- #1
geoduck
- 258
- 2
If you can think of an infinitismal transformation of fields that vanishes at the endpoints, then doesn't the action automatically vanish by the Euler-Lagrange equations?
For example take the Lagrangian:
L=.5 m v2
and the transformation:
x'(t)=x(t)+ε*(1/t2)
At t±∞, x'(±∞)=x(±∞), so the field x(t) doesn't change at the endpoints t=±∞ under this transformation. Since the transformation is infinitismal and the endpoints don't change, then the change in the action should be zero: hence there should be a conserved quantity, namely:
Q=mv*(1/t2)
But this quantity Q is not conserved for a free particle.
For example take the Lagrangian:
L=.5 m v2
and the transformation:
x'(t)=x(t)+ε*(1/t2)
At t±∞, x'(±∞)=x(±∞), so the field x(t) doesn't change at the endpoints t=±∞ under this transformation. Since the transformation is infinitismal and the endpoints don't change, then the change in the action should be zero: hence there should be a conserved quantity, namely:
Q=mv*(1/t2)
But this quantity Q is not conserved for a free particle.