Why two different Lagrangians to derive geodesic equations? - Physics Stack Exchange most recent 30 from physics.stackexchange.com 2019-08-19T12:26:27Z https://physics.stackexchange.com/feeds/question/183621 http://www.creativecommons.org/licenses/by-sa/3.0/rdf https://physics.stackexchange.com/q/183621 5 Why two different Lagrangians to derive geodesic equations? Peter4075 https://physics.stackexchange.com/users/4075 2015-05-12T18:52:44Z 2015-05-12T20:33:53Z <p>I'm trying (very early stages) to understand the derivation of the geodesic equation</p> <p>$$\frac{d^{2}x^{\alpha}}{d\lambda^{2}}+\Gamma_{\gamma\beta}^{\alpha}\frac{dx^{\beta}}{d\lambda}\frac{dx^{\gamma}}{d\lambda}=0$$ via Lagrangians and the Euler-Lagrange equations. I don't understand why some authors use the Lagrangian $$L\left(\dot{x}^{c},x^{c}\right)\equiv\frac{1}{2}g_{ab}\left(x^{c}\right)\dot{x}^{a}\dot{x}^{b}$$ (as in Foster and Nightingale's <em>A Short Course in General Relativity</em>, p60) and others use the Lagrangian $$L\left(\dot{x}^{\alpha},x^{\alpha}\right)\equiv\sqrt{-g_{\mu\nu}\left(x^{\alpha}\right)\dot{x}^{\mu}\dot{x}^{\nu}}$$ (as in Moore's <em>A General Relativity Workbook</em>, p90). I realise I must be confused here, but I can't why as they both seem to end up with the same geodesic equation.</p> https://physics.stackexchange.com/questions/183621/-/183640#183640 3 Answer by childofsaturn for Why two different Lagrangians to derive geodesic equations? childofsaturn https://physics.stackexchange.com/users/10762 2015-05-12T20:33:53Z 2015-05-12T20:33:53Z <p>Actually the equations of motion one ends up with are not manifestly the same:</p> <p>If I let $$L_1 = \sqrt{g_{\mu \nu} \frac{dx^{\mu}}{dt} \frac{dx^{\nu}}{dt}}$$ one finds that the Euler-Lagrange equation is</p> <p>$$\frac{d^2 x^{\mu}}{dt^2} + \Gamma^{\mu}_{\nu\sigma} \frac{dx^{\nu}}{dt} \frac{dx^{\sigma}}{dt} = f(t) \frac{dx^{\mu}}{dt},$$ for a suitable function $f(t)$ (you should try to derive this), whereas for $L_2 = (L_1)^2$, the equation is what you wrote above. So how are these two related, and why do they lead to the same solution? The answer has to do with the parameter used to describe the curve.</p> <p>In the case of $L_1$, the parameter is arbitrary. One can see that the action $S_1 = \int dt L_1$ is reparametrization invariant, namely if we take $t \rightarrow \lambda(t)$, $S_1$ doesn't change. This implies that the we are free to choose any parameter we like without affecting the physics. </p> <p>One can show that there is a particular $\lambda(t)$ for which the EOM of $L_1$ reduce to the EOM of $L_2$ (such a parameter is called an affine parameter). It's a very interesting exercise, and I leave you to work on the details.</p>