differential geometry - Geodesic Equation from variation: Is the squared lagrangian equivalent?

It is well known that geodesics on some manifold $M$, covered by some coordinates ${x_\mu}$, say with a Riemannian metric can be obtained by an action principle . Let $C$ be curve $\mathbb{R} \to M$, $x^\mu(s)$ be an affine parametrization of $C$. (Using same symbol for coordinates and parametrization here, but it's standard). The action yielding geodesics is:

$$S (C) =\int_CL ds$$ where $$L\equiv \sqrt{g_{\mu\nu} \dot{x}^\mu \dot{x}^\nu}$$

Now the popular text by Nakahara makes the claim that variation of

$$F \equiv \frac{L^2}{2}$$

Will yield the exact same action principle solutions. However, the reference above ONLY shows that:

$C$ solves euler Lagrange equation for $L$ $\implies$ $C$ solves euler Lagrange equation for $F$

And this can be shown through a direct brute force computation.

My question: Is the converse of the above true, and how does one go about proving it. In particular, how do we show that $F$ yields no extraneous solutions to the Euler Lagrange equations.

Answer

The upshot of this answer will be as follows: if a path satisfies the Euler-Lagrange equations for $L^2/2$, then it will satisfy the Euler-Lagrange equations for $L$, but the converse does not hold unless the path has affine parameterization.

Let $L = L(x, \dot x)$ be a lagrangian that is local function of only position and velocity, then a parameterized path $x(s) = (x^i(s))$ on $M$ is said to satisfy the Euler-Lagrange equations for $L$ provided

$$\begin{align} \frac{\partial L}{\partial x^i}(x(s), \dot x(s)) - \frac{d}{ds}\frac{\partial L}{\partial \dot x^i}(x(s), \dot x(s)) = 0 \end{align}$$ for all $i$ and for all $s$ is the domain of $x$.

Lemma 1. If $x$ satisfies the Euler-Lagrange equations for $L$, then the Beltrami-Identity holds for $x$:

$$\frac{d}{ds}L(x(s), \dot x(s)) = \frac{d}{ds}\left(\frac{\partial L}{\partial \dot x^i}\big(x(s), \dot x(s)\big)\cdot \dot x^i(s)\right)$$

for all $s$ in the domain of $x$.

Proof. Just do it. Note that, the proof hinges on the fact that $L$ is a local function of only $x$ and $\dot x$.

Lemma 2. If $L(x,\dot x) = \sqrt{g_{ij}(x)\dot x^i\dot x^j}$, then $L$ satisfies the following identity:

$$\frac{\partial (L^2/2)}{\partial \dot x^i}(x, \dot x) \dot x^i = L(x,\dot x)^2$$

Proof. Try this yourself too!

Corollary. If $L(x,\dot x) = \sqrt{g_{ij}(x)\dot x^i\dot x^j}$ and $x$ satisfies the Euler-Lagrange equations for $L^2/2$, then $x$ satisfies the Euler-Lagrange equations for $L$.

Proof. If $x$ satisfies the Euler-Lagrange equations for $L^2$, then Lemma 1 gives the following Beltrami identity (we use notational shorthand here -- all expressions should be evaluated on $x(s)$)

$$\frac{d(L^2/2)}{ds} = \frac{d}{ds} \frac{\partial (L^2/2)}{\partial \dot x^i}\cdot \dot x^i$$

On the other hand, evaluating both sides of Lemma 2 on $x(s)$, and taking the derivative of both sides with respect to $s$ gives

$$\frac{d}{ds} \frac{\partial (L^2/2)}{\partial \dot x^i}\cdot \dot x^i = \frac{d(L^2)}{ds}$$

Combining these facts shows that $d(L^2)/ds = 0$ which implies that $L^2$ is constant along $x(s)$ and therefore that $L$ is also constant along $x(s)$:

$$\frac{dL}{ds} = 0$$

Now, we separately notice that since $x$ satisfies the Euler-Lagrange equations for $L^2/2$, we have

$$\begin{align} 0 &= \frac{\partial(L^2/2)}{\partial x^i} - \frac{d}{ds} \frac{\partial (L^2/2)}{\partial \dot x^i} \\ &= L\left(\frac{\partial L}{\partial x^i} - \frac{d}{ds}\frac{\partial L}{\partial \dot x^i}\right) - \frac{dL}{ds}\frac{\partial L}{\partial \dot x^i} \tag{$\star$}\\ &= L\left(\frac{\partial L}{\partial x^i} - \frac{d}{ds}\frac{\partial L}{\partial \dot x^i}\right) \end{align}$$

and therefore as long as $L\neq 0$, we see that $x$ satisfies the Euler-Lagrange equations for $L$ as was desired.

The crucial point here is that because of the specific form of $L$, any path satisfying the Euler-Lagrange equation for $L^2/2$ has the nice property that $dL/ds = 0$ along the path. This allows one to kill the term in $(\star)$ which is the term that is the essential difference between the Euler-Lagrange equations for $L^2/2$ and the Euler-Lagrange equations for $L$.

However, if $x$ satisfies the Euler-Lagrange equations for $L$, then it is not necessarily the case that $dL/ds = 0$ along $x$, so in this case, one can't kill that term in $(\star)$, so it need not be a solution to the Euler-Lagrange equation for $L^2/2$.

Nonetheless, if $x$ is affinely parameterized, then it will automatically have the property that $L$ is constant along it, so it will automatically satisfy both Euler-Lagrange equations.

In fact, using parts of the computations above, it is not hard to show that

Proposition. Let $L(x, \dot x) = \sqrt{g_{ij}(x)\dot x^i\dot x^j}$. A path $x$ is an affinely parameterized geodesic if and only if is solves the Euler-Lagrange equations of both $L$ and $L^2/2$.

So the Euler-Lagrange equations of $L^2/2$ yield all affiniely parameterized geodesics, while the Euler-Lagrange equations of $L$ yield all geodesics, regardless of parameterization.