Derivatives of EM Four-Potential: Euler-Lagrange to $\nabla \times B$

[Gene Naden]

So the Euler-Lagrange equations give $\partial _\mu ( \partial ^\mu A^\nu - \partial ^\nu A^\mu ) = J^\nu$ with $B=\nabla \times A$. I want to convert this to $\nabla \times B - \frac{\partial E}{\partial t} = \vec{j}$. I reckon I am supposed to use the Minkowski metric to raise or lower indices, but am not sure how. I want to get from $(\partial ^\mu A^\nu - \partial ^\nu A^\mu )$ to $\nabla \times A$.

 

[PeterDonis]

I want to get from $(\partial ^\mu A^\nu - \partial ^\nu A^\mu )$ to $\nabla \times A$.

You can't, at least not if $\nabla \times A$ is supposed to contain all of the information; $(\partial ^\mu A^\nu - \partial ^\nu A^\mu )$ is a 4-vector expression and $\nabla \times A$ is a 3-vector expression, so the first contains information that the second does not.

Try writing out the components of $(\partial ^\mu A^\nu - \partial ^\nu A^\mu )$ explicitly, making sure to include $\partial_0$ and $A^0$. You should find that some of the components work out to $\nabla \times A$, where $A$ is a 3-vector; but there are also other components that mean something different. Once you figure out what those other components mean, you will have the solution to your problem.

 

[Orodruin]

You can't, at least not if ∇×A∇×A\nabla \times A is supposed to contain all of the information; (∂μAν−∂νAμ)(∂μAν−∂νAμ)(\partial ^\mu A^\nu - \partial ^\nu A^\mu ) is a 4-vector expression and ∇×A∇×A\nabla \times A is a 3-vector expression, so the first contains information that the second does not.

Not only is it an expression using 4-vectors instead of 3-vectors. It is an expression that describes the components of a (2,0) tensor while $\nabla \times \vec A$ is a 3-vector. Yoy can use the permutation symbol to relate the field tensor to the 3-vectors $\vec E$ and $\vec B$.

 

[vanhees71]

Of course one can convert from the Minkowski-covariant tensor notation to the non-covariant (1+3) notation (in a fixed (!) inertial reference frame). Indeed, the contravariant spatial components of the Faraday tensor is directly mapped one to one to the (1+3) Cartesian components $\vec{B}$. You have (latin indices run over the spatial indices only, i.e., $j \in \{1,2,3\}$ etc.):
$$F^{jk}=\partial^{j} A^{k} -\partial^{k} A^j=-\partial_j A^k + \partial_k A^j=\epsilon_{ikj} (\vec{\nabla} \times \vec{A})_i.$$
The usual difficulty is to keep in mind that in the (1+3) formalism
$$\vec{\nabla} = \vec{e}_j \frac{\partial}{\partial x^j}=\vec{e}_j \partial_j=-\vec{e}_j \partial^j.$$
One should also note that the notation is not easily made consistent since in the (1+3) formalism one usually writes all indices as lower indices, because in Cartesian components you have V_j=V^j, but in SR of course V_j=-V^j.

The difficulty is natural since the components $\vec{E}$ and $\vec{B}$ are vectors in the (1+3) formalism (i.e., their components behave as vector components under rotations in the fixed inertial frame), but they are not spatial components of four-vectors but in the 4-formalism are components of the antisymmetric Faraday tensor.

For completeness, here's the relation between the temporal-spatial Faraday tensor components with the (1+3) object $\vec{E}$ (electric field):
$$F^{j0}=\partial^j A^0-\partial^0 A^j=-\partial_j A^0-\partial_0 A^j=E_j,$$
i.e.,
$$\vec{E}=-\frac{1}{c} \dot{\vec{A}}-\vec{\nabla} A^0,$$
as is well-known from the (1+3) formalism of E-dynamics.