# Free Fall Conservation of Momentum

So I looked at the invariance of the Lagrangian under the Gallilei Transformations.
So for the free fall we have the Lagrangian
$$L = \frac{m}{2}\dot{z}^2 -mgz$$
Then I applied the transformation
$$x\mapsto x' = x+\delta x$$ which gives me the following transformed Lagrangian
$$L' = \frac{m}{2}\dot{z}^2-gmz-gm\delta z = L+(-gm\delta z)$$
as the Lagrangian can have an extra addition of a Constant that doesnt change the Euler Lagrange Equations which I also extra verified you get the usual $$z = \frac{1}{2}gt^2$$
So there is now the Noether theorem which states that when we have an symmetry aka. invariance under a transformation we have a conservation law according to my research if you have translation invariance you get the conservation of momentum.

Now my question is how do we have conservation of Momentum in free falling conditions am I missing something there.

You don't have translation invariance for the free fall in external gravitational potential, because as you calculated, $$\delta L = -gm \,\delta z \neq 0$$ and consequently you don't have momentum conservation.
• i am a bit confused i only add a constant $\delta z$, and according to my lectures the Lagrangian is Invariant if you only add a constant. Am i missunderstanding something here? – The Mastermage May 25 '19 at 23:16
• Noether theorem requires full invariance, without any constant added. Or at least invariance of the action $S = \int_{t_1}^{t_2} L dt$. Invariance up to a constant is not enough. Specifically it causes the momentum to not be constant, but grow linearly. – Adam Latosiński May 25 '19 at 23:23