Helicity integral in differential forms

In summary, by using the assumption that ##i^{*}\omega^{2}=0##, we can show that the tangential components of the vorticity vector are zero on the boundary, which then allows us to prove that the helicity integral is constant in time.
  • #1
spaghetti3451
1,344
33

Homework Statement



Let ##V^{3}(t)## be a compact region moving with the fluid.

Assume that at ##t=0## the vorticity ##2##-form ##\omega^{2}## vanishes when restricted to the boundary ##\partial V^{3}(0)##; that is, ##i^{*}\omega^{2}=0##, where ##i## is the inclusion of ##\partial V## in ##\mathbb{R}^{3}##.

(This does ##\textit{not}## say that ##\omega^{2}## itself vanishes, rather only that ##\omega({\vec{u}},{\vec{w}})=0## for ##\vec{u}##,##\vec{w}## tangent to ##\partial V^{3}(0)##.)

Then the ##\textbf{helicity}## integral ##\displaystyle{\int_{V(t)}{\vec{v}}\cdot{\vec{\omega}}\ dx\wedge dy\wedge dz}## can be constant in time.

Prove that the helicity integral is constant in time.

Homework Equations



The Attempt at a Solution



$$\frac{d}{dt}\int_{V(t)}{\vec{v}}\cdot{\vec{\omega}}\ dx\wedge dy\wedge dz$$
$$= \frac{d}{dt}\int_{V(t)}{\vec{v}}\cdot{\vec{\omega}}\ dx^{1}\wedge dx^{2}\wedge dx^{3}$$
$$= \frac{1}{o(x)\sqrt{g}(x)} \frac{d}{dt}\int_{V(t)} {\vec{v}}\cdot{\vec{\omega}}\ \text{vol}^{3}$$
$$= \frac{1}{o(x)\sqrt{g}(x)} \frac{d}{dt}\int_{V(t)} \nu^{1}\wedge \omega^{2}$$
$$= \frac{1}{o(x)\sqrt{g}(x)} \frac{d}{dt}\int_{V(t)} \nu^{1}\wedge {\rm d}\nu^{1}$$
$$= \frac{1}{o(x)\sqrt{g}(x)} \frac{d}{dt}\int_{V(t)} \nu\wedge {\rm d}\nu$$
$$= \frac{1}{o(x)\sqrt{g}(x)}\frac{1}{2} \frac{d}{dt}\int_{W(t)} (\nu\wedge {\rm d}\nu)+(\nu\wedge {\rm d}\nu)$$
$$= \frac{1}{o(x)\sqrt{g}(x)}\frac{1}{2} \frac{d}{dt}\int_{W(t)} (\nu\wedge {\rm d}\nu)-({\rm d}\nu\wedge \nu)$$
$$= \frac{1}{o(x)\sqrt{g}(x)} \frac{d}{dt}\int_{W(t)} {\rm d}(\nu\wedge \nu)$$
$$= \frac{1}{o(x)\sqrt{g}(x)} \frac{d}{dt}\oint_{\partial W(t)} \nu\wedge \nu$$
$$= 0.$$

I think this is wrong because I haven't used the assumption that at ##t=0## the vorticity ##2##-form ##\omega^{2}## vanishes when restricted to the boundary ##\partial V^{3}(0)##; that is, ##i^{*}\omega^{2}=0##, where ##i## is the inclusion of ##\partial V## in ##\mathbb{R}^{3}##.
 
Last edited:
Physics news on Phys.org
  • #2
I am not sure how to incorporate this assumption into my proof. Can someone please help me?

One way to incorporate this assumption is to use the fact that the vorticity 2-form can be written as ##\omega^{2} = \omega^{2}_{ij} dx^{i}\wedge dx^{j}##, where ##\omega^{2}_{ij}## are the components of the vorticity tensor. Then, using the fact that ##i^{*}\omega^{2} = 0##, we have ##i^{*}\omega^{2}_{ij} = 0## for all ##i,j##. This means that the tangential components of the vorticity tensor are zero on the boundary. Using this, you can show that the tangential components of the vorticity vector are also zero on the boundary. Then, you can use this fact in your proof to show that the helicity integral is constant in time.
 

1. What is the helicity integral in differential forms?

The helicity integral in differential forms is a mathematical concept used to calculate the total amount of helicity, or twisting, of a vector field in a given region. It is represented by integrating the exterior product of the vector field and its differential form over a closed surface.

2. How is the helicity integral related to fluid dynamics?

In fluid dynamics, the helicity integral is used to measure the amount of twisting and linking of fluid particles in a flow. It is closely related to the concept of vortex lines and is often used to study the behavior of turbulence and other complex flows.

3. Can the helicity integral be negative?

Yes, the helicity integral can have both positive and negative values. A positive value indicates a right-handed twist, while a negative value indicates a left-handed twist. The magnitude of the integral represents the strength of the twisting.

4. How is the helicity integral calculated?

The helicity integral is calculated by first expressing the vector field and its differential form in terms of their components. Then, the exterior product of these components is integrated over a closed surface using multivariable calculus techniques.

5. What are the applications of the helicity integral?

The helicity integral has many applications in mathematics, physics, and engineering. It is used to study fluid dynamics, plasma physics, and other areas of physics where the twisting and linking of vector fields play a crucial role. It is also used in topology to study the properties of knots and links.

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
571
  • Calculus and Beyond Homework Help
Replies
2
Views
2K
  • Calculus and Beyond Homework Help
Replies
9
Views
949
  • Calculus and Beyond Homework Help
Replies
1
Views
632
  • Calculus and Beyond Homework Help
Replies
9
Views
752
  • Calculus and Beyond Homework Help
Replies
6
Views
838
  • Calculus and Beyond Homework Help
Replies
7
Views
549
  • Calculus and Beyond Homework Help
Replies
3
Views
547
  • Calculus and Beyond Homework Help
Replies
5
Views
611
  • Calculus and Beyond Homework Help
Replies
2
Views
452
Back
Top