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I have the following formula in 3-Dimensions:

$$ \int_{\partial \omega} f(x,y,z) \vec{dS} = \int_{\omega} \nabla f dV$$

I want to write the above in the language of forms and derive it through stokes.

My attempt:

Let $\tilde{R}(u,v)$ be the position vector on the surface, we have the two tangent vectors as $\tilde{R}_u $ and $\tilde{R}_v$. As a form , I write as $r_u$ and $r_v$. The integral is rewritten as refer :

$$ \int_{\partial w} f(x,y,z) \frac{ r_u \wedge r_v}{|r_u \wedge r_v|} = \int_{\partial \omega} f(x,y,z) n$$

Where $n$ is the normal two form.

By stokes theorem:

$$ \int_{\partial w} fn = \int_{\omega} d(fn) = \int_{\omega} df \wedge n + f dn $$

Not sure how to proceed now. I am suppose the exterior derivative of the normal relates to the Riemann curvature tensor which I am nto so familiar about yet.

Maybe this will be helpful

tryst with freedom
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  • Do you know how to derive this formula from the usual divergence theorem $\int_{\partial\Omega}\mathbf{F}\cdot\mathbf{n},dS=\int_{\Omega}\text{div}(\mathbf{F}),dV$? (you have to use it three times first with the field $\mathbf{F}=(f,0,0)$ then with $(0,f,0)$ and finally with $(0,0,f)$, as mentioned in Kurt's answer below, and by my comments there) If yes, then the answer below essentially tells you how this form of the divergence theorem follows from the general Stokes theorem. – peek-a-boo May 16 '22 at 06:32
  • In fact... an equivalent way of stating the divergence theorem in $\Bbb{R}^n$ is that for all nice enough $f$ and $\Omega$, $\int_{\partial\Omega}f,n^i,dS=\int_{\Omega}\frac{\partial f}{\partial x^i},dV$, which is just the component expression of your equation. – peek-a-boo May 16 '22 at 06:35
  • Bruh that seems such a contrived way to do it and seems like the wrong way. If I am not mistaken, you're assuming a basis to do it, I thought in differential forms you could do these things without assuming any basis @peek-a-boo – tryst with freedom May 16 '22 at 07:05
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    You're integrating a vector-valued function, and integrals of vector-valued functions are defined as integrals of their components... Differential forms allow us to prove the general Stokes theorem. It is an extremely general theorem, which means it has a several special cases, and you can only appreciate its significance by extracting and putting together the various special cases. – peek-a-boo May 16 '22 at 07:39
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    Fellows ! Please check the edit of my answer. – Kurt G. May 16 '22 at 08:02

1 Answers1

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Classically we have two theorems in $\mathbb R^3$: Gauss' divergence theorem and Stokes' theorem. Your formula confuses me a bit as it seems to be neither one of them.

A better notation is probably $$\tag{1} \int_{\partial\omega}v=\int_\omega dv $$ where $$\tag{2} v=v_1\,(dy\wedge dz)+v_2\,(dx\wedge dz)+v_3\,(dx\wedge dy) $$ is a $2$-form. It looks like you are interested only in the cases $v=(f,0,0)$ or $v=(0,f,0)$ or $v=(0,0,f)$. (However I don't know what your notation $f\,\vec{dS}$ exactly means.)

The exterior derivative of $v$ is a $3$-form

\begin{align} dv&=(\partial_x v_1)\,(dx\wedge dy\wedge dz)-(\partial_y v_2)\,(dx\wedge dy\wedge dz)+ (\partial_z v_3)\,(dx\wedge dy\wedge dz)\\ &=\Big(\partial_x v_1-\partial_y v_2+\partial_z v_3\Big)(dx\wedge dy\wedge dz)\,. \end{align} Obviously, we now have a divergence: $$ \partial_x v_1-\partial_y v_2+\partial_z v_3=\nabla\begin{pmatrix}v_1\\-v_2\\v_3\end{pmatrix} $$ which gets integrated over the volume $\omega\,$. In other words, we have Gauss' theorem.

To cast Stokes' theorem into the form (1) note that this deals with a line integral over a closed curve $\gamma$ that is the boundary of a surface $\omega$. So here $v$ is a $1$-form

$$ v=v_1\,dx+v_2\,dy+v_3\,dz $$ and \begin{align} dv =&-(\partial_yv_3)\,(dy\wedge dz)+(\partial_z v_2)\,(dy\wedge dz)\\ &-(\partial_zv_1)\,(dx\wedge dz)+(\partial_x v_3)\,(dx\wedge dz)\\ &-(\partial_y v_1)\,(dx\wedge dy)+(\partial_x v_2)\,(dx\wedge dy). \end{align} Obviously, we now have a rotation $$ dv = (\nabla\times v)\cdot\begin{pmatrix}dy\wedge dz\\ dx\wedge dz\\ dx\wedge dy \end{pmatrix} $$ whose dot product gets integrated over the now two dimensional surface $\omega$.

The beauty of Elie Cartan's exterior calculus is that it automatically leads to the right expressions of divergence, resp. rotation.

peek-a-boo has kindly pointed out that your surface integral $\int_{\partial \omega}f\vec{dS}$ is that of a vector valued $2$-form whose components are those of $f$ multiplied with the unit outward normal vector $\vec{\boldsymbol{n}}$ to the surface. My favourite book on vector valued or tensor valued forms is Misner, Thorne and Wheeler Gravitation. I believe they would write this vector valued form as $$\tag{3} \boldsymbol{v}=\boldsymbol{v}_1\,(dy\wedge dz)+\boldsymbol{v}_2\,(dx\wedge dz)+\boldsymbol{v}_3\,(dx\wedge dy) $$ where each $\boldsymbol{v}_i$ is now a vector field in $\mathbb R^3$. In other words, each $\boldsymbol{v}_i$ has components $v_{ij}\,$: $$ \boldsymbol{v}_i=\begin{pmatrix}v_{i1}\\v_{i2}\\v_{i3}\end{pmatrix} $$ The form (3) is a stack of three forms of type (2) and clearly Gauss theorem yields a stack of three equations of type (1): $$ \int_{\partial \omega}\boldsymbol{v}=\int_\omega d\boldsymbol{v} $$ where the vector valued $3$-form has components

$$ d\boldsymbol{v}=\begin{pmatrix}\partial_x v_{11}-\partial_y v_{12}+\partial_z v_{13}\\\partial_x v_{21}-\partial_y v_{22}+\partial_z v_{23}\\\partial_x v_{31}-\partial_y v_{32}+\partial_z v_{33} \end{pmatrix}(dx\wedge dy\wedge dz)\,. $$ I.e., each component comes from the divergence of a vector field $$ \begin{pmatrix}v_{i1}\\-v_{i2}\\v_{i3}\end{pmatrix}\,. $$

Kurt G.
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    $f,d\vec{S}$ means the vector(bundle)-valued differential form $f,\vec{n},dS$, where $\vec{n}$ is the unit outward normal to the surface, and $dS=n\lrcorner dV$ is the surface element (i.e the 'volume' form of the surface). But you're right that this result does not generalize arbitrarily, because we can't really integrate vector-bundle-valued forms; only if they take values in a fixed vector space (here, we can imaging $n$ is $\Bbb{R}^3$ valued, so everything works out), and in this case, the result in question follows by a 3-fold application of the divergence theorem. – peek-a-boo May 15 '22 at 21:21
  • @peek-a-boo +1 and thanks ! – Kurt G. May 16 '22 at 04:04
  • Can you show how to do this please? This is a pretty big result in fluid dynamics and I can't I figure it out still – tryst with freedom May 16 '22 at 05:57
  • fdS is f times the normal vector multiplied by area of a patch – tryst with freedom May 16 '22 at 05:58
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    Well, I can't upvote twice, but just in case OP wants the same thing written in fancier notation: we can consider the $\Bbb{R}^p$-valued $(p-1)$-form $\boldsymbol{v}=\sum_{i=1}^p\underbrace{\left[\frac{\partial f}{\partial x^i}(\star dx^i)\right]}_{:=v_i},\mathbf{e}_i$ defined on a nice open set $\Omega\subset\Bbb{R}^p$ ($\star$ denotes the Hodge star). Then, we're just applying, as you mentioned, the vector-analogue of the generalized Stokes theorem here. – peek-a-boo May 16 '22 at 08:14
  • @peek-a-boo . Don't care about upvotes. Want to learn DG. Thanks a lot again ! – Kurt G. May 16 '22 at 08:15
  • I'm so confused now. The coefficient of the basis two forms can be vectors themself...??? I thought they were only scalars. – tryst with freedom May 16 '22 at 08:48
  • I have upvoted now. I think what you've written is infact what is needed.. the only problem is for me to understand it.. – tryst with freedom May 16 '22 at 08:50
  • Give tribute to peek-a-boo and be patient. I myself learned this stuff only a few months ago and I am almost retired :). – Kurt G. May 16 '22 at 08:56