Re: [Phys-l] Relativity Question

[John Denker <jsd@av8n.com>]

On 05/07/2007 11:46 AM, Tom Sandin wrote:

> It seems to me that the only potential energy that can be used is the 
> potential energy that belongs to the system being analyzed.

Yes.

  (Nitpickers might argue that "belongs to" is not a very
  precise way of saying it ... but in this case the meaning
  is clear so let's not worry about the wording.)

> In this case, the ball does not own the gravitational potential 
> energy, Rather, that potential energy belongs to the earth-ball 
> system.

Yes, the system is what counts.

> Ignoring frictional losses, that decrease in potential energy of the 
> earth-ball system shows up as an increase in the translational and 
> rotational kinetic energies of the ball.

That's narrowly true as stated ... but it blithely makes the
assumption that we can ignore losses, and that's not always
a good assumption.

For example, if we are observing the mass deficit associated with
the binding energy (in, say, a nuclear reaction) then the losses
are all-important;  if the binding energy of the system is not
lost to the system, no binding occurred.

> Of course, being a relativistic mass proponent, I say the mass of the 
> ball increases by an amount equal to its total kinetic energy 
> increase divided by c^2.

Whatever.

> As I understand the anti-relativistic mass viewpoint, the mass of the 
> ball increases by an amount equal to its rotational kinetic energy 
> divided by c^2 (because in a co-moving frame of reference the ball 
> has no translational kinetic energy but does have an increase in its 
> rotational kinetic energy). This increase is one example of a varying 
> "invariant" mass.

If that means what I think it means, it is wrong twice over.

*) First of all, the recent talk about variance of the invariant
mass is just a bunch of silly word games.  Next time, please resist
the temptation.  In this context, invariance means invariance with
respect to the Lorentz group, i.e. 4-dimensional rotations, including
boosts as well as spacelike rotations.

Even in non-relativistic geometry, we say that "THE" length of a
ruler is invariant with respect to rotations.  That doesn't mean it
is invariant to every imaginable manipulation;  for instance, it
is probably not invariant to beating on it with hammer and anvil.

The same goes for 4-D rotations:  "THE" length of the ruler is
invariant with respect to the choice of observer;  the length
has a physical reality independent of which observer(s), if
any, are looking at it.  OTOH there are lots of ways of changing
the ruler's length if you beat on it hard enough.

The same goes for mass:  It has a physical reality independent of
which observer(s), if any, are looking at it.  Changing observers
doesn't change the mass.  OTOH if you physically beat on the particle,
/that/ can change the mass.

*) Ironically, in the given example of allegedly varying mass, the
mass doesn't change at all.

Consider the less-than-general case of nearly lossless motion, as
exemplified by a comet in a highly eccentric orbit around the sun.
Because of the eccentricity, KE is constantly being converted to
PE and back again.  But the total energy E within the boundaries
of the system is constant, and the total 3-momentum ps within the
boundaries of the system is constant.  There is absolutely no
probably applying the formula

        m^2 = E^2 - ps^2

to the total E and total ps to find the total m ... which is
constant with respect to time.  It is *not* necessary to look
inside the black box, i.e. inside the boundaries of the
system, to ask how much of the energy is kinetic and how much
is potential at any given time.

As an example where this line of reasoning is put to good use,
see
  http://www.av8n.com/physics/bevatron.htm
Suppose we want to design an accelerator to produce antiprotons.
The question for today is, how much energy must the accelerator
supply?  There is an easy way to answer this question.  This
provides a wonderful illustration of the power of vectors in
general, and four-vectors in particular.   No math is required
beyond high-school ``Algebra I'' plus the rule for taking dot
products of 4-vectors.