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Derivatives of the Lagrangian in curved space

  1. Mar 16, 2015 #1
    Follow along at http://star-www.st-and.ac.uk/~hz4/gr/GRlec4+5+6.pdf and go to PDF page 9 or page 44 of the "slides." I'm trying to see how to go from the first to the third line. If we write the free particle Lagrangian and use q^i-dot and q^j-dot as the velocities and metric g_ij, how is it we cancel out the factor of 1/2 when when take the q^l-dot derivative? It looks like more than just a simple change of variables from j to l. If instead we took dL/dq^j-dot I'd expect to get 0.5m x g_ij x q^i-dot . I don't think I can assume I'm only taking derivatives when i=j, so how does this work?
     
  2. jcsd
  3. Mar 16, 2015 #2

    ShayanJ

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    ## \frac{\partial}{\partial \dot{q}^l}(\frac 1 2 g_{ij} \dot{q}^i \dot{q}^j)=[\frac 1 2 g_{ij}(\frac{\partial \dot{q}^i}{\partial \dot{q}^l} \dot{q}^j+\dot{q}^i \frac{\partial \dot{q}^j}{\partial \dot{q}^l})]=[\frac 1 2 g_{ij}(\delta^i_l \dot{q}^j+\dot{q}^i \delta^j_l)] =[\frac 1 2(g_{lj} \dot{q}^j+g_{il}\dot{q}^i)]=[\frac 1 2(g_{il} \dot{q}^i+g_{il}\dot{q}^i)]=g_{il}\dot{q}^i##
     
    Last edited: Mar 16, 2015
  4. Mar 17, 2015 #3
    Thank you a ton! That was clear and helpful.
     
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