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In problem 21 of Chapter 9 of Goldstein's "Classical Mechanics," 3rd edition, it is given that if the Hamiltonian $H(q, p, t) $ satisfies the scaling condition

$$H (q\lambda, p/\lambda, t\lambda^n) = \lambda^{-n} H(q, p, t)$$

then $$D = pq/n - tH$$ is a constant of motion. Does the quantity $D$ have any physical significance as a constant of motion?

Qmechanic
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1224physics
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2 Answers2

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Your transformation of coordinates is slightly incorrect, it should be $Q=\lambda q$, $p=\lambda P$, $t'=\lambda^2 t$ and only applies to hamiltonians of the form $H = p^n - aq^{-n}$.

By dimensional analysis D has units of angular momentum since $D \sim pq =mxv$. But I don't belive it has any immediate physical meaning. It is just the conserved charge of some sort of conformal symmetry.

Thomas Bastos
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  1. The quantity $$D~\equiv~ qp-ntH\tag{1}$$ is a constant of motion iff $$\begin{align} 0~\approx~&\frac{dD}{dt}~\approx~\{D,H\}+\frac{\partial D}{\partial t}\cr ~=~&p\frac{\partial H}{\partial p} - q\frac{\partial H}{\partial q}-nH - nt\frac{\partial H}{\partial t}. \end{align}\tag{2}$$ We see that $D$ is essentially the generator of a Hamiltonian vector field (HVF) in the form of an Euler vector field.

    For this reason $D$ must have dimension of angular momentum/action/$\hbar$.

  2. Eq. (2) is clearly equivalent to OP's homogeneity property $$ H(\lambda q,p/\lambda,\lambda^nt)~=~\lambda^{-n} H(q,p,t)\tag{3}$$ of the Hamiltonian function, i.e. $H$ is homogeneous function of weight $-n$ of the arguments $(q,p^{-1},t^{\frac{1}{n}})$.

    Eq. (3) can be view as a part of a dimensional analysis.

Qmechanic
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