Exterior differential

[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?

[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.

[Zurtex]

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

[hypermorphism]

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

[Zurtex]

Thanks a lot :smile: