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  How are closed 1-forms used in physics?

+ 3 like - 0 dislike
4133 views

Closed 1-forms are well studied in foliation topology, algebraic geometry, and theory of manifolds. What are examples of their most typical or most interesting applications in physics?

I do not mean exact 1-forms (roughly speaking, functions -- not interesting). I am interested in examples of applications of closed 1-forms that that are not exact.

My motivation is to mention several good examples in an introductory section of a mathematical paper on closed 1-forms to show their importance to physics, both classical and modern. So several good (typical, or interesting) examples suitable to be mentioned in such a section would do.

This post imported from StackExchange Physics at 2015-01-06 10:51 (UTC), posted by SE-user Irina
asked Jan 3, 2015 in Mathematics by Irina (75 points) [ no revision ]
This question is too broad - many areas of physics may be formulated with differential forms, and almost all of them will consequently deal with closed forms in particular.

This post imported from StackExchange Physics at 2015-01-06 10:51 (UTC), posted by SE-user ACuriousMind
why exactly are you interested in just closed 1-forms?

This post imported from StackExchange Physics at 2015-01-06 10:51 (UTC), posted by SE-user Phoenix87
@Phoenix87 I am a mathematician, I study closed 1-forms, because they have many specific mathematical properties. In the Preface to my papers I want to show their importance for physics, classical (mechanics, electrodynamics, crystallography?) and modern (cosmology and gravitation?). But I am not a physicist and I'm not sure which applications are most important to mention.

This post imported from StackExchange Physics at 2015-01-06 10:51 (UTC), posted by SE-user Irina
@ ACuriousMind I don't think the question is too broad: I mean precisely closed 1-forms --- not differential forms in general. And at the moment there is no answer :(

This post imported from StackExchange Physics at 2015-01-06 10:51 (UTC), posted by SE-user Irina
I edited the question to ask for good/best examples of applications and not for all applications. This allows for a reasonably short and specific answer.

This post imported from StackExchange Physics at 2015-01-06 10:51 (UTC), posted by SE-user Irina

3 Answers

+ 2 like - 0 dislike

1-forms appear prominently in thermodynamics, where they denote reversible changes in thermodynamic observables. Sometimes these are closed and then lead to conserved quantities.

answered Mar 3, 2015 by Arnold Neumaier (15,787 points) [ no revision ]
+ 1 like - 0 dislike

Many. Classical mechanics is essentially geometry. In the Hamiltonian formulation, the dynamics takes place on a cotangent bundle to a manifold, the configuration space $\Gamma$, known as the phase space $T^*\Gamma$. The tautological, or Poincaré 1-form $\theta$, leads through exterior derivative to the natural symplectic 2-form $\omega$ on the cotangent bundle $T^*\Gamma$, that is $\omega = \text d\theta$.

In Electrodynamics, the 4-potential $A$ can be viewed as a 1-form, and its exterior derivative $\text dA$ is the Faraday, or electromagnetic, tensor $F$, which describes both electric and magnetic fields and is linked to the 4-current 1-form $J$ through Maxwell's equations. For more on this subject see this answer.

For some other ideas in General Relativity see this other answer.

This post imported from StackExchange Physics at 2015-01-06 10:51 (UTC), posted by SE-user Phoenix87
answered Jan 3, 2015 by Phoenix87 (40 points) [ no revision ]
Thank you! The $\theta$ is not a closed form, otherwise $\omega=\mathrm{d}\theta$ were zero ($\omega$ is a closed form iff $\mathrm{d}\omega=0$). Your $\omega$ is closed but not a 1-form :-( It seems that the same holds for $A$ and $F$.

This post imported from StackExchange Physics at 2015-01-06 10:51 (UTC), posted by SE-user Irina
Yep, these are examples of forms rather than 1-forms, so this is why i was asking you for you interests in just closed ones.

This post imported from StackExchange Physics at 2015-01-06 10:51 (UTC), posted by SE-user Phoenix87
Perhaps fluid dynamics is another important example. Closed 1-forms describe irrotational flows under some circumstances, but i don't remember much at the moment

This post imported from StackExchange Physics at 2015-01-06 10:51 (UTC), posted by SE-user Phoenix87
+ 0 like - 0 dislike

Most notably, part of Maxwell's equations states that the Faraday 2-form is closed: $$dF=0$$ From this we can infer from Poincare's lemma that there exists a 1-form $A$ such that $dA=F$. In some elementary treatments $F$ is considered to be an exact form. But when considering magnetic monopoles is it important to treat it as a closed form because of the "locally" clause in the Poincare lemma.

A really trivial example is the following: let $g$ be an orthonormal metric. Then it is a closed 0-form $$dg=0$$ This is merely the equation for the antisymmetry of the spin connection on a Riemannian manifold with orthonormal metric.

Cohomology is used quite extensively in a little sector of physics called String Theory. I'm sure you know how important closed forms are for that. A really important closed form is the Kahler form: $$dJ=0$$

EDIT: Those weren't 1-forms. The curl operator is $\star d$. Thus a closed one-form is isomorphic to a vector that has zero curl! Some examples I can think of off the top of my head:

Take Faraday's law $\nabla\times\mathbf{E}+\dot{\mathbf{B}}=0$. Suppose the fields are static. Then $\dot{\mathbf{B}}=0$ and $\nabla\times\mathbf{E}=0$. If $\mathcal{E}=\mathbf{E}^\flat$ $$d\mathcal{E}=0$$

The same works for the Maxwell-Ampere law in a vacuum. Then the magnetic 1-form $\mathcal{B}=\mathbf{B}^\flat$ is closed $$d\mathcal{B}=0$$

Suppose the integral of some force $\mathbf{F}$ is path-independent. Work is defined by $$W_P=\int_P\mathbf{F}\cdot d\mathbf{x}$$ If $\mathcal{F}=\mathbf{F}^\flat$ then $$W_P=\int_P\mathcal{F}$$ The difference of work along two different paths vanishes ($P'-P$ is a closed curve which is the boundary of a surface $S$) $$W_{P'}-W_P=\int_{P'-P}\mathcal{F}=\int_S d\mathcal{F}=0$$ by Stokes' theorem. This implies for any conservative force $$d\mathcal{F}=0$$

This post imported from StackExchange Physics at 2015-01-06 10:51 (UTC), posted by SE-user 0celo7
answered Jan 3, 2015 by 0celo7 (50 points) [ no revision ]
Thank you! But Faraday form, Kahler form, even metrics are 2-forms. And I ask about a closed 1-form.

This post imported from StackExchange Physics at 2015-01-06 10:51 (UTC), posted by SE-user Irina
Oh snap, didn't notice that. I'll rack my brain!

This post imported from StackExchange Physics at 2015-01-06 10:51 (UTC), posted by SE-user 0celo7
Oh, so any irrotational vector field (in particular, electric or magnetic irrotational field) on a 3-manifold corresponds to a closed 1-form! Thank you. As for a conservative force -- it corresponds to an exact form, which is trivially closed, and thus is not very interesting. Except for irrotational vector fields -- they are so classic-- is there something relevant in modern physics?

This post imported from StackExchange Physics at 2015-01-06 10:51 (UTC), posted by SE-user Irina
I'm actually not sure if a conservative force is by definition exact. It really depends on definitions! A conservative force is path-independent. As I showed above, it implies that F is closed. By the converse of the gradient theorem (and certain properties of $\mathbb{R}^n$) we can find a potential $V$ such that $F=-\nabla V$ globally. It's a special case of the Poincare lemma!

This post imported from StackExchange Physics at 2015-01-06 10:51 (UTC), posted by SE-user 0celo7
Modern physics...That's hard. I can think of so many closed forms that are not 1-forms (Ricci form, NS-NS field strength, etc.) and tensor-valued forms... In fact, there are some tensor valued one-forms that are closed! That might be cheating, because we have to use the absolute exterior differential $D$. But here goes... The torsion 1-form is closed: $D\Theta^i=0$. Let $T_{\mu}=T_{\mu\nu}\theta^\nu$ be the energy-momentum 1-form. Then $DT_\mu=0$. Let $G_\mu=G_{\mu\nu}\theta^\nu$ be the Einstein 1-form. Then $DG_\mu=0$. I am looking furiously! More to come!

This post imported from StackExchange Physics at 2015-01-06 10:51 (UTC), posted by SE-user 0celo7
Here's a really good one! In the study of axisymmetric spacetimes it is convenient to introduce the complex-valued Ernst form on a two-dimensional Riemannian submanifold. On the vacuum part of the manifold (i.e. $\operatorname{Ric}=0$), this form is closed $d\mathcal{E}=0$. If you need more details, google first but ask if you don't find anything good. It's quite the complicated object actually.

This post imported from StackExchange Physics at 2015-01-06 10:51 (UTC), posted by SE-user 0celo7
"It really depends on definitions!": OK, I call a form exact if it is $\mathrm{d}f$ for some globally defined $f$. With this definition, it is a tautology that if the integral is path-independent, then the form is exact -- exactly as you show: $W(x)=\int_{x_0}^x{\mathcal{F}}$. (But see my next comment.)

This post imported from StackExchange Physics at 2015-01-06 10:51 (UTC), posted by SE-user Irina
However, if you change "Suppose the integral of some force $\mathbf{F}$ is path-independent" to "locally path-independent", that is, path-independent in any simply connected neighborhood, then we can find an interesting example. For example, I think the work of a magnet around a coil current is locally path-independent, but not globally: if you move the magnet through the coil, the work will be non-zero; does this give a closed but not exact form? A similar example could be the Kerr black hole, but I am not a specialist in this. Those are examples of what I am looking for.

This post imported from StackExchange Physics at 2015-01-06 10:51 (UTC), posted by SE-user Irina
Perhaps it makes sense to add new examples to the answer and not only to the comments. This will make your answer even better!

This post imported from StackExchange Physics at 2015-01-06 10:51 (UTC), posted by SE-user Irina
I'm not sure about the specific magnetic example you mentioned, but the magnetic field does not produce a conservative force. This is because you cannot put B in the form of a gradient. Alternatively, observe that the magnetic field couples to velocity in the Lorentz force.

This post imported from StackExchange Physics at 2015-01-06 10:51 (UTC), posted by SE-user 0celo7

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