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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 $$ L\equiv \sqrt{g_{\mu\nu} \dot{x}^\mu \dot{y}^\nu} $$

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.

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.

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 $$ L\equiv \sqrt{g_{\mu\nu} \dot{x}^\mu \dot{y}^\nu} $$

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 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.

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 $$ L\equiv \sqrt{g_{\mu\nu} \dot{x}^\mu \dot{y}^\nu} $$

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.