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In a paper I'm reading (https://arxiv.org/abs/1312.6120, equations 8-9) I've seen the following statement - given that $s$ is constant and $a,b$ are some dynamical variables obeying the following equations of motion:
$$\frac{d}{dt}a=b(s-ab), \qquad \frac{d}{dt}b=a(s-ab)\tag{1}$$
then, as this is going down the gradient of the function $$E(a,b)=\frac{1}{2}(s-ab)^2\tag{2}$$ and there's the scaling symmetry $$a\to\lambda a, \qquad b\to\lambda^{-1} b,\tag{3}$$ as per Noether's theorem the quantity $a^2-b^2$ is conserved. While it is easy to manually check that this is true, how is Noether's theorem used here? I haven't managed to formulate a Lagrangian or derive a useful first-order version of the theorem. Is there some generic Noether-analogous result for cases like these?

Qmechanic
  • 201,751
Yonatan Gideoni
  • 21

1 Answers1

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  1. Yes, OP is right: One can strictly speaking not apply Noether's theorem without an action formulation. (It is of course easy to verify that $a^2-b^2$ indeed is an integral of motion (IOM), as OP already mentioned.)

  2. Let us for fun try to cook up an action formulation.

    • First note that we can recast OP's EOMs (1) as Hamilton's equations $$ \dot{a}~=~\{a,H\}, \qquad \dot{b}~=~\{b,H\} \tag{A}$$ with Hamiltonian $$H~=~\frac{1}{2}(b^2-a^2)\tag{B}$$ and fundamental Poisson bracket $$ \{a,b\}~=~s-ab,\tag{C}$$ cf. this Phys.SE post.

    • It is no coincidence that the Hamiltonian $H$ is (a function of) OP's IOM, since for a 2D phase space, this is essentially unique.

    • The corresponding symplectic 2-form is $$ \omega~=~\frac{1}{s-ab}\mathrm{d}b\wedge \mathrm{d}a~=~\mathrm{d}\theta,\qquad \theta~=~-\ln|s-ab|\mathrm{d}\ln |a|, \tag{D}$$ cf. e.g this Phys.SE post.

    • The corresponding Hamiltonian Lagrangian becomes $$ L_H~=~ -\frac{1}{a}\ln|s-ab|\dot{a} - \frac{1}{2}(b^2-a^2),\tag{E} $$ cf. e.g this Phys.SE post.

    • It is straightforward to check that the corresponding Euler-Lagrange (EL) equations are OP's EOMs (1).

    • However, the OP's transformation (3) is not a symmetry of the action (E). This can essentially be traced back to the fact that the Hamiltonian $H$ is not invariant under OP's transformation (3).

    • Since $L_H$ has no explicit time dependence, Noether's theorem implies that $H$ is conserved.

Qmechanic
  • 201,751