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--- math121b:sample-m1 [2020/02/21 23:23]
pzhou
+++ math121b:sample-m1 [2026/02/21 14:41] (current)
@@ Line 82: / Line 82: @@
 The divergence of $V$ is
-$$ div(V) = H^{-2} \d_u (H^2) = 2 \cosh(u)/\sinh(u) $$
+$$ div(V) = H^{-2} \d_u (H^2) = \frac{2 \cosh(u)\sinh(u)}{ \sinh^2(u) + \sin^2(v) }$$
+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.

Lecture Notes