- #1
Isaek_
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- TL;DR
- Struggling to match the E-L equations from an assumed Lagrangian
I decided recently that I would work through Einstein's original paper on the foundation of GR and I've recently hit a roadblock. It starts with Einstein's Lagrangian for an empty space-time,
$$ \begin{align} \mathcal{L}= g^{\mu \nu} \Gamma^{\alpha}_{\nu \beta} \Gamma^{\beta}_{\mu \alpha} \tag{47a} \end{align}, $$
which, Einstein shows, gives the two quantities:
$$ \begin{align*} \frac{\partial \mathcal{L}}{\partial g^{\mu\nu}} = - \Gamma^{\alpha}_{\mu \beta} \Gamma^{\beta}_{\nu \alpha}; \\
\frac{\partial \mathcal{L}}{\partial (\partial_\sigma g^{\mu\nu})} = \Gamma^{\sigma}_{\mu \nu}. \tag{48} \end{align*} $$
I can follow the maths as given in the paper to arrive at this result, but I am confused by the first step. Namely, ##\mathcal{L}## is a function of ##(g^{\mu\nu}, \partial_\sigma g^{\mu\nu})##, yet it is differentiated wrt ## (g^{\mu\nu}, \Gamma^{\beta}_{\nu\alpha}).## But isn't ##\Gamma## a function of both ##(g^{\mu\nu}, \partial_\sigma g^{\mu\nu})##? When I differentiate wrt to these, I get the following quantities:
$$ \begin{align*}
\mathcal{L}(g^{\mu \nu}, \partial_\rho g^{\mu \nu}) &= g^{\sigma \tau} \Gamma^{\alpha}_{\sigma \beta} \Gamma^{\beta}_{\tau \alpha}; \\
\delta \mathcal{L} &= \left[ \Gamma^{\alpha}_{\sigma \beta} \Gamma^{\beta}_{\tau \alpha} + 2g^{\sigma\tau}\frac{\partial \Gamma^\alpha_{\sigma\beta}}{\partial g^{\mu\nu}} \right]\delta(g^{\mu\nu}) + 2\left[g^{\sigma\tau}\Gamma^{\alpha}_{\sigma\beta}\frac{\partial \Gamma^{\beta}_{\tau\alpha}}{\partial(\partial_\rho g^{\mu\nu})} \right]\delta(\partial_\rho g^{\mu\nu});\\
\frac{\partial \mathcal{L}}{\partial g^{\mu\nu}} &= 3\Gamma^{\alpha}_{\mu \beta} \Gamma^{\beta}_{\nu \alpha}; \\
\frac{\partial \mathcal{L}}{\partial (\partial_\sigma g^{\mu\nu})} &= \Gamma^{\sigma}_{\mu \nu}.
\end{align*}
$$
I guess my main question is, why are we disregarding the ## g^{\mu\nu}## dependency in ##\Gamma##? I trace the difference to the second term in the first square bracket: if its sign were different, I would obtain the same answer! Thanks.
(Here's a link to the paper for ease of access ^^: https://en.wikisource.org/wiki/The_Foundation_of_the_Generalised_Theory_of_Relativity#§_15._Hamiltonian_Function_for_the_Gravitation-field._Laws_of_Impulse_and_Energy.)
$$ \begin{align} \mathcal{L}= g^{\mu \nu} \Gamma^{\alpha}_{\nu \beta} \Gamma^{\beta}_{\mu \alpha} \tag{47a} \end{align}, $$
which, Einstein shows, gives the two quantities:
$$ \begin{align*} \frac{\partial \mathcal{L}}{\partial g^{\mu\nu}} = - \Gamma^{\alpha}_{\mu \beta} \Gamma^{\beta}_{\nu \alpha}; \\
\frac{\partial \mathcal{L}}{\partial (\partial_\sigma g^{\mu\nu})} = \Gamma^{\sigma}_{\mu \nu}. \tag{48} \end{align*} $$
I can follow the maths as given in the paper to arrive at this result, but I am confused by the first step. Namely, ##\mathcal{L}## is a function of ##(g^{\mu\nu}, \partial_\sigma g^{\mu\nu})##, yet it is differentiated wrt ## (g^{\mu\nu}, \Gamma^{\beta}_{\nu\alpha}).## But isn't ##\Gamma## a function of both ##(g^{\mu\nu}, \partial_\sigma g^{\mu\nu})##? When I differentiate wrt to these, I get the following quantities:
$$ \begin{align*}
\mathcal{L}(g^{\mu \nu}, \partial_\rho g^{\mu \nu}) &= g^{\sigma \tau} \Gamma^{\alpha}_{\sigma \beta} \Gamma^{\beta}_{\tau \alpha}; \\
\delta \mathcal{L} &= \left[ \Gamma^{\alpha}_{\sigma \beta} \Gamma^{\beta}_{\tau \alpha} + 2g^{\sigma\tau}\frac{\partial \Gamma^\alpha_{\sigma\beta}}{\partial g^{\mu\nu}} \right]\delta(g^{\mu\nu}) + 2\left[g^{\sigma\tau}\Gamma^{\alpha}_{\sigma\beta}\frac{\partial \Gamma^{\beta}_{\tau\alpha}}{\partial(\partial_\rho g^{\mu\nu})} \right]\delta(\partial_\rho g^{\mu\nu});\\
\frac{\partial \mathcal{L}}{\partial g^{\mu\nu}} &= 3\Gamma^{\alpha}_{\mu \beta} \Gamma^{\beta}_{\nu \alpha}; \\
\frac{\partial \mathcal{L}}{\partial (\partial_\sigma g^{\mu\nu})} &= \Gamma^{\sigma}_{\mu \nu}.
\end{align*}
$$
I guess my main question is, why are we disregarding the ## g^{\mu\nu}## dependency in ##\Gamma##? I trace the difference to the second term in the first square bracket: if its sign were different, I would obtain the same answer! Thanks.
(Here's a link to the paper for ease of access ^^: https://en.wikisource.org/wiki/The_Foundation_of_the_Generalised_Theory_of_Relativity#§_15._Hamiltonian_Function_for_the_Gravitation-field._Laws_of_Impulse_and_Energy.)
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