Conditions for Lagrangian Conserved Quantity A(q,p)=qp

[zetafunction]

Homework Statement

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

Homework Equations

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

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).

 

[guguma]

Homework Statement

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

Homework Equations

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

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).

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 dependent and the right term is time dependent 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.