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 geodesic equation can be derived from the Euler-lagrange equations only. I tried to pluh in the Lagrangian $$ L= \frac{1}{2} g_{\mu\nu}\frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda}$$

in $$\frac{d}{d\lambda}\frac{\partial L}{\partial(dx^\mu/d\lambda)} = \frac{\partial L}{\partial x^\mu} $$

but I am running into derivation problems and the corresponding chain rule. May you please help me out here, I have to understand how does the geodesic equation get derived from Euler-Lagrange equations. Thank you very much in advance!!

  • $\begingroup$ I am still not very good at deriving these stuff with the metric. I searched in a lot of books for a detailed procedure but couldn't find it using euler-lagrange equation. I only found using variational principle. I would really appreciate it if you could supply me with a name of a book that does it. $\endgroup$ – PhilosophicalPhysics Sep 26 '14 at 20:18
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    $\begingroup$ There's a derivation of the geodesic equation starting from an Euler-Lagrange equation in the Geodesics as curves of stationary interval Wikipedia article section. Click the "Show" link there for details. $\endgroup$ – Red Act Sep 26 '14 at 20:46
  • $\begingroup$ Thank you very much, I went through the steps but I have a simple question as the author went from step 6 to step 7. So we had on the very left side of step 6 $$g_{\mu\nu }\frac{\partial x^\mu}{\partial \tau}\frac{\partial x^\nu}{\partial \tau}$$, what he did to reach step 7 is to divide by the latter term. As far as I can tell, he changed $$x^\mu , x^ \nu to x^\alpha , x^\beta$$ and thus it canceled with the term on the right side. My question is do we have the right to change indices to indices already there during division? As far as I know, during multiplication we have to chose diff ones $\endgroup$ – PhilosophicalPhysics Sep 27 '14 at 9:11
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    $\begingroup$ Related question with square root Lagrangian: physics.stackexchange.com/q/94348/2451 and links therein. $\endgroup$ – Qmechanic Sep 27 '14 at 14:18

Let us do the RHS first. This just gives us a derivative on the metric: $$\frac{\partial L}{\partial x^\lambda}=\frac{1}{2}\partial_\lambda g_{\mu\nu}\dot x^\mu\dot x^\nu$$ The first derivative on the LHS is essentially a derivative of a square, thus $$\frac{\partial L}{\partial \dot x^\lambda}=g_{\mu\lambda}(x(\lambda))\dot x^\mu$$ where we have made the dependence of $g$ on $\lambda$ clear for the next step. Now we differentiate with respect to the curve parameter: $$\frac{\mathrm{d}}{\mathrm{d}\lambda}[g_{\mu\lambda}(x(\lambda))\dot x^\mu]=\partial_\nu g_{\mu\lambda}\dot x^\mu\dot x^\nu+g_{\mu\lambda}\ddot x^\mu=\frac{1}{2}\partial_\nu g_{\mu\lambda}\dot x^\mu\dot x^\nu+\frac{1}{2}\partial_\mu g_{\nu\lambda}\dot x^\mu\dot x^\nu+g_{\mu\lambda}\ddot x^\mu$$ where in the last step we split the first term apart and rearranged indices. Putting it all together, we obtain $$g_{\mu\lambda}\ddot x^\mu=-\frac{1}{2}\left(\partial_\nu g_{\mu\lambda}+\partial_\mu g_{\nu\lambda}-\partial_\lambda g_{\mu\nu}\right)\dot x^\mu\dot x^\nu=-\Gamma_{\lambda\mu\nu}\dot x^\mu\dot x^\nu$$ where in the last step we used the definition of the Christoffel symbols with three lower indices. Now contract with the inverse metric to raise the first index and cancel the metric on the LHS. We obtain $$\ddot x^\lambda=-\Gamma^\lambda{}_{\mu\nu}\dot x^\mu\dot x^\nu$$ as was to be shown.


Here is one way to derive the geodesic equations from the Euler-Lagrange equations. First consider a natural Lagrangian system $(M,L)$, where $L\in C^\infty(TM)$. Let $g$ be a Riemannian metric. Suppose in our mechanical system the net force is zero. That is, the Lagrangian is just equal to the kinetic energy, $$L(p,V_p)=\frac{1}{2}mg_p(V_p,V_p)$$In particular, if $\gamma:[a,b]\to M$ is representing a particle of mass $m$ then its kinetic energy is $\frac{1}{2}mg_{\gamma(t)}(\gamma^\prime(t),\gamma^\prime(t))$. The Euler-Lagrange equations are derived by finding the critical points of the action $$\mathcal A(\gamma)=\int_{\gamma(t)}g_{\gamma(t)}(\gamma^\prime(t),\gamma^\prime(t))dt.$$ A standard fact from Riemannian geometry is that the critical points of this functional (the length functional) are geodesics.

Now back to your question of deriving the geodesic equation from the Euler-Lagrange equations. I believe this works, although it might not be the idea you had in mind.

Consider the setup above, a Lagrangian system $(M,L)$ where $(M,g)$ is a Riemannian manifold. Let $(x^1,\dots,x^n)$ denote a coordinate chart on $M$. Let $(x^1,\dots,x^n,v^1,\dots,v^n)$ and $(x^1,\dots,x^n,\xi_1,\dots,\xi_n)$ denote the induced charts on $TM$ and $T^\ast M$ respectively. In these coordinates, $L=K=\frac{1}{2}mg_{ij}v^iv^j\in C^\infty(TM)$. Using the Legendre transform we can consider the Hamiltonian system $(T^\ast M,H)$ where $H=\frac{\partial L}{\partial v^i}$ is the induced Hamiltonian. In coordinates, this means $H=\frac{1}{2}g^{ij}\xi_i\xi_j$ where $g^{ij}$ is the inverse matrix of $g_{ij}$. Let $\gamma(t)=(x(t),\xi(t))$ be a curve in $T^\ast M$. In order to satisfy Hamilton's equations, you need that $$\dot x^k=\frac{\partial H}{\partial \xi_k} \ \ \ \text{ and } \ \ \ \dot\xi_k=-\frac{\partial H}{\partial x^k}.$$ You can check that $\frac{\partial H}{\partial\xi_k}=g^{kj}\xi_j$ and $-\frac{\partial H}{\partial x^k}=-\frac{1}{2}\frac{\partial g^{ij}}{\partial x^k}\xi_i\xi_j$. Using this and the fact that $\xi_k=g_{ak}\dot x^a$, we can plug this into the second of Hamiltons equations to get that $$\dot\xi_k=\frac{\partial g_{ak}}{\partial x^q}\dot x^a\dot x^q+g_{ak}\ddot x^a=-\frac{1}{2}\frac{\partial g^{ij}}{\partial x^k}g_{ia}g_{jp}\dot x^a\dot x^p$$Since $-\frac{\partial g^{ij}}{\partial x^k}g_{ia}g_{jp}=\frac{\partial g_{ap}}{\partial x^k}$, the above equation becomes $$\frac{\partial g_{ak}}{\partial x^q}\dot x^a\dot x^q+g_{ak}\ddot x^a=\frac{1}{2}\frac{\partial g_{ap}}{\partial x^k}\dot x^a\dot x^p.$$ Re-arranging and simplifying, we see that $\gamma(t)$ satisfies the Hamilton equations if and only if $$\ddot x^b=-\frac{1}{2}g^{kb}\left(\frac{\partial g_{ak}}{\partial x^p}\dot x^a\dot x^p+\frac{\partial g_{pk}}{\partial x^a}\dot x^a\dot x^p-\frac{\partial g_{ap}}{\partial x^k}\dot x^a\dot x^p\right).$$This is precisely the geodesic equation. Your claim now follows from the fact that motions in the Lagrangian system $(M,L)$ (i.e. curves satisfying Euler-Lagrange equations) are in correspondence to motions in the Hamiltonian system $(T^\ast M,H)$ (i.e. curves satisfying the Hamilton equations).

I should also point out that this last equation is the final line of the proof from the link in the comments, once you recall the definition of Christoffel symbols.

  • $\begingroup$ Also, with your above notation, $\frac{ d x^\mu}{d\lambda}$ is equal to $v^\mu$ in what I wrote above. $\endgroup$ – JonHerman Oct 10 '14 at 15:37

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