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Suppose we have a ball of mass $m$ in the Earth's gravitational field ($g=const.$). Equation of motion reads as: $$ ma = -mg $$ From here we can conclude that we have translational symmetry of the form $x(t) \to x(t) + const$ (we are working in only 1D). However, we cannot see this symmetry from the Lagrangian: $$ L = \frac{mv^2}{2} - mgx $$ because the linear term "breaks" this symmetry. Moreover, we also do not have the corresponding conserved quantity (as far as I can see).

Does this mean that we can have symmetries in the Newtonian sense (transformations that map solutions to other solutions) that are not present in the Lagrangian?

RedGiant
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3 Answers3

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Well, it seems to me that under a translation $x(t) \to x(t) + c$, the Lagrangian goes to $$\mathcal{L} \to \mathcal{L}' = \frac{1}{2}m v^2 - mg(x+c) = \mathcal{L} -mgc.$$

So yes, the Lagrangian may appear to be different, however since it only shifts by a constant, these two Lagrangians ($\mathcal{L}$ and $\mathcal{L}'$) are equivalent and produce the same Euler-Lagrange Equations. Indeed, more generally, two Lagrangians are equivalent if their difference is a total time derivative. i.e. $\mathcal{L}$ and $$\mathcal{L}' = \mathcal{L} + \frac{\text{d}f}{\text{d}t}$$ are equivalent for any $f(t)$.

Philip
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You can do an integration by parts on the last term (and discard the resulting boundary term) to yield an action with equivalent EOMs:$$ S'= \int \left( \frac{1}{2} m \dot{x}^2 + m g t \dot{x} \right) \, dt $$ In this context, the symmetry $x \to x + C$ is obvious at the level of the Lagrangian. Moreover, the Euler-Lagrange equations become $$ \frac{d}{dt} \left( m \dot{x} + m g t \right) = \frac{\partial \mathcal{L}'}{\partial \dot{x}} = 0 $$ and thus the quantity $m \dot{x} + m g t$ is a constant of the motion. Specifically, it's the initial momentum of the particle.

(This seems "cheap", somehow, and I'm not 100% sure whether it's a legitimate move. Comments are welcome.)

Michael Seifert
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  1. The (infinitesimal) translation $$\delta x~=~\epsilon$$ changes OP's Lagrangian with a total time-derivative $$\delta L~=~mg \epsilon~=~ \frac{d}{dt}(mg \epsilon t).$$ This is known as a quasi-symmetry. Noether's theorem does also hold for quasi-symmetries.

  2. Concerning symmetries of action vs. EOM, see also e.g. this related Phys.SE post.

Qmechanic
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