Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I want to understand the Stokes -theorems deeper. I am trying to understand the operation from

$$\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}} {\partial t}$$

to

$$\oint_{\partial \Sigma} \mathbf{E} \cdot d\boldsymbol{\ell} = - \int_{\Sigma} \frac{\partial \mathbf{B}}{\partial t} \cdot d\mathbf{A}$$

relating to Maxwel-Faraday's law here.

You have there surface-integral, closed line-integral, curl and the rate of change. Explain the transition from the differential from into the integral form.

Bonus points and Puzzles

  1. Notation? I am unsure whether $\int$ is just physical convention, instead of $\int\int$ for the surface integral like the Kelvin-Stokes -theorem $\oint_{\Gamma} \mathbf{F}\, d\Gamma = \iint_{\mathbb{S}} \nabla\times\mathbf{F}\, d\mathbb{S}$, the two forms of notation confuses me greatly -- mathematical notation and physical notation?
share|improve this question
add comment

2 Answers 2

up vote 2 down vote accepted

If I undertstand your query correctly, all you really need to complete the thought are Stoke's and Gauss' Theorems: $$ \iint_{\partial E} \vec{F} \cdot d\vec{S} = \iiint_E \nabla \cdot \vec{F} dV $$ $$ \int_{\partial M} \vec{F} \cdot d\vec{r} = \iint_{M} \nabla \times \vec{F} \cdot d\vec{S}$$ Here we suppose that $E$ is a simple solid region with boundary $\partial E$ and $M$ is a simply connected surface with boundary $\partial M$. The boundaries must be consistent with the interiors of the integration regions. This means $\partial E$ has outward pointing normal whereas a trip around $\partial M$ finds the interior of $M$ always on the left of your journey.

Begin with $\nabla \times \vec{E} = -\frac{\partial \vec{B}}{\partial t}$ and integrate over some surface $M$ at time $t$. Apply Stokes' theorem to convert the flux integral of the curl to the line integral of the boundary: $$ \iint_{M} -\frac{\partial \vec{B}}{\partial t} \cdot d\vec{S}=\iint_{M} \nabla \times \vec{E} \cdot d\vec{S}=\int_{\partial M} \vec{E} \cdot d\vec{r} = \mathcal{E}_{induced}$$ Moreover, the term on the l.h.s. of the above can be expressed as $\frac{\partial}{\partial t} \iint_{M} \vec{B} \cdot d\vec{S} = \partial_t \Phi_B$. We thus derive the integral form of Faraday's Law; the voltage induced around a closed loop is proportional to the change in the magnetic flux through the loop.

share|improve this answer
    
Actually, I didn't need Gauss' theorem for your post, if you had asked how $Q_{enclosed}/\epsilon_o=\Phi_E$ arises from $\nabla \cdot \vec{E} = \rho/\epsilon_o$ then we would need Gauss' integral theorem to derive the integral form of Gauss' law. –  James S. Cook Sep 4 '12 at 2:27
    
I see in the Wikipedia they left out many integral signs between area -integrals so confusing things. Thank you +1! –  hhh Sep 4 '12 at 10:18
    
glad to help. These are some of my favorite equations :) –  James S. Cook Sep 4 '12 at 11:24
    
...by the theorems can I deduce Maxwell equations in Integral form from the Maxwell equations in differential form? –  hhh Sep 4 '12 at 14:58
add comment

You need to use the theories below. The notation with integrals between the two cases in KS -threorem look different: $\int\int_S$ and $\int_\Sigma$. I think they are still meaning the same thing. The $\partial\Sigma$ means the boundary of the manifold $\Sigma$.

I am unable to get inside this theorem so I am unable to explain how to get from the differential from into the integral form, anyway below some working, perhaps someone could continue here.

Trials

I. Suppose I integrate $\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}} {\partial t}$ with respect to the area so

$$\int_\Sigma\left(\nabla\times \bf E \right)\cdot d\bf A=-\int_\Sigma \frac{\partial \mathbf{B}} {\partial t}\cdot d\bf{A}:=RHS.$$

II. Suppose I integrate it with respect to the line so

$$\oint_{\partial \Sigma} \left(\nabla\times \mathbf{E}\right) \cdot d\boldsymbol{\ell} = \oint_{\partial \Sigma} \left(-\frac{\partial \mathbf{B}} {\partial t}\right) \cdot d\boldsymbol{\ell}$$

err no LHS emerging.

III. Some other way? I am clearly not getting sides equal like that, at least easily.

Theories

  1. Stokes theorem $\int_{\partial \Omega} w =\int_\Omega dw$ (sthing about boundaries and manifolds here).

  2. Kelvin-Stokes -theorem $$\oint_{\Gamma} \mathbf{F}\, d\Gamma = \iint_{\mathbb{S}} \nabla\times\mathbf{F}\, d\mathbb{S}$$ (source) or $$\int_{\Sigma} \nabla \times \mathbf{F} \cdot d\mathbf{\Sigma} = \oint_{\partial\Sigma} \mathbf{F} \cdot d \mathbf{r}$$ (source).

Perhaps related here.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.