
    Last edited: Nov 27, 2013
  11. Nov 27, 2013 #10

    BruceW

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    I don't understand which bit you don't understand. There's 2 steps:
    k.E = divE
    divE = 0
    and therefore k.E = 0 (which is the result that you were asked to get to). So which step are you having trouble with?

    edit: so what I mean is there's two independent steps, then you combine them to get the result.
     
    Last edited: Nov 27, 2013
  12. Nov 27, 2013 #11
    Are you trolling me?

    I already explained several times I do not know if or why k.E = divE. I said k.E=divE only in the case k.E =0, but I do not know if k.E=0.
     
  13. Nov 27, 2013 #12

    BruceW

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    It kinda threw me off when you said the tricky bit is why k.E=0 (since this is the entire problem). But anyway, I understand what you mean now. You know that divE=0 But you don't know why k.E=divE, right?
    You can derive this from the equation of a plane wave in free space. So unfortunately, you still need to remember the equation for a plane wave in free space. But I think that is easier to remember, and you can derive a lot of other stuff by just remembering this. So give it a go. Use the plane wave solution for the electric field in the two expressions k.E and divE and you'll see that once you do the calculation, they are equal to each other :)

    edit: I guess that also, since the wave is monochrome, it is not too difficult to derive the form of the plane wave from the wave equation. And you can get the wave equation from Maxwell's equations in free space. So it is possible to derive the answer from Maxwell's equations, along with a few assumptions. I'm guessing that you won't be required to do this in exam... It depends on what is the maximum amount of marks for the question, right?
     
    Last edited: Nov 27, 2013
  14. Nov 27, 2013 #13

    vela

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    When you said, "Only link between divE and k I can think of, is that k.E should equal to divE," I think it was natural for us to assume you knew how to calculate div E.

    At this point, it's not really clear what you know and what you don't know. All you keep saying is you don't know how to do the problem. What's the expression for E for a plane wave? Do you know that? If not, you need to figure that out first. It should be in your book or notes. If you do, you need to take its divergence. Do you not know how to calculate the divergence of a vector field? If so, show us your best attempt and we can see where you're going astray.
     
  15. Nov 27, 2013 #14
    Okay, thanks bruce. I'm sorry for snapping at you, what I wrote could easily be misinterpreted. What I meant was, that k.E will equal divE if and only if k.E=0. And yes, I do remember the plane-wave function.

    Anyway, I took divE, but I only get ##\nabla . \vec{E} = - sin(\vec{k}.\vec{r}-\omega t) \cdot \vec{E_0} \cdot \nabla (\vec{k}.\vec{r})##

    where ##\vec{E} = \vec{E_0} cos(\vec{k}.\vec{r}-\omega t)##
     
    Last edited: Nov 27, 2013
  16. Nov 27, 2013 #15

    BruceW

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    that's it dude, that's it. You're most of the way there now. p.s. I was wrong before when I said the two expressions E.k and divE are equal. (obviously, since divE has a sine, while E.k has a cosine). But the sine and cosine don't matter for showing that E.k=divE=0 for this special case of a monochrome plane wave.

    Anyway, keep going. you're getting there. Now, calculate
    [tex]\vec{E_0} \cdot \nabla (\vec{k}.\vec{r})[/tex]
    Then you will see pretty soon how this is related to E.k

    edit: well, I wasn't wrong to say E.k and divE are equal in this case (since they're both zero). But yeah, the part of them that evaluates to zero is the same (even before evaluating them both as zero), which is the important bit.
     
  17. Nov 27, 2013 #16
    So ##\vec{r}=[a_i x,a_j y, a_k z]## (why?), where the a_x,a_y and a_z are constants? And in that case, there is one ##a## for each of the partial derivatives of ##r##: ##r_x=[a_i,0,0]##, ##r_y = [0,a_j,0]##, ##r_z=[0,0,a_k]##? So I just insert this into the expression I found and out rolls something similar to E.k?


    I appreciate the effort and all of that, but I would prefer it if you could just show me the link directly?
     
    Last edited: Nov 27, 2013
  18. Nov 27, 2013 #17

    BruceW

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    hehe, you'll get there. uh, I'm not sure why you say ##\vec{r}=[a_i x,a_j y, a_k z]## Surely ##\vec{r}## is simply given by: ##\vec{r} = (x,y,z)##, right? The vector ##\vec{k}## will be some constant vector though. So anyway, you need to calculate ##\vec{E_0} \cdot \nabla## and then use this on ##\vec{k} \cdot \vec{r}##
     
  19. Nov 27, 2013 #18
    OK then I get it. I was mixing r up. thanks.

    finita la commedia
     
  20. Nov 27, 2013 #19

    BruceW

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    nice work, "the farce is over", eh? only until you start the next physics problem :) hehe. life is farcical anyway, right?
     
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