Stoke's Theorem - Curl of a Magnetic Field

[Astrum]

Griffiths's proof of Ampère's Law was probably one of the ugliest things I've seen. All that product rule, integration by parts and what not, really could have brought tears to the eyes of ANY man.

I mean, have you LOOKED at this? GAAAAH it's terrible, really, terrible.

The simple statement that he presents on pg. 221 - 222 is much more elegant. But one thing confuses me.

How is $$\int \vec{J} \cdot d \vec{a} = \int (\nabla \times \vec{B}) \cdot d \vec{a}$$

I see that this is an application of Stoke's Theorem, so I guess I'm asking for a clarification of what Stoke'es Theorem actually SAYS, and why it makes sense.

I'm not sure if this belongs in the classical physics section, or the math section.

 

[WannabeNewton]

Take a look at references online or a textbook because Stokes' theorem can be explained in a more coherent manner through such mediums. For example: http://ocw.mit.edu/courses/mathemat...007/video-lectures/lecture-31-stokes-theorem/ or p.92 of Purcell's electromagnetism text (3rd ed.)

 

[Astrum]

Take a look at references online or a textbook because Stokes' theorem can be explained in a more coherent manner through such mediums. For example: http://ocw.mit.edu/courses/mathemat...007/video-lectures/lecture-31-stokes-theorem/ or p.92 of Purcell's electromagnetism text (3rd ed.)

I'll take a look. And perhaps I'll get my hands on Purcell's, since you keep telling us how awesome of a text it is, I'll have to check it out for myself

 

[jtbell]

I see that this is an application of Stoke's Theorem, so I guess I'm asking for a clarification of what Stoke'es Theorem actually SAYS, and why it makes sense.

See section 1.3.5 of Griffiths, pages 34-36:

The fundamental theorem for curls, which goes by the special name of Stokes' Theorem, states that

$$\int{(\vec \nabla \times \vec v) \cdot d \vec a} = \oint {\vec v \cdot d \vec l}$$

$\vec v$ is any vector field. In this particular application, $\vec v = \vec B$. The integral on the left is a surface integral, taken over some surface. The integral on the right is a line integral around the boundary of that surface.

It's useful, although somewhat tedious, to verify Stoke's Theorem by calculating the integrals explicitly for a few simple cases, as in Example 1.11 and Problem 1.33.

 

[atyy]

The equation in the OP is not an application of Stokes's theorem. In Stokes's theorem a quantity integrated over the boundary is related to a quantity integrated over the space enclosed by the boundary.

In the OP, one takes the differential form of Ampere's law in the absence of a changing electric field ∇ X B =J and integrates it.

The differential form of Ampere's law is given in the right column of the Wikipedia link http://en.wikipedia.org/wiki/Maxwell's_equations. It is in getting from the equations in the right column to the equations in the left column that one uses Stokes's theorem - in the equation in the left column you see the integral on the left and right hand side of the equations are integrated over different things - one being the boundary denoted ∂Ʃ and the other being the area enclosed by the boundary Ʃ.

 

[jtbell]

The equation in the OP is not an application of Stokes's theorem.

No; however, the right-hand side is the result of an application of Stokes' Theorem.

Griffiths starts with Ampere's Law in the form

$$\oint {\vec B \cdot d \vec l} = \mu_0 I_{enc}$$

On the left side, he applies Stokes' Theorem. On the right side, he applies the relationship between current and current density

$$I_{enc} = \int {\vec J \cdot d \vec a}$$

The result is the equation in the OP, with left and right sides switched.

 

[atyy]

@jtbell, yes, I agree.