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Re: [Phys-l] Action

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.