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*From*: John Denker <jsd@av8n.com>*Date*: Thu, 07 Jan 2010 17:15:05 -0700

On 01/07/2010 01:48 PM, Josh Gates asked:

is there an alternative way to calculate K and U (.5mv^2 and

mgy) here?

That's quite an open-ended question. The wise-guy answer would

be yes, there are always alternative ways. Here's a stab at

a more helpful answer: Here's how I would attack it:

*) We have a series of (t,x,y) points. Such points can be called

_events_ in spacetime.

*) We assume that t is monotonic and can be used to parameterize

progress along the path.

*) We assume that between events, x varies as a linear function

of time.

*) We assume that between events, y varies as a linear function

of time. This is of course not exactly true, but if the step

size is small enough the results should come out OK. (So here

is a nontrivial alternative: keep track of t, x, y, xdot, and

ydot and use a second-order model for the positions.)

*) Another alternative, not trivially simple but not ridiculous

either: Assume that KE(t) and PE(t) are piecewise linear

functions of t. (This is incompatible with the previous

assumptions; it's an "instead-of" not an "also".)

*) For each interval (the interval between two adjacent events)

do the integral analytically. Integrating a low-order polynomial

is easy.

If those aren't the sort of alternatives you were looking for,

please ask a more specific question.

As others have pointed out, the usual setup is to specify the

initial (t,x,y) and the final (t,x,y) and leave everything else

including initial angle and initial energy to be determined on

the fly. This is not the only way of doing things; obviously

the particle knows how to get from t to t+dt without knowing

the final destination.

- The bit about cons. of E was, I thought, a requirement.

Actually you don't need to put it in as a prerequisite; it

will come out as a consequence.

The general principle here is:

"The Lagrangian knows all and tells all."

In particular, given the Lagrangian you can choose a variable

and then just turn the crank to find whatever momentum is

canonically conjugate to that variable. You can also find

the equations of motion (using the principle of least action).

You can also find the Hamiltonian and show that the equations

of motion conserve it.

In contrast, given just the Hamiltonian, when you choose a

variable you don't have any automatic way of finding the

conjugate momentum, so you can't proceed.

To repeat: The Lagrangian will give you the Hamiltonian and

not vice versa.

Also the fact that the Lagrangian density is well behaved

with respect to special relativity should be a clue that

there's something special about it.

**References**:**[Phys-l] Action***From:*Josh Gates <jgates@tatnall.org>

**Re: [Phys-l] Action***From:*John Mallinckrodt <ajm@csupomona.edu>

**Re: [Phys-l] Action***From:*Josh Gates <jgates@tatnall.org>

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