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When we write the Lagrangian $\mathcal{L}=\frac{1}{2}m\dot{x}^2-U(x)$, where $U$ is the potential energy, we are assuming that the mass $m$ is constant, the only variables being the velocity $\dot{x}$ and position $x$. What can be done to determine the equation of motion of the particle in case the mass is changing?

I know that we cannot simply use the formula $$\dot{p}=m\ddot{x}+\dot{m}\dot{x},$$ with $p=\frac{\partial \mathcal{L}}{\partial \dot{x}}$, because it isn't Galilean invariant and the system is not closed, so some other procedure must be used.

Perhaps the method of Lagrange multipliers may be used? Or via a non-standard Lagrangian that somehow reproduces the equation of motion given here?

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If you just give the mass an explicit time-dependence, $$L = \frac12 m(t) \dot{x}^2 - U(x)$$ then the Euler-Lagrange equation is $$\frac{d}{dt} (m \dot{x}) = \dot{m} \dot{x} + \dot{m} \ddot{x} = - \frac{dU}{dx}.$$ It's unclear to me why you think "we simply cannot use" this result. It isn't Galilean invariant, but once you let $m(t)$ have arbitrary time-dependence, the action isn't Galilean invariant either.

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  • $\begingroup$ It is because of this answer that I saw physics.stackexchange.com/q/53980 $\endgroup$ – Don Al Feb 16 at 1:51
  • $\begingroup$ @DonAl It depends on what one means by a "variable mass" system. If you mean a system that can exchange mass with an environment, then the caveats there apply. But if you just mean some isolated system with a mass term in the Lagrangian which can change explicitly with time (as in your question), it's much simpler and what I said applies. $\endgroup$ – knzhou Feb 16 at 1:54
  • $\begingroup$ Note, however, that this technique does not reproduce the rocket equation that was linked to by the OP. $\endgroup$ – Michael Seifert Feb 16 at 12:34

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