0
$\begingroup$

I am currently trying to show that the form $$\omega_{a,b,c}=\frac{(x-a)dy\wedge dz+(y-b)dz\wedge dx+(z-c)dx\wedge dy}{[(x-a)^2+(y-b)^2+(z-c)^2]^{3/2}}$$ is closed.

So I read some of the answers on the site, and made $r = [(x-a)^2+(y-b)^2+(z-c)^2]^{1/2}$ and made $f = \frac{1}{r^3}$, and $$\alpha = (x-a)dy\wedge dz + (y-b)dz\wedge dx + (z-c)dx\wedge dy$$ So to find $\partial\omega_{a,b,c}$, I showed that $$\partial\omega_{a,b,c} = \partial(f\alpha) = \partial f\wedge\alpha+f\partial\alpha$$But when I found $\partial\alpha$, I calculated it as:$$\partial\alpha=d(x-a)\wedge dy\wedge dz + d(y-b)\wedge dz\wedge dx + d(z-c)\wedge dx\wedge dy$$ Now I also get $$\partial f = \frac{-3}{r^5}((x-a)dx+(y-b)dy+(z-c)dz)$$ So I tried to find $\partial f\wedge\alpha$, which led to:$$\frac{-3}{r^5}(r^2(dx\wedge dy\wedge dz))$$ So somewhere my maths is going awry, and I can't seem to figure out why. Any help would be greatly appreciated!

  • 1
    Your work is just fine, just simplify: $\partial f\wedge \alpha = -\frac{3}{r^3}dx\wedge dy\wedge dz$ and $fd\alpha = \frac{3}{r^3}dx\wedge dy\wedge dz$, so $\partial f\wedge \alpha + f d\alpha = 0$. – 2017-01-25

1 Answers 1

0

you make everything right. $df\wedge \alpha=-\frac{3}{r^3}dx\wedge dy\wedge dz $. On the other hand $f d\alpha=\frac{3}{r^3}dx\wedge dy\wedge dz $. So $d(f\alpha)=0$ as you wanted to prove.