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In summary, the conversation discusses the use of differential forms, specifically in the context of Gauss's Law. It is mentioned that 0 can be considered a 0-form, and there is a discussion on the notation and meaning of d\omega=0. The use of musical isomorphism is also mentioned. However, there is a discrepancy in the notation used in the steps, and it is clarified that it should be d(\star \vec E ^\flat)=\frac{\rho}{\epsilon_0} dV.

- #1

davi2686

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if i have [itex]\int_{\partial S} \omega=0[/itex] by stokes theorem [itex]\int_{S} d \omega=0[/itex], can i say [itex]d \omega=0[/itex]? even 0 as a scalar is a 0-form?

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- #2

ShayanJ

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Consider [itex] d\omega=x^3 dx [/itex] integrated over [itex] S=(-a,a) [/itex]. The integral gives zero but the integrand is zero in only one point of the region of integration. So this is a counterexample to [itex] \int_S d\omega=0 \Rightarrow d\omega=0 [/itex].

- #3

davi2686

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Shyan said:

Consider [itex] d\omega=x^3 dx [/itex] integrated over [itex] S=(-a,a) [/itex]. The integral gives zero but the integrand is zero in only one point of the region of integration. So this is a counterexample to [itex] \int_S d\omega=0 \Rightarrow d\omega=0 [/itex].

thanks, but have no problem with 0 is a 0-form and [itex]d\omega[/itex] a k-form? so can i work with something like [itex]d\omega=4 [/itex]?

- #4

HallsofIvy

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I have no idea what "[itex]d\omega= 4[/itex]", a differential form equal to a number, could even **mean**. Could you please explain that?

- #5

davi2686

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HallsofIvy said:

I have no idea what "[itex]d\omega= 4[/itex]", a differential form equal to a number, could even

mean. Could you please explain that?

my initial motivation is in Gauss's Law, [itex]\int_{\partial V} \vec{E}\cdot d\vec{S}[/itex]=[itex]\int_V \frac{\rho}{\epsilon_0}dV[/itex], i rewrite the left side with differential forms, [itex]\int_{\partial V} \star\vec{E}^{\flat}=\int_V \frac{\rho}{\epsilon_0}dV[/itex] which by the Stokes Theorem [itex]\int_{V} d(\star\vec{E}^{\flat})=\int_V \frac{\rho}{\epsilon_0}dV\Rightarrow d(\star\vec{E}^{\flat})=\frac{\rho}{\epsilon_0}[/itex], if i don't make something wrong in these steps, in left side we get a n-form and right side a 0-form, and that i don't know if i can do.

- #6

ShayanJ

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davi2686 said:

thanks, but have no problem with 0 is a 0-form and [itex]d\omega[/itex] a k-form? so can i work with something like [itex]d\omega=4 [/itex]?

Its correct that 0 is a 0-form but by a zero 1-form we actually mean [itex] \omega= 0 dx [/itex] and write it as [itex] \omega= 0[/itex] when there is no chance of confusion.

davi2686 said:

my initial motivation is in Gauss's Law, [itex]\int_{\partial V} \vec{E}\cdot d\vec{S}[/itex]=[itex]\int_V \frac{\rho}{\epsilon_0}dV[/itex], i rewrite the left side with differential forms, [itex]\int_{\partial V} \star\vec{E}^{\flat}=\int_V \frac{\rho}{\epsilon_0}dV[/itex] which by the Stokes Theorem [itex]\int_{V} d(\star\vec{E}^{\flat})=\int_V \frac{\rho}{\epsilon_0}dV\Rightarrow d(\star\vec{E}^{\flat})=\frac{\rho}{\epsilon_0}[/itex], if i don't make something wrong in these steps, in left side we get a n-form and right side a 0-form, and that i don't know if i can do.

You missed something. You should have written [itex] d(\star \vec E ^\flat)=\frac{\rho}{\epsilon_0} dV [/itex].(What's [itex] \flat [/itex] anyway?)

- #7

davi2686

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You missed something. You should have written [itex] d(\star \vec E ^\flat)=\frac{\rho}{\epsilon_0} dV [/itex].

Thanks i did not know that.

What's [itex] \flat [/itex] anyway?

That is musical isomorphism [itex]\flat:M \mapsto M^*[/itex], in fact i understand it works like a lower indice, [itex]\vec{B}^{\flat}[/itex] give me a co-variant B or it related 1-form.

## Related to Can \( d \omega = 0 \) Be Concluded from \( \int_{\partial S} \omega = 0 \)?

## 1. What is the difference between k and p forms?

The main difference between k and p forms is their mathematical properties. K forms represent objects that can be described using vectors and have a fixed orientation in space, while p forms represent objects that can be described using planes and have a variable orientation in space.

## 2. How does equality between k and p forms affect physics?

Equality between k and p forms is essential in physics, particularly in the study of electromagnetism and relativity. It allows for a better understanding of the relationship between space and time, and helps in the development of mathematical models to describe physical phenomena.

## 3. Can you give an example of a physical concept that uses both k and p forms?

Maxwell's equations, which describe the behavior of electric and magnetic fields, use both k and p forms. The electric field is represented by a k form, while the magnetic field is represented by a p form.

## 4. How does the concept of equality between k and p forms relate to differential geometry?

Equality between k and p forms is a fundamental concept in differential geometry, as it helps to define the geometric structures of a space. It is also used to define the concept of curvature, which is crucial in understanding the behavior of objects in curved spaces.

## 5. Is there a practical application for the concept of equality between k and p forms?

Yes, there are many practical applications for this concept, particularly in the fields of physics and engineering. It is used in the development of mathematical models for electromagnetic devices, such as antennas and transformers, and in the study of fluid dynamics and general relativity.

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