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In Hamiltonian statement the Noether theorem is read as follows. Consider a system with the Hamiltonian function $$H=H(z),\quad z=(p,x),\quad p=(p_1,\ldots,p_m),\quad x=(x^1,\ldots,x^m)$$ and the phase space ##M,\quad z\in M.## Assume that this system has a one parametric group of symmetry ##z\mapsto g^s(z)##. This group is generated by a Hamiltonian ##F=F(z):##
$$\frac{d g^s(z)}{ds}=v_F( g^s(z)),\quad g^0(z)=z,$$
here ##v_F## stands for a Hamiltonian vector field that corresponds to the Hamiltonian function ##F(z)##.
By definition, "group of symmetry" means $$H(g^s(z))=H(z),\quad \forall s.\qquad (1).$$
Then Noether says: ##F## is a first integral for the system with the Hamiltonian ##H##. To prove this trivial observation one just should differentiate (1) in s and put s=0.
My question is little bit philosophical. The point is that the assertion of Noether's theorem is the least of what we can actually extract from the symmetry group given. Indeed,
assume that for some point ##z_0## one has ##dF(z_0)\ne 0## . Then in a neighborhood of this point there are local canonical coordinates ##P,X## such that in these new coordinates the function ##F## has the form ##F=X^1.##
This fact is proved in [Olver, P. J. (1986), Applications of Lie Groups to Differential Equations, Graduate Texts in Mathematics, 107, Springer] but there is a much simpler proof by means of the generating functions.
The coordinates ##(P,X)## can be constructed explicitly provided the group ##g^s## is given .Then by the Noether theorem in the coordinates ##(P,X)## we get
$$\{F,H\}=0=\frac{\partial H}{\partial P_1}.$$
So that the function ##H## does not depend on ##P_1## and the coordinate ##X^1=const## is a first integral. We obtain a Hamiltonian system with ##m-1## degrees of freedom ##H=H(P_2,\ldots,P_m,const,X^2,\ldots,X^m)##. Observe that the first integral by itself does not give possibility to reduce the system explicitly to one with ##m-1## degrees of freedom.
So my question is why do physics textbooks never mention this consequence from the presence of symmetry group?
$$\frac{d g^s(z)}{ds}=v_F( g^s(z)),\quad g^0(z)=z,$$
here ##v_F## stands for a Hamiltonian vector field that corresponds to the Hamiltonian function ##F(z)##.
By definition, "group of symmetry" means $$H(g^s(z))=H(z),\quad \forall s.\qquad (1).$$
Then Noether says: ##F## is a first integral for the system with the Hamiltonian ##H##. To prove this trivial observation one just should differentiate (1) in s and put s=0.
My question is little bit philosophical. The point is that the assertion of Noether's theorem is the least of what we can actually extract from the symmetry group given. Indeed,
assume that for some point ##z_0## one has ##dF(z_0)\ne 0## . Then in a neighborhood of this point there are local canonical coordinates ##P,X## such that in these new coordinates the function ##F## has the form ##F=X^1.##
This fact is proved in [Olver, P. J. (1986), Applications of Lie Groups to Differential Equations, Graduate Texts in Mathematics, 107, Springer] but there is a much simpler proof by means of the generating functions.
The coordinates ##(P,X)## can be constructed explicitly provided the group ##g^s## is given .Then by the Noether theorem in the coordinates ##(P,X)## we get
$$\{F,H\}=0=\frac{\partial H}{\partial P_1}.$$
So that the function ##H## does not depend on ##P_1## and the coordinate ##X^1=const## is a first integral. We obtain a Hamiltonian system with ##m-1## degrees of freedom ##H=H(P_2,\ldots,P_m,const,X^2,\ldots,X^m)##. Observe that the first integral by itself does not give possibility to reduce the system explicitly to one with ##m-1## degrees of freedom.
So my question is why do physics textbooks never mention this consequence from the presence of symmetry group?
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