3

I want to understand how Stoke's theorem shows that the integral form of Faraday's law: $$\int_{c(A)} E dr = -\frac{1}{c} \frac{d}{dt} \int_A B ds$$ ($A$ is a surface and $c(A)$ its boundary curve) is equivalent to its differential form, i.e. the Maxwell equation $$\operatorname{rot}E + \frac{1}{c} \frac{d}{dt} B = 0.$$ Now, by applying Stokes theorem we can exchange the line integral on the left by the surface integral $$\int_A \operatorname{rot} E dA.$$ Then $$\int_A \operatorname{rot} E dA + \frac{1}{c} \frac{d}{dt} \int_A B dA = \int_A \left(\operatorname{rot} E + \frac{1}{c} \frac{d}{dt} B\right) \cdot dA$$ (what is the rigorous explanation that we are allowed to exchange differentiation by time with the integration here?). If the integrand is zero (i.e. the Maxwell equation holds) then this integral is zero (i.e. Faraday's law in integral form holds). But how do we argue the other way around? Why does it follow here from integral = zero that the integrand = zero?

Thanks.

Vincenzo Tibullo
  • 10,955
Mekanik
  • 1,761

2 Answers2

1

As to your first question, about when it's OK to interchange the order of differentiation and integration, see the Leibniz integral rule - it basically states that if $f(x,y)$ is continuous over a region on the form $[x_0,x_1]\times[y_0,y_1]$, then

$\frac{\partial}{\partial x} \int_{y_0}^{y_1} f(x,y) dy = \int_{y_0}^{y_1} f_x(x,y) dy$

for $x$ in this region. For proofs and more general statements, see the wikipedia page on the Leibniz integral rule.

For your second question, about the integration, this is related to the surface $A$, and the fact that your conclusions hold for any choice of $A$. In other words, we can make $A$ arbitrarily small, thus requiring the integrand to be identically zero everywhere.

If you need further proof of this, think about what happens if the integrand is not zero everywhere. Then there is (at least) a small region somewhere that contributes to the integral, and another small region somewhere else which cancels this contribution. But since our choice of $A$ is arbitrary, we can chose $A$ so that only one of the regions is included; the contributions don't cancel, and the integral value is nonzero, which contradicts Faraday's law in integral form.

Tomas Aschan
  • 473
0

the integral form of Faraday's law is $$ \int_{\partial A} \vec E \cdot \mathrm{d}\vec s = -\int\frac{\partial \vec B}{\partial t} \cdot \mathrm{d}\vec A $$

The form $$ \int_{\partial A} \vec E \cdot \mathrm{d}\vec s = -\frac{\mathrm{d}}{\mathrm{d}t}\int\vec B \cdot \mathrm{d}\vec A $$ is usually wrong, if there is a temporal change of A, because $$ \frac{\mathrm{d}}{\mathrm{d}t} \int_A \vec{B}\cdot \mathrm{d}\vec{A}=\int_A \frac{\partial \vec{B}}{\partial t} \cdot \mathrm{d}\vec{A} + \oint_{\partial A} (\vec{B} \times \vec{u}) \cdot \mathrm{d}\vec{s} + \int_A (\nabla \cdot \vec{B}) \cdot \vec{u} \cdot \vec{d}\mathbf{A}. $$ with $\vec u$ being the (local) speed of the area $A$ and its contour line $\partial A$, respectively. The term with $\nabla \cdot \vec B$ does not need to be considered here, because of the Maxwell equation $\nabla \cdot \vec B = 0$.

Note that the field $\vec E$ needs to be measured from the same frame of reference as all other variables. This means in particular, that an (ideal) conductor moving in space with speed $\vec v$ has an $\vec E$ field of $\vec E=-\vec v \times \vec B$ in the laboratory frame, whereas an observer sitting on the conductor measures an $\vec E'$ field of $\vec E'=\vec 0$ (the prime ' marks a change of the frame of reference). Many physicists think that the wrong equation is right, because they confuse the different frames of reference.

You will find more information in: http://de.wikipedia.org/wiki/Elektromagnetische_Induktion#.C3.9Cbergang_von_der_differentiellen_Form_zur_Integralform

and

https://en.wikipedia.org/wiki/Classical_electromagnetism_and_special_relativity

You might want to use google translate to translate the German text.

Best regards

guest
  • 1