math121b:02-10

$$\gdef\div{\text{div}} \gdef\vol{\text{Vol}} \gdef\b{\mathbf} \gdef\d{\partial}$$

We first finish up the derivation of Laplacian last time. See the lecture note 2020-02-07, Friday. Then, we review three concepts $df, \nabla f, \nabla \cdot \b V$ ($\nabla \times \b V$ is a bit special for $\R^3$).

Then, we will follow Boas 10.8 and 10.9, to reconcilliate the math notation and physics notations.

In Cartesian coordinate, the differential of a function $f$ is $$ df = \sum_i \frac{\d f }{\d x_i} dx_i. $$

In general coordinate $(u_1, \cdots, u_n)$, the differential of a function $f$ is $$ df = \sum_i \frac{\d f }{\d u_i} d u_i. $$

You can specify the differential of a function directly: $df$ at a point $p \in \R^n$ is a linear function on $T_p \R^n$, $df(p) \in (T_p \R^n)^*$. It does the following $$ df(p) : \b v_p \mapsto \b v_p(f), \quad \b v_p \in T_p \R^n $$ where $\b v_p(f)$ is the directional derivative of $f$ along $\b v_p$.

In Cartesian coordinate, the gradient of a function is $$ \gdef\grad{\text{ grad } } \grad f = \sum_i \frac{\d f}{\d x_i} \frac{\d }{\d x_i}. $$

In general coordinate, the gradient of a function is more complicated $$ \grad f = \sum_{i,j} g^{ij} \frac{\d f}{\d u_i} \frac{\d }{\d u_j}, $$ where $g^{ij}$ is the entry of the inverse matrix of the matrix $[g_{kl}]$. And it just happens that, for Cartesian coordinate, $g^{ij} = \delta_{ij}$.

Note that $g_{ij}$ and $g^{ij}$ depends on the coordinate system. $$ g_{ij} = g(\frac{\d}{\d u_i}, \frac{\d}{\d u_j}), \quad g^{ij} = g^*(d u_i, d u_j). $$ Beware that $\nabla u_i \neq \frac{\d}[\d u_i}$.

** Notation ** $$ \nabla f = \grad f.$$

Let $\R^n$ be the flat space, with standard coordinates $(x_1, \cdots, x_n)$.

Let $\b V$ be a vector field on $\R^n$, that is, for each point $p \in \R^n$, we specify a tangent vector $$ \b V(p) = \sum_{i=1}^n V^i(p) \d_i \in T_p \R^n. $$ We require that $V^i(p)$ varies smoothly with respect to $p$.

The divergence of $\b V$ is a function on $\R^n$, $$ \div(\b V) = \sum_{i=1}^n \d_i( V^i ) $$ recall that $V^i$ is a function on $\R^n$, and $\d_i$ is taking the partial derivative with respect to $x_i$.

** Notation ** $$ \nabla \cdot \b V = \div (\b V).$$

**What does divergence mean?** Geometrically, it measure the relative change-rate of the volume of an infinitesimal cube situated at point $p$. Suppose $\Phi^t: \R^n \to \R^n$ is the flow generated by $\b V$ (every point moves as dictated by $\b V$). And let $C= C(p, \epsilon)$ be a cube of side-length $\epsilon$, center at $p$.
Then, we have the geometrical interpretation as
$$ \div(\b V) = \lim_{\epsilon \to 0} \frac{1}{\vol(C)} \frac{d \vol(\Phi^t(C))}{dt} \vert_{t=0}. $$
That is why, if $S \subset \R^n$ is an open domain, we can compute the change-rate of the volume of $S$ by
$$ \frac{d \vol(\Phi^t(S)) } {dt}\vert_{t=0} = \int_{S} \div(\b V)(\b x) \, d \vol(\b x). $$

** In curvilinear coordinate. **
The formula for computing the divergence is the following, suppose $\b V = \sum_i V^i \frac{\d }{\d u_i}$, then
$$ \div (\b V) = \sum_{i=1}^n \frac{1}{\sqrt{|g|}} \frac{\d (\sqrt{|g|} V^i)}{\d u_i} $$

The reason we have the above formula is that , for any compactly supported function $\varphi$ ^{1)}, we have
$$ \int_{\R^n} (\nabla \cdot \b V)\, \varphi\, \sqrt{|g|} du_1\cdots d u_n = \int_{\R^n} \b V \cdot (\nabla \varphi)\, \sqrt{|g|} du_1\cdots d u_n $$

For Cartesian coordinate, we have basis vectors $\b i, \b j, \b k$.

For spherical coordinate, we have **unit** basis vectors $\b e_r, \b e_\theta, \b e_\phi$, and corresponding coordinate basis vectors $\b a_r, \b a_\theta, \b a_\phi$ (not unit length). These $\b a_n$ corresponds to our coordinate basis tangent vectors:
$$ \b a_r = \frac{\d }{\d r}, \quad \b a_\theta = \frac{\d }{\d \theta}, \cdots $$

See Example 2 on page 523, for how to consider a general curvilinear coordinate $(x_1, x_2, x_3)$ on $\R^3$.

The notation $d \b s$ corresponds to
$$ \sum_{i=1}^n \frac{\d }{\d x_i} \otimes d x_i \in (T_p \R^n) \otimes (T_p \R^n)^*. $$
An element $T$ in $V \otimes V^*$ can be viewed as a linear operator $V \to V$, by inserting $v \in V$ to the second slot of $T$. In this sense $d \b s$ is the identity operator on $T_p \R^n$. You might have seen in Quantum mechanics the bra-ket notation $1 = \sum_n | n \rangle \otimes \langle n |$ ^{2)} It is the same thing, where $| n \rangle \in V$ forms a basis and $ \langle n | \in V^*$ are the dual basis.

** Orthogonal coordinate system (ortho-curvilinear coordinate) **, the matrix $g_{ij}$ is diagonal, with entries $h_i^2$ (not to be confused with our notation for dual basis). This is
the case we will be considering mainly.

Suppose we have orthogonal coordinate system $(x_1, x_2, x_3)$, and **unit** basis vectors $\b e_i$, we have
$$ \b a_i = \frac{\d }{\d x_i} = h_i \b e_i. $$

Given a vector field $V$, we write its component in the basis of $\b e_i$ (warning! this is not our usual notation, we usual write with basis $\frac{\d }{\d x_i}$) $$ \b V = \sum_i V^i \b e_i $$.

Try to do problem 1.

An important property is the “Leibniz rule” $$ \b \nabla \cdot (f \b V) = \b \nabla f \cdot \b V + f \b \nabla \cdot \b V.$$

To compute the curl, we note the following rule $$ \b \nabla \times (f \b V) = \b \nabla(f) \times \b V + f \b \nabla \times \b V $$ and $$ \b \nabla \times \nabla f = 0 $$.

In the ortho-curvilinear coordinate, we can use the above rule to get a formula for the curl. I will not test on the curl operator in the orthocurvilinear case.

a compactly supported function on $\R^n$ is a function that vanishes outside a sufficently large ball.

$\otimes$ sometimes omitted as usual in physics.

math121b/02-10.txt · Last modified: 2020/02/22 18:03 by pzhou