Line integral in n dimension(Stokes' theorem)

1. May 26, 2012

youngurlee

I have shown by my intuition that if a good field g(2th or more differentiable) in n dimension satisfies
$\frac{∂g_{i}}{∂x_{j}}-\frac{∂g_{j}}{∂x_{i}}$=0 for all i,j,
then $\oint$g$\cdot$dl=0,
hence there exist a scalar function $\phi$ such that
$\frac{∂\phi}{∂x_{i}}=g_{i}$ for all i.

I want to know what in general will the ring integral $\oint$g$\cdot$dl be,
can it be written as a surface integral, as in Kelvin–Stokes theorem?

2. May 26, 2012

algebrat

wouldn't it be the surface integral of the flux of the curl of g over a surface with boundary the curve?

$\int_S\nabla\times\textbf{g}\cdot d\textbf{A}$

also keep in mind you're only guaranteed a potential locally. A spiral staircase has no local curl, but a global potential cannot be defined.

3. May 26, 2012

youngurlee

but in n dimension, what then will $\nabla\times \textbf{g}$ be? is it also an n dimensional vector? what is it's form? moreover, if $d\textbf{A}$ is a vector, shouldn't it has n elements, then what are these elements?
consider in 4D, if the elements take the form dxi$\wedge$ dxj (imitate the 3D case), wouldn't then $d\textbf{A}$ owns 6 elements, how could then $d\textbf{A}$ be a vector in 4 dimension?

but Kelvin–Stokes theorem doesn't seem to care about local or global, it just tells that a field with no curl will yields a zero ring integration, and hence owns a potential.
could you explain your idea, please? thanks!

4. May 26, 2012

algebrat

Sure, I can't assume that these things are defined for n-dimensions. So two issues, what is the integral, and is there a potential:

1. I have to appeal to $\int_{\partial M}\omega=\int_Md\omega$.

The left hand side, is in your case g integrated around the boundary. The dot product comes from the pullback of g using the inclusion map of the boundary to manifold with boundary. Strictly speaking, I'm not sure which textbook or source would lead you from the pull back to a dot product, in arbitrary n-dimensions, so I'm not the only one here messing with the definitions. But I thought what you wrote and mixed together was fine.

So, I shouldn't be writing the cross product in n-dimensions, and I'm not even supposed to use curl unless there's a metric. But if you take it to mean $\sum\int_S\frac{\partial f_i}{\partial x_j}-\frac{\partial f_j}{\partial x_i}dx_idx_j$, then we were headed in the right direction, because that's what the abstract stuff would lead to.

2. The potential. If the exterior derivative (in our case, should be okay to call it curl) is zero all along the surface, than the integral is zero.

If all integrals of curl are zero, than we have a potential, but you never said anything about all integrals, so based off one integral I can't guarantee there is a potential. There are counterexamples.

5. May 26, 2012

algebrat

dA is the sum of three 2-forms, which is still a 2-form, just like if you add 3 vectors, it's still a vector. It's really the whole integrand $d\omega$ which is the 2-form, some linear combinations of 2-forms $dx_i\wedge dx_j$.

For the potential issue: to build some topology intuition, ignore the integral for a moment. If the derivative is zero, then there is a local potential. If you move around and the derivative is remains zero, then you can extend that potential. But if you travel around a hole extending the potential, then you could get two values for your potential function, which is not a function. like the spiral staircase. So it's not enough for an integral to be zero once, it's got to be zero repeatedly, and all over the place, and once again as you extend the potential in this way, you can't extend the potential around a hole.

If Kelvin-Stokes theorem doesn't treat this, please reference where you saw this.

6. May 26, 2012

youngurlee

Thank you very much.
I am not a maths student, I can't really understand what stuffs like $\int_{\partial M}\omega=\int_Md\omega$ mean. I saw it in analysis books, I know that it is a very
generalized case on Stokes' theorem, from which many cute identities in vector integral can be deduced.

my question came to me when I read Dirac' s "Principles of QM"(page 92, Eq(34)), while he affirms a potential for a vector field like g.
I also guessed the ring integral may yields to $\sum\int_S\frac{\partial f_i}{\partial x_j}-\frac{\partial f_j}{\partial x_i}dx_idx_j$, but I failed to proof that.
If it really holds and can be deduced from $\int_{\partial M}\omega=\int_Md\omega$, I would not bother myself vexing about it(since I don't have that maths basics), but I don't know whether it is so. (could you tell me if it is the case?)
Yes, that's true, I forgot to say that "g has that property for all points in space".

7. May 26, 2012

algebrat

$$\oint_C\textbf{g}\cdot d\textbf{l}=\int_{\partial M}\omega=\int_Md\omega$$
$$\omega=g_1dx^1+\cdots+g_ndx^n$$
$d\omega=dg_1\wedge dx^1+\cdots+dg_n\wedge dx^n=(\partial_1g_1dx^1+\cdots+\partial_ng_1dx^n) \wedge dx^1+\cdots+(\partial_1g_1dx^1+\cdots+\partial_ng_1dx^n) \wedge dx^n$

But at this point, I have to apologize, I'm going to tap out and let you check the signs on these, I'm guessing you know a little about how the signs flip on those wedges, and you should know dx wedge dy, in the integral is just dxdy for the normal integral. Also those partial g things mean partial of g sub i with respect to x sub j. Try the n=2 case and n=3 case to make sure it works. But remember, this is definitely the generalization of green's and curl theorem, not divergence or gradient theorem, since we went from 1-forms to 2-forms.

8. May 26, 2012

youngurlee

Yes, I guess I get the idea.
If as you said, signs flip over wedge, that is$$dx\wedge dy=-dy\wedge dx$$, and if your prescription about $d\omega$ is all right, then it works out well in n=3. so it seems that my problem is totally solved, for then $$\int_Md\omega=0$$, and in general, $$\oint_C\textbf{g}\cdot d\textbf{l}=\int_Md\omega=∫∫ Ʃ_{i>j}(\frac{∂g_{i}}{∂x_{j}}-\frac{∂g_{j}}{∂x_{i}})dx_{i}\wedge dx_{j}$$
thank you!

9. May 26, 2012

algebrat

Okay, but I don't know about the i>j guess. I'm not even sure it's true for n=2. if you do it for n=4, you may have the pattern (use x,y,z,w to speed up the algebra, then switch back to x_i to find the rule). n=3 would show a certain cyclic behavior, 3<2<1<3<2 i'm guessing, but I doubt that same pattern would work for n=4, so you gotta try n=4 at least.

10. May 26, 2012

youngurlee

The exterior derivative defined by $d\omega=dg_1\wedge dx^1+\cdots+dg_n\wedge dx^n=(\partial_1g_1dx^1+\cdots+\partial_ng_1dx^n) \wedge dx^1+\cdots+(\partial_1g_1dx^1+\cdots+\partial_ng_1dx^n) \wedge dx^n$ is all right (I found it in a lecture document), so according to $$\int_{\partial M}\omega=\int_Md\omega$$, which is to be true for all k-form and its exterior derivative in virtue of Stokes' theorem, in our case the 1-form corresponds to the line integral element, and its exterior derivative corresponds to what I have written in the integrand,
my guess with the result $$\oint_C\textbf{g}\cdot d\textbf{l}=∫∫ Ʃ_{i>j}(\frac{∂g_{i}}{∂x_{j}}-\frac{∂g_{j}}{∂x_{i}})dx_{i}\wedge dx_{j}$$ must be right, while i>j just follows from the reorder of sum and the identity $$dx\wedge dy=-dy\wedge dx$$.

11. May 26, 2012

youngurlee

actually the order of $$dx_{i}\wedge dx_{i}$$ isn't of much importance, and hence the cyclic behavior or some other. one can always get a desired order, since the coefficients of the basic 2-forms can be made opposite in step.

12. May 26, 2012

youngurlee

Oh, I am wrong, I messed up Stokes' theorem.
Stokes theorem applies to (n-1)-form in n D, not for all k-forms with k<n.
It seems that my problem remains unsolved.

13. May 26, 2012

youngurlee

How could you apply Stokes' theorem in n D for 1-form but not (n-1)-form?
Is it valid for all k-forms, where k<n?

14. May 26, 2012

algebrat

No, that is not an issue.

We related 1 and 2 forms in n dimensions. We integrated a 1-form $\omega$ along the boundary $\partial M$, which was a 1-manifold. This was equal to integrating the exterior derivative of the 1-form $d\omega$, which gives us a 2-form, along a 2-manifold $M$. So 1-form meant it was sum of differentials of coordinates, $dx^i$, and 2-forms are sum of single wedge of differentials of coordinates, $dx^i\wedge dx^j$. A k-form could be expressed as sum of $k$ wedges of coordinate differentials, $dx^{i_1}\wedge\cdots\wedge dx^{i_k}$. If our $k$-manifold M is coordinatized in degree $n$, then we have that many coordinate differentials to consider, $\{dx^1,\dots,dx^n\}$. So $M$ might be a $k$-manifold, locally immersed in ℝ$^n$, then $1\le k\le n$, and $\omega$ is a $k-1$-form.

While you can't immerse an $n+1$-manifold in ℝ$^n$, you can take the exterior derivative of an $n$-form. It's not hard to show that it is zero.

Last edited: May 26, 2012
15. May 27, 2012

youngurlee

You're right, I saw this claim in "Multivariate Calculus" by Alder(see the attached).
I know not the detail, but I feel that the validity of such a claim is just what I want.

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