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HubertP
HubertP is offline
#5
Nov23-12, 06:33 AM
P: 9
I fully agree with your example. The Lagrangian is valid, it describes isolated system (no explicit time dependence - potential only depends on distance between system components and not the absolute positions). In such case it's obvious that translation will not change the Lagrangian.
Your example made me aware of one important thing, though (which, in fact, addresses my doubts)! Even though we have certain freedom in constructing the Lagrangian (we can always add total time derivative of function of positions and time to the Lagrangian and end up with the same trajectory), we DO WANT to construct the Lagrangian for isolated system in its simplest form, e.g. independent (explicitly) from time and absolute positions, which is probably equivalent to saying we construct it in an inertial frame of reference.

To show an example of the systems I was thinking about consider:
[itex]\mathcal{L} = \frac{1}{2}m \dot{x}^2 + \dot{x}x[/itex]
It can be shown that even though the translation changes the Lagrangian, it does not alter the equations of motion - original trajectory after translation is still valid (minimizes the action), because the change in the action due to translation is (by [itex]\delta S[/itex] I mean how much the action differs between translated and original system for the same trajectory, assuming the translation is small):
[itex]\delta \mathcal{S} = \int_{t1}^{t2} \delta \mathcal{L} dt = \int_{t1}^{t2} \bigg( \frac{d \mathcal{L}}{dx} \delta x + \frac{d \mathcal{L}}{d \dot{x}} \delta \dot{x} \bigg) dt[/itex]
Now, since [itex]\delta \dot{x}[/itex] is 0 the second term in integral is zero, too, so we can write:
[itex]\delta \mathcal{S} = \delta x \int_{t1}^{t2} \frac{d \mathcal{L}}{dx} dt = \delta x \int_{t1}^{t2} \dot{x} dt = \delta x \int_{t1}^{t2} \frac{dx}{dt} dt = \delta x \bigg(x \bigg|_{t1}^{t2} \bigg)[/itex]
This is constant and will vanish when variating the trajectory.
So, the Lagrangian has different form after translation, yet it yields the same equations of motion. However, as I just realized, it does not describe isolated system (at least not in an inertial frame), due to explicit dependence on position in the second term. It might be equivalent to some isolated system expressed in non-inertial frame (I'm not sure). Anyway - it doesn't matter. You helped me understand this. Now I think I know why we say that translation does not change the Lagrangian when deriving conservation laws. It's because the original Lagrangian does not depend explicitly on positions and time. Thanks!