Re: Derivative of KE

["John S. Denker" <jsd@MONMOUTH.COM>]

At 10:04 PM 10/16/01 -0400, Tim O'Donnell wrote:
>  KE = 1/2mv^2 equation and take the derivative I
> believe I get  mv which is momentum (p).  Now clearly
> there has to be a relationship between KE and p simply
> because we are talking about the same mass and velocity.
> As I understand derivatives, if I was graphing the  delta
> KE of an object with respect to time, the slope of said
> graph would be the momentum of the object at that time.

That's about half right.  Do not confuse the following:
  -- The derivative of total energy w.r.t time is power.  That's an
interesting quantity, but not to be confused with momentum.
  -- The derivative of the Lagrangian w.r.t velocity is the
momentum.  That's a horse of a different color, and very, very interesting
in its own right.

In the case of a free particle, the Lagrangian is equal to the kinetic
energy and to the total energy, so in this case it is no accident that the
derivative of the KE w.r.t velocity is the momentum.  No accident indeed!!!!

More generally, the Lagrangian is KE - PE, in contrast to the plain old
energy which is KE + PE.  The Lagrangian has dimensions of energy, but it
is _not_ "the energy" and does not obey a local conservation law the way
energy does.

The Lagrangian is a wonderful creature.  Given the Lagrangian you can find
the energy -- but the converse is not true!  The Lagrangian knows all,
tells all.

As an example of the power of the Lagrangian, let's try to figure out the
analogy between a mechanical oscillator (mass on a spring) and an
electrical oscillator (inductor and capacitor).  You know there is a
quantum [position, momentum] uncertainty relation for the mechanical
oscillator.  The question is, what is the corresponding uncertainty
relation for the electrical oscillator?  What electrical quantity
corresponds to position?  What is the dynamically conjugate momentum?  The
Lagrangian will tell us!

Here's the drill:
 -- Choose some relevant variable to be the "coordinate".  For this
example, I choose the charge Q.  Once we have chosen a coordinate, the rest
is just turning the crank;  the formalism takes care of everything.
 -- Write down the energy in the capacitor in terms of the chosen
coordinate;  in this case it is 1/2 Q^2 / C.
 -- Write down the energy in the inductor in terms of the SAME
variable.  That must be 1/2 L (dQ/dt)^2.
 -- Whatever piece of the energy depends on the chosen coordinate (Q) must
be the potential energy.  Whatever piece of the energy depends on (dQ/dt)
must be the kinetic energy.  Therefore the Lagrangian is 1/2 L (dQ/dt)^2 -
1/2 Q^2 / C.
 -- Differentiate the Lagrangian w.r.t velocity (dQ/dt).  The answer is L
(dQ/dt) which is just LI which is just the flux in the inductor.
 -- Conclude that flux is dynamically conjugate to charge.  Hence we will
have a [charge, flux] uncertainty relation.  The correspondence is
        charge  <-->  position
        flux    <-->  momentum.

Homework:
  1a) In the foregoing, find the electrical quantity that plays the role
corresponding to mass.
  1b) Similarly, find the electrical quantity that plays the role
corresponding to spring constant.

2) Repeat the foregoing steps, except use flux (not charge) as the chosen
coordinate (generalized position).  Turn the crank.  Find the dynamically
conjugate quantity (generalized momentum).
  2a) Identify the electrical quantity that plays the role corresponding
to mass.
  2b) Identify the electrical quantity that plays the role corresponding
to spring constant.

Hint:  Answer 2a is not the same as answer 1a.

Last but not least:  Given the Lagrangian, you can derive the full equation
of motion via the principle of least action.  The Lagrangian knows all,
tells all.

Notice:  Limited warranty:  The foregoing has amazingly great but not
unlimited generality.  Special handing is required if the kinetic energy
contains funny stuff such as second (or higher) derivatives of the
coordinate.  You also need to think a bit when generalizing from
point-particles to fields.  Check ye olde classical mechanics book to get
the details.