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Conserved quantity

  1. Jul 17, 2009 #1
    1. The problem statement, all variables and given/known data

    what condition must satisfy the potential so a Lagrangian [tex] m \dot q \dot q - V(q) [/tex]

    has as a conserved quantity [tex] A(q,p)=qp [/tex]


    2. Relevant equations

    [tex] A(q,p)=qp [/tex] [tex] m \dot q \dot q - V(q) [/tex]

    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 [tex] v(cq)=V(q)c [/tex] for any constant 'c'

    the other attempt to solution is this, since 'A' is a conserved quantity then the Poisson brackets should vanish so [tex] {A,H}=0 [/tex] using the definition of Poisson bracket i should get an ODe for the potential V(q).
     
  2. jcsd
  3. Jul 18, 2009 #2
    Interesting problem, I tried the poisson brackets got a solution check it out if it makes sense to you.

    [tex]\frac{\partial}{\partial q}A \frac{\partial}{\partial p}H - \frac{\partial}{\partial p}A \frac{\partial}{\partial q}V = 0[/tex]

    [tex]p\frac{\partial}{\partial p}H - q \frac{\partial}{\partial q}V = 0[/tex]

    [tex]m\dot q\frac{\partial m\dot q^2}{\partial m\dot q} - q \frac{\partial}{\partial q}V = 0[/tex]

    [tex]2\left(\frac{\partial q}{\partial t}\right)^2 = q \frac{\partial}{\partial q}V [/tex]

    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 [tex]C[/tex]

    [tex]C = q \frac{\partial}{\partial q}V[/tex]

    which leaves us with

    [tex]V(q) = C\ln(q) + C_{1}[/tex]

    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
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