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How to explain the overdetermination of Maxwell's equations

  1. Mar 16, 2015 #1
    Maxwell's equations seem overdetermined, in that they involve six unknowns (the three components of E and B) but eight equations (one for each of the two Gauss's laws, three vector components each for Faraday's and Ampere's laws). (The currents and charges are not unknowns, being freely specifiable subject to charge conservation.)

    The usual explaination is: It can be proven that any system satisfying Faraday's law and Ampere's law automatically also satisfies the two Gauss's laws, as long as the system's initial condition does. By introducing dummy variables characterizing these violations, the four equations become not overdetermined after all. (from wiki)

    I think this explaination is not correct. For example, how to explain electrostatic field by this explaination? [tex]\nabla \times{E} = 0, \nabla \cdot E=\rho [/tex]
    There are four equations and three unknowns.


    Maybe it perhaps can be explained in this way.




    Typesetting is so bad. The docx file is uploaded.
     

    Attached Files:

    Last edited: Mar 16, 2015
  2. jcsd
  3. Mar 17, 2015 #2
    In algebra, there are some vectors Yi (i=1,…,n) in a vector space. When there are coefficients (ci), not all zero, such that [tex]\sum\limits_{i = 1}^n {{c_i}{Y_i} = 0} [/tex]; The vectors Yi (i=1,…,n) are linearly dependent. Now I generalize the definition of linearly dependence in differential equations.


    There is a linearly partial differential equation:

    [tex](1)\left\{ {\begin{array}{*{20}{c}}
    {\sum\limits_{ij}^{} {a_{ij}^{\left( 1 \right)}\frac{{\partial {Z_j}}}{{\partial {x_i}}}} + {f_1} = 0}\\
    {\sum\limits_{ij}^{} {a_{ij}^{\left( 2 \right)}\frac{{\partial {Z_j}}}{{\partial {x_i}}}} + {f_2} = 0}\\
    \vdots \\
    {\sum\limits_{ij}^{} {a_{ij}^{\left( n \right)}\frac{{\partial {Z_j}}}{{\partial {x_i}}}} + {f_n} = 0}
    \end{array}} \right.[/tex]

    And we make [tex]{Y_k} = \sum\limits_{ij}^{} {a_{ij}^{\left( k \right)}\frac{{\partial {Z_j}}}{{\partial {x_i}}}} + {f_k}[/tex]

    The definition of linear dependence in differential equations is:
    ① there are coefficients (ci), not all zero, such that [tex]\sum\limits_k^{} {{c_k}{Y_k}} \equiv 0[/tex]
    ② there are coefficients (cij), not all zero, such that [tex]\sum\limits_{kl}^{} {{{c'}_{kl}}\frac{{\partial {Y_k}}}{{\partial {x_l}}} \equiv 0} [/tex]

    If the condition①or ②, or both conditions (①&②) are satisfied, then Eqs. (1) are thought as linearly dependent.

    The difference between this definition and the one in algebra is: we take one (or more) differentiation on Yk in differential equation ( condition ②).


    As we all know there are two curl equations in Maxwell equation. And

    [tex]\nabla \cdot \nabla \times {\bf{E}} \equiv 0[/tex]

    the above equation is satisfied with the condition ②. Therefore, there are two differential linearly dependent equations, and the number of independent equations are six (8-2=6), which are equal to the number of the unknowns.

    vixra:1503.0063
     
    Last edited: Mar 17, 2015
  4. Mar 17, 2015 #3
    a. ∇⋅D =ρ
    b. ∇⋅B=0
    c. ∇×E + ∂B/∂t =0
    d. ∇×H - ∂D/∂t = J

    In the 4 Maxwell's equation above you can easily deduce b from c ( by taking the divergence of c) which are sometimes referred to as the first pair of Maxwell's eq. You can deduce a from d ( by taking the divergence of d) which are referred to as the second pair of Maxwell's Eq. So b and c and a and d are not independent.
     
  5. Mar 17, 2015 #4

    PeterDonis

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    That tells you that ##\nabla \cdot \left( \partial B / \partial t \right) = 0##, or, exchanging time and space derivatives, ##\partial / \partial t \left( \nabla \cdot B \right) = 0##. But that does not imply ##\nabla \cdot B = 0##. It only implies that ##\nabla \cdot B## is not a function of time.

    That tells you that ##\partial / \partial t \left( \nabla \cdot D \right) = \nabla \cdot J##; but that is not the same as equation a.

    The "pairs" are not because one of the pair can be derived from the other. They are because each "pair" is actually the expression of a single Lorentz covariant equation; in Lorentz covariant form, there are only two Maxwell Equations, not four.
     
  6. Mar 18, 2015 #5

    PeterDonis

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    Another way of putting this is that the two Gauss's Laws are constraint equations, not evolution equations; they tell you what kind of initial conditions you have to have, not how a given set of initial conditions evolves in time.

    It's also worth noting that, if you write the fields in terms of a vector and scalar potential, i.e., ##E = - \partial A / \partial t - \nabla \phi##, ##B = \nabla \times A##, then the two source-free Maxwell equations (b and c) are automatically satisfied. (In Lorentz covariant form, this is just one equation, ##dF = 0##, which is an identity if ##F = dA##.) So yet another way of describing what's going on is that there are actually only four independent variables, not six (three vector potential components and one scalar potential), and four equations (a and the three components of d) linking those four variables to the source (assumed known). (Again, in Lorentz covariant form, there is just one equation with four components, which can be written schematically as ##\nabla \cdot F = J##.)
     
  7. Mar 18, 2015 #6
    PeterDonis et. al.

    Regarding your critique of my comment. I did not go through the full explanation of the issue at hand. Yes the div B is at this point not necessarily equal to zero but if we assume that B was, can or will be zero at some time at any point in space we then are allowed to ( must) set div B=0. Likewise for Div D which equals ρ +C where C is a constant and a function of coordinates. If at anytime we can assume the D and ρ did, are, or will become zero then C being independent of time must be zero. and div D = ρ. Note that the two dependent equations = 0 and are also referred to as the homogeneous pair and the two ≠ 0 are referred to as the inhomogeneous pair. This begs the question If there are only two Maxwell's Eqs why do we continue to tell student that there are four?
     
  8. Mar 18, 2015 #7

    Delta²

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    I believe one crucial error is that the current and charge density are not unknowns. This is not always true, for example in antenna and transmission line related problems everything is considered to be unknown, the only known thing is the voltage input at the feed point of the antenna or the transmission line.
     
  9. Mar 18, 2015 #8

    PeterDonis

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    But what justifies this assumption? There are certainly solutions of Maxwell's equations for which this is not the case.

    Again, what justifies this assumption? There are solutions for which it is not the case.

    Because EM is usually taught in ordinary "3+1" notation, i.e., physical quantities are expressed using 3-vector formalism as functions of time. Relativistically, this corresponds to choosing a particular inertial frame and expressing everything in terms of quantities relative to that frame. In that formalism, there are four equations: two have one component (they are scalar equations), and two have three components (they are 3-vector equations), for a total of eight components.

    In Lorentz covariant formalism, we do not choose a particular reference frame; we write equations in a form that is valid in any frame. In that form, there are two equations, each with four components, for a total of eight components. So the total number of components, formally speaking, is the same.

    Of course, once we start looking at what the components actually express, we find that there are not eight independent degrees of freedom. But that's a separate question.
     
  10. Mar 18, 2015 #9
    Drat!!!! I meant t if we assume that div B was, can or will be zero at some time at any point in space we then are allowed to ( must) set ( or keep) div B=0
     
  11. Mar 18, 2015 #10

    PeterDonis

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    Yes, this is true, and it means, as I said in a previous post, that ##\nabla \cdot B = 0## is really a constraint on initial conditions, not an evolution equation.
     
  12. Mar 18, 2015 #11
    Just curious, if we had magnetic monopoles could we reduce it to one equation?
     
  13. Mar 18, 2015 #12
    everyone,
    now we talk some simpler things, while not talking Maxwell equations.
    Electrostatic fields equations
    [tex](1) \nabla \cdot E = \rho ,\nabla \times E = 0[/tex]
    There are four eqs, and three unknowns. How to explain it?

    Here we only talk Eq. (1), while NOT talking the following
    [tex]
    \begin{array}{l}
    first,\nabla \times E = 0 \Rightarrow E = - \nabla \varphi \\
    then,\nabla \cdot E = \rho \Rightarrow {\nabla ^2}\varphi = - \rho
    \end{array}
    [/tex]
     
  14. Mar 18, 2015 #13

    PeterDonis

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    No. In fact, adding magnetic monopoles would increase the number of independent degrees of freedom.
     
  15. Mar 18, 2015 #14

    PeterDonis

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    Well, one way would be to ignore your next statement:

    If you view the electrostatic field ##E## as the gradient of the scalar potential, there is only one degree of freedom (the potential) and one equation (Gauss's Law). The other equation is an identity.

    Physically, since you have restricted yourself to a static situation, you should expect any equation with a time derivative in it to be empty of actual content. So there really is only one equation that tells you anything useful, and therefore one degree of freedom is the correct number.
     
  16. Mar 23, 2015 #15

    Demystifier

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    Maxwell equations are not overdetermined, because they are differential equations. For instance, a first-order differential equation for one variable x(t) does not have only one solution, but an infinite number of solutions. A unique solution is determined if you also specify the initial condition x(0). Similarly, Maxwell equations for 6 variables have 6 equations with first time derivatives, so for a unique solution you need 6 initial conditions. The 2 Maxwell equations which do not contain time derivatives are the so-called constraint equations, which serve to constrain 2 initial conditions. In this way you still have freedom to choose arbitrarily 6-2=4 initial conditions.

    These 4 initial conditions can also be interpreted as 2 physical degrees of freedom. Namely, Maxwell equations can also be written as second-order differential equations for 4 potentials. Each second-order differential equation requires 2 initial conditions, implying that each degree of freedom requires 2 initial conditions. Therefore the 4 initial conditions correspond to 2 degrees of freedom.
     
  17. Mar 23, 2015 #16

    Delta²

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    Guys what do you think about the statement in the OP that "current and charge density are not unknows, being freely specifiable subject to charge conservation".

    In my opinion they are not freely specifiable they both depend on the electric field. This follows from the definition of current density J=ρυ and either Ohm's law J=σE or from the fact that the electric field exerts force on the charged particles affecting their velocity , and thus the current density and charge density(from continuity equation).
     
  18. Mar 25, 2015 #17
    I think that the title of this thread is not good enough.
    Maybe the following is better: How to explain the overdetermination of Electrostatic equations
    4 equations and 3 unknowns
    [tex]div E=\rho, curl E=0[/tex]
     
  19. Mar 25, 2015 #18

    PeterDonis

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    This was already answered in post #14. The correct count for electrostatics is 1 equation (Gauss's Law) and 1 unknown (the scalar potential). The other three equations (curl E = 0) are identities and do not contain any additional information.
     
  20. Mar 25, 2015 #19
    I know what you mean. [tex]div E=\rho, curl E=0 <=> \nabla \nabla \phi=-\rho[/tex]
    However, which one is dependent in four electrostatics equations?
    The Gauss law? or the one of curls?
    And what are the algebraic structure of the four equations?
     
  21. Mar 25, 2015 #20

    PeterDonis

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    The curl equation (all 3 components) is an identity if ##E = - \nabla \phi##. So it contains no information. The only equation that contains information is Gauss's law, which, as you say, becomes ##\nabla^2 \phi = - \rho## and determines the scalar potential. Since the scalar potential is the only degree of freedom in the electrostatic case, everything works out fine.

    I'm not sure what you mean by this. There is one equation and one degree of freedom, as above.
     
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