1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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}##
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Maxwell equations in terms of potentials
  1. Maxwell's Equations (Replies: 3)

Loading...