Newton Lagrange Hamilton formalism

[Ali]

Are Newoton's Eqs. Lagrange formalism and Hamilton Equations the same
and equal? Or one is more general than another?

[Rock Brentwood]

On Nov 28, 8:09*am, Ali <ph.questi...@gmail.com> wrote:
> Are Newoton's Eqs. Lagrange formalism and Hamilton Equations the same
> and equal? Or one is more general than another?

They're of overlapping generality.

First, Newton's laws pertain to a specific type of composite system --
one composed of discrete "corpuscles", each described solely by its
trajectory with no internal degrees of freedom or parameters other
than its mass. The 2nd order system resulting from the force law is,
therefore, not the most general possible 2nd order system that you can
have. Some of the possibilities excluded might be describable by
Lagrangians or Hamiltonians.

Going the other way, for definiteness, consider only Lagrangians for
2nd order systems. So, the Lagrangian L is a function L(q,v) of the
configuration coordinates q = (q1,q2,...,qN) and configuration
velocities v = (v1,v2,...,vN). A general 2nd order system will take
the form dv/dt = a(q,v); v = dq/dt, where a = (a1,a2,...,aN) is the
"acceleration", as a function of q and v.

In order for this to be a Lagrangian system, an extra set of
conditions -- the Helmholz conditions -- need to be imposed. That's
because the Euler-Lagrange equations f = dp/dt are linked to partial
differential equations of the form p = dL/dv, f = dL/dq. So, there's a
requirement of "integrability" -- the order of differentiation should
commute. So, making the indices explicit, you have
m_{ab} = dp_a/dv^b, where p_a = dL/dv^a.
Therefore m_{ab} = m_{ba} follows from the requirement that the d/dv's
commute, when applied to L.

It also follows, applying the same argument to the coefficients of
inertia matrix m, that
dm_{ab}/dv^c = dm_{ac}/dv^b.

You then have that dp_a/dq^b = df_b/dv^a, where f_b = dL/dq^b. This is
meant to work in conjunction with the equation of motion, so
dp_a/dq^b = d/dv^a (d/dt (p_b)).
The total time derivative, explicitly in terms of the partial
deriatives, is:
d/dt = d_t + v d/dq + a(q,v) d/dv
where d_t is the partial derivative with respect to t.

So, the condition can be reduced further, by pulling out the "d/dt",
making use of the commutator
[d/dv^a, d/dt] = [d/dv^a, d_t + v d/dq + a(q,v) d/dv] = d/dq^a + da/
dv^a d/dv.
The result is
d/dv^a (d/dt p_b) = d/dt (dp_b/dv^a) + dp_b/dq^a + da/dv^a dp_b/dv.
So, the result is an equation for dm_{ba}/dt. The equation involves
the combination
S_{ab} = dp_a/dq^b - dp_b/dq^a.
So, since S_{ab} is anti-symmetric in a,b and m_{ab} is symmetric, the
equation can be split into an anti-symmetric part (an equation for S_
{ab}) and a symmetric part (an equation for dm_{ab}/dt that doesn't
involve S_{ab}).

Finally, the commuting of the d/dq's on L gives you the condition df_a/
dq^b = df_b/dq^a, or after applying the Euler-Lagrange equations,
d/dq^b (d/dt p_a) = d/dq^a (d/dt p_b).
The d/dt can be pulled out by applying the commutator
[d/dq^a, d/dt] = da/dq^a d/dv.
And the result is a condition that gives you an equation for dS_{ab}/
dt.

The conditions can be stated in terms of the "acceleration" function
a, or in terms of the "force" function F_a = sum_b (m_{ab} a^b). The
more usual formulation (and the one that generalizes to field theory)
is the second one. It eliminates any explicit reference to an
"acceleration" function.

To go from Hamiltonian <-> Lagrangian requires solving the Legendre
transform. This inverts the relation p <-> v. The requirement for
going from v -> p is that the gradient m = dp/dv not be singular --
that is, the coeffcients of inertia should be non-singular (no "zero
mass modes"). The requirement for going from p -> v is that the
inverse matrix W = dv/dp not be singular (no "infinite mass modes").

Most systems are constrained and have singular m's or W's. So, this
leads into the Dirac/Bergmann method for handling constrained systems.

[Oh No]

Thus spake Rock Brentwood <markwh04@yahoo.com>
>> Are Newoton's Eqs. Lagrange formalism and Hamilton Equations the same
>> and equal? Or one is more general than another?
>
>They're of overlapping generality.
>
>First, Newton's laws pertain to a specific type of composite system --
>one composed of discrete "corpuscles", each described solely by its
>trajectory with no internal degrees of freedom or parameters other than
>its mass. The 2nd order system resulting from the force law is,
>therefore, not the most general possible 2nd order system that you can
>have. Some of the possibilities excluded might be describable by
>Lagrangians or Hamiltonians.
[[Mod. note -- The usual term in physics for such a "corpuscle"
is "point particle".
-- jt]]
>
>Going the other way, for definiteness, consider only Lagrangians for
>2nd order systems. So, the Lagrangian L is a function L(q,v) of the
>configuration coordinates q = (q1,q2,...,qN) and configuration
>velocities v = (v1,v2,...,vN). A general 2nd order system will take the
>form dv/dt = a(q,v); v = dq/dt, where a = (a1,a2,...,aN) is the
>"acceleration", as a function of q and v.

Let us add Maxwell's equations to Newton's laws, as they were not known
at the time of Newton. So we can also have charges and electromagnetic
potentials, as well as gravitational.

Now, in what way is the Lagrangian formulation distinct from a set of N
corpuscles described by its trajectory and having no parameters other
than mass and charge?

Regards

--
Charles Francis
moderator sci.physics.foundations.
charles (dot) e (dot) h (dot) francis (at) googlemail.com (remove spaces and
braces)

http://www.teleconnection.info/rqg/MainIndex

[Rock Brentwood]

Oh No wrote:
> Let us add Maxwell's equations to Newton's laws, as they were not known
> at the time of Newton. So we can also have charges and electromagnetic
> potentials, as well as gravitational.

Good luck with that. To even define what all that means is still
largely an open problem, though recent work has appeared in the
literature on how to consistently define field theory with point-like
sources (there's trouble even line-like sources, as Maxwell himself
made clear). The problem, in fact, also appeared in one of the MAA
monthly's back in the 1980's under the open problem challenge.

[Oh No]

Thus spake Rock Brentwood <markwh04@yahoo.com>
>Oh No wrote:
>> Let us add Maxwell's equations to Newton's laws, as they were not known
>> at the time of Newton. So we can also have charges and electromagnetic
>> potentials, as well as gravitational.
>
>Good luck with that. To even define what all that means is still
>largely an open problem, though recent work has appeared in the
>literature on how to consistently define field theory with point-like
>sources (there's trouble even line-like sources, as Maxwell himself
>made clear). The problem, in fact, also appeared in one of the MAA
>monthly's back in the 1980's under the open problem challenge.
>
I am happy that that isn't really a problem within classical
electrodynamics. We know very well that "point-like" in classical
mechanics means using an approximation at greater scales than one needs
to take quantum effects into account.

Regards

--
Charles Francis
moderator sci.physics.foundations.
charles (dot) e (dot) h (dot) francis (at) googlemail.com (remove spaces and
braces)

http://www.teleconnection.info/rqg/MainIndex

[Rock Brentwood]

Oh No wrote:
> Let us add Maxwell's equations to Newton's laws, as they were not known
> at the time of Newton. So we can also have charges and electromagnetic
> potentials, as well as gravitational.

Good luck with that. To even define what all that means is still
largely an open problem, though recent work has appeared in the
literature on how to consistently define field theory with point-like
sources (there's trouble even line-like sources, as Maxwell himself
made clear). The problem, in fact, also appeared in one of the MAA
monthly's back in the 1980's under the open problem challenge.