Feynman's derivation of Stokes' theorem

[hobz]

While reading the Feynman lectures, I stumbled upon a passage which I do not quite understand.

 

In Vol II, 3-9, Feynman is telling us about Stokes' theorem.

 

He writes:

 

[math]

\oint \vec{C} \cdot d\vec{s} = C_x(1) \Delta x + C_y(2)\Delta y -C_x(3) \Delta x - C_y(4)\Delta y

[/math]

and he looks at

[math]

[C_x(1) -C_x(3)]\Delta x

[/math]

and writes:

"You might think that to our approximation the difference is zero. That is true to the first approximation. We can be more accurate, however, and take into account the rate of change of [math]C_x[/math]. If we do, we may write

[math]

C_x(3)=C_x(1)+\frac{\partial C_x}{\partial y} \Delta y

[/math]

"If we included the next approximation, it would involve terms in [math](\Delta y)^2[/math], but since we will ultimately think of the limit as [math]\Delta y \rightarrow 0[/math], such terms can be neglected."

 

How can the theorem be correct if he neglects terms?

Why doesn't the neglecting logic apply to the first order derivative? (The derivative term vanishes when we let [math]\Delta y \rightarrow 0[/math])

I thought [math]dy=\frac{\mathrm{d}y}{\mathrm{d}x}dx[/math] without any other terms.

[Mr Skeptic]

Nothing is negligible compared to zero, so you have to include your smallest term if otherwise you get zero. For very small x, the square of x is often negligible compared to x, often even if there is other stuff in the term.

[hobz]

Interesting. Does "have to" correspond to some theorem in calculus?

[Mr Skeptic]

Maybe. All I'm saying is that your largest term generally can't be ignored as negligible.

[hobz]

Considering what he is arriving at, it seems a bit informal to include one term and neglect others.

I thought Stokes' theorem was more formal/rigid than what is explained here.

[Cap'n Refsmmat]

Hmm. I took vector calculus last semester, so I might be able to help, but I'm not familiar with how this is written:

 

[math]

\oint \vec{C} \cdot d\vec{s} = C_x(1) \Delta x + C_y(2)\Delta y -C_x(3) \Delta x - C_y(4)\Delta y

[/math]

 

Could you explain what [math]C_x(1)[/math] and so on represent?

[hobz]

Yes. Feynman has a nice drawing in his book which I didn't bother to include at first.

 

I have scanned it and attached it.

 

 

[math]C_x(1)[/math] is the tangential component of the vector field [math]C[/math] (denoted [math]1[/math] on the illustration.)

[Cap'n Refsmmat]

I think Feynman is taking advantage of the fact that at arbitrarily small distances, any function looks linear. Does he present this as an actual proof of Stokes' Theorem or just as an explanation of why it works?

[Mr Skeptic]

Considering what he is arriving at, it seems a bit informal to include one term and neglect others.

 

It's always perfectly reasonable to ignore negligible terms. So long as the closer you get to zero you can get the other terms to be arbitrarily much smaller than the first, they're safe to ignore.

 

I thought Stokes' theorem was more formal/rigid than what is explained here.

 

Famous theorems have all kinds of proofs all over the place. If you want a more formal proof, feel free to google for one.

[hobz]

I think Feynman is taking advantage of the fact that at arbitrarily small distances, any function looks linear. Does he present this as an actual proof of Stokes' Theorem or just as an explanation of why it works?

 

From the tone of his lectures, the latter seems more correct.

 

It's always perfectly reasonable to ignore negligible terms. So long as the closer you get to zero you can get the other terms to be arbitrarily much smaller than the first, they're safe to ignore.

 

It seems like the second term is included to prevent the whole thing from being zero.

If the third term is discarded, then it is really more of an approximation, although it is stated as an equation.

 

So why not keep the third term when keeping the second term?

[Mr Skeptic]

Because it's negligible. If you have limit of 1+x as x-->0, you can ignore the x, as it is negligible. If you get limit of 1+x+x^2 as x-->0, you can ignore both the x and x^2. If you have limit of 1+x+1/x as x-->0, now you can ignore the 1 and the x. It's not a matter of "is equal now" but of "will be equal later".

[hobz]

True.

 

But by that logic, [math]\lim_{\Delta x \rightarrow 0}\frac{\partial y}{\partial x}\Delta x = 0[/math], just as [math]\lim_{\Delta x \rightarrow 0}\frac{\partial^2 y}{\partial x^2}(\Delta x)^2 = 0[/math], which is discarded for its negligibility.

[hobz]

I would agree with Feynman, if he didn't mention the higher order approximations.

 

Wouldn't the first order derivative alone reveal how much the field [math]C[/math] changes along [math]\Delta y[/math]?

[Cap'n Refsmmat]

Only if its rate of change is constant.

[hobz]

So it (Stokes theorem) is at best a good approximation?

[Cap'n Refsmmat]

No, I think there are more rigorous proofs available. Here's one:

 

http://higheredbcs.wiley.com/legacy/college/hugheshallett/0471484822/theory/hh_focusontheory_sectionm.pdf