- #1
Pentaquark5
- 17
- 2
I've recently read in a textbook that a geodesic can be defined as the stationary point of the action
\begin{align}
I(\gamma)=\frac{1}{2}\int_a^b \underbrace{g(\dot{\gamma},\dot{\gamma})(s)}_{=:\mathcal{L}(\gamma,\dot{\gamma})} \mathrm{d}s \text{,}
\end{align}
where ##\gamma:[a,b]\rightarrow M## is a differentiable curve. Thus,
\begin{align}
\mathcal{L}(x^\mu ,\dot{x}^\nu)=\frac{1}{2}g_{\alpha \beta}(x^\mu)\dot{x}^\alpha \dot{x}^\beta\text{.}
\end{align}
How exactly does the Lagrangian in ##(2)## follow from ##(1)##?
\begin{align}
I(\gamma)=\frac{1}{2}\int_a^b \underbrace{g(\dot{\gamma},\dot{\gamma})(s)}_{=:\mathcal{L}(\gamma,\dot{\gamma})} \mathrm{d}s \text{,}
\end{align}
where ##\gamma:[a,b]\rightarrow M## is a differentiable curve. Thus,
\begin{align}
\mathcal{L}(x^\mu ,\dot{x}^\nu)=\frac{1}{2}g_{\alpha \beta}(x^\mu)\dot{x}^\alpha \dot{x}^\beta\text{.}
\end{align}
How exactly does the Lagrangian in ##(2)## follow from ##(1)##?