- #1
lync4495
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Hello,
I'm beginning to learn GR and ran across the following in the beginning of Stephani's book.
In deriving the equations of motion for a force-free particle, we have the Lagrangian:
[tex]L = \frac{m}{2} g_{\alpha \beta} \dot{x}^\alpha \dot{x}^\beta[/tex]
To construct the equations of motion we need
[tex]\frac{\partial L}{\partial \dot{x}^\nu} = m g_{\alpha \nu} \dot{x}^\alpha[/tex]
and
[tex]\frac{\partial L}{\partial x^\nu} = L_{,\nu} = \frac{m}{2} g_{\alpha \beta, \nu} \dot{x}^\alpha \dot{x}^\beta[/tex]
Then from the Lagrange equations we get
[tex]g_{\alpha \nu }\ddot{x}^\alpha + g_{\alpha \nu, \beta} \dot{x}^\alpha \dot{x}^\beta - \frac{1}{2}g_{\alpha \beta, \nu}\dot{x}^\alpha \dot{x}^\beta =0[/tex]
I don't understand where the second term in the last equation comes from.
I'm beginning to learn GR and ran across the following in the beginning of Stephani's book.
In deriving the equations of motion for a force-free particle, we have the Lagrangian:
[tex]L = \frac{m}{2} g_{\alpha \beta} \dot{x}^\alpha \dot{x}^\beta[/tex]
To construct the equations of motion we need
[tex]\frac{\partial L}{\partial \dot{x}^\nu} = m g_{\alpha \nu} \dot{x}^\alpha[/tex]
and
[tex]\frac{\partial L}{\partial x^\nu} = L_{,\nu} = \frac{m}{2} g_{\alpha \beta, \nu} \dot{x}^\alpha \dot{x}^\beta[/tex]
Then from the Lagrange equations we get
[tex]g_{\alpha \nu }\ddot{x}^\alpha + g_{\alpha \nu, \beta} \dot{x}^\alpha \dot{x}^\beta - \frac{1}{2}g_{\alpha \beta, \nu}\dot{x}^\alpha \dot{x}^\beta =0[/tex]
I don't understand where the second term in the last equation comes from.