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D'Alembertian question again (sorry)

  1. Mar 30, 2007 #1


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    Sorry to cluter forum with previous thread. I cannot work out the letex code


    in above pdf i do not understand the transition from eq 32 to 33.

    That is, I don't understand the presence of the last i*psi/c term in the new 4-vector potential. SHouldn't it just be a= (A1,A2,A3,A4) to get eq29?
    Last edited by a moderator: Apr 22, 2017
  2. jcsd
  3. Mar 30, 2007 #2
    You understand 4 vectors? [itex]\phi[/itex] is the electric potential & A is the magnetic potential. They're trying to combine the two.
  4. Mar 30, 2007 #3


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    But there is no "A_4" in equation 29 so what do you mean?

    No, their equation 33 contains both eqs 29 AND 30!! (with the four-vector [itex] j_\mu[/itex] defined in eq 27)
    Last edited by a moderator: Apr 22, 2017
  5. Mar 31, 2007 #4


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    I understand the intention not the method. so I guess I don't understand 4-vectors!

    my problem is this:

    if square=(d2/dx1,d2/dx2,d2/dx3,-1/c^2 d2/dx4) (eqn 31)

    & the combined vector & scalar potential 4-vector a:

    a=(A1,A2,A3,i*psi/c) (eqn 32)

    Then surely:

    square*a=laplacian A - i/c^3 d^2 (scalar)/dt^2

    but instead they have (correctly) square*a = mu*j = mu(J1,J2,J3,icp)

    why? I cant see how that relates to eq 29 & 30?
  6. Mar 31, 2007 #5
    As nrqed said, you're combining eqs 29 & 30 to form a 4 vector. J and A in eq 29 range as {1,2,3} & eq 30 becomes {4}.

    Compare 29 and 30 with 33. Factor out [itex]\mu_0[/itex] from 30 & you have [itex]-\rho c^2[/itex] .. which is why you need to divide by c in 32 to make the 4D [itex]j_\mu[/itex] that you see in 27.
    Last edited: Mar 31, 2007
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