Area element in polar co's/trouble with diff. forms

[speeding electron]

So I'm fiddling about with formulae...trying things out...consolidating...as you do...and something didn't work - why?

The infinitesimal area element in polar coordinates dA = r*d(theta)*dr . Agreed? Now in the same coordinate system $$x = r \cos{\theta} ; y = r \sin{\theta}$$. Taking differentials, we have:
$$dx = dr \cos{\theta} - r \sin{\theta} d\theta ; <br /> dy = dr \sin{ \theta} + r \cos{\theta} d\theta$$.
Multiplying together we get dx dy, the area element in cartesian coordinates, and hopefully equal to r*d(theta)*dr, but
$$dx dy = {dr}^2 \cos{\theta} \sin{\theta} - d\theta^2 r^2 \sin{\theta} \cos{\theta} + \cos^2{\theta} r d\theta dr - \sin^2{\theta} r d\theta dr$$.
Neglecting squares of differential forms, we have that:
$$dx dy = \cos{2\theta} r d\theta \dr$$
Not what I was hoping, you see. So could someone explain? It's probably down to some assumption I'm making, but I don't know what it is. All I can think of is that either the area elements are different in the different coordinates, or that since we are considering the product of differentials the squared forms cannot be ignored.

P.S. can someone refer me to somewhere that'll show me where I can type Greek letters without reverting to Latex? Thanks.

 

[robphy]

Try $dx \wedge dy$

 

[speeding electron]

Looks like I'm going to have to learn a bit more linear algebra. For now, could you explain why multiplying the good ol' way doesn't work? Thanks.

 

[Tide]

I think it's because part of the dA arises from the unit vectors in the x and y directions varying as you generate your dr and $d\theta$.

 

[Wong]

The algebra for forming area elements doesn't work that way.

Think about this. When you are finding the area element $$dxdy$$, what you are finding is in fact *the area formed by the vector $$dx\vec i$$ and $$dy\vec j$$*. Using your calculation, (intuitively), we should have $$dx \vec i = dr \cos{\theta}\vec{e_{r}} - r \sin{\theta} d\theta\vec{e_{\theta}}$$ and $$dy\vec{j} = dr \sin{ \theta}\vec{e_{r}} + r \cos{\theta} d\theta\vec{e_{\theta}}$$. Then the problem is, given two vectors $$\vec{a}$$ and $$\vec{b}$$, how can we find the area spanned by them? The answer is $$\vec{a}\times\vec{b}$$. So to find the area element in dr and $$d\theta$$ what you need to do is cross $$dx\vec{i}$$ with $$dy\vec{j}$$. (Since cross product only exists in three dimension, you may like to “artificially” add in another dimension z.)

You may compare this with your original method. You will discover the algebra is different from that of your method. In fact, as robphy said, the algebra with the transformation of area elements is the same as the algebra of forms

 

[HallsofIvy]

In particular "dx dx" makes no sense. Multiplication of "differential forms" is anti-symmetric- that's that $dx \wedge dy$ that robphy mentioned. $x= rcos(\theta)$ so $dx= cos(\theta)dr- r sin(\theta)d\theta$ and $y= rsin(\theta)$ so $dy= sin(\theta)dr+ rcos(\theta)d\theta$. Multiplying those (carefully keeping the "order" straight): $dxdy= sin(\theta)cos(\theta)dr^2+ rcos^2(\theta)drd\theta- rsin^2(\theta)d\thetadr- r^2sin^2(\theta)d\theta^2$.

Since multiplication of differential forms is anti-symmetric, $dr^2= 0, d\theta^2= 0$ and $d[theta]dr= -drd[\theta]$. Putting those into the formula, $dxdy= rcos^2(\theta)drd\theta+ rsin^2(\theta)drd\theta= r dr d\theta$


You can also do it in terms of the "Jacobian", the determinant of the derivatives, or as a cross product of vectors as Wong said.

 

[speeding electron]

OK, thanks for that. Forming the cross product in polar coordinates (i.e. treating $\vec{e_{\theta}}$ and $\vec{e_{r}}$as basis vectors) gives the right answer, but in cartesian, thought the multiplication is of course still anti-symmetric, dx dy is still, well, dx dy multiplied my way or -dx dy, depending on the order. Why will the cross product only work in the appropriate coordinate system?
I now remember reading in a brief section on differential forms (they weren't 'within the book's scope') that in a differential form, any odd number of permutations reverses the sign of the form, and an even number produces the same form (does the Levi-Civita tensor come in here?). One more question: why then, in a double integral calculation, can you integrate with respect to either variable first? Thanks for your time.

 

[Hurkyl]

One more question: why then, in a double integral calculation, can you integrate with respect to either variable first?

Because when you switch the order of integration, you are reversing both the order of the differentials and the orientation of the region. Both of these contribute a negative sign to the integral, which cancel out.