math121b:sample-m1
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math121b:sample-m1 [2020/02/22 18:05] pzhou [Partial Solution] |
math121b:sample-m1 [2020/02/23 16:56] (current) pzhou |
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| The divergence of $V$ is | 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. | 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. | ||
math121b/sample-m1.1582423538.txt.gz · Last modified: 2020/02/22 18:05 by pzhou
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