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K^2 is offline
Nov23-12, 07:44 AM
Sci Advisor
P: 2,470
Actually, it's slightly more complicated. You didn't just add potential. You've modified the kinetic term.

[tex]p = \frac{\partial L}{\partial \dot{x}} = m\dot{x} + x[/tex]

So momentum of your particle is position-dependent. But lets take a look at total energy.

[tex]H = \dot{x}p - L = \frac{1}{2}m\dot{x}^2[/tex]

It looks like a Hamiltonian of a free particle. Of course, we need it in terms of momentum, so that's going to get messy.

[tex]H = \frac{p^2}{2m} - \frac{xp}{m} +\frac{2x-x^2}{2m}[/tex]

So it's a particle whose kinetic energy is function of p and x in a potential.

In other words, you did not simply create a different Lagrangian with same trajectory. You came up with a completely new set of physics where under a particular potential, you end up with the same equations of motion as free particle under normal physics. There is no surprise, then, that in general, the conservation laws are going to be very different with these new physics.