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Maxwell's equations specify two vector and two scalar (differential) equations. That implies 8 components in the equations. But between vector fields $\vec{E}=(E_x,E_y,E_z)$ and $\vec{B}=(B_x,B_y,B_z)$, there are only 6 unknowns. So we have 8 equations for 6 unknowns. Why isn't this a problem?

As far as I know, the answer is basically because the equations aren't actually independent but I've never found a clear explanation. Perhaps the right direction is in this article on arXiv.

Apologies if this is a repost. I found some discussions on PhysicsForums but no similar question here.

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It isn't a problem because two of the eight equations are constraints and they're not quite independent from the remaining six.

The constraint equations are the scalar ones, $$ {\rm div}\,\,\vec D = \rho, \qquad {\rm div}\,\,\vec B = 0$$ Imagine $\vec D=\epsilon_0\vec E$ and $\vec B=\mu_0\vec H$ everywhere for the sake of simplicity.

If these equations are satisfied in the initial state, they will immediately be satisfied at all times. That's because the time derivatives of these non-dynamical equations ("non-dynamical" means that they're not designed to determine time derivatives of fields themselves; they don't really contain any time derivatives) may be calculated from the remaining 6 equations. Just apply ${\rm div}$ on the remaining 6 component equations, $$ {\rm curl}\,\, \vec E+ \frac{\partial\vec B}{\partial t} = 0, \qquad {\rm curl}\,\, \vec H- \frac{\partial\vec D}{\partial t} = \vec j. $$ When you apply ${\rm div}$, the curl terms disappear because ${\rm div}\,\,{\rm curl} \,\,\vec V\equiv 0$ is an identity and you get $$\frac{\partial({\rm div}\,\,\vec B)}{\partial t} =0,\qquad \frac{\partial({\rm div}\,\,\vec D)}{\partial t} =-{\rm div}\,\,\vec j. $$ The first equation implies that ${\rm div}\,\,\vec B$ remains zero if it were zero in the initial state. The second equation may be rewritten using the continuity equation for $\vec j$, $$ \frac{\partial \rho}{\partial t}+{\rm div}\,\,\vec j = 0$$ (i.e. we are assuming this holds for the sources) to get $$ \frac{\partial ({\rm div}\,\,\vec D-\rho)}{\partial t} = 0 $$ so ${\rm div}\,\,\vec D-\rho$ also remains zero at all times if it is zero in the initial state.

Let me mention that among the 6+2 component Maxwell's equations, 4 of them, those involving $\vec E,\vec B$, may be solved by writing $\vec E,\vec B$ in terms of four components $\Phi,\vec A$. In this language, we are left with the remaining 4 Maxwell's equations only. However, only 3 of them are really independent at each time, as shown above. That's also OK because the four components of $\Phi,\vec A$ are not quite determined: one of these components (or one function) may be changed by the 1-parameter $U(1)$ gauge invariance.

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    $\begingroup$ Lubosh, $\vec{E},\vec{B}$ are expressed via 6 time and space derivatives of $\phi$ and $\vec{A}$; that is why there is an ambiguity in potentials. $\endgroup$ – Vladimir Kalitvianski Jan 27 '12 at 18:48
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    $\begingroup$ Dear Vladimir, I have answered your question in detail. Again. There's a 1-parameter ambiguity in the 4 potentials - the U(1) gauge invariance - because locally in spacetime, the 4 potentials are only constrained by 3 equations, curl H = $j+\partial D / \partial t$. The fourth equation with currents, ${\rm div}\,\, D=\rho$, isn't independent: its time derivative follows from the previous three. The remaining 3+1 equations for $B,E$ are satisfied automatically if $B,E$ are expressed in terms of the 4 potentials, they're Bianchi identities. $\endgroup$ – Luboš Motl Jan 27 '12 at 18:54
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    $\begingroup$ Is Lubos or Lubosh more appropriate for the "hatted s"? $\endgroup$ – Nikolaj-K Jan 28 '12 at 12:00
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    $\begingroup$ Dear @Vladimir, ${\rm div}\,\vec D=\rho$ is a constraint for $\vec D$, surely in the technical sense. It is not a purely algebraic constraint; if it were, then it would be solvable and one could just erase some components of $\vec D$ immediately. Instead, it contains spatial derivatives. But this difference only allows the overall $\vec D$ in space to move by constant: independent of space. At individual points of the initial slice, the presence of derivatives is irrelevant for the counting and there's 1 constraint per point (except for one point in space) just like if it were an algebraic one $\endgroup$ – Luboš Motl Jan 31 '12 at 7:16
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    $\begingroup$ Dear @Nick, Lubos or Lubosh is surely easier to write and I don't get insulted. Many people, even those outside Central and Eastern Europe where š may be typed on keyboard, are actually able to write the character in a second or so, by copy-and-paste etc., so it's not a huge sacrifice if they write it correctly. But yes, š is pronounced as sh. $\endgroup$ – Luboš Motl Jan 31 '12 at 7:18
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I) Let us just for fun generalize OP's question to $n$ spacetime dimensions, and check how the counting of eqs. and degrees of freedom (d.o.f.) work out in this general setting. We shall use Lubos Motl's answer as a template for this part. Also we shall use a special relativistic $(-,+,\ldots,+)$ notation with $c=1$, where $\mu,\nu\in\{0,\ldots,n-1\}$ denote spacetime indices, while $i,j \in\{1,\ldots,n-1\}$ denote spatial indices. Maxwell eqs. are the following.

  1. Source-free Bianchi identities: $${\rm d}F~=~0 \qquad\qquad \Leftrightarrow \qquad\qquad \sum_{\rm cycl.~\mu,\nu,\lambda} d_{\lambda} F_{\mu\nu} ~=~0, \qquad\qquad F~:=~\frac{1}{2} F_{\mu\nu}~ {\rm d}x^{\mu} \wedge {\rm d}x^{\nu}.$$ Here $$\left(\begin{array}{c} n \cr 3\end{array}\right) {\rm~Bianchi~identities} ~=~ \left(\begin{array}{c} n-1 \cr 3\end{array}\right) {\rm~constraints}~+~ \left(\begin{array}{c} n-1 \cr 2\end{array}\right) {\rm~dynamical~eqs.} $$ $$~=~ ({\rm No~magnetic~monopole~eqs.})~+~ ({\rm Faraday's~law}). $$

  2. Maxwell eqs. with source terms: $$ d_{\mu}F^{\mu\nu}~=~-j^{\nu} .$$ Here $$n {\rm~source~eqs.}~=~1 {\rm~constraint} ~+~ (n-1) {\rm~dynamical~eqs.}$$ $$~=~({\rm Gauss'~law}) ~+~ ({\rm Ampere's~law~with~displacement~term}).$$

We have used the terminology that a dynamical eq. contains time derivatives, while a constraint does not. So the number of dynamical eqs. is

$$ \left(\begin{array}{c} n-1 \cr 2\end{array}\right)~+~(n-1)~=~ \left(\begin{array}{c} n \cr 2\end{array}\right),$$

which precisely matches

$${\rm the~number~} \left(\begin{array}{c} n \cr 2\end{array}\right) {\rm~of~} F_{\mu\nu} {\rm~fields}$$ $$~=~\left(\begin{array}{c} n-1 \cr 2\end{array}\right){~\rm magnetic~fields~} F_{ij} ~+~(n-1) {\rm~electric~fields~}F_{i0} .$$

Maxwell eqs. with source terms imply the continuity eq.

$$ d_{\nu}j^{\nu} ~=~-d_{\nu}d_{\mu}F^{\mu\nu}~=~0,\qquad\qquad F^{\mu\nu}~=~-F^{\nu\mu},$$

so one must demand that the background sources $j^{\nu}$ obey the continuity eq.

For consistency, the time derivative of each of the constraints should vanish. In the case of the no-magnetic-monopole-eqs., this follows from Faraday's law. In the case of Gauss' law, this follows from the modified Ampere's law and the continuity eq.

II) The previous section (I) made the counting in terms of the $\left(\begin{array}{c} n \cr 2\end{array}\right)$ field strengths $F_{\mu\nu}$. In terms of the $n$ gauge potentials $A_{\mu}$, the counting goes as follows. The Bianchi identities are now trivially satisfied,

$$F~=~{\rm d}A\qquad\qquad A~:=~A_{\mu}~ {\rm d}x^{\mu}. $$

There are still the $n$ Maxwell eqs. with source terms

$$ (\Box\delta^{\mu}_{\nu}-d^{\mu}d_{\nu})A^{\nu}~=~-j^{\mu} , \qquad\qquad \Box~:=~d_{\mu}d^{\mu}. $$

There is a single gauge d.o.f. because of gauge symmetry $A \to A + {\rm d}\Lambda$ and $F \to F$. If one gauge-fixes using the Lorenz gauge condition

$$d_{\mu}A^{\mu}~=~0, $$

the Maxwell eqs. become $n$ decoupled wave equations

$$ \Box A^{\mu}(x)~=~-j^{\mu}(x). $$

By a spatial Fourier transformation, these become decoupled linear second-order ODEs with constant coefficients,

$$ (d^2_t+\vec{k}^2) \hat{A}^{\mu}(t;\vec{k})~=~\hat{j}^{\mu}(t;\vec{k}) , $$

which, starting from some initial time $t_0$, may be solved for all times $t$, cf. OP's question. [One should check that the solution

$$\hat{A}^{\mu}(t;\vec{k}) ~=~\int {\rm d} t^{\prime} ~G(t-t^{\prime};\vec{k})~\hat{j}^{\mu}(t^{\prime};\vec{k}), \qquad\qquad (d^2_t+\vec{k}^2)G(t-t^{\prime};\vec{k})~=~\delta(t-t^{\prime}),$$

satisfies the Lorenz gauge condition. This follows from the continuity eq.]

III) It is interesting to derive the complete solution $\tilde{A}^{\mu}(k)$ in $k^{\nu}$-momentum space without gauge-fixing. The Fourier-transformed Maxwell eqs. read

$$M^{\mu}{}_{\nu}~\tilde{A}^{\nu}(k)~=~\tilde{j}^{\mu}(k), \qquad\qquad M^{\mu}{}_{\nu}~:=~k^2\delta^{\mu}_{\nu} -k^{\mu}k_{\nu}. $$

To proceed one must analyze the matrix $M^{\mu}{}_{\nu}$ for fixed $k^{\lambda}$. There are three cases.

  1. Constant mode $k^{\mu}=0$. Then the matrix $M^{\mu}{}_{\nu}=0$ vanishes identically. Maxwell eqs. are only possible to satisfy if $\tilde{j}^{\mu}(k=0)=0$ is zero. The gauge potential $\tilde{A}_{\mu}(k=0)$ is not restricted at all by Maxwell eqs., i.e., there is a full $n$-parameter solution.

  2. Massive case $k^2\neq 0$. The matrix $M^{\mu}{}_{\nu}$ is diagonalizable with eigenvalue $k^2$ (with multiplicity $n-1$), and eigenvalue $0$ (with multiplicity $1$). The latter corresponds to a pure gauge mode $\tilde{A}^{\mu}~\propto~k^{\mu}$. The complete solution is a $1$-parameter solution of the form $$\tilde{A}^{\mu}(k) ~=~\frac{\tilde{j}^{\mu}(k)}{k^2}~+~ik^{\mu}\tilde{\Lambda}(k).$$ Apart from the source term, this is pure gauge.

  3. Massless case $k^2=0$ and $k^{\mu}\neq 0$. The matrix $M^{\mu}{}_{\nu}$ is not diagonalizable. There is only eigenvalue $0$ (with multiplicity $n-1$). Maxwell eqs. are only possible to satisfy if the source $\tilde{j}^{\mu}(k)=\tilde{f}(k)k^{\mu}$ is proportional to $k^{\mu}$ with some proportionality factor $\tilde{f}(k)$. In that case Maxwell eqs. become $$ -k_{\mu}\tilde{A}^{\mu}(k)~=~\tilde{f}(k). $$ Let us introduce an $\eta$-dual vector$^1$ $$k^{\mu}_{\eta}~:=~(-k^0,\vec{k})\qquad {\rm for}\qquad k^{\mu}~=~(k^0,\vec{k}).$$ Note that $$k_{\mu}~k^{\mu}_{\eta}~=~(k^0)^2+\vec{k}^2$$ is just the Euclidean distance square in $k^{\mu}$-momentum space. The complete solution is an $(n-1)$-parameter solution of the form $$\tilde{A}^{\mu}(k) ~=~-\frac{k^{\mu}_{\eta}}{k_{\nu}~k^{\nu}_{\eta}}\tilde{f}(k) ~+~ik^{\mu}\tilde{\Lambda}(k)~+~\tilde{A}^{\mu}_{T}(k).$$ The term proportional to $k_{\mu}$ is pure gauge. Here $\tilde{A}^{\mu}_{T}(k)$ denote $n-2$ transversal modes, $$k_{\mu}~\tilde{A}^{\mu}_{T}(k)~=~0, \qquad\qquad k_{\mu}^{\eta}~\tilde{A}^{\mu}_{T}(k)~=~0. $$ The $n-2$ transversal modes $\tilde{A}^{\mu}_{T}$ are the only propagating physical d.o.f. (electromagnetic waves, photon field).

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$^1$ Longitudinal and timelike polarizations are in the massless case proportional to $k^{\mu}\pm k^{\mu}_{\eta}$, respectively.

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  • $\begingroup$ Why do the transversal modes satisfy $k_{\mu}^{\eta}~\tilde{A}^{\mu}_{T}(k)~=~0$? Maxwell equs. demand $k_{\mu}~\tilde{A}^{\mu}_{T}(k)~=~0$. Where does this extra constraint come from? $\endgroup$ – image Jan 13 '17 at 17:06
  • $\begingroup$ It may be viewed as part of the definition of what transversal means. $\endgroup$ – Qmechanic Jan 13 '17 at 17:10
  • $\begingroup$ 1. But there is no actual constraint that tells me that they have to be transversal? 2. Also: What is the argument for $k^2 = 0$ if not the Lorenz-gauge Maxwell equs. $\Box A^\mu = 0$? $\endgroup$ – image Jan 13 '17 at 17:12
  • $\begingroup$ 1. No, there is not. 2. I assume you are referring to Section III: The massless case $k^2=0$ is one possibility. There's also a massive case $k^2\neq 0$. $\endgroup$ – Qmechanic Jan 13 '17 at 17:21
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Equations are written for any time $t$ and there is no need to "prove" their validity at any time. These equations are the experimental laws and are, of course, consistent at any time. The constraints are imposed here not to the fields, but to the electric and magnetic charges. The charges do not have sources/sinks so the derived equations like $\partial\rho/\partial t + \rm div \vec{j}=0$ say namely that and are called the charge conservation laws. (They are are an experimental fact.) The charge conservation laws do not determine the charge dynamics; for the latter the "mechanical" equations exist. In case of one elementary charge $q$, its conservation means its time-independence: $\frac{dq}{dt}=0$ which is not usually written as an additional equation, but used as its solution $q=const$ in the "mechanical" equations.

So you have six equations for fields and two as conservation laws for charges.

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  • $\begingroup$ Can't we see the following two views as equivalent: 1. There are six equations for the fields and there are two additional constraints on the charges, i.e. the conservation laws. 2. There are six dynamical equations for the evolution of the fields and the two additional boundary conditions which need to be satisfied at the initial instant of time by all physically real electromagnetic fields.? $\endgroup$ – Dvij Mankad Nov 4 '18 at 16:54
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Maxwell's equations are indeed redondant, if one works with the normal variables the redundecies are eliminated. A very clear discussion is found into:

Photons and atoms: introduction to quantum electrodynamics Claude Cohen-Tannoudji, Jacques Dupont-Roc, Gilbert Grynberg

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  • $\begingroup$ Why is this an improvement on Lubos and Qmechanic's fine answers? $\endgroup$ – Ron Maimon Jul 18 '12 at 2:19
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The Maxwell's equations do not over-determine the electric and magnetic fields. This becomes clearer if we rewrite the four Maxwell's equations into one using geometric algebra: $$(c^{-1}\partial_t + \vec\nabla)(\vec E + i\zeta\vec H) = \zeta(\rho c +\vec j)$$, where the vector products follow the Pauli identity $\vec a\vec b = \vec a\cdot\vec b + i\vec a\times \vec b$ . In principle, we can invert the Maxwell's equation to solve for the electromagnetic field $\vec E + i\zeta\vec H$, by applying boundary conditions.

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  • $\begingroup$ You can write in TeX by enclosing the texcode with dollar signs (inline), or double dollars. $\endgroup$ – Manishearth Feb 7 '12 at 14:03
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Here's a related question I always throw at students. In free space you can convert Maxwell's equation into 2 vector Helmholtz equations, one for E and one for B. So how come they are decoupled? It would seem that we could calculate E and B separately. The clue is that in free space, to have non-zero fields at all you must specify some boundary conditions. And the boundary conditions must be consistent with Maxwell's equations. So the transverse, coupled nature of the fields comes from the BC and is propagated into free space.

Incidentally, for finite beams the E and B fields need not be transverse to one another (i.e., there are longitudinal fields). This makes working with finite beams a lot harder than working with unphysical plane waves.

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This is easy to see if you use the Maxwell equations to arrive at the decoupled, inhomogeneous wave equations for the fields, $$\begin{split}\Box \vec{E} &= - \mu_0 \frac{\partial \vec{J}}{\partial t} - \vec{\nabla} \frac{\rho}{\varepsilon_0},\\ \Box \vec{B} &= \mu_0\vec{\nabla}\times \vec{J}, \end{split} $$ with $\Box \equiv \frac{1}{c^2} \frac{\partial^2}{\partial t^2} - \nabla^2$ the dalembertian. This derivation requires the use of all of Maxwell's equations and a solution exists and is uniquely defined if we use appropriate boundary conditions, therefore Maxwell's equations are not independent.

A hint for their dependence of one on another is the Helmholtz theorem, provided that the sources are localized. According to the theorem, a field is uniquely defined if both its divergence and its curl are known, i.e. the theorem defines 3 functions from 4 dependent equations.

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