Ampere's Law with Maxwell's correction is equivelant to Ampere's Law?

[Question Man]

Is it true that Ampere's Law with Maxwell's correction is equivelant to Biot-Savart Law?
Under what assumptions?

 

[vanhees71]

Biot-Savart holds for stationary fields, where Maxwell's displacement current doesn't play a role, i.e., you have the two magnetostatic equations (here for simplicity I neglect medium effects, i.e., use the vacuum equations in Heaviside-Lorentz units)

\vec{\nabla} \times \vec{B}=\frac{\vec{j}}{c}, \quad \vec{\nabla} \cdot \vec{B}=0.

From the second equation, which says that there are no magnetic charges, we see that the magnetic field is a pure solenoidal field, i.e., there is a vector potential, \vec{A} such that

\vec{B}=\vec{\nabla} \times \vec{A}.

For a given magnetic field, the vector potential is only determined up to the gradient of a scalar field, and thus we can choose a constraint on the potential. In the so called Coulomb gauge one assumes

\vec{\nabla} \cdot \vec{A}=0.

Plugging now this ansatz into the first equation, which is Ampere's Law, we get (in Cartesian coordinates!)

\vec{\nabla} \times (\vec{\nabla} \times \vec{A})=\vec{\nabla} (\vec{\nabla} <br /> \cdot \vec{A})-\Delta \vec{A}=-\Delta \vec{A}=\frac{\vec{j}}{c}.

Now this looks like the equation of electrostatics for each Cartesian component of the vector potential. From this we get immediately the solution in terms of the Green's function of the Laplacian:

\vec{A}(\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x}&#039; \frac{\vec{j}(\vec{x}&#039;)}{4 \pi c |\vec{x}-\vec{x}&#039;|}.

Taking the curl of this solution, directly yields the Biot-Savart Law,

\vec{B}(\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x}&#039; \vec{j}(\vec{x}&#039;) \times \frac{\vec{x}-\vec{x}&#039;}{4 \pi c |\vec{x}-\vec{x}&#039;|^3} .

 

[Question Man]

Muchas Gracias!