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Homework Help: Given a plane wave characterized by Ex and By

  1. Aug 2, 2012 #1
    1. The problem statement, all variables and given/known data

    Given a plane wave characterized by Ex , By , propagating in the positive z-direction,

    [itex]E[/itex] [itex]=[/itex] [itex]E[/itex][itex]0[/itex][itex]sin[/itex][[itex]\frac{2\pi}{\lambda}[/itex][itex](z-ct)[/itex]][itex]\widehat{x}[/itex]

    show that it is possible to take scalar potential ϕ = 0 . Find a possible
    vector potential A for which the Lorentz gauge is satisfied.

    2. Relevant equations

    i)[itex]E[/itex]= [itex]-∇[/itex][itex]\varphi[/itex]

    ii)[itex]∇E[/itex] = [itex]\frac{\rho}{ε}[/itex]

    iii)[itex]B[/itex] = [itex]∇×A[/itex]

    iv)[itex]∇A[/itex] = [itex]\frac{-1}{c^2}[/itex][itex]\frac{\delta\varphi}{\delta t}[/itex]

    3. The attempt at a solution

    So basically I am thinking combining the above first two equations which results in the laplace operator but I am not sure how this ties into phi equating to 0. Then there is equation 4 the Lorenz gauge and I assume values of A and phi are plugged in to see if it balances. But its the first step i am not sure on. How to show the scalar potential can be zero..

    [itex]-∇^2[/itex][itex]\varphi[/itex] = [itex]\frac{-\rho}{\epsilon}[/itex]
  2. jcsd
  3. Aug 2, 2012 #2


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    For any vector potential and scalar potential combination to be valid, the resulting fields must simultaneously satisfy all of Maxwell's equations, so I would say that the other 2 Maxwell's equations would also be "relevant equations" for this problem. :wink:

    Also, the divergence of a field [itex]\mathbf{E}[/itex] is usually written as [itex]\mathbf{\nabla} \cdot \mathbf{E}[/itex], not [itex]\mathbf{\nabla}\mathbf{E}[/itex] (which is how one writes the gradient of a vector field, which is a 2nd rank tensor).

    Now, in general you have [itex]\mathbf{E}= -\mathbf{ \nabla } \varPhi -\frac{ \partial \mathbf{A} }{ \partial t}[/itex] (this should also have been listed as a relevant equation :wink:). What do you get for [itex]\mathbf{A}[/itex] when you choose [itex]\varPhi = 0 [/itex]? Does it satisfy the Lorentz Gauge? Does the resulting [itex]\mathbf{B}[/itex]-field, together with the given [itex]\mathbf{E}[/itex]-field satisfy all of Maxwell's equations?
  4. Aug 3, 2012 #3
    Yeah I knew the correct way to write divergence but couldn't see the middle dot symbol anywhere haha :) thanks.

    Okay that makes sense, there are a list of four maxwell equations that can be manipulated and I need to make sure all are satisfied when I pick values of A and phi. The question states that I have to do this with phi equating to zero.

    Sooo plug chug and see what happens? Finishing with my equation 'iv)'?


    [itex]E[/itex] = [itex]E[/itex]0[itex]sin[/itex][[itex]\frac{2π}{λ}[/itex][itex](z−ct)]xˆ[/itex]

    [itex]E = -∇\Phi -\frac{\delta A}{\delta t}[/itex] NB. Set '[itex]\Phi[/itex]' = 0 so;

    [itex]E = -\frac{\delta A}{\delta t}[/itex] and then

    [itex]-∫E.\delta t = A[/itex] and then

    [itex] B = ∇×A, ∇.B = 0 = ∇.(∇×A)[/itex] which will test if B is correct and finally both values into Lorenz gauge to equal 0?
    Last edited: Aug 3, 2012
  5. Aug 3, 2012 #4
    If my understanding is correct then after I have took the negative integral of E and then found the curl of it I get a cosine function in the y-axis as I would expect but the following divergence of this function would give me a scalar function instead of the zero I am looking for?

    Never mind no it doesn't haha. I think I have solved it. Thank you.
    Last edited: Aug 3, 2012
  6. Aug 3, 2012 #5


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    No problem, the [itex]\LaTeX[/itex] code for it is just "\cdot". And for the partial derivative symbol, it is "\partial".

    Not quite sure what [itex]-∫E.\delta t = A[/itex] means (how does one integrate with respect to a partial differential?), don't you mean [itex]\mathbf{A} = -\int \mathbf{E} dt[/itex]?

    Other than that it looks like you've probably done everything correctly (you didn't actually post what you got for [itex]\mathbf{A}[/itex] or [itex]\mathbf{B}[/itex], so I can't be 100% sure). Just 2 things that you should keep in mind:

    (1) Make sure you check all 4 of Maxwell's equations - you made no mention of checking the curl and divergence of [itex]\mathbf{E}[/itex]

    (2) Is your choice of [itex]\mathbf{A}[/itex] unique? What about the "constant" of integration, are there restrictions placed on it by Maxwell's equations or the Lorentz gauge condition?
  7. Aug 4, 2012 #6
    Thank you for the correct Latex code and I did indeed mean [itex] A = - ∫Edt [/itex].

    For [itex]A[/itex] I got [itex]\frac{\lambda}{2\pi}[/itex][itex]E[/itex][itex]0[/itex][itex]cos[\frac{2\pi}{\lambda}(z-ct)]\widehat{x}[/itex]

    For [itex]B[/itex] I got [itex]-E[/itex][itex]0[/itex][itex]sin[\frac{2\pi}{\lambda}(z-ct)]\widehat{y}[/itex]

    And thank you for all your help I shall check all of Maxwell's equations and integration constants.

    Kind Regards Hilly
  8. Aug 4, 2012 #7


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    You are missing a factor of [itex]\frac{1}{c}[/itex], and you'll want to recheck the sign of B.

    You're welcome! And remember the integration "constant" is not just a scalar in this case, and can depend on position (just not time).
  9. Aug 4, 2012 #8
    I am in desperate need of a warning before posting :), the factor of [itex]\frac{1}{c}[/itex] I do have in my answer, I merely forgot it in this post :)whoops!

    And double checking the sign of [itex]B[/itex] I came to the same negative sine hmm.. Am I going wrong somewhere?

    The operator [itex]∇×A[/itex] ends up as equating to [itex]\frac{\partial{Ax}}{\partial{z}}[/itex]..

    1)everything before the cosine has no variable z so is treated as constant
    2)the function inside cosine is [itex]\frac{2z\pi}{\lambda} - \frac{2ct\pi}{\lambda}[/itex]
    3)so second part has no z and first part has derivative [itex]\frac{2\pi}{\lambda}[/itex]
    4)cosine changes to negative sine?
  10. Aug 4, 2012 #9


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    Sorry, I meant A (and subsequently B)
  11. Aug 4, 2012 #10
    Right you are, :) thank you again.
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