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Independent equations of maxwell equations

  1. Aug 23, 2011 #1
    Maxwell equations are 8.

    we need them to determine electric and magnetic field.
    the component of magnetic and electric field are 6.

    in linear algebra say for solving a system of equation with n free variable, we need n equations to solving it uniquely.

    here arises a question and that, why do we need 8 Maxwell equations to determine 6 field component?

    with a little attention to wave equation, we can understand they are 6 equations to determine component of field!

    with deep thinking to Maxwell equation, can be realized curl and divergence of field appear because of Helmholtz theorem, so the about question in general is true for Helmholtz theorem.
     
  2. jcsd
  3. Aug 23, 2011 #2

    Hootenanny

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    Well, there are actually only 4 "different" Maxwell equations. The integral equations can be transformed to the differential equations and vice-versa.

    The number of equations required to determine particular fields depends on the problem at hand. One can make use of various symmetries and/or boundary conditions as well as conservation laws. It is simply wrong to say that we need to use all of Maxwell's equations to determine a given field. Sometimes we can use just one.
     
  4. Aug 23, 2011 #3
    no i think take it mistake.

    we have four Maxwell equations that two of them are vector equation which each of them has three equations. each of curl equations.


    i speak in general case not at special issue with specific symmetry.
     
  5. Aug 23, 2011 #4

    Dale

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    There is also the potential formulation of Maxwell's equations. That formulates Maxwell's equations in terms of a scalar potential equation (1 equation in 1 unknown) and a vector potential equation (3 equations in 3 unknowns). That seems to match well the number of equations and unknowns, but then you can do a gauge transformation and get a different set of potentials which describe the exact same fields.
     
  6. Aug 24, 2011 #5
    This is an excellent question. With only six unknowns and 8 equations in Maxwell's equations, you would think that two of the equations are either over-specifying the problem and will lead to contradictions, or are linearly dependent on the other equations and are therefore useless.

    The answer lies in the fact that there are derivatives in these equations (in the curl and divergence operators) and derivatives destroy information in a sense. The simplest example is if I have the equation df/dx = 3, I cannot uniquely determine the function f. It is f = 3x + C. Yes, a boundary condition can help determine C in this particular case, but when you go to curl operators, boundary condition are not enough. You can think of the curl and divergence operators as orthogonal operators so that their dot product is zero. What this means is that any general vector field has a curling part, a diverging part, and a residual part. To uniquely solve for the vector, we must write down an equation determining its curling part, its diverging part, and its residual part. The divergence equations determine the diverging part of the E and B vector fields in Maxwell's equations, the curl equations determine the curling part of E and B, and the boundary conditions determine the residual part.

    By the way, transforming Maxwell's equations into the potentials formulation does not reduce the number of equations (it does reduce the number of useful equations), you still have some equations sitting idly by, such as the Gauge equation.
     
  7. Aug 24, 2011 #6
    Strictly speaking, there are always 8 Maxwell equations (when taking vector components separately), and they must all be satisfied no matter how simple the problem. If the problem is an electrostatics problem so that B = 0 and dE/dt = 0, then most of Maxwell's equations reduce down to 0 = 0. True, these equations are not entirely useful, but strictly speaking, they still exist and they still must be satisfied.
     
  8. Aug 24, 2011 #7

    Hootenanny

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    Fair point.

    Although in actually working with the two vector equations it is usually easier (in my experience) to treat them as vector equations, rather than trying to deal with each component separately, given that we are dealing with curls.

    I do take you point, however.
     
  9. Aug 24, 2011 #8
    i can't your answer chrisbaird

    i ask why do we have 8 equations for determining 6 component of field in ME rather than we have 6 equations in Wave Equations

    but you say
    The answer lies in the fact that there are derivatives in these equations (in the curl and divergence operators) and derivatives destroy information in a sense

    WE also consist of vector operation, but they are 6
     
  10. Aug 24, 2011 #9

    Born2bwire

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    I think chrisbaird hit the nail on the head here. In many situations the equations can be shown to have redundancy in them but I do not believe that this is so in general. When you have derivatives, you add extra degrees of freedom in the solution because you lose information (such as the constants when doing a first order differentiation). So to uniquely determine the final solution, you need additional information in the form of boundary conditions.

    However, Maxwell's Equations implicitly contain the boundary conditions. Take a look at this simple derivation of the boundary conditions:

    http://www.amanogawa.com/archive/docs/EM5.pdf

    The boundary conditions are all derived by inspection of Maxwell's Equations. So there must be additional information or constraints imbedded into the equations and this could account for the fact that we have eight equations for six unknowns.
     
  11. Aug 25, 2011 #10
    The wave equations do not completely specify the fields. They are essentially only kinematic. The original Maxwell equations are also dynamic. For example, derive plane waves in vacuum using the wave equations. Now try deriving the relative directions of the E and B fields using only the wave equations. You can't do it. You have to go back to the original Maxwell equations to derive that for plane waves in vacuum, the E field vector, the B field vector, and the wavevector are all mutually perpendicular (in other words the plane is transverse.) This is only one example showing that because the fields wave equations consist of only 6 equations instead of the original 8 in Maxwell's equations, they are incomplete.
     
  12. Aug 25, 2011 #11
    Let's be careful. You can derive from Maxwell's equations boundary condition equations linking two adjacent regions, but that is not what I meant. I meant that an actual boundary value must be supplied externally in addition to the Maxwell's equations to uniquely determine the fields. This is the case with any differential equations. Even if the region of interest is unbounded (infinite), we still have to mathematically provide a boundary condition to get a unique solution. The typical one is that the field strengths are finite (or even zero) out at infinity. But this is not why Maxwell's equations contain 8 instead of 6 equations, this is just an explanation of why boundary values are needed. The reason you need 8 equations is because the curl operator does not uniquely determine a field's solution, it only determines the curling part. Put another way, the divergence of the curl is always zero, so the solution to ∇.B = 0 is not some unique function B = Bdiverging but is the infinite set of functions B = Bdiverging + ∇xBcurling where we can choose any Bcurling in the world and the original equation will still be satisfied. So there must be another equation to determine the curling part of the field.

    BOTTOM LINE: The concept of 6 linearly independent equations being necessary and sufficient to determine 6 unknowns is only true in linear algebra. When you use curl and divergence operators, you have moved beyond linear algebra.
     
  13. Aug 25, 2011 #12

    Born2bwire

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    Yeah I forgot. In the homogeneous isotropic case we still need a single boundary condition which can be provided in the form of the Sommerfeld Radiation Condition. This is an externally derived boundary condition although it is derived from simple deduction.
     
  14. Aug 25, 2011 #13

    marcusl

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    Nicely explained, chrisbaird.
     
  15. Aug 28, 2011 #14
    You can see a paper.

    http://arxiv.org/abs/1002.0892
    A discussion to contradiction of the number of unknown variables and the number of Maxwell equations
     
  16. Aug 28, 2011 #15

    vanhees71

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    I've only read the abstract, and from that I'd not waste my time to read the paper.

    Maxwell's equations can be grouped into two parts, the homogeneous and the inhomogeneous equations. Let's start with the homogeneous ones. These are Faraday's Law of Induction and the absence of magnetic poles (I'm using Heaviside-Lorentz units with [itex]c=1[/itex]):

    [tex]\partial_t \vec{E} + \vec{\nabla} \times \vec{B}=0, \quad \vec{\nabla} \cdot \vec{B}=0.[/tex]

    From these equations it follows from Helmholtz's fundamental theorem of vector calculus that there exists a scalar and a vector potential such that

    [tex]\vec{E}=-\partial_t \vec{A}-\vec{\nabla} \Phi, \quad \vec{B}=\vec{\nabla} \times \vec{A}.[/tex]

    Now, [itex](\vec{E},\vec{B})[/itex] is the physically observable electromagnetic field, and the potentials are auxiliary quantities to identically fulfill the homogeneous Maxwell equations. Now, for a given em. field, the potentials are determined only up to a scalar field since if [itex]\Phi[/itex] and [itex]\vec{A}[/itex] define this em. field, also the new fields

    [tex]\Phi'=\Phi+\partial_t \chi, \quad \vec{A}'=\vec{A}-\vec{\nabla} \chi[/tex]

    lead to the same em. field. To make the solution for the potentials at a given em. field unique, we have to impose a constraint on the four potentials, e.g., the Lorenz-gauge condition,

    [tex]\partial_t \Phi+\vec{\nabla} \cdot \vec{A}=0.[/tex]

    Thus we have not four but only three independent potentials to describe the em. field.

    Now, we have also the inhomogeneous Maxwell equations, namely the Ampere-Maxwell Law and Gauss's Law:

    [tex]\partial \vec{B}-\partial_t \vec{E}=\vec{j}, \quad \vec{\nabla} \cdot \vec{E}=\rho.[/tex]

    In Lorenz gauge this leads to the wave equations,

    [tex]\Box \Phi=\rho, \quad \Box \vec{A}=\vec{j},[/tex]

    and the charge distribution and current density have to obey the consistency condition

    [tex]\partial_t \rho+\vec{\nabla} \cdot \vec{j}=0[/tex]

    such that the Lorenz-gauge condition is valid, i.e., gauge invariance dictates that the electric charge is a conserved quantity. This constraint also immediately follows from the original gauge invariant form of the inhomogeneous Maxwell equations. The em. field at the presence of charges thus has three independent field-degrees of freedom, i.e., the four potentials with one gaug-fixing condition.

    For free fields, however, one gauge-fixing constraint is not enough. Using the Lorenz-gauge condition, for free fields one has still the freedom to do a restricted gauge transformation with

    [tex]\Box \chi=0[/tex]

    This additional gauge freedom can be used to impose another constraint, i.e., the radiation-gauge condition [itex]\Phi=0[/itex], i.e., we then have the two conditions

    [tex]\Phi=0, \quad \vec{\nabla} \cdot \vec{A}=0.[/tex]

    Thus, for free fields, we only have two independent field-degrees of freedom. Writing the em. fields in terms of plane-wave modes one finds that the free em. field is transverse, i.e., for each wave vector [itex]\vec{k}[/itex] there are two linearly independent plane-wave solutions with [itex]\vec{A} \perp \vec{k}[/itex]. An em. wave has only two, not three, independent polarization states.

    The deeper reason for that is the special structure of the massless realizations of the Poincare symmetry of Minkowski space in relativistic quantum field theory.
     
  17. Aug 28, 2011 #16
    sadegh. Maxwell's equations consist of 8 equations and 10 unknowns, not just 6 unknowns for the fields.

    Ten unknowns comes from including the charge and current density amoung the number of unknowns. There is one unknown for charge density and 3 unknowns for current density.

    However, the number of unknowns can be reduced by 2 by introducing the 4-vector potential, A. The magnetic and electric fields are not completely independent, so 6 is too many.

    The Faraday tensor, F=dA (see Wikipedia), replaces the 3 electric field components and 3 magnetic field components with 4 independent quantities giving us 8 equations and 8 unknowns.
     
    Last edited: Aug 28, 2011
  18. Aug 29, 2011 #17
    The charge and current density are not unknowns. If they are unknowns, then everything is unknown and there is nothing to solve.

    The Faraday tensor has 16 components (it is a 4 by 4 matrix), but many of these components are redundant. It only contains 6 independent elements, the different spatial components of the E and B fields.
     
  19. Aug 29, 2011 #18
    Going back to the OP:

    Below, I will refer to both the microscopic and the macroscopic Maxwell equations; if anybody is unfamiliar with these terms, please see the first two tables here: http://en.wikipedia.org/wiki/Maxwell's_equations.

    The OP implicitly talks about the microscopic Maxwell equations. At first glance, indeed there seem to be eight scalar equations for only six scalar unknowns; however, this is actually not the case. The reason is that the two divergence equations are not independent of the two curl equations, i.e. the former can be derived from the latter. Simply take the divergence of Ampere's law and you get Gauss' law. Similarly, take the divergence of Faraday's law and you get the law of no magnetic monopoles. Since the two divergence equations are simply consequences of the two curl equations, we find that there are actually only six independent scalar equations for the six unknown scalar field quantities.

    For the macroscopic Maxwell equations, by similar arguments, you wind up with six independent scalar equations for twelve (!) scalar unknowns. The remedy is that the constitutive relations give you six more independent scalar equations.

    Hope that clears everything up!
     
  20. Aug 30, 2011 #19
    chrisbaird.

    OK. better said, there are 8 variables and 8 scalar equations.

    however, you’re taking the question to another level. For this it’s a lot simpler after having converted maxwell’s equations to differential forms. It’s far more obvious and natural.

    charge density, current density and the fields are completely specified by the 4 vector potential for all spacetime points.

    the electric and magnetic fields are specified from F=dA. F is the exterior derivative of A. Using J=d*F, the electric charge and current density are specified.** So there are really only 4 independent variables. J=d*F obtains the 4 maxwell equations and d2J=0 4 more. However, the last is a mathematical identity saying there are no magnetic charges or currents.

    going backwards, beginning with the charge and current density, is harder. Integration introduces constants of integration. Physically, these are boundary conditions or gauges. I’m not as comfortable with the integral forms, though the resultant gauging is fairly obvious once you have it.

    to get F when given J, schematically, *F = ∫J + θ. The gauge θ is a 4x4 tensor with 6 independent terms.

    (*F)μν dxμdxν = (∫Jρμν dxρ) dxμ dxν + θμν dxμdxν

    with the gauge included, the differential equations are J = d(*F + θ), where dθ = 0.

    so, in the typical problems usually encountered where charge and current densities are the "knowns", to get the fields knowing the charge and currents, 6 additional variables need to be specified by setting a gauge or giving boundary condtions.

    also, A = ∫F + φ, where φ is a dual vector, so 4 additional independent terms arise.

    Aν dxν = (∫Fμν dxμ) dxν + φν dxν

    F = d(A+φ), where dφ = 0.

    This second gauge is not encountered in the usual problems. It's just known that A is regaugable and probably cannot be directly measured to an exact value, but known only up to a gauge offset.

    ** reference: exterior derivative, Hodge dual, and “all exact forms are closed”.

    For inclusion of magnetic monopoles, 'A' can have complex valued entries, but things are a little more convoluted. F is not only complex but, as I recall, regauges itself, i.e.: *iF regauges F.
     
    Last edited: Aug 30, 2011
  21. Sep 7, 2011 #20
    This has been bothering me. Doesn't this self-gauging put sevier restraints on magnetic monopole fields? With a little thought, inclusion of magnetic charge seems to place constaints on boundary conditions. This seems a little odd as to how this could occur. This could lead to a no-go theorem for classical magnetic charge, couldn't it?
     
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