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Re: [Phys-l] Action

The action integral is to be taken over time (t),not x.

You are probably employing dt =dx/v_x and assuming that's OK since v_x is constant in time. But this is assured true only for the true path, not for neighboring paths, which are to be included.

Bob Sciamanda
Physics, Edinboro Univ of PA (Em)

From: "John Denker" <>
Sent: Thursday, January 07, 2010 11:29 AM
To: "Forum for Physics Educators" <>
Subject: Re: [Phys-l] Action

On 01/07/2010 05:32 AM, Josh Gates wrote:

Here's how I'm trying to do it (which apparently has one or more flaws):

- I made a spreadsheet, with the columns x, y, v, KE, PE, E, K-U
* x increments in .1 m steps from 0 to 5m
* y is a function of x, defining the path
* v is root(v_i^2-2gy), satisfying cons. of E
* KE and PE are defined in the ordinary way
* E is there to check my formulas, verifying cons. of E
* I average all of the K-U entries to give something similar to the action

Since the x steps are all the same, integrating K-U dx and dividing by
the total delta x should give me the same thing that the average does (I
think). It occurs to me now that there's a problem with paths that go
straight up at any point, but I'm willing to work with that later. My
current issue is that there are other parabolic paths that give a lower
K-U average than the correct path.

Anyone see where I went awry?

The general approach seems reasonable. However, some
opportunities for improvement include:

1) You gave very explicit definitions for all quantities
*except* K and U. It would simplify the discussion if
we had comparably explicit definitions for those, too.

2) Action is the total, not the average.

3) Action is the integral _dt_ so there would never be
any problem with vertical segments.

Also: There is a scintillating discussion of the topic
in Feynman volume I. Well worth re-reading.
Forum for Physics Educators