CLASSICAL THEORY in .NET framework

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CLASSICAL THEORY
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changed, Eq. (1-13). Indeed, ex could be chosen as the parameter characterizing the transformation. If we also have rotational invariance, consider an infinitesimal transformation of angle bex around the axis 0:
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q ---+ q
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+ bex 0
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Pn (0 x qn) h
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The very same formula M(2, 1) = P bq - H bt Ii leads to
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Pn (0 x qn) 12
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Since 0 is arbitrary we therefore find the conservation of total angular momentum as (1-91)
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Of course, if we only have rotational invariance around an axis the corresponding component of angular momentum will be the only one conserved. In summary, when the dynamical problem admits a symmetry the stationary actions /(2, 1) and /(2', I') are equal, where primes denote transformed boundary conditions. Whenever the transformations form a continuous group we obtain a conservation law by differentiating with respect to the group parameters. However, symmetries need not be continuous. Such is the case for parity, time reversal, etc. For instance, in the latter case we have /(q2' t2; qb tl) = /(ql' - tl; Q2, - t2) where boundary conditions are interchanged and time is reversed, corresponding to P2 +-> - Pl' Of course, this in variance does not lead to a conservation law. We finally cast the above considerations in a form suitable for later generalizations. To be specific we return, for instance, to rotational invariance. Under an infinitesimal rotation q goes into Rq = q + bex 0 x q. The Lagrange function is assumed invariant when R is time independent. However, the leastaction principle allows us to choose variations around the stationary path Q(t) of the form
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bq(t) = bex(t) 0 x Q(t)
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(1-92)
provided bex(t2) = bex(tl) = O. Using the in variance of L when bex is constant it follows that
M(Q) _ d oL _ 0 bex(t) - - dt Ob&(t) -
Hence the conserved quantity is proportional to
(1-93)
since oijjOb& = Oqi/Obex. In the present case this is, of course, the component of angular momentum along the direction o.
QUANTUM FIELD THEORY
For time translations the above method is to be used with care. If the infinitesimal parameter Drx is constant, initial and final times are displaced. We set Dq = Drx(t)q, Dq = Drxij + Daq with Drx(td = Drx(t2) = O. In the neighborhood of the real trajectory
o= M=
dt [ Drx (OL qoq dt
+ ij OL) + Da OL] --;--;- q
oq oq oq OL] ot
f2 dt Drx
[~(L _ q o~) _
Invariance means generally that oL/ot vanishes, in which case energy H more generally (d/dt)H = -oL/ot.
pq - L is conserved;
It may occur that the equations of motion are invariant but not the Lagrange function. Under an infinitesimal time-independent transformation, L is modified through the addition of a total derivative in time. In other words, oL/oE>a = (d/dt)cp and (d/dt)(oL/obri. - cp) = 0 for a time-dependent E>a. Thus we find a conserved quantity which is not oL/oE>a any more and explicitly depends on time.
As an example, the dynamics of a particle moving under a constant force is translation invariant but momentum is, of course, not conserved. The Lagrange function is L = j-mq2 + q' F(t). Under a translation Da(t), DL = F Da = (d/dt) St dt' F(t')' Da(t'). The integral of the motion is therefore mq - S:o dt' F(t') = constant in agreement with naive expectations. Note that even for a constant F this is explicitly time dependent. Physically it is, of course, impossible to create such a force throughout all space as required by translational invariance.
Extending these relations to infinite systems will not create difficulties. We shall distinguish two types of symmetries. The first one will correspond to geometrical transformations of space and time under which the lagrangian 2'(x) will go into 2'(x'), where x' is the transformed point (Sec. 1-2-2). The second type will leave the lagrangian invariant and will be called internal (Sec. 1-2-3). Symmetries play such a fundamental role that we shall devote Chap. 11 to a deeper study.
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