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4

The problem is that they have too many solutions if the gauge is not fixed. Imagine you have some initial values, and want to solve it on the computer. Then you have to solve the equations for the next time step given the values for the previous one. But you have to compute the values for, say, four variables but have only three equations. You somehow ...


5

First, the gauge invariance means that the solutions $A_\mu(x^\alpha)$ are not unique. For every solution, the gauge transformations of it are solutions, too. That may be a problem because sometimes we want to have specific values of $A_\mu(x^\alpha)$ that answer a physical question. Second, we sometimes gauge fix because the equations simplify. For ...


12

The equations are entirely equivalent, as can be proven using Gauss' and Stokes' theorems. The integral forms are most useful when dealing with macroscopic problems with high degrees of symmetry (e.g. spherical or axial symmetry; or, following on from comments below, a line/surface integrals where the field is either parallel or perpendicular to the ...


5

Your confusion lies in failing to recognize that they are exactly the same equations. Take for example Gauss's law $$ \vec \nabla \cdot \vec E = \dfrac{\rho}{\epsilon_0}$$ You can see that there $\rho$ is the charge distribution, and in general can be a funcion of the position. Now consider a volume $V$, you can just integrate the density to obtain the ...


3

As is written here the two remaining equations follow from the Bianchi identity which says that the anti-symmetrized derivative is zero, ie. $$ \partial_{[a} F_{bc]} = \partial_{a} F_{bc}+\partial_{b} F_{ca}+\partial_{c} F_{ab} = 0 $$ (remember the $F_{\mu\nu}$ is antisymmetric itself!)


3

All this tells you is that the fields satisfies both the inategral and the differential equations. The two are related by the mathematical identities called the divergence theorem and Stokes' theorem. So which do you apply? Well, which ever one you want! If you run into an integral, you use the integral form, and if you're ever asked for the divergence or ...


2

If you define $C:q\rightarrow -q$, $P:(x,y,z)\rightarrow (-x,-y,-z)$ and $T:t\rightarrow -t$, then all the Maxwell eaquations are invariant under $C$, $P$, $T$ or any combination of them. To see this you just have to notice how the transformations act on sources and coordinates. The charge conjugation acts non trivially as \begin{align} C\rho&=-\rho,\\ ...


1

As a general note, asking notation questions without providing a reference to the original occurrence (from which we'd be able to infer the context) is an excellent recipe for an unanswerable question. In this particular case, though, it's pretty clear that it refers to the unit basis vector in the $z$ direction, $$\hat e_z=(0,0,1).$$ It arises in this ...


0

As long as $\mathbf{B}$ is a continuous (once-differentiable) function, when you look at small enough sizes, $\mathbf{B}$ has a Taylor series, the first term of which is a constant. As you let the loop size shrink, only the constant term matters. But then $\int_S \mathbf{B}\cdot d\mathbf{a}\rightarrow \mathbf{B}\int_S d\mathbf{a}\rightarrow 0$ since the area ...


0

Maxwell's equations, in their microscopic form as formulated by Lorentz, are the standard postulates of electrodynamics. From them all electromagnetic formulas and properties can be derived. The symmetry of these equations relative to spatial translation implies, as follows from Noether's theorem, the law of conservation of linear momentum of the charges ...


0

Start by taking the curl of Maxwell's third equation (for vacuum), and substituting $\vec{B}=\mu_0\vec{H}$ one can obtain, $$ \nabla^2.\vec{E} = \mu_0 \epsilon_0 \frac{\partial^2}{\partial t^2} \vec{E} $$ Similarly, by taking curl of Maxwell's fourth equation, substituting $$ \nabla\times\vec{E} = -\frac{\partial}{\partial t}\mu_0\vec{H} $$ one can ...


1

People are not understanding your question. I think you want someone to verify explicitly that the fields produced in your special case do (or don't) obey the generally valid wave equation. After all, the field produced does not look like a wave. A general solution of the wave equation for a disturbance traveling in the $x$ direction is ${\bf{E}}({\bf ...



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