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Cross product on undivided to components vector

  1. Jul 19, 2011 #1
    he question asks to find the ratio between E0 and B0 and the ratio between w and k
    ?

    E and B are on the x-y plane
    they are given as verctors without any components dividing
    no x direction part,y direction part,z direction part
    eq153d.jpg
    so when when i use the maxwell equations
    [TEX]\nabla\times E=-\frac{{dB}}{dt}[/TEX]
    [TEX]\nabla\times B=\mu_{0}\varepsilon_{0}\frac{dE}{dt}[/TEX]

    i cant do the cross product of B or E
    because i dont know what to put in the tererminant whch is calculating the cross product


    i tried to make some components by my self
    but i get so many variables
     
  2. jcsd
  3. Jul 19, 2011 #2

    hunt_mat

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    In general you can't. You just have yo leave it in the form that you have it. However, there are some things that you can do:
    [tex]
    \mathbf{E}=\varphi (t,z)\mathbf{E}_{0}(x,y)
    [/tex]
    and likewise for B and there are general rules for this. That is the only way that I can see.
     
  4. Jul 19, 2011 #3
    what next?
     
  5. Jul 19, 2011 #4

    hunt_mat

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    So you should have some vector identities which cover it:
    so for example:
    [tex]
    \nabla\times (\varphi\mathbf{E})=\varphi\nabla\times\mathbf{E}+\nabla\varphi\times\mathbf{E}
    [/tex]
    and the like. As I said, there is not much you can do.
     
  6. Jul 19, 2011 #5
    its a very inportant questiom
    i dont know what to do next with those identities
    could youwrite down how youwould solved it
    ?
     
  7. Jul 19, 2011 #6

    hunt_mat

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    [itex]\varphi =e^{ikz-i\omega t}[/itex], so [itex]\nabla\varphi =ike^{ikz-i\omega t}\mathbf{k}[/itex]

    I will do one of maxwell's equations for you, you can do the rest and we will see where to go from there.
    [tex]
    \nabla\times\mathbf{E}=-\frac{\partial\mathbf{B}}{\partial t}
    [/tex]
    In our case:
    [tex]
    \nabla\times\mathbf{E}=e^{ikz-i\omega t}\nabla\times\mathbf{E}_{0}+ik\mathbf{k}\times \mathbf{E}_{0}
    [/tex]
    and also:
    [tex]
    \frac{\partial\mathbf{B}}{\partial t}=-i\omega \mathbf{B}_{0}e^{ikz-i\omega t}
    [/tex]
    and the equation becomes:
    [tex]
    \nabla\times\mathbf{E}_{0}+ik\mathbf{k}\times \mathbf{E}_{0}=i\omega\mathbf{B}_{0}
    [/tex]
     
    Last edited: Jul 19, 2011
  8. Jul 19, 2011 #7
    you made a type mistake
    iside the formula its E_0
    the hole formula is for E
    first question:
    [tex]\varphi\nabla\times\mathbf{E_0}[/tex]
    means dot product of [tex]\varphi[/tex] with [tex]\nabla\times\mathbf{E_0}[/tex]
    but i dont have components of them to do thedot product.
    and i dont know how to find
    [tex]\nabla\times\mathbf{E_0}[/tex]


    thing that might help is
    that the B field and E field are perprendicular(90 degree angle between them)
    so there dot product is 0

    ??
     
  9. Jul 19, 2011 #8

    hunt_mat

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  10. Jul 19, 2011 #9
    i dont think so
    those two are the ones i solved similar question before

    how to continue your idea
    ?
     
    Last edited: Jul 19, 2011
  11. Jul 19, 2011 #10

    hunt_mat

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    By continuing my idea! You saw the sort of thing I did in my post, I gave you a reference for the vector identities. I will do one more only, [itex]\nabla\cdot \mathbf{B}=0[/itex]:
    [tex]
    \nabla\cdot (\mathbf{B}_{0}e^{ikz-i\omega t})=e^{ikz-i\omega t}\nabla\cdot \mathbf{B}_{0}+\mathbf{B}_{0}\cdot\nabla (e^{ikz-i\omega t}) =e^{ikz-i\omega t}\nabla\cdot \mathbf{B}_{0}+ik \mathbf{k}\cdot \mathbf{B}_{0}e^{ikz-i\omega t}
    [/tex]
    Which shows that:
    [tex]
    \nabla\cdot \mathbf{B}_{0}+ik \mathbf{k}\cdot \mathbf{B}_{0}=0
    [/tex]
     
  12. Jul 19, 2011 #11
    ok i wil do gauss law
    [tex]\nabla\cdot (\mathbf{E}_{0}e^{ikz-i\omega t})=\frac{\rho}{\epsilon_0}[/tex]

    but still i have to find [tex]\nabla\cdotE[/tex]
    and [tex]\nabla\cdot B[/tex]
    how to do that
    ?
     
  13. Jul 19, 2011 #12

    hunt_mat

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    So expand the left hand side, what do you get?
     
  14. Jul 19, 2011 #13
    [tex]\nabla\cdot (\mathbf{E}_{0}e^{ikz-i\omega t})=\frac{\rho}{\epsilon_0}[/tex]
    i wil expand like you did on B

    but still i have to find [tex]\nabla\cdot E_0[/tex]
    and [tex]\nabla\cdot B_0[/tex]
    how to do that
    ?
     
    Last edited: Jul 19, 2011
  15. Jul 19, 2011 #14

    hunt_mat

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    Do you have any sources of charge in your problem? I don't think that you have as you're looking for a wave propagating in the z direction. You still haven't expanded the left hand side of your equation as I did.

    Once we have all the reduced equations, ten we will take another look at where we can go from there.
     
  16. Jul 19, 2011 #15
    when you expanded you wrote kk in some place
    why didnt you write k^2
    ?
     
  17. Jul 19, 2011 #16

    hunt_mat

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    They are different k's. one is a vector direction and the other is effectively a wave number. Keep going.
     
  18. Jul 19, 2011 #17
    it takes me alot of time to write on latex
    so here is a photo of the gauss equaation you said to develop
    2uo18cn.jpg
    what is the next step?
     
  19. Jul 19, 2011 #18

    hunt_mat

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    I think what we're doing here is looking for normal modes and therefore we can set all sources to zero, this will simplify you answer somewhat. You should also be careful about cross products
     
  20. Jul 19, 2011 #19
    ok
    but i dint know how to solve it
    i followed your lead so far
    do you know how to finish it?
     
  21. Jul 19, 2011 #20

    hunt_mat

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    You have one more to do, Ampere's law, what does that reduce to?

    You can't get a ratio of two vectors, so I am not too sure what they're referring to here but I can help you derive a dispersion relation (and equation giving [itex]\omega[/itex] as a function of k.
     
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