Homework 2 Problem 7

Solution by: Andy Bosiljevac
Problem:
Given:

(1)
\begin{align} \mu = 13d\theta \wedge d\phi \end{align}

Compute Div(X)

Solution:

(2)
\begin{align} \mathcal{L}_X \mu = (\text{div}_\mu X) \mu = d(X\overset \lrcorner \, \, \mu) + X \overset \lrcorner \, \, d\mu \end{align}

Starting in parts,

(3)
\begin{align} d\mu = d(13d\theta)\wedge d\phi + (-1)^k 13d\theta \wedge d(d\phi) = 0 \end{align}

, so

(4)
\begin{align} X \overset \lrcorner \, \, d\mu = 0. \end{align}

Next,

(5)
\begin{align} X \overset \lrcorner \, \, \mu = X \overset \lrcorner \, \, (13 d\theta \wedge d\phi) \end{align}
(6)
\begin{align} = 13(X \overset \lrcorner \, \, d\theta) \wedge d\phi - 13d\theta \wedge (X \overset \lrcorner \, \, d\phi) \end{align}
(7)
\begin{align} =13 \langle X_\theta \frac{\partial}{{\partial \theta}} + X_\phi \frac{\partial}{{\partial \phi}}, d\theta \rangle \wedge d\phi - 13d\theta \wedge \langle X_\theta \frac{\partial}{{\partial \theta}} + X_\phi \frac{\partial}{{\partial \phi}}, d\phi \rangle \end{align}
(8)
\begin{align} =13 X_\theta \wedge d\phi - 13d\theta \wedge X_\phi \end{align}

Now, taking the differential of $X \overset \lrcorner \, \, \mu$

(9)
\begin{align} d(X \overset \lrcorner \, \, \mu) = d(13 X_\theta \wedge d\phi - 13d\theta \wedge X_\phi) \end{align}
(10)
\begin{align} = 13d(X_\theta \wedge d\phi) - 13d(d\theta \wedge X_\phi) \end{align}
(11)
\begin{align} = 13\frac{\partial X_\theta}{{\partial \theta}} d\theta \wedge d\phi + 13d\theta \wedge \frac{\partial X_\phi}{{\partial \phi}} d\phi \end{align}
(12)
\begin{align} = (\frac{\partial X_\theta}{{\partial \theta}} + \frac{\partial X_\phi}{{\partial \phi}}) 13d\theta \wedge d\phi = Div_\mu (X) \mu \end{align}

Therefore,

(13)
\begin{align} Div_\mu (X) = \frac{\partial X_\theta}{{\partial \theta}} + \frac{\partial X_\phi}{{\partial \phi}}. \end{align}

Scott agrees! Note that for the particular $X$ defined in Problem 4, the divergence therefore equals zero.

page revision: 29, last edited: 10 Apr 2007 18:33