Conserved quantity

1. Jul 17, 2009

zetafunction

1. The problem statement, all variables and given/known data

what condition must satisfy the potential so a Lagrangian $$m \dot q \dot q - V(q)$$

has as a conserved quantity $$A(q,p)=qp$$

2. Relevant equations

$$A(q,p)=qp$$ $$m \dot q \dot q - V(q)$$

3. The attempt at a solution

since we have the conserved quantity A(q,p)=qp [/tex] i believe that a condition for the potential is to be scale-invariant $$v(cq)=V(q)c$$ for any constant 'c'

the other attempt to solution is this, since 'A' is a conserved quantity then the Poisson brackets should vanish so $${A,H}=0$$ using the definition of Poisson bracket i should get an ODe for the potential V(q).

2. Jul 18, 2009

guguma

Interesting problem, I tried the poisson brackets got a solution check it out if it makes sense to you.

$$\frac{\partial}{\partial q}A \frac{\partial}{\partial p}H - \frac{\partial}{\partial p}A \frac{\partial}{\partial q}V = 0$$

$$p\frac{\partial}{\partial p}H - q \frac{\partial}{\partial q}V = 0$$

$$m\dot q\frac{\partial m\dot q^2}{\partial m\dot q} - q \frac{\partial}{\partial q}V = 0$$

$$2\left(\frac{\partial q}{\partial t}\right)^2 = q \frac{\partial}{\partial q}V$$

Now left term is time dependant and the right term is time independant for this DE to hold for all time both of them must be equal to a constant say $$C$$

$$C = q \frac{\partial}{\partial q}V$$

which leaves us with

$$V(q) = C\ln(q) + C_{1}$$

does this make sense to you? I saw no one gave this one a shot so I tried, but I am not entirely sure.

Last edited: Jul 18, 2009