Does the Lagrangian $L(x,y,\dot{x},\dot{y})=\frac{1}{2}m(\dot{x}^2-\dot{y}^2)$ have any physical significance?

Question

Let us consider the example of a free particle moving in 2D plane. Certainly, if we take the Lagrangian to be $$L_1(x,y,\dot{x},\dot{y})=\frac{1}{2}m(\dot{x}^2+\dot{y}^2),\tag{1}$$ it produces the correct EoM i.e. $$\ddot{x}=0,\qquad \ddot{y}=0.$$ On the other hand, if we take an inequivalent Lagrangian of the form $$L_2(x,y,\dot{x},\dot{y})=\frac{1}{2}m(\dot{x}^2-\dot{y}^2),\tag{2}$$ then too, the same (correct) EoM are reproduced. Here by inequivalent I mean that $L_2-L_1$ is not equal to $dF/dt$ where $F(x,y,t)$ is a function of coordinates and time.

Does it mean that both the Lagrangians are valid descriptions of the system? Note that, the Hamiltonian obtained from $L_1$ and $L_2$ are $$H_1=\frac{1}{2m}(p_x^2+p_y^2), \quad H_2=\frac{1}{2m}(p_x^2-p_y^2)$$ respectively. Clearly, $H_1$ represents the total energy of the system while $H_2$ does not. So I want to know whether $L_2$ is a valid description of a free particle moving in 2D plane. If not, does $L_2$ have any physical significance at all?

Answer 1

Here’s one interpretation (not motion in a 2D Euclidean plane, but rather a Lorentzian one): instead of calling the coordinates $x,y$, call them $x,t$ respectively, and if you call the parameter $\lambda$ instead, then we have $L_2(t,x,\dot{t},\dot{x})=\frac{m}{2}(-\dot{t}^2+\dot{x}^2)$, which can be viewed as the Lagrangian associated to the Lorentzian metric $\eta=m(-dt^2+dx^2)$ on $\Bbb{R}^{1+1}$. So, the solutions Euler-Lagrange equations associated to $L_2$ are the (affinely parametrized) geodesics in the Minkowski spacetime $(\Bbb{R}^{1+1},\eta)$, which are simply straight lines in the usual affine sense.

The associated Hamiltonian is then $H_2=\frac{1}{2}\eta^{ab}p_ap_b=\frac{1}{2m}(-p_t^2+p_x^2)$.

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Edit: Conserved Quantities

In general, suppose $Q$ is a configuration space and that for $i\in\{1,2\}$ we are given:

• a Lagrangian $L_i$
• $\mathcal{S}_i$ the set of solutions of the ELE associated to $L_i$
• $\mathcal{C}_i$ the set of conserved quantities associated to $L_i$ (i.e the set of smooth functions $f:TQ\to \Bbb{R}$ such that for all $\gamma\in \mathcal{S}_i$, we have that $f\circ\gamma’$ is constant).

Then, it is trivial to check that $\mathcal{S}_2\subset\mathcal{S}_1\implies \mathcal{C}_1\subset \mathcal{C}_2$. In particular, if $\mathcal{S}_1=\mathcal{S}_2$ then $\mathcal{C}_1=\mathcal{C}_2$.

So, in the present case, we have that $M:= x\dot{y}-y\dot{x}$ is a conserved quantity for $L_1$ and $L_2$ as we can see in a variety of ways.

• a direct calculation.
• the usual rotational symmetry argument applied to $L_1$ tells us the angular momentum $xp_y-yp_x$ is conserved, hence by the “general” theorem from above, it follows that once we express this as functions of positions and velocities, $x\dot{y}-y\dot{x}$ is conserved for both.
• apply the Lorentz invariance of $L_2$ tells us that $xp_y+yp_x$ is conserved for $L_2$, but really back in terms of positions and velocities this means $-x\dot{y}+y\dot{x}$ is the conserved quantity for $L_2$, and hence by the general theorem above, it is also conserved for $L_1$.

Part of OP’s confusion in the comments might have been due to $p_y$ in both cases actually being negatives of each other, so although $M:=x p_y- yp_x$ and $\tilde{M}:=xp_y+yp_x$ are different functions (on the cotangent bundle $T^*(\Bbb{R}^2)$), they’re the same function of positions and velocities (super precisely, if $\Bbb{F}L_i:T(\Bbb{R}^2)\to T^*(\Bbb{R}^2)$ are the fiber-derivatives/Legendre transforms, then $(\Bbb{F}L_1)^*(M):= M\circ \Bbb{F}L_1$ and $(\Bbb{F}L_2)^*(\tilde{M}):= \tilde{M}\circ \Bbb{F}L_2$ are, up to an overall minus sign the same function on $T(\Bbb{R}^2)$, namely $\pm (x\dot{y}-y\dot{x})$).

All of this is to say that conserved quantities cannot distinguish $L_1,L_2$; neither mathematically nor physically (because both have the same set of ELE solutions hence the same conserved quantities). Although if one naively expresses the conserved quantities in terms of momenta, then it may look like the system has different conserved quantities, but that’s due purely to the fact that the fiber-derivatives/Legendre transforms of $L_1,L_2$ are different.

Answer 2

1. OP's Lagrangian $L_2$ has the form of the Lagrangian $$L(x,t,\dot{x},\dot{t},e)~=~\frac{\dot{x}^2 - \dot{t}^2}{2e}\tag{A}$$ for a massless relativistic point-particle in 1+1D Minkowski spacetime in the gauge $e=1$, cf. e.g. this Phys.SE post.

2. We note that OP's Lagrangian $L_2$ is not of the form kinetic energy minus potential energy. In fact the $y$-variable (if it is not a gauge DOF) is a (bad) ghost. At the classical level, this is not a problem per se.

   However, (bad) ghosts may make it difficult to build a unitary quantum theory. (In OP's case this is still not a problem since OP's theory is free: The $x$ and $y$ sectors don't talk to each other, so an opposite sign convention for the $y$-sector is in principle fine.)

3. In complex coordinates $z=x+iy$, OP's Lagrangians (1) & (2) take the form $$\begin{align} L_1(z,\dot{z},t)~=~&\frac{m}{2}|\dot{z}|^2 ,\cr L_2(z,\dot{z},t)~=~&\frac{m}{2}{\rm Re}(\dot{z}^2), \end{align}\tag{B}$$ with the same equations of motion (EOM) $$ \ddot{z}~\approx~0 .\tag{C}$$

4. The fact that $L_1$ and $L_2$ are not proportional (modulo total time derivative terms) is related to the fact that rotations $z\to e^{i\theta}z$ are a symmetry of EOM (C) but not a quasisymmetry of $L_2$, cf. e.g. this Phys.SE post.

5. If we rotate the coordinate system $z\to e^{-\frac{i\pi}{4}}z$, then OP's pair (B) becomes exactly example 3 in my Phys.SE answer here.