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I Maxwell equations in terms of potentials

  1. Apr 22, 2016 #1
    These are Maxwell´s equations in potential formulation:
    29ed508c553583e1477fc6a3681158fe.png
    621ac837ee1d1d742fcf5ed3f73c033a.png

    2φ = DIV(grad(φ)) . Am I right?
    2A = ROT(ROT(A))=ROT(B)=grad(DIV(A))-Laplace(A) . Am I right?
    In coulomb gauge in every point and at any time DIV(A)=[PLAIN]https://upload.wikimedia.org/math/4/4/1/44131cc26bd9db464d0edb7459ccca84.png. [Broken] Am I right?
    Where could I find Maxwell´s equations in terms of potentials without vector operator?

    How must ROT (same as curl) be generalized to make the equations describe EM-field in D-dimensional space equally with these equation
    ##\begin{cases}
    & \sum_{i=1}^D(\frac{\partial E_i}{\partial x_i})=\rho \frac{1}{{\epsilon_0}} \\
    & \frac{\partial E_a}{\partial t}=\sum_{i=1}^D(\frac{\partial B_{[i;a]}}{\partial x_i})-J_a \\
    & \frac{\partial B_{[a;b]}}{\partial t}=\frac{\partial E_b}{\partial x_a}-\frac{\partial E_a}{\partial x_b}\\
    & \frac{\partial B_{[a;b]}}{\partial x_c}+\frac{\partial B_{[b;c]}}{\partial x_a}+\frac{\partial B_{[c;a]}}{\partial x_b}=0
    \end{cases}##
    ,which are in terms of E and B?

    φ is electripotentialfield.
    E is electricvectorfield.
    A is magneticpotentialvectorfield.
    B is magneticvectorfield.
    ρ is electriccharge density.
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Apr 22, 2016 #2
    Yes
    No. Look here: https://en.wikipedia.org/wiki/Vector_Laplacian
    Yes
    You can do this yourself. Just take what you have above and replace with partial derivatives. Also, this might be useful
    https://en.wikipedia.org/wiki/Covariant_formulation_of_classical_electromagnetism

    Your last question is a little weird. Are interested in a 2D subset (x,y) of 3D space (x,y,z) where z is a symmetry?
     
  4. Apr 22, 2016 #3
    It seems to confirm, that ROT(ROT(A))=grad(DIV(A))-Laplace(A) .
    also this http://m.wolframalpha.com/input/?i=curl+(curl+A)&x=0&y=0 seems to confirm, that ROT(ROT(A))=grad(DIV(A))-Laplace(A) . Or does ∇2 A in the second equation note Laplace(A) not ROT(ROT(A))?

    I am interested about theoretical physical system with D-dimensional space. Understanding physical meaning of such system is not necessary to answer the question.
     
  5. Apr 22, 2016 #4
    ##\nabla^2 A## means the same thing as Laplace(A).

    How many dimensions do you want to work in?
     
  6. Apr 23, 2016 #5
    D dimensions. Generalization must include variable D.
     
  7. Apr 23, 2016 #6
    Now as I understand there are different Laplacians for vectorfield and scalarfield it makes more sense.

    So are these equations in lorenz gauge
    ##\begin{cases}
    \frac{\partial^2 A_{D_1}}{\partial t^2}=J_{D_1}\cdot k_E \cdot \pi \cdot c^2 \cdot 4+\sum_{i=1}^D(\frac{\partial^2 A}{\partial x_i \cdot \partial x_{D_1}}) \cdot c^2-ROT(ROT(A)) \cdot c^2\\
    \frac{\partial^2 \phi}{\partial t^2}=\rho \cdot k_E \cdot \pi \cdot 4+\sum_{i=1}^D(\frac{\partial^2 \phi}{\partial x_i^2})\\
    \frac{\partial \phi}{\partial t}=\sum_{i=1}^D(\frac{\partial A_i}{\partial x_i}) \cdot c^2\\
    F_E=q\cdot(\sum_{i=1}^D(v_i \cdot \frac{\partial A_i}{\partial x_{D_1}})-\frac{\partial A_{D_1}}{\partial t}-\frac{\partial \phi}{\partial x_{D_1}})
    \end{cases}##
    ?

    But what must ROT(ROT(A)) equal to in D-dimensional space so that the Maxwell equations were equal with these
    ##\begin{cases}
    & \sum_{i=1}^D(\frac{\partial E_i}{\partial x_i})=\rho \frac{1}{{\epsilon_0}} \\
    & \frac{\partial E_a}{\partial t}=\sum_{i=1}^D(\frac{\partial B_{[i;a]}}{\partial x_i})-J_a \\
    & \frac{\partial B_{[a;b]}}{\partial t}=\frac{\partial E_b}{\partial x_a}-\frac{\partial E_a}{\partial x_b}\\
    & \frac{\partial B_{[a;b]}}{\partial x_c}+\frac{\partial B_{[b;c]}}{\partial x_a}+\frac{\partial B_{[c;a]}}{\partial x_b}=0
    \end{cases}##
    equation?
    I also want the equation system to be formed as short and easy as possible. What gauge you recommend me to choose? Any ideas how to simplify the equations?
     
    Last edited: Apr 23, 2016
  8. Apr 23, 2016 #7
    I have hypothesis, that Laplace(A) must be generalized to be equal to ## 2\cdot\frac{\partial^2 A_{D_1}}{\partial x_{D_1}^2}-\sum_{i=1}^D(\frac{\partial^2 A_{D_1}}{\partial x_i^2})## .
    Can anybody check if it is correct?
     
  9. Apr 24, 2016 #8
    You know what the electromagnetic tensor is? You should use a tensor formulation if you want to generalize to different dimensions. I asked what dimension you wanted to work in because it affects the answer. Cross product only really works in 3 or 7 dimensions. You might want to use the Levi-Civita symbol.
     
  10. Apr 24, 2016 #9
    I got the equations in E and B formulation from Maxwell equations in electromagnetic tensor formulation.

    The equations should not contain vector products on operators, but partial derivatives and sums. D must be a variabe, that can be equal to any natural number.
     
    Last edited: Apr 24, 2016
  11. Apr 30, 2016 #10
    I have got my answer. Correct equation system is:
    ##\begin{cases}
    \frac{\partial^2 A_{D_1}}{\partial t^2}=J_{D_1}\cdot k_E \cdot \pi \cdot 4+\sum_{i=1}^D(\frac{\partial^2 A_{D_1}}{\partial x_i^2}) \cdot c^2\\
    \frac{\partial^2 \phi}{\partial t^2}=\rho \cdot k_E \cdot c^2 \cdot \pi \cdot 4+\sum_{i=1}^D(\frac{\partial^2 \phi}{\partial x_i^2})\\
    \frac{\partial \phi}{\partial t}=\sum_{i=1}^D(\frac{\partial A_i}{\partial x_i}) \cdot c^2\\
    F_E=q\cdot(\sum_{i=1}^D(v_i \cdot \frac{\partial A_i}{\partial x_{D_1}})-\frac{\partial A_{D_1}}{\partial t}-\frac{\partial \phi}{\partial x_{D_1}})
    \end{cases}##
     
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