# Exterior differential

1. Nov 10, 2005

### Zurtex

Hey, having a little problem with a question because I am a little to unsure how to do it, I've been asked to calculate some exterior differentials, would this be write:

$$d\left(r^3 (\cos (4 \theta) dr - r \sin (4\theta) d\theta)\right)$$

Simplified:

$$d\left(r^3 \cos (4\theta) dr - r^4 \sin (4\theta) d\theta\right)$$

Goes to:

$$2r^3 dr \left(-4 \sin (4\theta) d\theta dr - 4r^3 dr \cos (4\theta) d\theta d\theta \right) = 0$$

Am I even slightly right?

2. Nov 10, 2005

### hypermorphism

Note that if p and q are k-forms, then $dp\wedge dq = -dq\wedge dp$ which leads to $dp\wedge dp = 0$ and the differential follows the product rule. Thus we get:
$$d\left(r^3 \cos (4\theta) dr - r^4 \sin (4\theta) d\theta\right)$$
$$=-4r^3\sin(4\theta)d\theta\wedge dr - 4r^3\sin(4\theta)dr\wedge d\theta$$
= 0.
You could have saved a bit of work by noting that your expression is just
$$d(\frac{r^4\cos(4\theta)}{4})$$
since the differential of a differential is zero.

3. Nov 10, 2005

### Zurtex

When say exterior differentiating $r^3 \cos (4\theta) d\theta$ Does only dr get generated and not an extra $d\theta$?

4. Nov 10, 2005

### hypermorphism

That's due to the anticommutativity of the wedge product. The full calculation is:
$$d(r^3 \cos (4\theta) d\theta) = d(r^3\cos(4\theta))\wedge d\theta + (-1)^{1}r^3\cos(4\theta)\wedge d(d\theta))$$
$$= (3r^2 \cos(4\theta) dr - 4r^3 \sin(4\theta )d\theta ) \wedge d\theta - 0$$
$$= (3r^2 \cos(4\theta)) dr\wedge d\theta - (4r^3 \sin(4\theta )) d\theta\wedge d\theta$$
$$= (3r^2 \cos(4\theta)) dr\wedge d\theta - 0$$
Just too much to type out. if you have a k-form where k is greater than or equal to one, you can decrease your drudge work by just ignoring partials with respect to differentials already present in the form and noticing existing differentials.
All you really need is the anticommutativity, from which can be derived $dp\wedge dp = 0$, the generalized product rule, and d(dp)=0 for any form p.

Last edited: Nov 10, 2005
5. Nov 12, 2005

Thanks a lot