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### Lagrangian, metric and geodesics: obtention of equation of motion in GR [duplicate]

I would like to check my global understanding on G.R: how we determine the trajectories of particles. Everything starts from the equations of Einstein. We have to find a metric that satisfies ...
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### Geodesic equation from Euler - Lagrange

There are several ways to derive the geodesic equation. One of which is the variational method which I seemed to understand it because it was written in great details. Then it was mentioned that the ...
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### Geodesics equations via variational principle

I would like to recover the (timelike) geodesics equations via the variational principle of the following action: $$\mathcal{S}[x] = -m \int d\tau = -m \int \sqrt{-g_{\mu\nu}\,dx^{\mu}\,dx^{\nu}}$$ ...
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I'm trying (very early stages) to understand the derivation of the geodesic equation $$\frac{d^{2}x^{\alpha}}{d\lambda^{2}}+\Gamma_{\gamma\beta}^{\alpha}\frac{dx^{\beta}}{d\lambda}\frac{dx^{\gamma}}{... 2answers 640 views ### From Euler-Lagrange equation to non-affine geodesic equation I have some problems to demonstrate the non-affine geodesic equation from Euler-Lagrange's equations. I start defining the square root Lagrangian$$L=\sqrt{ g_{ij}(x) \dot{x}^i \dot{x}^j},$$but then ... 1answer 747 views ### Geodesic equation from the proper time integral This is something that has been bothering me for a little while. The usual procedure that I've seen is to write the proper time as the line integral$$\tau=\int_\gamma d\tau$$along some curve \gamma... 2answers 492 views ### Do the same equations of motion imply the same Lagrangians? [duplicate] If two Lagrangian (densities) \mathcal{L} give the same equations of motion, are they equivalent? 2answers 318 views ### Why can we choose affine parameterization? In general relativity when deriving the geodesic equation$$\ddot{x}^\mu + \Gamma^\mu_{\alpha\beta}\dot{x}^\alpha\dot{x}^\beta = 0\tag{1}$$from the action$$S = \int d\tau \sqrt{|g_{\mu\nu} \dot{x}...
In Relativity the Lagrangian of a free particle is \begin{align} \mathcal L=\sqrt{g_{ab}\frac{dx^a}{d\tau}\frac{dx^b}{d\tau}}\end{align} Since $\mathcal L$ is parameterization invariant we can always ...