solutions for problems on vector analysis and differential forms


Assume that 4 vectors in \(\mathbb{R}^3\) satisfy \(A+B+C+D\)=0.
$$ A\times B-B\times C+C \times D-D\times A$$

Solution: By the anti-symmetry \(-B\times C = C\times B\)
and \(-D\times A = A\times D\) we can rewrite the expression as
$$ A\times B+C\times B+C \times D+A\times D$$
Using the distributive law of the cross product, we have
$$ (A+C)\times B+(C+A) \times D$$
and again
$$ (A+C)\times (B+D) =** $$
However, as \(F=(A+C) = -(B+D)\) we get
$$ **= F\times (-F) = 0 $$

Exercise: Integrate \(dx/y\) over the circle (oriented counterclockwise)

Solution: utilize a counter clockwise parametrization
$$ x(t)=R\cos(2\pi t),~~~~~y(t)=R\sin(2\pi t),~~~~ t\in [0,1) $$
The differential becomes
$$ \frac{dx}{y} = \frac{ -2\pi R\sin(2\pi t) dt}{R\sin(2\pi t)}= -2\pi dt $$
(except two removable singularities at \(t=0,\frac{1}{2}\)).
The overall integral would then be
$$ \int_{C} \frac{dx}{y} = \int_{0}^1 (-2\pi dt) = -2\pi $$

An alternative approach is to integrate over \(x\) and use the explicit
form of \(y=\pm \sqrt{R^2-x^2}\) (and account for integration direction):
\underbrace{\int_{R}^{-R} \frac{dx}{\sqrt{R^2-x^2}} dx}_{
\text{Upper half integral} \\
\text{Note: right to left direction}
\end{array}} +
\underbrace{\int_{-R}^{R} \frac{dx}{(-\sqrt{R^2-x^2})} dx}_{
\text{Lowed half integral}}
= -2\int_{-R}^{R} \frac{dx}{\sqrt{R^2-x^2}}
=** $$
Then, by substitution \(x = -\cos(\pi t) \) with end-points \(t=0,1\) we get
$$ ** = -2\int_{0}^1 \frac{\pi R\sin(\pi t)}{R\sin(\pi t)} dt = -2\pi $$

Exercise: let
\(\omega = dx\wedge dy + ds \wedge dt + du\wedge dv \). Find \(\omega^3\).

Solution: This is a \(2\)-form on \(\mathbb{R}^6\), so \(\omega^3\)
would be a \(6\)-form on \(\mathbb{R}^6\), hence a multiple of the volume form:
$$ \omega^3 = cdx \wedge dy \wedge du \wedge dv \wedge ds \wedge dt $$
So that we just need to find \(c\).

First compute the \(4\)-form \(\omega^2\):
\omega\wedge\omega &= (dx\wedge dy + ds \wedge dt + du\wedge dv )\wedge
(dx\wedge dy + ds \wedge dt + du\wedge dv ) \\
= &(dx\wedge dy)\wedge (dx\wedge dy) +
(dx\wedge dy)\wedge (ds \wedge dt) +
(dx\wedge dy)\wedge (du\wedge dv) \\
& +(ds \wedge dt)\wedge (dx\wedge dy)
+(ds \wedge dt)\wedge (ds \wedge dt)
+(ds \wedge dt)\wedge (du\wedge dv) \\
& +(du\wedge dv)\wedge (dx\wedge dy)
+(du\wedge dv)\wedge (ds \wedge dt)
+(du\wedge dv)\wedge (du\wedge dv) \\ \end{align*} $$
We can get rid of parenthesis since the wedge product is also associative
(in contrast to the cross product ). The diagonal elements vanish
since \(dx\wedge dx = ds\wedge ds = du\wedge du =0\), and the
off diagonal elements are equal as \(2\)-forms have commutative wedge
\(\alpha^\wedge \beta = (-1)^4 \beta\wedge \alpha \)
so that
\omega^2 &= 2\left[ (dx\wedge dy)\wedge (ds \wedge dt)
+ (dx\wedge dy)\wedge (du\wedge dv)
+ (ds \wedge dt)\wedge (du\wedge dv) \right]
\end{align*} $$

Then by the same rules we get
\begin{align*} \omega^3 =
\omega \wedge \omega^2 &= 2\left[ dx\wedge dy + ds \wedge dt + du\wedge dv\right] \wedge \left[(dx\wedge dy)\wedge (ds \wedge dt)
+ (dx\wedge dy)\wedge (du\wedge dv)
+ (ds \wedge dt)\wedge (du\wedge dv) \right] \\
&= 2 \left[(dx\wedge dy)\wedge (du \wedge dv) \wedge (ds\wedge dt) +
(ds\wedge dt)\wedge(dx\wedge dy)\wedge (du \wedge dv) +
(du \wedge dv) \wedge (dx\wedge dy)\wedge (ds\wedge dt) \right] \\
& = 6 dx\wedge dy\wedge du \wedge dv \wedge ds\wedge dt
\end{align*} $$

No comments yet.

Leave a Reply