Conference rusure::math

The following text discusses the application of Hamilton's principle (cf
note 618.18-26).  Hamilton's principle is more widely used, especially in 
advanced mechanics, than Newton's laws.  Indeed, Hamilton's principle finds wide 
use not only in advanced mechanics, but in thermodynamics and quantum theory as 
well.  One further note, now that I understand what you mean by invariant, 
Hamilton's principle is no more invariant than Newton's Laws.

Now, for the purposes of discussion, let's use Hamilton's principle to derive 
Newton's equations of motion for a particle under the influence of a 
gravitational field (W/O REFERENCE TO NEWTON'S LAW).  O.K? Here we
go...

Let

(1)        L = K - P
        
(1a)       Where K = Kinetic Energy = 1/2 mv^2
(1b)             P = Potential Energy = F(x,y,z) = mgz
                     (Note:  P is represented as a function of position, in
                             this example, only the position wrt z.  Also,
                             see Footnote 1, below)
Since

(2)        V^2 = (ds/dt)^2 = [(dx/dt)^2 + (dy/dt)^2 + (dz/dt)^2]

            "              = [x'^2 + y'^2 + z'^2]
                        
By substituting (2) into (1a), then (1a) and (1b) into (1) we obtain

(3)     L = 1/2 m[x'^2 + y'^2 +z'^2] - mgz

Expressed in this way, L is sometimes referred to as the Lagrangian of
the system.  Hamilton's principle, mathematically stated is that the
value of the integral of L over some time period (i.e., between T(1) and T(2)),
is STATIONARY.  Mathematically, STATIONARY is a term commonly applied to points 
on some curve, F(x), whose derivative F'(x) = 0 at that point.  Such points are 
either minima, maxima, or points of inflection.

In general, we seek to find a STATIONARY function, I, such that

                      T(2)
                I = INT   L(x',y',z',x,y,z)dT
                      T(1)

To solve this integral such that I is STATIONARY requires one to employ 
variational calculus.  Variational calculus is very straightforward but is a 
real pain to show on DEC terminals (see footnote 2, below).  What you end up
with after a some straightforward algebraic manipulation is Euler's equation
(see below).  Rather than derive Euler's equation for you, though, let me try to 
describe the main concepts behind the mathematics, instead:

You all remember maxima and minima (aka extremal) problems from calculus and 
diff. eq.  In these problems, the task was to find the max/min POINTS of a given 
function.  Variational calculus enable us to find stationary FUNCTIONS (i.e., 
whole sets of points that might represent, for example, the shortest path 
between two points on a sphere).

For example, suppose you want to find a function that represents the LEAST
expenditure of energy in pushing a particle from point 'A' to point 'B'.
There are an infinite number of paths over which one could push the particle.
Variational calculus allows us to find an expression for the FUNCTION that
describes the one that will result in the least expenditure of energy.

If you understand this notion, that of finding min/max functions rather
than min/max points, it is fairly easy to show (although not on a DEC
terminal) the derivation of Euler's equation,
                _   _ 
             d |P[L] |    P[L]     
(4)         ---|---- |  - ---- = 0      (* Euler's Equation *)
             dt|P[x']|    P[x]     
                -   -                   
Where
        P  = the partial differentiation operator
        x' = the derivative of x w.r.t. time
        L  = the Lagrangian (See equation (3))
        
To derive Newton's equations of motion, one sets up the Euler equation for each 
of the three dimensions in the cartesian space, {x,y,z}.  That is, we set up
a system of homogeneous, linear, differential equations of the form

(5a)        Euler (x',x) = 0        ...This is a shorthand form of (4)
        
(5b)        Euler (y',y) = 0

(5c)        Euler (z',z) = 0
        
To solve these equations, we must compute the partial derivatives and substitute
them into the appropriate Euler equations (5a,b,or c).  Thus, from (1b) we have

        F[x,y,z] = mgz         ...is the potential energy function
        
Such that 

(6a)     P[F]/P[x] = 0        ...Partial of F wrt x
(6b)     P[F]/P[y] = 0        ...Partial of F wrt y
(6c)     P[F]/P[z] = mg       ...Partial of F wrt z

Similarly we compute the partial derivatives of L wrt x',y',and z' to obtain

(7a)     P[L]/P[x'] = m[dx/dt] = mx'    ...Partial of L wrt x'
(7b)     P[L]/P[y'] = m[dy/dt] = my'    ...Partial of L wrt y'
(7c)     P[L]/P[z'] = m[dz/dt] = mz'    ...Partial of L wrt z'

Substituting (7a,b,c) into their respective Euler equations (5a,b,c) we obtain

                       d[mx']
(8a)    Euler (x',x) = ------ = 0, or x' = constant     ...constant velocity
                         dt
                         
                       d[my']
(8b)    Euler (y',y) = ------ = 0, or y' = constant     ...constant velocity
                         dt

                       d[mz']
(8c)    Euler (z',z) = ------ + mg = 0, or z'' = -g     ...acceleration due to
                         dt                                gravity.
                         
These are just the familiar equations obtained from Newton's law;  They say
that in a gravitational field near the surface of the earth, the horizontal
velocity (in x and y directions) is constant, and the vertical acceleration
is (-g).

The advantage of this method is not really brought out in simple
examples of this type.  But in more complicated situations it may be simpler
to find one scalar function, L, to which to apply the Euler equation, than
to find six functions (the components of the two vectors, force and
acceleration).  One example of the utility of this approach is to derive
the equations of motion from Newton's laws for spherical coordinates.
It's trivial using Hamilton's principle.  It's extremely tedious using
Newton's laws, especially for the acceleration components.

FOOTNOTE 1:

The gravitational potential energy function, V = mgz, where z is the
"up and down" axis can be found in any elementary physics text.  In
words, it simply says that the gravitational potential energy is proportional
to the distance between a particle of mass, m, and the center of mass.
Or in simple cases, the height above the surface of the earth.

FOOTNOTE 2:

If any of you are upset by this hand-waving, I'll understand.  Here are
some references (I recommend the first, highly).

1. Boas, Mary L.,  Mathematical Methods in the Physical Sciences, John Wiley
   & Sons, 2nd edition, 1983.  See Chapter 9, "Calculus of Variations"
   
2. Bradbury, Ted C., Mathematical Methods with Applications to Problems
   in the Physical Sciences, John Wiley & Sons, 1984.  See chapter 10,
   "The Calculus of Variations"

Note:  This is for the heavies among you.  It's very rigorous, but
very good if you can slog through it.

3. Feynman, Leighton, Sands, The Feynman Lectures on Physics, Addison-
   Weysley, 6th edition 1977.  See chapter 19, Vol II, "The Principle
   Of Least Action"
   
Note: A very nice job of describing (Euphemism for hand-waving, much
like this note) the applicability of this technique.  Quite readable.

To carry the discussion 621.1 a little further:  One can use
Hamilton's principle to derive the equations of simple harmonic motion
just by noting that the potential energy function of a system
executing SHM is...

        P = 1/2 Kx^2
        
By using this Potential Energy function, instead of P = mgz (in the case
of a graviational field) you'll end up with the equations of SHM. That is,

        d^2[x]
      m ------  = - Kx   ...Force is proportional to the displacement.
         dt^2
         
Try it...

/mtp

    In one sense, a the equivalence of a variational statement of a
    problem and a differential equation statement is like the difference
    between a boundry value problem and an initial value problem.

    Suppose you can solve a given boundry value problem (for example,
    by expanding in an appropriate space of eigenfunctions).  If the
    solution is smooth enough, you could then get the initial conditions
    at one of the boundries and now get the same solution as an initial
    value problem.  The latter would be like the 'shooting method' for
    solving a boundry problem, where you try various initial conditions
    at one end and try and hit the other boundry.

    The beauty of variational prinicples is that you are often not interested
    in the path itself, but only in some global property (such as the
    capacitance per unit length of a microstrip transmission line).
    It also leads to computationally effective ways of solving problems
    via the finite element method (Galerken method, Ritz method,
    moment methods, and so forth).

    To rigorously probe the limits of applicability of this, one gets into
    some hairy mathematics (Sobolev spaces and the like).

    A few references I'm familair with:

	"Classical Mechanics" by Goldstein - very good elementry exposition
	without assuming you know all sorts of stuff about manifolds, etc.

	"Feynman Lectures on Physics" - everyone should look through
	these sometime.  I think Vol II has a chapter on least action.

    Also, any book on quantum mechanics will use Hamiltonians, since
    nuclear 'forces' are conservative.

    - Jim

    Re .1:
    
    How does that indicate a tendency for the difference between kinetic
    and potential energies to become minimized?
    
    You define "stationary" for a point, but then refer to a "stationary
    function".  What is the relationship between them?
    
    You also describe the value of the integral of L over some time period
    as stationary.  How does this relate to the definition of "stationary"
    that F'(x) = 0.  In other words, what are x and F for the value of the
    integral of L over some time period?
    
    
    				-- edp 

        You also describe the value of the integral of L over some time period
    as stationary.  How does this relate to the definition of "stationary"
    that F'(x) = 0.  In other words, what are x and F for the value of the
    integral of L over some time period?

>   I'm not sure I completely understand your question, but I'll give it
>   a shot:  First, when we note that F'(x) = 0, we say that the point, x,
>   is stationary, not the function F'(x).

>   For a function, F(x,x') to be stationary it must meet the following
>   condition.  That is, The derivative of the partial derivative of F wrt x' 
>   minus the partial derivative of F wrt to x, must be equal to zero!
>   In otherwords, the function that is derived by setting Euler's equation
>   to zero is, by definition, a stationary function.

>   To prove that to you would require me apply variational calculus, which
>   I'm unwilling to do given that these proofs are readily found in the
>   texts referenced at the end of note .1

    Hope this helps,
    
    /mtp
    

    Consider a simple one dimensional problem of a particle moving from
    point x2 to x2 in time t1 to t2; these are the boundry conditions.

    We now assume that the path of 'least action' is x(t) and ask what
    happens to the action (integral of the Lagrangian over the interval)
    when we perturb the path by some admissable e(t) (no Dirac deltas, etc).

    We have the kinetic energy which is a function of velocity,

	KE = .5*m*(dx(t)/dt)^2

    and the potential energy as some function of position,

	PE = V(x)

    The action is defined as the functional

	S = int(t1,t2) KE-PE dt

    The claim is that the perturbation e(t) will vanish over the interval
    if x(t) satisfies Newtons law.

    Assume e(t) is small so we can expand (d(x+e)/dt) and V(x+e)
    in series in the neighborhood of x neglecting second order terms
    in e:

	V(x+e) = V(x) + e*V'(x) + ...

	(d(x+d)/dt)^2 = (dx/dt)^2 + 2(dx/dt)(de/dt) + (square ignored)

    Now the action is (to first order in e)

	S = INT(t1,t2) .5*m(dx/dt)^2 - V(x) + m*(dx/dt)(de/dt) - e*V'(x) dt

    So this is of the form S + dS, 

	dS = INT(t1,t2) m*(dx/dt)(de/dt) - e*V'(x) dt

    Now we integrate by parts (one of those wonderfully obscure seeming moves)

	dS = m*dx/dt*(e(t2) - e(t1)) - INT(t1,t2) e*(m*(dx^2/dt^2) + V'(x) dt

    Note that e(t1), e(t2) are both zero by the boundry conditions, so
    perturbation in the action functional is just the last integral

	dS = INT(t1,t2) e(t)*(m*(dx^2/dt^2) + V'(x)) dt

    Now, we are allowing e(t) to be arbitrary over the interval, and for
    dS to vanish for any admissible e(t) then we must have

	m*(dx^2/dt^2) = -dV(x)/dx

    since force is the gradient gravitational potential we recover F = m*a
    in this case...

    - Jim

    Nice job, Jim.  I just wasn't motivated to wrestle with ASCII
    characters.  BTW, with a slightly more rigorous approach one ends
    up with Euler's equation.  Having Euler's equation is like not
    having to build a hammer everytime you want to hammer a nail. Anyway,
    I liked your derivation.
    
    /Mike

    It's been a long time since I took a mechanics course (using
    Goldstein), but I remember that the use of the Langrangian rather
    than Newton's laws made it MUCH easier to include constraint equations
    into your equations of motion (constraints like "this object is
    constrained to move upon the surface of a sphere").