# Why two different Lagrangians to derive geodesic equations?

I'm trying (very early stages) to understand the derivation of the geodesic equation

$$\frac{d^{2}x^{\alpha}}{d\lambda^{2}}+\Gamma_{\gamma\beta}^{\alpha}\frac{dx^{\beta}}{d\lambda}\frac{dx^{\gamma}}{d\lambda}=0$$ 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 A Short Course in General Relativity, 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 A General Relativity Workbook, 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.

• It is not unusual for different Lagrangian's to yield same equation of motion. Possible duplicate of physics.stackexchange.com/q/149082 – zzz May 12 '15 at 18:58
• Do you really want to derive geodesic equation with Lagrangians? I think it's simpler if you derive and practice differential geometric concepts in simple sphere. You should first understand the covariant derivative; when it's zero you defined parallel transport; when you parallel transport a vector towards the direction in points to, you'll follow a geodesic and that yields the geodesic equation, quite intuitively. Basically the equation says how do you need to change the coordinates of your velocity vector to remain on a straight path. – Calmarius May 12 '15 at 22:34
• Technically speaking, the square-root Lagrangian and the non-square-root Lagrangian lead to slightly different EL eqs. As is well-known, the solutions are for both eqs. (parametrized) geodesics, but where the former eq. produces all parametrized geodesics, the latter eq. only produces affinely parametrized geodesics, cf. e.g. my Phys.SE answer here. – Qmechanic May 13 '15 at 5:10
• @Qmechanic - thanks. How come Valter Moretti's derivation here (which is over my head) starts with the square root Lagrangian and ends with (what I assume is) the affinely parametrized geodesic equations, ie my first equation? – Peter4075 May 13 '15 at 13:09
• His derivation (v6) assumes this at some point. – Qmechanic May 13 '15 at 13:38

Actually the equations of motion one ends up with are not manifestly the same:

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

$$\frac{d^2 x^{\mu}}{dt^2} + \Gamma^{\mu}_{\nu\sigma} \frac{dx^{\nu}}{dt} \frac{dx^{\sigma}}{dt} = f(t) \frac{dx^{\mu}}{dt},$$ 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.

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.

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.

• Thanks. I will try to derive the non-affine geodesic equations you give. In the meantime do you have a link to this derivation? I'm still puzzled as to how Moore starts with the square root Lagrangian and ends up with the affine geodesic equations. – Peter4075 May 13 '15 at 13:17
• I'm not familiar with the source you're using, but the clearest (IMO) way to get the affine geodesic equations from the square root Lagrangian, is to first get the non-affine equations, and then switch your arbitrary parameter to an affine one. As far as having a source for the derivation, I don't know of any which derives the non-affine equation directly (although it's not difficult). Carroll in his notes starts with the square-root Lagrangian and then changes to an affine parameter in the middle of his derivation. Look at page 15 in: preposterousuniverse.com/grnotes/grnotes-three.pdf – childofsaturn May 13 '15 at 14:17
• Thanks, but I just cannot see how he gets $d\tau$ as the denominator after making the affine parameter substitution (Equation 3.52). – Peter4075 May 13 '15 at 15:48
• One simply applies the chain rule for that, if I understand your confusion correctly. – childofsaturn May 13 '15 at 17:41
• Any chance of a hint as to what that chain rule looks like? I cannot see how he substitutes (3.52) into (3.51) to get the equation at the bottom of page 69. Thank you for your patience. – Peter4075 May 13 '15 at 18:33