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--- math121b:sample-m1 [2020/02/19 09:04]
pzhou
+++ math121b:sample-m1 [2020/02/23 16:56] (current)
pzhou
@@ Line 15: / Line 15: @@
   * Find the metric tensor $g$ in terms of $u,v$ (or equivalently, the line element $ds^2$),
   * The volume element.
-  * The expansion of $\d_u$ and $\d_v$ using $\d_x$ and $\d_y$ (or equivalently, the line element $ds^2$).
+  * The expansion of $\d_u$ and $\d_v$ using $\d_x$ and $\d_y$.
   * (bonus) If a function on $\R^2$ is given by $f(u,v) = 2u + 3v$, find its gradient expressed using basis vectors $\d_u, \d_v$.
   * (bonus) Find the divergence of the vector field $V = \d_u$.
@@ Line 23: / Line 23: @@
 $$ div(V^i \d_i) = \frac{1}{\sqrt{g}} \d_i(\sqrt{g} V^i) $$
 where we used Einstein summation convention, and $\d_i = \frac{\d}{\d u_i}$, $u_1 = u, u_2 = v$.
+====== Partial Solution ======
+. Draw the picture to see. You should draw a parallogram with $v$ along the diagonal, and two sides parallel to $e_1,e_2$.
+. Since $V$ is 3 dim, suffice to show that $f_i$ are linearly independent. That is, if we have
+$$ 0 = c_1 f_1(t) + c_2 f_2(t) + c_3 f_3(t) $$
+for all $t$, then we need to show $c_i=0$. Plug in $t=2,3,5$ to conclude this is indeed so.
+Same thing for the next problem, suppose we try to find the expansion
+$$ f(t) = c_1 f_1(t) + c_2 f_2(t) + c_3 f_3(t) $$
+then we can get $c_i$ by plug-in $t=2,3,5$.
+You may worry about: what about $t$ equal to other values? Well, since we have an equation about degree two polynomials, if it holds at 3 points, then it holds for all $t$.
+. This is good place to introduce another trick, the relation about 2-dim orthogonal coordinate and holomorphic functions.
+Suppose $z = f(w)$ is a holomorphic function, if we write $z = x+ i y$, $w = u + iv$, we get Cauchy-Riemmann equation
+$$ \begin{cases}
+\frac{\d x}{\d u} = \frac{\d y}{\d v} \cr
+\frac{\d x}{\d v} = - \frac{\d y}{\d u}
+\end{cases}
+$$
+Hence, we have
+$$ g(\frac{\d }{\d u} ,\frac{\d }{\d u}) =  \left( \frac{\d x}{\d u} \right)^2 + \left( \frac{\d y}{\d u} \right)^2 = \left( \frac{\d y}{\d v} \right)^2 + \left( - \frac{\d x}{\d v} \right)^2= g(\frac{\d }{\d v} ,\frac{\d }{\d v})$$
+and
+$$ g(\frac{\d }{\d u} ,\frac{\d }{\d v}) = \frac{\d x}{\d u} \frac{\d x}{\d v} + \frac{\d y}{\d u} \frac{\d y}{\d v} = - \frac{\d x}{\d u} \frac{\d y}{\d u} + \frac{\d y}{\d u} \frac{\d x}{\d u} = 0 $$
+Hence, we have orthogonal coordinate, and furthermore, the length of the two basis vectors are the same
+$g_{uu} = g_{vv}$.
+In this problem, we have
+$$ x + i y = \cosh(u + iv) $$
+so the above method applies. Another examples is
+$$ y = uv , x = (u^2 - v^2)/2 $$
+this is from
+$$ x + i y = (u + iv)^2/2.$$
+Of course, one can do the problem without using the above trick. One then do
+$$dx = ... du + ... dv, \quad dy = ... du + ... dv$$
+then plug into
+$$ ds^2 = dx^2 + dy^2 $$
+to express $ds^2$ using $du$ and $dv$. You should get in the end
+$$ ds^2 = (\cosh^2(u)- \cos^2(v)) (du^2 + dv^2) $$
+Or, the factor can be written in different ways, since
+$$ \cosh^2(u)- \cos^2(v) = \sinh^2(u) + \sin^2(v) .$$
+Let $H^2 = \sinh^2(u) + \sin^2(v)$, then we have
+$$ g = [g_{ij}] = \begin{pmatrix} H^2 & 0 \cr 0 & H^2 \end{pmatrix} $$
+So we have $g_{ij} = \delta_{ij} H^2, \quad g^{ij} = H^{-2} \delta_{ij}$.
+The volume element is
+$$ \sqrt{g} du dv =  (\sinh^2(u) + \sin^2(v)) du dv$
+The expansion is
+$$ \d_u = \d_u(x) \d_x + \d_u(y) \d_y = ... $$
+The gradient is
+$$ grad(f) = H^{-2} \d_u(f) \d_u +  H^{-2} \d_v(f) \d_v =  H^{-2} (2  \d_u +  3 \d_v) $$
+The divergence of $V$ is
+$$ div(V) = H^{-2} \d_u (H^2) = \frac{2 \cosh(u)\sinh(u)}{ \sinh^2(u) + \sin^2(v) }$$
+More details: $V = 1 \d_u + 0  \d_v$ so $V^1 = 1$, $V^2=0$, and $\sqrt{g} = \sqrt{H^4} = H^2$. Plug in the divergence formula will yield the answer.
+In Boas' notation, we have $e_1 = \d_u / \| \d_u \| = \d_u / H$, so
+$$V = \d_u = H e_1$$
+and $V^1_{Boas} = H$, also $h_1 = h_2 = H$. Using that Boas formula, we have
+$$div V = \frac{1}{h_1h_2} \d_u( h_2 V^1_{Boas}) = H^{-2} \d_u (H^2) $$
+yield the same answer.

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