Stokes' theorem and surface integrals

[dRic2]

TL;DR Summary

Compute the integral of the "curl" of the vector field $A_i(x_i)$ over a 2-dimensional surface.

Hi,

So my goal is to compute the integral of the "curl" of the vector field $A_i(x_i)$ over a 2-dimensional surface. Following a physics book that I am reading, I introduce the antisymmetric 2-nd rank tensor $\Omega_{ij}$, defined as:
$$\Omega_{ij} = \frac {\partial A_i}{\partial x_j} - \frac {\partial A_j}{\partial x_i}$$
then, from what I gather reading the Wikipedia page (https://en.wikipedia.org/wiki/Surface_integral), if the surface can be parametrized by a set of 2 variables $u$ and $v$ I should be able to write the surface element as $dudv$ times the determinant of the jacobian:
$$ \frac {\partial x_i}{\partial u} \frac {\partial x_j}{\partial v} - \frac {\partial x_j}{\partial u} \frac {\partial x_i}{\partial v}$$
Putting all together I get:
$$\int_S \left( \frac {\partial A_i}{\partial x_j} - \frac {\partial A_j}{\partial x_i} \right) \left( \frac {\partial x_i}{\partial u} \frac {\partial x_j}{\partial v} - \frac {\partial x_j}{\partial u} \frac {\partial x_i}{\partial v} \right) dudv$$
Using the chain rule, I can simplfy the expression to yield:
$$ \int_S \left( \frac {\partial A_i}{\partial v} \frac {\partial x_i}{\partial u} - \frac {\partial x_i}{\partial v} \frac {\partial A_i}{\partial u} - \frac {\partial A_j}{\partial u} \frac {\partial x_j}{\partial v} + \frac {\partial x_j}{\partial u} \frac {\partial A_j}{\partial v} \right) dudv = 2 \int_S \left( \frac {\partial A_i}{\partial v} \frac {\partial x_i}{\partial u} - \frac {\partial x_i}{\partial v} \frac {\partial A_i}{\partial u} \right) dudv
$$
But I am pretty sure there should not be a 2 in the last equation. Is my reasoning wrong?

Thanks
Ric

 

[Eclair_de_XII]

I think you should check your distributions from the second-to-last line. From what I got, the terms on the last line should all be distinct.

This is assuming that your reasoning is correct, though. Algebraically, your final answer does not look right.

 

[dRic2]

In the last line $u$ and $v$ were swapped. I corrected the OP for clarity.

 

[pasmith]

Summary:: Compute the integral of the "curl" of the vector field $A_i(x_i)$ over a 2-dimensional surface.

Hi,

So my goal is to compute the integral of the "curl" of the vector field $A_i(x_i)$ over a 2-dimensional surface. Following a physics book that I am reading, I introduce the antisymmetric 2-nd rank tensor $\Omega_{ij}$, defined as:
$$\Omega_{ij} = \frac {\partial A_i}{\partial x_j} - \frac {\partial A_j}{\partial x_i}$$
then, from what I gather reading the Wikipedia page (https://en.wikipedia.org/wiki/Surface_integral), if the surface can be parametrized by a set of 2 variables $u$ and $v$ I should be able to write the surface element as $dudv$ times the determinant of the jacobian:
$$ \frac {\partial x_i}{\partial u} \frac {\partial x_j}{\partial v} - \frac {\partial x_j}{\partial u} \frac {\partial x_i}{\partial v}$$
Putting all together I get:
$$\int_S \left( \frac {\partial A_i}{\partial x_j} - \frac {\partial A_j}{\partial x_i} \right) \left( \frac {\partial x_i}{\partial u} \frac {\partial x_j}{\partial v} - \frac {\partial x_j}{\partial u} \frac {\partial x_i}{\partial v} \right) dudv$$
Using the chain rule, I can simplfy the expression to yield:
$$ \int_S \left( \frac {\partial A_i}{\partial v} \frac {\partial x_i}{\partial u} - \frac {\partial x_i}{\partial v} \frac {\partial A_i}{\partial u} - \frac {\partial A_j}{\partial u} \frac {\partial x_j}{\partial v} + \frac {\partial x_j}{\partial u} \frac {\partial A_j}{\partial v} \right) dudv = 2 \int_S \left( \frac {\partial A_i}{\partial v} \frac {\partial x_i}{\partial u} - \frac {\partial x_i}{\partial v} \frac {\partial A_i}{\partial u} \right) dudv
$$
But I am pretty sure there should not be a 2 in the last equation. Is my reasoning wrong?

Thanks
Ric

There are $n(n-1)$ possible off-diagonal values of $(i,j)$. The way you have set your integral up, you only need the $\frac12 n(n-1)$ terms where $i < j$. After changing variables to $u$ and $v$ your integrand is symmetric in $i$ and $j$; hence your answer is twice as large as it should be.

 

[dRic2]

Oh I see. Since $dx_i \wedge dx_j = - dx_j \wedge dx_i$ I was counting the same surface twice!

Even though I don't know differential geometry at all, I read on Wikipedia (https://en.wikipedia.org/wiki/Differential_form) that the "differential 2-form" (whatever that is...) is defined for i<j in order to give the correct surface element. I didn't understand most of the article but it seems to me that it says the same thing that you said. Thanks!

 

[pasmith]

Oh I see. Since $dx_i \wedge dx_j = - dx_j \wedge dx_i$ I was counting the same surface twice!

Even though I don't know differential geometry at all, I read on Wikipedia (https://en.wikipedia.org/wiki/Differential_form) that the "differential 2-form" (whatever that is...) is defined for i<j in order to give the correct surface element. I didn't understand most of the article but it seems to me that it says the same thing that you said. Thanks!

Note that $$\int \partial_jA_i\,dx_i \wedge dx_j = \int (\partial_2A_1-\partial_1A_2)\,dx_1 \wedge dx_2 + \int (\partial_3A_2-\partial_2A_3)\,dx_2 \wedge dx_3 + \int (\partial_3A_1-\partial_1A_3)\,dx_1 \wedge dx_3$$