Re: Newton Vs Lagrange Vs Hamilton

From: Neil Price (infiniteirresponsability_at_hotmail.com)
Date: 03/27/05


Date: Sun, 27 Mar 2005 08:56:15 +0000 (UTC)

vonnyn@hotmail.com (Vonny N) wrote in message news:<df799de5.0503251424.2f339940@posting.google.com>...
> What is the relationship between:
>
> 1. Newtonian Mechanics
> 2. Lagrangian Mechanics
> 3. Hamiltonian Mechanics

1. --> 2. In case of conservative forces, one can study a Lagrangian
of the form L = T - V with V the potential energy. This is equivalent
to the original Newtonian setup. There are, however, Lagrangians
which are not of the form L = T - V and therefore are not
representable in Newtonian form. (This should be taken with a grain
of salt).

2. --> 3. To construct a Hamiltonian from a given Lagrangian, one
studies a map called the Legendre transform (this expresses the p's as
derivatives of the Lagrangian). In order to do this construction, one
has to invert the Legendre transformation, so the transition to the
Hamiltonian formalism only works when this map is actually invertible.

If you know that p = dL/dv, it's easy to see that invertibility is
equivalent to the matrix d^2 L/dv^2 being invertible. Now, this is
always the case for a Lagrangian of mechanical type (one of the form L
= T - V, with T the kinetic energy).

The above should again be taken with a grain of salt. Even if the
Lagrangian is degenerate, there are more sophisticated ways of going
to the Hamiltonian formalism. In this case, some constraints will
arise and these will determine a subset of the phase space. Take the
example where the Lagrangian does not depend on a certain velocity
coordinate v_0. Then the associated momentum p_0 = dL/dv_0 will be
identically zero.

Furthermore, the above treatment only goes through in the case of
first-order theories, i.e. depending on x, v, but not the derivatives
of v. In that case, one can still make sense of a lot of these
things, but it is all a lot more involved. It certainly is no mere
extension of the first order case.

> Can they be declared to be *rigorously* mathematically equivalent?
>
> Does it suffice to say that they are reformulations of each other,
> each well-adapted to a particular class of problems?

The prevailing mathematical point of view is that Lagrangian mechanics
takes place on the tangent bundle of the configuration space, whereas
the Hamiltonian formalism is dealt with on the cotangent bundle.
There are various intrinsic objects defined on these bundles, and the
equations of motion can be expressed by use of them.

As to which formulation to use, I think it mostly depends (in the
regular case) on the formulation of the problem. Sometimes one is
preferred over the other: in the Lagrangian picture one has a
"variational principle"; i.e. the equations of motion are derived by
extremizing a certain functional. This can sometimes be exploited,
e.g. in the construction of numerical schemes which preserve (some of)
the geometrical content of the problem.

On the other hand, Hamiltonian mechanics is more directly amenable to
techniques from symplectic geometry, e.g. symplectic reduction,
definitions involving integrability, etc.
 
> If so, can we describe reasonably clearly which type of problems are
> more easily represented and/or solved using each version of mechanics?
>
> Finally, is there a book on classical mechanics which treats all three
> of these approaches to mechanics, along with their
> inter-relationships, with mathematical rigour? (I'm afraid I don't
> much like Goldstein's widely cited book).
>

See the book "foundations of mechanics" by Abraham and Marsden or
"introduction to mechanics and symmetry" by Marsden and Ratiu. These
make use of a lot of differential geometry so they might not be very
easy going on a first reading (especially FoM), but they tell you just
about everything you'll ever want to know on this topic. There's also
the book by V.I. Arnold "Mathematical Methods of Classical Mechanics",
which is very good.

N.



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