Consider Lagrangian $$ L_1= q\dot{\alpha}+\alpha^2 $$
The Euler-Lagrange equations for $q$ and $\alpha$ read $$ \dot{q}=2\alpha $$ $$ \dot{\alpha}=0 $$ These two equations can be combined, leading to: $$ \ddot{q}=0 $$ $$ \alpha=\frac{\dot{q}}{2} $$ One can thus observe that $\alpha(t)$ is proportional to $\dot{q}(t)$ and, in this sense, I say that $\alpha$ plays an auxiliary role.

Consider also Lagrangian $$ L_2=\dot{q}^2 $$ The Euler-Lagrange equation for $q$ simply reads: $$ \ddot{q}=0 $$ So, Lagrangian $L_1$ and Lagrangian $L_2$ produce the same motion equation for $q(t)$.

If one is interested just in the motion equation for $q(t)$, is it possible to prove the equivalence of $L_1$ and $L_2$? In other words, is it possible to make the auxiliary role of $\alpha$ in Lagrangian $L_1$ explicit? Do $L_1$ and $L_2$ belong to some (suitably defined) "equivalence class"?


OP's 2 Lagrangians $L_1$ & $L_2$ are (up to normalization and total time derivative terms) just $$\overbrace{L_H~=~ p\dot{q}-\underbrace{\frac{p^2}{2m}}_{\text{Hamiltonian}}}^{\text{Hamiltonian Lagrangian}^1}\qquad\stackrel{p~\approx~ m\dot{q}}\longrightarrow\qquad \overbrace{L~=~\frac{m}{2}\dot{q}^2}^{\text{Lagrangian}}$$ for a free non-relativistic 1D particle, respectively, and therefore equivalent.


$^1$ In general, the Hamiltonian Lagrangian $L_H=p\dot{q}-H$ is the Lagrangian for Hamiltonian theories. Its Euler-Lagrange (EL) equations are Hamilton's equations, see e.g. H. Goldstein, Classical Mechanics, section 8.5. The Lagrangian and Hamiltonian formulations are related via a Legendre transformation, see e.g. this Phys.SE post.

  • $\begingroup$ I've deleted a number of obsolete comments and/or responses to them. $\endgroup$
    – David Z
    May 7 '20 at 21:20

Yes you can! Indeed, observe that the two Lagrangians basically give rise to the same action (modulo boundary terms and an overall normalization which do not impact the equations of motion). First of all, write

$$ S_{1}=\int_{t_{i}}^{t_{f}}dt\ \left\{q\frac{d\alpha}{dt}+\alpha^{2}\right\}=[q\alpha]^{t_{f}}_{t_{i}}-\int_{t_{i}}^{t_{f}}dt\ \alpha(\dot{q}-\alpha) $$

Here $\alpha$ appears explicitly as a Lagrange multiplier, and the fixed-extrema functional derivative of $S_{1}$ with respect to $\alpha$ reads

$$ \frac{\delta S_{1}}{\delta\alpha}=-\dot{q}+2\alpha $$

Solving the equation for $\alpha$ by replacing $\alpha=\dot{q}/2$ in $S_{1}$ you obtain

$$ S_{1}\to\frac{1}{2}\,[q\dot{q}]^{t_{f}}_{t_{i}}-\int_{t_{i}}^{t_{f}}dt\ \frac{\dot{q}^{2}}{4}=\frac{1}{2}\,[q\dot{q}]^{t_{f}}_{t_{i}}-\frac{S_{2}}{4} $$

Now, with respect to fixed-extrema variations (i.e. the ones you need in order to derive the equations of motion) the boundary term $[q\dot{q}]^{t_{f}}_{t_{i}}/2$ is a constant and can be neglected. The factor of $-1/4$ in front of $S_{2}$ also does not impact the equations of motion. Therefore, denoting with $\cong$ the congruence modulo additive constants and multiplicative normalizations,

$$ S_{1}\cong\frac{1}{2}\,[q\dot{q}]^{t_{f}}_{t_{i}}-\frac{S_{2}}{4}\cong-\frac{S_{2}}{4}\cong S_{2} $$

where the first $\cong$ comes from having integrated out the Lagrange multiplier $\alpha$.


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