Differential Form: Closed/Exact

In summary, the differential form ##\omega## is closed because it contains the partials and the equation from class says that it is if the function ##g : \mathbb{R}^n \to \mathbb{R}## is such that the derivative is equal to the differential. The domain of ##\omega## is also R^3 because it contains the domains of all three terms.
  • #1
Karnage1993
133
1
1 & 3
I have this differential form:

##\omega = F_1 dx + F_2 dy + F_3 dz##

And I concluded that ##\omega## is closed because I calculated the partials and found out that ##\displaystyle \frac{\partial F_i}{\partial x_j} = \frac{\partial F_j}{\partial x_i}##.

Also, ##F_1## contains only ##x,y## terms, ##F_2## contains ##x,y,z## terms and ##F_3## only ##y,z## terms.

So according to an equation from class, the Domain of ##\omega## = Domains of ##F_1 \cap F_2 \cap F_3 = \mathbb{R}^2 \cap \mathbb{R}^3 \cap \mathbb{R}^2 = \mathbb{R}^2##.

Here's where I'm confused. How is the domain of ##\omega = \mathbb{R}^2##? The differential form contains all 3 parameters so I don't see how it can be. Also, would the ##g## also have domain ##\mathbb{R}^2##? This domain problem is preventing me from concluding that ##\omega## is exact. Once I figure out it's exact, then I can carry out the computations to find ##g(x,y,z)##.

Homework Equations


Definition of exact is:

Let ##\omega## be a first order differential form in ##\mathbb{R}^n##. If ##\omega = dg##, for some ##g : \mathbb{R}^n \to \mathbb{R}##, then ##\omega## is said to be exact.

exact ##\Rightarrow## closed
 
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  • #2
What's this equation from class? I suspect you are all worried about nothing. F(x,y,z)=x+z still has domain R^3 even though there is no y in it. You can't even really add things that have different domains.
 
  • #3
Can you also post the text of your problem, so we can judge what you wrote. Thanks!
 
  • #4
So does that mean something like ##F_1 dx## = ##(2x + xy^2)dx## would mean that the domain of ##F_1## is ##\mathbb{R}^3## as well?

The equation is really from an example. It just says that the domain of ##\omega## = Domain ##F_1## intersect Domain ##F_2## intersect Domain ##F_3##. I suspect this is only for ##\mathbb{R}^3##.

EDIT: Here's the problem:

"For each of the following differential forms ##\omega## determine if there exists a function ##g## such that ##\omega## = ##dg##."

The specific ##\omega## I'm working on is pretty long but it is exactly as I described it with the specific parameters I mentioned.
 
  • #5
Karnage1993 said:
So does that mean something like ##F_1 dx## = ##(2x + xy^2)dx## would mean that the domain of ##F_1## is ##\mathbb{R}^3## as well?

The equation is really from an example. It just says that the domain of ##\omega## = Domain ##F_1## intersect Domain ##F_2## intersect Domain ##F_3##. I suspect this is only for ##\mathbb{R}^3##.

EDIT: Here's the problem:

"For each of the following differential forms ##\omega## determine if there exists a function ##g## such that ##\omega## = ##dg##."

The specific ##\omega## I'm working on is pretty long but it is exactly as I described it with the specific parameters I mentioned.

Yes, I think it's implicit that the domain of all three terms is R^3.
 
  • #6
Oh, okay, that makes everything easier. Thanks for the help!
 

FAQ: Differential Form: Closed/Exact

1. What is the definition of a closed differential form?

A closed differential form is a type of mathematical expression that describes how a quantity changes over different directions and at different points in space. It is called "closed" because the total amount of change in the quantity is equal to zero, meaning that the form is independent of the path taken to calculate the change.

2. How does a closed differential form differ from an exact differential form?

An exact differential form also describes how a quantity changes, but it is "exact" because the form is not only independent of the path taken, but also its value can be calculated by taking the gradient of a scalar function. In other words, an exact form is the gradient of a potential function.

3. Can a closed differential form be exact?

Yes, a closed differential form can be exact. This means that the form is both closed and exact, and its value can be calculated by taking the gradient of a scalar function. In this case, the form is called a "harmonic form."

4. What are some real-world applications of closed and exact differential forms?

Closed and exact differential forms have many practical applications in science and engineering, such as in electromagnetism, fluid mechanics, and thermodynamics. They are also used in mathematical models for describing physical phenomena, such as the movement of fluids or the behavior of electric fields.

5. How are closed and exact differential forms related to Stokes' Theorem?

Stokes' Theorem is a fundamental theorem in vector calculus that relates the integral of a closed differential form over a surface to the integral of the corresponding exact form over the boundary of that surface. In other words, it shows the connection between closed and exact forms and how they can be used to evaluate integrals over different regions in space.

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