A Why electromagnetic tensor (Faraday 2-form) is exact? (and not closed)

AI Thread Summary
The discussion centers on the nature of the electromagnetic field tensor (Faraday 2-form) F, questioning why it is considered exact (F=dA) rather than merely closed. It acknowledges that while all exact forms are closed, not all closed forms are exact, raising the issue of the conditions under which F is defined as exact. The Poincaré lemma indicates that F can be expressed as the differential of a 1-form A locally, but this does not guarantee global exactness. The example of a potential vortex illustrates a scenario where F is closed globally but not exact due to the topology of the space involved. This highlights the complexities in the relationship between closed and exact forms in electromagnetic theory.
phoenix95
Gold Member
Messages
81
Reaction score
23
Following from Wikipedia, the covariant formulation of electromagnetic field involves postulating an electromagnetic field tensor(Faraday 2-form) F such that
F=dA
where A is a 1-form, which makes F an exact differential form. However, is there any specific reason for expecting F to be exact? Could it be the case that in general, F is a closed differential form, but by virtue of the Poincare lemma we define F to be this way?
 
Physics news on Phys.org
That's just the homogeneous Maxwell equations, ##\mathrm{d} F=0##. In Ricci-calculus notation that's
$$\partial_{\mu} ^{\dagger} F^{\mu \nu}=\partial_{\mu} \frac{1}{2} \epsilon^{\mu \nu \rho \sigma} F_{\rho \sigma}=0.$$
The Poincare lemma tells you that (at least locally) ##F=\mathrm{d} A## or, in Ricci notation,
$$F_{\mu \nu}=\partial_{\mu} A_{\nu}-\partial_{\nu} A_{\mu}.$$
 
vanhees71 said:
That's just the homogeneous Maxwell equations, ##\mathrm{d} F=0##. In Ricci-calculus notation that's
$$\partial_{\mu} ^{\dagger} F^{\mu \nu}=\partial_{\mu} \frac{1}{2} \epsilon^{\mu \nu \rho \sigma} F_{\rho \sigma}=0.$$
The Poincare lemma tells you that (at least locally) ##F=\mathrm{d} A## or, in Ricci notation,
$$F_{\mu \nu}=\partial_{\mu} A_{\nu}-\partial_{\nu} A_{\mu}.$$
Thanks for the reply. I understood that. But as much as I know, not all closed forms are exact (although all exact forms are closed). So is there a specific reason why we always write F=dA? In other words, just because it is closed why do we expect it to be exact?

In your answer, you wrote F=dA at least locally right? So am I right in saying that the differential 2-form F, in general, is not exact globally (although we both agree that F has to be closed globally)?
 
Well, there are examples like the "potential vortex", where you have a multiply connected region, where you have ##\text{curl} \vec{B}=0## everywhere except along an arbitrary infinite line (e.g., along the ##3##-axis of a Cartesian coordinate system) and
$$\vec{B}=\frac{C}{x^2+y^2} \begin{pmatrix}-y \\x \\ 0 \end{pmatrix},$$
which has
$$\int_{K} \mathrm{d} \vec{r} \vec{B}=2 \pi C N$$
for any closed curve ##K##, which winds ##N## times around the ##z##-axis.
 
Thread 'Motional EMF in Faraday disc, co-rotating magnet axial mean flux'
So here is the motional EMF formula. Now I understand the standard Faraday paradox that an axis symmetric field source (like a speaker motor ring magnet) has a magnetic field that is frame invariant under rotation around axis of symmetry. The field is static whether you rotate the magnet or not. So far so good. What puzzles me is this , there is a term average magnetic flux or "azimuthal mean" , this term describes the average magnetic field through the area swept by the rotating Faraday...
Back
Top