1

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?

knzhou
  • 101,976
SRS
  • 26,333

1 Answers1

4

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
knzhou
  • 101,976
  • My question is whether the non-invariance of the EOM under $t\to -t$ implies non-invariance of the Lagrangian? @knzhou – SRS Mar 22 '18 at 15:11
  • @SRS I am using the contrapositive: "not A implies not B" is logically equivalent to "B implies A". – knzhou Mar 22 '18 at 15:12
  • 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 – SRS Mar 22 '18 at 15:15
  • @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 $\dot{q} = 0$. – knzhou Mar 22 '18 at 15:17
  • So your point is that only under time-reversal both the EOM and the action have the same behaviour. Am I getting it correct? @knzhou – SRS Mar 22 '18 at 15:22
  • @SRS The point is that symmetries of the action give symmetries of the EOM, which is logically equivalent to saying that asymmetries of the EOM imply asymmetries of the action. However, the converse/inverse of these statements is not true. – knzhou Mar 22 '18 at 15:30
  • Regarding your example of adding $t$ to the Lagrangian, what if we disregard total derivatives in the Lagrangian? – innisfree Mar 22 '18 at 22:08
  • 1
    @innisfree How about $L = t \dot{q}^2$? The equation of motion is $\ddot{q} + \dot{q} = 0$. – knzhou Mar 22 '18 at 22:19