Difference between revisions of "Aufgaben:Problem 9"
(Created page with "==Solution== '''a)''' $$\nabla_k g_{ij} = \partial_k g_{ij} - g_{nj} \Gamma^n_{kj} - g_{in}\Gamma^n_{kj} =$$ $$ = \partial_k g_{ij} - g_{nj}(\frac{1}{2} g^{np}(\partial_k g_...") |
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again we used the symmetry of \( g_{ij} \) in the first bracket. | again we used the symmetry of \( g_{ij} \) in the first bracket. | ||
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+ | '''b)''' My idea is to transform \( g^{ij} \nabla_i(\partial_j f) \) into \(\Delta f \). From there we have to show \( Lg(f) = \Delta(f) \) as in notes_new at pg 82. | ||
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+ | The first part should be: | ||
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+ | $$ g^{ij} \nabla_i \partial_j + \underbrace{ \nabla_i g_{ij}}_\text{= 0, from a)} = g^{ij} \nabla_i \partial_j + \underbrace{( \nabla_i g_{ij}) \partial_j}_\text{= 0} = g^{ij} \nabla_i \partial_j + \underbrace{( \nabla_i g^{ij})\partial_j}_\text{= 0} = \nabla_i (g^{ij} \partial_j) = \Delta$$ |
Revision as of 14:32, 14 June 2015
Solution
a)
$$\nabla_k g_{ij} = \partial_k g_{ij} - g_{nj} \Gamma^n_{kj} - g_{in}\Gamma^n_{kj} =$$ $$ = \partial_k g_{ij} - g_{nj}(\frac{1}{2} g^{np}(\partial_k g_{ip} + \partial_i g_{kp} - \partial_p g_{ki})) - g_{in}(\frac{1}{2} g^{nq}(\partial_k g_{jq} + \partial_j g_{kq} - \partial_q g_{kj})) =$$
from the symmetry of \( g_{ij} \) and from: \( g_{ki}g^{ij} = g^{ji}g_{ik} = \delta^i_k \) we obtain:
$$ = \partial_k g_{ij} - \frac{1}{2} \delta^p_j(\partial_k g_{ip} + \partial_i g_{kp} - \partial_p g_{ki}) - \frac{1}{2} \delta^q_i(\partial_k g_{jq} + \partial_j g_{kq} - \partial_q g_{kj}) = $$
$$ = \partial_k g_{ij} - \frac{1}{2} (\partial_k g_{ij} + \partial_i g_{kj} - \partial_j g_{ki}) - \frac{1}{2} (\partial_k g_{ji} + \partial_j g_{ki} - \partial_i g_{kj}) = $$
$$ = \partial_k g_{ij} - \frac{1}{2} (\partial_k g_{ij} + \partial_k g_{ji}) - \frac{1}{2} (\partial_j g_{ki} - \partial_i g_{kj} + \partial_i g_{kj} - \partial_j g_{ki} ) = 0 $$
again we used the symmetry of \( g_{ij} \) in the first bracket.
b) My idea is to transform \( g^{ij} \nabla_i(\partial_j f) \) into \(\Delta f \). From there we have to show \( Lg(f) = \Delta(f) \) as in notes_new at pg 82.
The first part should be:
$$ g^{ij} \nabla_i \partial_j + \underbrace{ \nabla_i g_{ij}}_\text{= 0, from a)} = g^{ij} \nabla_i \partial_j + \underbrace{( \nabla_i g_{ij}) \partial_j}_\text{= 0} = g^{ij} \nabla_i \partial_j + \underbrace{( \nabla_i g^{ij})\partial_j}_\text{= 0} = \nabla_i (g^{ij} \partial_j) = \Delta$$