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.

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.

After thinking about this a bit I realize the question was due to a rather silly misconception,the converse was not mentioned in the usual references because it's trivial from the proof of the usual direction. For future reference I will include a quick version here.

$$F = L^2/2$$ $$EL[F] \equiv \frac{d}{ds}(\frac{\partial}{\partial \dot{x^\mu}}F)- \frac{\partial}{\partial x^\mu}F \\ = \frac{d}{ds}(\frac{\partial L}{\partial \dot{x^\mu}} L)- \frac{\partial L}{\partial x^\mu}L\\ =\left(\frac{d}{ds}\left(\frac{\partial L}{\partial \dot{x^\mu}}\right) -\frac{\partial L}{\partial x^\mu}\right)L+ \frac{\partial L}{\partial \dot{x^\mu}} \frac{dL}{ds}\\$$

For the geodesic equation (of interest here) it is always possible to choose parametrization where $\frac{dL}{ds}=0$, in such parametrizations the last term vanishes and we have:

$$EL[F] = EL [L]$$ Hence the equations of motion yielded are equivalent.

1. The question formulation seems to talk about affine parametrizations before applying the principle of stationary action.

2. In the context of Riemannian$^1$ geometry, an affine parametrization of a (not necessarily geodesic) curve means by definition that the arc-length $s$ and the curve parameter $\lambda$ are affinely related $s=a\lambda +b$.

3. However, it is not always possible to maintain that all virtual paths [that satisfy pertinent boundary conditions (BCs)] are parametrized affinely with the same common parameter $\lambda$. Note in particular, that the BCs refer to the same initial and final values $\lambda_i$ and $\lambda_f$ for all paths.

4. Hence an a priori requirement of affine parametrization of virtual paths must be abandoned.

5. Also any partial a priori requirement of affine parametrization would be illogical, since we imagine that we do not know the geodesics on beforehand. In fact, that's what we intend to use the principle of stationary action to find.

6. Define the Lagrangian $$\tag{1}L_0(x,\dot{x})~:=~ g_{ij}(x) \dot{x}^i \dot{x}^j~\geq~0,$$ where dot means differentiation wrt. the parameter $\lambda$.

7. Now the square-root action $$\tag{2}S[x]~=~\int_{\lambda_i}^{\lambda_f}\! d\lambda \sqrt{L_0}$$ is invariant under reparametrizations of $\lambda$, so the stationary solutions for (2) with pertinent BCs will be all geodesics satisfying the BCs, independent of parametrization.

8. On the other hand, the stationary solutions for the non-square-root action $$\tag{3}S_0[x]~=~\int_{\lambda_i}^{\lambda_f}\! d\lambda ~L_0$$ with pertinent BCs will only be all affinely parametrized geodesics satisfying the BCs.

9. In more detail, because $L_0$ in the action (3) does not depend explicitly of $\lambda$, one may show (e.g. by use of Noether's theorem) that the energy function $$\tag{4}h~:=~p_i\dot{x}^i-L_0~=~L_0, \qquad p_i~:=~\frac{\partial L_0}{\partial \dot{x}^i}~=~2g_{ij}(x) \dot{x}^j,$$ is conserved on-shell, i.e. $$\tag{5}\dot{L}_0~\approx~ 0.$$ [This argument (5) does not work for the reparametrization-invariant action (2), where the energy function $h=0$ vanishes identically.]

10. (Let us assume for simplicity that $L_0$ is not zero on-shell, and leave the case where $L_0$ is zero on-shell as an exercise to the reader.) Because of eq. (5), a solution of the EL eqs. for $L_0$ is also a solution to the EL eqs. for $\sqrt{L_0}$: $$\tag{6} \frac{d}{d\lambda}\frac{\partial \sqrt{L_0}}{\partial \dot{x}^i}~\stackrel{(4)}{=}~\frac{d}{d\lambda}\frac{p_i}{2\sqrt{L_0}}~\stackrel{(5)}{\approx}~\frac{\dot{p}_i}{2\sqrt{L_0}}~\stackrel{\text{EL eq.}}{\approx}~\frac{1}{2\sqrt{L_0}}\frac{\partial L_0}{\partial x^i}~=~\frac{\partial \sqrt{L_0}}{\partial x^i}.$$

11. Conversely, a solution of the EL eqs. for $\sqrt{L_0}$ is not necessarily a solution to the EL eqs. for $L_0$. However, if we endow the solution of the EL eqs. for $\sqrt{L_0}$ with an affine parametrization, then we can obtain eq. (5), and reverse the argument (6), so that it is also a solution of the EL eqs. for $L_0$.

--

$^1$ In pseudo-Riemannian geometry, with mixed signature, the situation is more complicated, cf. e.g. this Phys.SE post. Moreover, there the argument of the square-root $\sqrt{L_0}$ may become negative.

$^2$ Terminology and Notation: Equations of motion (EOM) means Euler-Lagrange (EL) equations. The words on-shell and off-shell refer to whether EOM are satisfied or not. The $\approx$ symbol means equality modulo EOM.

I think there is some sort of bug near the end of the first answer (joshphysics), possibly in the variational step which involves integration by parts. Unless some further assumptions were used. Otherwise the implied conclusion seems to be that the EL eqns for any functional with integrand $F(x(s),\dot{x}(s))$ are generally the same as for the functional with integrand $F^2$.

[A simple counterexample is $F = x(s)$, though I realise this is getting away from geodesics. Then the EL eqn for $F^2$ would be $x=0$ whereas the EL eqn for $F$ itself would be $1=0$.]

Anyway, getting back to the original question, often the words "Cauchy-Schwarz" appear when relating $\int f^2$ to $\int f$.

Ignoring that, and looking just at EL equations, consider minimizing/maximizing the two functionals $\int F^2 ds$ and $\int F ds$ (with the same parameter and bcs), where $F = F(x(s),\dot{x}(s))$. When are these two problems equivalent (or at least, one implies the other)?

The EL eqns for the two cases are generally different (cf counterexample above), but one condition for equivalence is $dF/ds = 0$ (when evaluated on a solution to either of the two EL eqns).

We also have the Beltrami identities for each case, which are first integrals of the EL eqns (cf energy conservation): for $F$ we have $\dot{x}_i\frac{\partial F}{\partial \dot{x}_i} - F = c_1$ and for $F^2$ we have $\dot{x}_i\frac{\partial (F^2)}{\partial \dot{x}_i} - F^2 = c_2$ (*).

Restricting to the problem of geodesics on some surface, we have an integrand of the form $F = \sqrt{g_{ij}(x)\dot{x}^i\dot{x}^j}$ and this satisfies $\dot{x}_i \frac{\partial F}{\partial \dot{x}_i} = F$ (identically), and also satisfies $\dot{x}_i \frac{\partial (F^2)}{\partial \dot{x}_i} = 2F^2$ (+).

Let's suppose we have a solution to the EL eqn for $F^2$, and thus to the Beltrami identity $(*)$ for $F^2$. Then (*) and (+) together give $F^2 = c_2$ i.e. $F$ is constant when evaluated on our solution.
[this is essentially point 9 of Qmechanics's answer] Hence the equivalence condition mentioned above ($dF/ds = 0$) is satisfied, and so the EL eqn for $F$ is also satisfied.

The converse doesn't hold in general since the Beltrami identity for such an $F$ is satisfied identically (with $c_1 = 0$), so does not give anything "extra".

• You're right, I made an error. I'm glad you caught that. It turns out that I did, in fact, neglect a certain term containing $dL/ds$ in integrating by parts, and I find, exactly as you indicate, that the EL problems are equivalent provided $dL/ds = 0$. Thanks for the careful read -- I will edit soon. – joshphysics Jun 17 '15 at 18:16