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

In summary: In this case, ##\vec{B}## is not an exact differential form, but it is closed and its divergence is zero.
  • #1
phoenix95
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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?
 
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  • #2
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,
$$F_{\mu \nu}=\partial_{\mu} A_{\nu}-\partial_{\nu} A_{\mu}.$$
 
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  • #3
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)?
 
  • #4
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.
 
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FAQ: Why electromagnetic tensor (Faraday 2-form) is exact? (and not closed)

What is the electromagnetic tensor (Faraday 2-form)?

The electromagnetic tensor, also known as the Faraday 2-form, is a mathematical representation of the electromagnetic field in differential geometry. It combines the electric and magnetic fields into a single antisymmetric tensor, allowing for a more unified and elegant description of electromagnetism in the context of special relativity.

Why is the electromagnetic tensor considered exact?

The electromagnetic tensor is considered exact because it can be expressed as the exterior derivative of the electromagnetic potential 1-form. In mathematical terms, if \( A \) is the electromagnetic potential 1-form, then the electromagnetic tensor \( F \) is given by \( F = dA \). This relationship implies that \( F \) is exact, as it is derived from another differential form.

What does it mean for a differential form to be exact?

A differential form is said to be exact if it can be expressed as the exterior derivative of another differential form. In other words, a form \( \omega \) is exact if there exists another form \( \alpha \) such that \( \omega = d\alpha \). Exact forms are always closed, meaning their exterior derivative is zero.

Is the electromagnetic tensor closed?

Yes, the electromagnetic tensor is closed. A differential form is closed if its exterior derivative is zero. Since the electromagnetic tensor \( F \) is the exterior derivative of the potential \( A \) (i.e., \( F = dA \)), taking the exterior derivative of \( F \) gives \( dF = d(dA) \), which is always zero due to the properties of the exterior derivative. Thus, \( F \) is closed.

Why is the distinction between exact and closed important in electromagnetism?

The distinction between exact and closed forms is important because it provides insight into the underlying structure of the electromagnetic field. Exact forms imply the existence of a potential from which the field can be derived, while closed forms ensure that certain conservation laws, such as Gauss's law for magnetism (which states that there are no magnetic monopoles), are satisfied. This duality helps in understanding the mathematical and physical properties of the electromagnetic field.

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