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

[davi2686]

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

 

[ShayanJ]

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

 

[davi2686]

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

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

 

[HallsofIvy]

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

 

[davi2686]

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

my initial motivation is in Gauss's Law, \int_{\partial V} \vec{E}\cdot d\vec{S}=\int_V \frac{\rho}{\epsilon_0}dV, i rewrite the left side with differential forms, \int_{\partial V} \star\vec{E}^{\flat}=\int_V \frac{\rho}{\epsilon_0}dV which by the Stokes Theorem \int_{V} d(\star\vec{E}^{\flat})=\int_V \frac{\rho}{\epsilon_0}dV\Rightarrow d(\star\vec{E}^{\flat})=\frac{\rho}{\epsilon_0}, 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.

 

[ShayanJ]

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

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

my initial motivation is in Gauss's Law, \int_{\partial V} \vec{E}\cdot d\vec{S}=\int_V \frac{\rho}{\epsilon_0}dV, i rewrite the left side with differential forms, \int_{\partial V} \star\vec{E}^{\flat}=\int_V \frac{\rho}{\epsilon_0}dV which by the Stokes Theorem \int_{V} d(\star\vec{E}^{\flat})=\int_V \frac{\rho}{\epsilon_0}dV\Rightarrow d(\star\vec{E}^{\flat})=\frac{\rho}{\epsilon_0}, 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 d(\star \vec E ^\flat)=\frac{\rho}{\epsilon_0} dV.(What's \flat anyway?)

 

[davi2686]

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

Thanks i did not know that.

What's \flat anyway?

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