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Maxwell's Equations - Tensor form 2

  1. Jan 22, 2016 #1
    1. The problem statement, all variables and given/known data
    The Lorenz gauge ∂Φ/∂t + ∇. A = 0 enables the Maxwell equations (in terms of potentials) to be written as two uncoupled equations;
    2Φ/∂t2 - ∇2Φ = ρ 1 and

    2A/∂t2 - ∇2A = j 2

    The tensor version using the Lorenz gauge
    is, i am told,
    μ∂μ Aα = jα 3

    expanded this is: ∂2Φ/∂t2 - ∇2A = jα 4

    where Jα is the 4-current; ρ + J

    2. Relevant equations

    If one adds 1 and 2 one gets two terms not included in 4. Namely;

    2A/∂t2 - ∇2Φ My query is, what happened to these terms when we go to 3 (or 4). Have I misunderstood the transition from 1 and 2 to the tensor forms?


    3. The attempt at a solution

    If one tries to cancel these terms, or equate their sum to zero, one sees that the first is a vector and the second a scalar. Secondly; ∂2A/∂t2 = ∂E/∂t which > 0 for a time varying field. And ∇2Φ = ∇. E = div E which also is not zero unless no charges are present, which they can be in 3.

    Any enlightenment would be greatly appreciated.
     
  2. jcsd
  3. Jan 22, 2016 #2

    Orodruin

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    No, this is not correct. Expanded you get four different equations, one for each value of alpha.
     
  4. Jan 25, 2016 #3
    Please could you tell me where can I find out how to do this? I understood from my texts that ∂μμ acting on a tensor was equivalent to ∂2 /∂t2 - ∇2 the d'Alambert operator. How do I get the other values of alpha?
     
  5. Jan 25, 2016 #4

    Orodruin

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    Yes, this is not what was incorrect. What was incorrect is that you let the different terms act on different components of the 4-potential.
     
  6. Jan 25, 2016 #5
    Thanks but I'm still not sure what's wrong. I proved the EM tensor by writing all the terms out long hand i.e I did ∂αAβ - ∂βAα . By constructing two matrices and subtracting to get a resultant matrix from which I identified the elements of Fαβ the EM tensor. Should i take ∂μAα which I think results in a 16 element square matrix and then let ∂μ act on that to get the correct elements for jα ? I am not conversant with tensor algebra yet you see.
     
  7. Jan 25, 2016 #6
    Can anyone actually show me what I am doing wrong here!
     
  8. Jan 25, 2016 #7

    Orodruin

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    So you wrote
    Can you go step by step to show how you obtained this from
    In particular, there is a free index in this equation. Which value did you put it to?
     
  9. Jan 26, 2016 #8
    Hi Orodruin, actually this morning I did what I said above; first I expanded ∂μAν i.e. (∂t, ∂x,∂y,∂z )(Φ, Ax, Ay, Az) and got the 16 element tensor. Then I let ∂μ act on that tensor and got the 16 element tensor in 2nd derivatives; with all the time derivatives in the 1st column. The top line is ∂2tΦ - ∂2xΦ etc, i.e. ≡ ∂2Φ/∂t2 - ∇2Φ and in the remaining array if one assumes the terms in ∂2iAj = zero (i ≠ j) for the mixed spacial derivatives; then one obtains the 3 spacial vector components as you said I should. That is ∂2tAx - ∂2x, and the same for y and z.

    Adding gives me the equation: ∂2A/∂t2 - ∇2A

    If this is right
    (and please say it is!) then i get back to the two equations 1 and 2 I started with and all is accounted for. NB I should have used superscripts for the spacial comps of A perhaps.
     
  10. Jan 26, 2016 #9
    The last sentence in paragraph one above should read, ∂2tAx - ∂2Ax, and same for y and z components of A (spacial vector).
     
  11. Jan 27, 2016 #10
    Sorry, line 2, I meant 4 element column tensor/matrix (in 61 terms - 4 on each row) in 2nd derivatives. Anyway I am satisfied that my workings are correct.
     
  12. Jan 28, 2016 #11

    ChrisVer

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    the equation: [itex] (Something~ done) X^a = Y^a[/itex] is not 1 equation... it's as many equations as are the values of a (a is not a summed/null index)... say that a=0,1,2,3...then it's 4 equations, which once you expand read:
    [itex] (Something~ done) X^0 = Y^0[/itex]
    [itex] (Something~ done) X^1 = Y^1[/itex]
    [itex] (Something~ done) X^2 = Y^2[/itex]
    [itex] (Something~ done) X^3 = Y^3[/itex]
    As for example the [itex]\nabla^2 \vec{X} = \vec{Y}[/itex] is 3 equations and not 1.

    your 3->4 transition is incorrect...in fact you destroyed a free index which is an unhealthy thing to do everywhere but especially in relativity. And you can also notice that from the fact that your equation 4 adds a scalar ∂2Φ/∂t2 with a vector ∇2A:nb). What would you say as a result to the [itex]2t^2 + \begin{pmatrix}2t \\ \sin t \\ u(t) \end{pmatrix}[/itex]? and of course equates it to the components of a 4 vector (which comes from destroying the free index)
     
    Last edited: Jan 28, 2016
  13. Jan 28, 2016 #12
    I have now understood where I went wrong (actually I said that I know you can't equate a scalar to a vector). Multiplying out the tensors long hand I found my error. I don't really get what these free indices are yet.
     
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