It is well known from Noether's Theorem how from continuous symmetries in the Lagrangian one gets a conserved charge which corresponds to linear momentum, angular momentum for translational and rotational symmetries and others.
Is there any elementary argument for why linear or angular momentum specifically (and not other conserved quantities) are conserved which does not require knowledge of Lagrangians? By elementary I mean, "if this is not so, then this unreasonable thing occurs".
Of course, we can say "if we want our laws to be the same at a different point in space then linear conservation must be conserved", but can we derive mathematically the expression for the conserved quantity without using the Lagrangian?
I want to explain to a friend why they are conserved but he doesn't have the background to understand the Lagrangian formalism.
Answer
The answer is yes, the essence of Noether's theorem for linear and angular momentum can be understood without using the Lagrangian (or Hamiltonian) formulation, at least if we're willing to focus on models in which the equations of motion have the form mn¨xn=Fn(x1,x2,...)
(This answer still uses math, but it doesn't use Lagrangians or Hamiltonians. An answer that doesn't use math is also possible, but it would be wordier and less convincing.)
The inputs to Noether's theorem are the action principle together with a (continuous) symmetry. For a system like (1), the action principle can be expressed like this: Fn(x1,x2,...)=−∇nV(x1,x2,...).
First consider linear momentum. Suppose that the model is invariant under translations in space. In the context of Noether's theorem, this is a statement about the function V. This is important! If we merely assume that the system of equations (1) is invariant under translations in space, then conservation of momentum would not be implied. (To see this, consider a system with only one object subject to a location-independent force.) What we need to do is assume that V is invariant under translations in space. This means V(x1+c,x2+c,...)=V(x1,x2,...)
Now consider angular momentum. For this, we need to assume that V is invariant under rotations. To be specific, assume that V is invariant under rotations about the origin; this will lead to conservation of angular momentum about the origin. The analogue of equation (5) is ∑nxn∧∇nV(x1x2,...)=0
No comments:
Post a Comment