A Noether and the derivative of the Action

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The discussion centers on the relationship between the Action and Noether's theorem, highlighting that the Action has units of Energy·time or Momentum·position. The derivative of the Action with respect to time yields Energy, while the derivative with respect to spatial coordinates yields Momentum, reinforcing the connection to conserved quantities. It is noted that if the Lagrangian is independent of time or a spatial coordinate, energy and momentum are conserved, respectively, which is not seen as coincidental. The question of whether the derivative of the Action with respect to phase could relate to electric charge is raised but met with confusion, emphasizing the established link between symmetries and conservation laws. Overall, the discussion underscores the fundamental principles of classical mechanics as articulated by Noether's theorem.
nemuritai
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I know that the Action has units Energy·time or Momentum·position. A second fact is that the derivative of the action with respect to time is Energy and similar with momentum-position, consistent with a units ie. dimensions check.Is it a coincidence that both are Noether conserved quantities? For example will the derivative of the Action with respect to phase be electric charge?
 
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Angular momentum is a conserved quantity, but action is not though the both has same dimension.
 
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nemuritai said:
I know that the Action has units Energy·time or Momentum·position. A second fact is that the derivative of the action with respect to time is Energy and similar with momentum-position, consistent with a units ie. dimensions check. Is it a coincidence that both are Noether conserved quantities?
If the Lagrangian is independent of time then energy is conserved. And, if it is indepenent of a spatial coordinate, then momentum in that direction is conserved. I don't see that as a coincidence.
nemuritai said:
For example will the derivative of the Action with respect to phase be electric charge?
I don't understand this question.
 
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It's of course not a coincidence, because if the Lagrangian doesn't depend on some coordinate ##q## (but only its generalized velocity ##\dot{q}##), then the canonical momentum is conserved for the solutions of the equation of motion, i.e., it is a conserved quantity:
$$p_q=\frac{\partial L}{\partial \dot{q}} \; \Rightarrow \; \dot{p}_q = \frac{\partial L}{\partial q}=0.$$
This canonical momentum is the generator of a symmetry (Noether's theorem).
 
For the quantum state ##|l,m\rangle= |2,0\rangle## the z-component of angular momentum is zero and ##|L^2|=6 \hbar^2##. According to uncertainty it is impossible to determine the values of ##L_x, L_y, L_z## simultaneously. However, we know that ##L_x## and ## L_y##, like ##L_z##, get the values ##(-2,-1,0,1,2) \hbar##. In other words, for the state ##|2,0\rangle## we have ##\vec{L}=(L_x, L_y,0)## with ##L_x## and ## L_y## one of the values ##(-2,-1,0,1,2) \hbar##. But none of these...

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