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Re: [Phys-l] Force and potential



Thanks - as soon as I saw your Eqn [4] it became obvious where I went astray. I deal with this same issue in thermodynamics (specifying what is being held constant), yet I let myself fall into that trap here.

Cheers,

Bob



________________________________________
From: phys-l-bounces@carnot.physics.buffalo.edu [phys-l-bounces@carnot.physics.buffalo.edu] on behalf of John Denker [jsd@av8n.com]
Sent: Thursday, April 05, 2012 4:24 PM
To: Forum for Physics Educators
Subject: Re: [Phys-l] Force and potential

On 04/05/2012 09:13 AM, LaMontagne, Bob wrote:

When I posted the musings about the problem (an illustration of
blindly following the math instead of the physics), I probably should
not have focused on the position (1,1) at the rim of the bowl (and
the infinity that results).

OK, let's consider points below the rim.

The nonsensical result even applies below the rim.

I'm not sure what result is being called "nonsensical". See
below for one possibility.

The potential generating force is gravity (not the contact
force) which only acts along the y direction. [1]

That's one way to formulate the problem. Let's call that the
one-potential formulation.

However, it is also possible to formulate the problem in terms
of a "force of constraint" which is associated with its own
very-steep-sided potential. Let's call this the two-potential
formulation. [2]

Yet - the blind use of
the expression F_x = -dU/dx gives an x component of force related to
the potential. The nonsense occurs from folding in the constraint y =
1 - sqrt(1-x2), not from any problem with the expression F_x = -dU/dx
per se.

Let's think about that some more. I'm guessing you prefer
the one-potential formulation, so I will adopt that for now.
That's fine with me. (If I have guessed wrong, please re-ask
the question.)

Let's look more closely at the expression F_x = -dU/dx.

I think the "d" operators there are not quite correct. I think
it is important to re-write that, to clearly indicate that it
involves a /partial/ derivative:

-∂U
F_x = ------ [3]
∂x

and at that point all sorts of red flags and alarms should go
off in your brain, because the conventional notation for partial
derivatives is bug-bait. It practically begs to be misunderstood.
A partial derivative is a /directional/ derivative, and the
notation in equation [3] does not sufficiently specify the
direction. Whenever you see a partial derivative, you should
immediately ask "at constant WHAT?" except perhaps in the rare
cases where the direction is obvious from context ... and even
when it is obvious it doesn't hurt to make it explicit anyway.
So, let's rewrite [3] as

-∂U |
F_x = ------| [4]
∂x |y

and meanwhile of course

-∂U |
F_y = ------| [5]
∂y |x

Now we can easily see what the problem is: In the one-dimensional
formulation of the problem, there is no motion "at constant y" or
"at constant x" (except on a set of measure zero). The only thing
that matters in this formulation is

-∂U |
F_θ = ------| [6]
∂θ |r

in polar coordinates.

=============

Here are several ways of analyzing the situation:

1) The easiest way is to start the analysis by immediately
switching to polar coordinates, and writing U as a function
of r and θ. Use the constraint to declare r to be a constant
Differentiate U with respect to only remaining variable θ.
You now have an equation of motion in terms of θ.

===

2) I suspect you want to treat it as a vector problem. We can do
that, but there are a couple of details that must be attended to.

Let's think of -dU as a vector. A gradient vector. Formally it
is a one-form (as opposed to a pointy vector). For details on
what I mean by this, see:
http://www.av8n.com/physics/thermo-forms.htm#sec-def-forms

In all generality, the chain rule says we can write:

-∂U | -∂U |
-dU = ------| dy + ------| dx [7]
∂y |x ∂x |y

provided the indicated partial derivatives exist. If U is the
gravitational potential, then the second term on the RHS of
equation [7] vanishes.

The vector dy can also be expanded. In all generality:

∂y | ∂y |
dy = ------| dθ + ------| dr [8]
∂θ |r ∂r |θ

provided the indicated partial derivatives exist (which they
always do, for ordinary polar coordinates).

If we have a constraint that says dr=0 then we can throw away
the second term on the RHS of equation [8]. Note that the
reason here is quite different from the reason for dropping
the second term in equation [7].

If we choose to use [8] to eliminate dy from [7], we get

-∂U | ∂y |
-dU = ------| ------| dθ [9]
∂y |x ∂θ |r

where both derivatives on the RHS are easy to evaluate
(assuming U contains only gravitational terms). This once
again gives us an equation of motion in terms of the single
variable θ,

===

3) If you really want to treat it as a two-dimensional motion
problem, with an x-force and a y-force, you can calculate F_θ
from equation [6] or equation [9] and then project out the
x-component and y-component of the force.

===

4) If you want to describe the force of constraint in terms of
a potential, that gives an elegant "unified" view of the problem.
Working out the details is possible but nasty. I'm not going to
do it. Not today anyway. It involves lots of infinities, because
an ideal constraint corresponds to a potential with infinitely
steep infinitely tall sides.
_______________________________________________
Forum for Physics Educators
Phys-l@carnot.physics.buffalo.edu
https://carnot.physics.buffalo.edu/mailman/listinfo/phys-l
_______________________________________________
Forum for Physics Educators
Phys-l@carnot.physics.buffalo.edu
https://carnot.physics.buffalo.edu/mailman/listinfo/phys-l