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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1. Why is the electromagnetic tensor considered to be an exact 2-form?

The electromagnetic tensor, also known as the Faraday 2-form, is considered to be an exact 2-form because it can be expressed as the exterior derivative of a 1-form. This means that it can be written as the gradient of a scalar potential, known as the electromagnetic potential. This property is important in mathematical physics as it allows for the formulation of Maxwell's equations, which describe the behavior of electromagnetic fields.

2. What does it mean for the electromagnetic tensor to be exact?

When we say that the electromagnetic tensor is exact, it means that it satisfies the condition of being closed. In other words, the exterior derivative of the electromagnetic tensor is equal to zero. This is a mathematical property that is used to describe the behavior of differential forms, which are used to represent physical quantities in mathematical physics.

3. How does the exactness of the electromagnetic tensor relate to Maxwell's equations?

The exactness of the electromagnetic tensor is directly related to Maxwell's equations. This is because the exactness property allows us to express the electromagnetic tensor as the exterior derivative of a 1-form, which is used in the formulation of Maxwell's equations. These equations describe the behavior of electric and magnetic fields and their interactions with charged particles.

4. Can the electromagnetic tensor be closed and not exact?

Yes, it is possible for the electromagnetic tensor to be closed and not exact. This means that the exterior derivative of the electromagnetic tensor is equal to zero, but it cannot be expressed as the exterior derivative of a 1-form. In other words, there is no scalar potential that can be used to represent the electromagnetic tensor as an exact 2-form. This situation can occur in certain non-trivial topologies or when considering non-linear theories of electromagnetism.

5. What are the implications of the exactness of the electromagnetic tensor?

The exactness of the electromagnetic tensor has important implications in the study of electromagnetism and mathematical physics. It allows for the formulation of Maxwell's equations, which are essential for understanding the behavior of electromagnetic fields and their interactions with matter. It also allows for the development of mathematical techniques, such as differential forms, that are used to describe physical quantities in a concise and elegant manner.

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