415(6): Angular momentum of m theory
Many thanks again! The angular momentum from the Lagrangian theory is obtained from Eq. (20) of Note 415(3) , this gives: p bold = m gamma r dot bold = partial lagrangian / partial r bold
and
L = r bold x p bold
Using the Lagrangian defined in this way guarantees that the right p is defined, and therefore the right L. So the lagrangian variable r bold gives the same angular momentum as the kinematic result. It would be interesting to experiment as you suggest with the Lagrange variables r / sqrt m(r) and phi, to see of these give the right L. It is possible to just use r bold (the complete vector) as the Lagrange variable. As in Ryder "Quantum Field Theory" it is also possible to use four vectors as the Lagrange variable. The weakness of the Lagrange method is that the Lagrange variables have to be guessed, as you know.
ote 415(6): Angular momentum of m theoryTo: Myron Evans <myronevans123>
The evaluation of the Lagrangian prduces the constant of motion
,
see eq. (o25) of the protocol I just sent over. This does not contain the function m(r) outside of the generalized gamma factor. This seems to be different from the result in note 415(6). I wonder if this has to do with the interpretation of r coordinate. Is this directly observable or do we have to take r/sqrt(m(r)) ? Then the Lagrange coordinate r has to be transformed to give the observable radius coordinate.
Horst