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I know how to derive the formula, but, in view of freedom, the left hand side $E$ has 3 components while the right hand side provides 1 constraint from $\varphi$ and 3 from $A$. I mean, why does a 3 variables unknown vector needs 4 constraints to determine? Is there any abundance or dependence?

And I have no idea what the relevant theorems about connection between solutions of equations and freedom are? (Does "freedom" mean number of unknown variables minus constraints or equations?)

Let me put it another way, in physics, $E$ is uniquely determined by two potentials, namely scalar $\varphi$ and vector $A$. But why do we need 4 components to determine only 3 unknowns?

chridam
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Shuchang
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  • not $(x,y,z)$ and $t$? I don't get what you mean. – eccstartup Aug 31 '13 at 03:01
  • $\varphi$ is a function of all of the spacetime variables just like $E$ and $A$. I'm not sure wha tyou're asking. – Cameron Williams Aug 31 '13 at 03:03
  • I think the Physics site will be a better match for this question. – Maesumi Aug 31 '13 at 03:15
  • @Cameron I'm mainly asking about freedom and constraints. Please see my complement in the last paragraph. – Shuchang Aug 31 '13 at 03:22
  • @eccstartup Not $(x,y,z,t)$ but $(\varphi,A_1,A_2,A_3)$ – Shuchang Aug 31 '13 at 03:22
  • If the magnetic potential is defined as: the magnetic field is given by $B = \nabla\times A$, then I think the correct way of writing the electric field is $E = -\nabla \phi - \frac{\partial A}{\partial t}$ – Shuhao Cao Aug 31 '13 at 03:27
  • @ShuhaoCao Cao Yes, but whatever, this doesn't affect the problem – Shuchang Aug 31 '13 at 03:30
  • @ShuchangZhang For me the relation is more like to determine $\phi$ and $A$ from $E$ and $B$, not the reverse. If $\phi$ and $A$ are given, expression known, then $E$ is determined anyway, there is no differential equation on $E$. – Shuhao Cao Aug 31 '13 at 04:25

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A better way to look at this is that the four potential $A$ gives the Faraday by exterior differentiation in $\mathbb{R}^4$; $dF=A$. Then the sources are coupled to the Hodge dual by $d *F = \mu_o *J$ where $J$ is the four-current. In particular, $A = \phi dt + \omega_{\vec{A}}$ and $J = \rho dt+ \omega_{\vec{J}}$ and $\omega_{\vec{F}} = F_1dx+F_2dy+F_3dz$ is the work-form correspondence of vectors and forms. Here I denote $\rho = dQ/dV$ the charge-density and $\phi$ the electric potential and $\vec{A}$ the vector potential. The exterior derivatives above encode the curl and divergence since: $$ d \omega_{\vec{F}} = dt \wedge \frac{\partial \vec{F}}{\partial t}+\Phi_{\nabla \times \vec{F}} $$ and $\Phi_{\vec{F}} = F_1 dy \wedge dz+F_2dz \wedge dx+F_3 dx \wedge dy $. Or, more to the point: $$ F=dA = d\phi \wedge dt + \Phi_{\nabla \times \vec{A}} = \omega_{\nabla \phi} \wedge dt +dt \wedge \frac{\partial \vec{A}}{\partial t}+ \Phi_{\nabla \times \vec{A}} = -dt \wedge \omega_{\vec{E}}+\Phi_{\vec{B}}$$ I might be off by a sign in what follows, but basically the Hodge dual flips work and flux forms $$ *F= -dt \wedge \omega_{\vec{B}}+\Phi_{\vec{E}} $$ and so $d*F = \mu_o *J$ gives a curl and divergence for $\vec{E}$ and $\vec{B}$ whereas the $dF=0$ gives a curl and divergence for the $\vec{B}$ and $\vec{E}$. These four equations are of course Maxwell's Equations. The asymmetry between the equations is an experimental fact (so far as we know) there is no magnetic charge. Look up "complexion" in Misner Thorne and Wheeler for more on how the equations modify in the presence of magnetic charge and current.

Ok, so let's do some counting in view of all this. We have 6 functions of spacetime which are coupled under Maxwell's Equations. These are $E_1,E_2,E_3,B_1,B_2,B_3$. These are derived from the four-potential which has 4 functions of spacetime; $\phi, A_1,A_2,A_3$. Even so, there is freedom in the choice of the potential. The choice of gauge allows us to shift $A$ by $dg$ since $d(A+dg)=dA$ so $A+dg$ generates the same physical fields (ignoring Aharanov Bohm...).

In any event, I don't think it's right to think of $\vec{E}$ and $\vec{B}$ as independent variables. They're not. We must respect Maxwell's Equations so they're coupled and when you write the potential and vector potential you fix both the electric and magnetic field.

Perhaps there is something to your question, but I don't understand it yet.

James S. Cook
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  • Thanks. Now my question becomes since electromagnetic field tensor $F$ has 6 independent components, intuitively we need 6 equations to determine them uniquely, right? However, Maxwell's equation $$dF=0, dF=\mu_0J$$ gives $(4+4)=8$ equations. That means number of equations is greater than number of unknowns. Generally, this system doesn't admit a solution unless some dependencies involved. – Shuchang Aug 31 '13 at 05:29
  • @ShuchangZhang the number of independent components is not usually the number of equations. For example, $A^T=-A$ for $3 \times 3$ matrices gives $6$ independent equations and the dimension of the antisymmetric subspace is $3$.Honestly, I'm not sure where our counting is supposed to begin. I mean, what do you take as the starting point? In my hypothetical, the backdrop is the 9-dimensional set of $3 \times 3$ matrices. In your counting problem, what is the starting point? What space are we imposing conditions upon? – James S. Cook Sep 01 '13 at 00:55