Noether's theorem is based on the assumption that the Lagrangian is independent of position/time/angle/etc. Does this mean it doesn't prove, for example, conservation of momentum in a gravitational field, where the potential energy of the Lagrangian does depend on position? If so, is there some adaptation of Noether's theorem, or new theorems altogether, that justify conservation laws in non-uniform spatial/temporal fields?

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    $\begingroup$ Hint: The gravitational potential depends only on relative positions. $\endgroup$ – Danu Apr 12 '15 at 10:08

You cite the example of a (Newtonian) gravitational potential which, naively, seems to depend on the position of the bodies under consideration. However, a little more contemplation is warranted here.

Consider the following: When we ask a basic question about the gravitational potential energy of some body relative to Earth, do we have to specify where Earth and this body are located in the universe? No, we don't. This should arouse the suspicion that something slightly more subtle is going on.

The reason why we don't have to specify such things is that the gravitational potential only depends on the relative positions of the two gravitationally interacting bodies: $\lvert\vec{r}_1-\vec{r}_2\rvert$. The gravitational potential energy is given by $$U=-G\frac{m_1m_2}{\lvert\vec{r}_1-\vec{r}_2\rvert} $$ The quantity $\lvert\vec{r}_1-\vec{r}_2\rvert$ is (by definition) invariant under any transformation that preserves the distances between points. This includes both rotations and translations. Thus, Noether's theorem applied to the Lagrangian for two gravitationally interacting bodies, given by

$$L=\frac{1}{2}\left(m_1v_1^2+m_2v_2^2\right)+G\frac{m_1m_2}{\lvert\vec{r}_1-\vec{r}_2\rvert} $$

does actually show that momentum (both linear and angular) is conserved.

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  • $\begingroup$ Oh, right! Of course! Sorry to make you have to write that out beyond the original hint. Are there any situations that can occur in reality where Noether's theorem doesn't apply? (The theorem is new to me.) $\endgroup$ – perilousGourd Apr 14 '15 at 6:30

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