Suppose the Euler-Lagrange equation of a system $$\frac{\partial L}{\partial q}=\frac{d}{dt}\Bigg(\frac{\partial L}{\partial \dot{q}}\Bigg)$$ is known to be not invariant under the discrete transformation $t\to -t$. It's given. Also assume that nothing is known about the functional form of the Lagrangian $L(q,\dot{q},t)$.

Is it possible to infer from that information whether the Lagrangian (or action) will be invariant under $t\to -t$ or not?


Your question asks if T-asymmetry of the equations of motion implies T-asymmetry of the Lagrangian. That's equivalent to asking if T-symmetry of the Lagrangian implies T-symmetry of the equations of motion.

The answer is yes, by the principle of least action. Specifically, T-symmetry of the Lagrangian implies the paths $q(t)$ and $q(-t)$ have the same action. Then $q(t)$ is a stationary point of the action if and only if $q(-t)$ is, so if $q(t)$ solves the equation of motion, so does $q(-t)$.

To clarify questions in the comments, very explicitly, the facts are

  • symmetries of the action imply symmetries of the EOM
  • asymmetries of the EOM imply asymmetries of the action
  • symmetries of the EOM don't imply symmetries of the action
  • asymmetries of the action don't imply asymmetries of the EOM
  • $\begingroup$ My question is whether the non-invariance of the EOM under $t\to -t$ implies non-invariance of the Lagrangian? @knzhou $\endgroup$ – SRS Mar 22 '18 at 15:11
  • $\begingroup$ @SRS I am using the contrapositive: "not A implies not B" is logically equivalent to "B implies A". $\endgroup$ – knzhou Mar 22 '18 at 15:12
  • $\begingroup$ To understand your point let me ask you the following. Would also say that if an EOM is invariant under $t\to t+t_0$ (time-translation), the Lagrangian will also be invariant under $t\to t+t_0$? @knzhou $\endgroup$ – SRS Mar 22 '18 at 15:15
  • $\begingroup$ @SRS No, not at all. For a very simple example, let $L = \dot{q}^2 + t$. The equation of motion is time translation invariant, as it is simply $\dot{q} = 0$. $\endgroup$ – knzhou Mar 22 '18 at 15:17
  • 1
    $\begingroup$ @innisfree How about $L = t \dot{q}^2$? The equation of motion is $\ddot{q} + \dot{q} = 0$. $\endgroup$ – knzhou Mar 22 '18 at 22:19

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