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The Vector Integral page on the Wolfram mathworld website lists as eq.(4) the following vector integral identity:

If $$\mathbf{F}:=\mathbf{c}\,F,$$ then $$\int_{C}F\,ds=\int_{S}d\mathbf{a}\times\nabla\mathbf{F}.$$

It seems clear from context that $\mathbf{c}$ is supposed to be a constant vector, though the page doesn't explicitly list this as a condition.

Try as I might, I can't wrap my head around this integral. In fact, it looks like patent nonsense to me, since on the LHS we have a scalar line integral (that is, the line-element $ds$ is a scalar) of a scalar field resulting in a scalar value, while the integral on the RHS looks to me like it must be a 2nd rank tensor.

I'd be surprised if this is an error on Wolfram's part that's somehow gone unnoticed all this time, so I can only assume that I have a fundamental misunderstanding somewhere that's preventing me from parsing this formula correctly. Can someone please set me straight?


UPDATE:

My thanks to @TylerHG for helping me confirm that this is indeed a typo on Wolfram's part. Instead of closing this question, I'd like to ask a follow-up question prompted by the falsehood of the identity proposed above. Given a scalar field $f(\mathbf{r})$ and a surface $\Sigma$ with boundary $\partial\Sigma$, does there exist any kind of Stokes' theorem analogue for the scalar line integral,

$$\oint_{\partial\Sigma}f\,\mathrm{d}\ell =\,???$$

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    $\begingroup$ Yea it is a typo. That F should be a scalar and the ds should be a vector. Look at this mathworld.wolfram.com/CurlTheorem.html $\endgroup$ – ClassicStyle Jun 13 '14 at 20:23
  • $\begingroup$ @TylerHG Well I'll be... Thank you for restoring my sanity. $\endgroup$ – David H Jun 13 '14 at 20:36
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    $\begingroup$ The cross product of a vector and the gradient of a vector...nooooooo! @DavidH $\endgroup$ – ClassicStyle Jun 13 '14 at 20:38
  • $\begingroup$ @TylerHG You can imagine upsetting this was to me. :o $\endgroup$ – David H Jun 13 '14 at 20:58
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For any type of Stokes-like result to hold, you have to be able to interpret $\int_{\partial\Sigma} f ds$ as the line integral of a vector field over the curve $\partial\Sigma$, and indeed, you can write $$ \int_{\partial \Sigma} f ds = \int_{\partial \Sigma} (f \mathbf{t}) \cdot d\mathbf{r}, $$ where $\mathbf{t}$ denotes the unit tangent vector field on $\partial\Sigma$. For Stokes's theorem to apply, you have to be able to extend $f \mathbf{t}$ to a vector field on all of $\Sigma$; since $f$ is (presumably) defined on some neighbourhood of $\Sigma$ in $\mathbb{R}^3$ anyway, the entire problem is in extending the unit tangent vector field $\mathbf{t}$ to a vector field $\tilde{\mathbf{t}}$ on all of $\Sigma$. If such a vector field $\tilde{\mathbf{t}}$ exists, then, and only then, can you apply Stokes's theorem to conclude that $$ \int_{\partial \Sigma} f ds = \int_{\partial \Sigma} (f \tilde{\mathbf{t}}) \cdot d\mathbf{r} = \int_\Sigma \nabla \times (f\tilde{\mathbf{t}}) \cdot d\mathbf{S} = \int_\Sigma \left( \nabla f \times \tilde{\mathbf{t}} + f \nabla \times \tilde{\mathbf{t}} \right) \cdot d\mathbf{S}, $$ i.e., that $$ \int_{\partial \Sigma} f ds = \int_\Sigma \left( \nabla f \times \tilde{\mathbf{t}} \right) \cdot d\mathbf{S} + \int_\Sigma f \left(\nabla \times \tilde{\mathbf{t}}\right) \cdot d\mathbf{S}. $$ If $\mathbf{t}$ cannot be extended to a vector field on $\Sigma$, or equivalently, if $\mathbf{t}$ isn't the restriction to $\partial\Sigma$ of some vector field defined on all of $\Sigma$, then you're out of luck. However, in any reasonable situation, $\mathbf{t}$ can be extended to a vector field on an open neighbourhood of $\partial\Sigma$ in $\mathbb{R}^3$, in which case you can use bump functions to get an extension $\tilde{\mathbf{t}}$ of $\mathbf{t}$ that vanishes outside that neighbourhood.


There is exactly one other possibility, which is to use the divergence theorem for surfaces: $$ \int_{\partial \Sigma} \mathbf{F} \cdot (\mathbf{t}\times\mathbf{n}) ds = \int_\Sigma \operatorname{div}_\Sigma(\mathbf{F}) dS, $$ where $\mathbf{t}$, as before, is the unit tangent vector field on $\partial\Sigma$, $\mathbf{n}$ is the unit normal vector field on $\Sigma$, and $\operatorname{div}_\Sigma$ denotes the divergence on $S$, i.e., $$ \operatorname{div}_\Sigma(\mathbf{F}) := \mathbf{n} \cdot \nabla \times (\mathbf{n} \times \mathbf{F}). $$ Then, since $$ \int_{\partial\Sigma} fds = \int_{\partial\Sigma} f(\mathbf{t}\times\mathbf{n}) \cdot (\mathbf{t}\times\mathbf{n}) ds, $$ it follows that if $\mathbf{t}\times\mathbf{n}$ extends to a vector field $\mathbf{v}$ on $\Sigma$, then you can apply the divergence theorem for surfaces to conclude that $$ \int_{\partial\Sigma} fds = \int_{\partial\Sigma} f\mathbf{v} \cdot (\mathbf{t}\times\mathbf{n}) ds = \int_\Sigma \operatorname{div}_\Sigma(f\mathbf{v}) dS. $$ This, however, gives you nothing new; on the one hand, if $\mathbf{t}$ does indeed extend to $\tilde{\mathbf{t}}$, then you can set $\mathbf{v} = \tilde{\mathbf{t}} \times \mathbf{n}$, whilst on the other, if $\mathbf{v}$ exists, then you can check, using the BAC-CAB formula, that $$ \tilde{\mathbf{t}} := \mathbf{n} \times \mathbf{v} $$ correctly extends $\mathbf{t}$.

Since Stokes's theorem and the divergence theorem for surfaces are the only two ways to rewrite, in the language of vector analysis, the abstract Stokes's theorem for $1$-forms and $2$-dimensional submanifolds with boundary, i.e., $$ \int_{\partial\Sigma} \omega = \int_\Sigma d\omega, $$ in the special case where the ambient space is $\mathbb{R}^3$, this really now exhausts all the obvious possibilities for a Stokes-like result.

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  • $\begingroup$ Very informative! This is actually a question that's nagged at me on and off for at least a few years. Good to finally have my answer. :) $\endgroup$ – David H Jun 13 '14 at 23:44

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