Equality between k and p forms

[davi2686]

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?

 

[ShayanJ]

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].

 

[davi2686]

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]?

 

[HallsofIvy]

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?

 

[davi2686]

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.

 

[ShayanJ]

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.

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?)

 

[davi2686]

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.