Re: [Phys-l] Conservation of energy in nuclear reactions

[John Denker <jsd@av8n.com>]

On 05/02/2009 06:43 PM, Robert Cohen wrote:

> Consider an astronaut dropping a rock on the moon.  While the rock is
> falling, how does the kinetic energy of the moon/rock/astronaut SYSTEM change?
> How does the rest energy of the moon/rock/astronaut SYSTEM change?  Are there
> any other forms of energy that are changing?

There are at least two ways of answering the question.  It's
not obvious which was intended ... and indeed I suspect that
the ambiguity may be the main reason for asking the question.

1) The 4-momentum P of the overall "SYSTEM" is not changed by
 dropping the rock.  It's a conserved quantity.
 -- Therefore the overall mass M of the SYSTEM is not changed
  since it depends on P•P.
 -- Therefore in any chosen inertial frame, the overall energy
  P_0 is not changed.
 -- In such a frame, P_s is not changed, where P_s is the 3-momentum
  i.e. the spatial part of P.
 -- In such a frame, the overall kinetic energy (P_s)^2 / (2M) is 
  not changed.

These results are so obvious that it's hard to believe there could
ever be a question about it.

That's how I prefer to look at it.  Nice and simple.

2) Things are slightly more interesting if we cut open the black box
 and look inside.  We write P = Q + R where P is the overall 4-momentum
 of the system, R is the 4-momentum of the rock, and Q is the 4-momentum
 of the other material objects, i.e. moon + astronaut.

 -- As previously established, dropping the rock has no effect on P.
 -- Dropping the rock causes a change in R ... and a less obvious
  but still important change in Q.
 -- In an inertial frame comoving with the astronaut, dropping the
  rock allows the KE of the rock to increase.  So there is some KE
  "within" the SYSTEM.  However, this is not properly considered KE
  "of" the overall SYSTEM.  We see that there is tremendous potential
  for confusion about what is meant by "the KE".  This confusion has
  nothing to do with relativity or with nuclear reactions;  it goes
  back to the most elementary Day One definition of KE.
     http://www.av8n.com/physics/kinetic-energy.htm
 -- Dropping the rock does not change the mass (i.e. rest energy)
  of the rock ... and also does not change the mass (i.e. rest energy)
  of the astronaut+moon subsystem.
 -- We now calculate the rest mass of the Q+R combination.  In particular, 
  the negative of mass squared is (Q+R)•(Q+R) = Q•Q + R•R + 2Q•R.  That's 
  interesting, because I was taught when I was 5 years old that the mass 
  of a combined system was equal to the sum of the masses of the component 
  parts, and we have just proved that this is not true.  The correction 
  term, i.e. the cross term 2Q•R, is small in ordinary rock-dropping
  experiments, but may well be significant in the analogous nuclear 
  reactions.
 -- We have also stumbled onto the fact that the overall system cannot
  be written as P=Q+R.  In particular, it should be clear that we cannot
  have P•P = Q•Q + R•R + 2Q•R because P•P is constant, Q•Q is constant,
  R•R is constant, but the Q•R cross term is not constant while the rock
  is dropping.  We simply must write P=Q+R+G where G is the 4-momentum
  of the gravitational field.

So, the primary sources of possible confusion are:
 -- Mass is not additive.  The mass of the compound system is the mass 
  of the constituent parts _plus_ a term representing the binding energy.
 -- Changes in binding energy are immeasurably small for ordinary chemical
  reactions and ordinary rock-dropping.  That's true as far as I know, but 
  the changes seem just barely beyond the reach of experiments, and I would 
  not be the least surprised to read that some enterprising experimentalist
  managed to pull off such a measurement.
 -- Changes in binding energy are routinely measured for nuclear reactions.
 -- KE within a system is not properly considered KE "of" the overall
  system.
 -- Gravitational potential energy is not properly considered as "in" or
  "of" the rock, even if/when it is for all practical purposes determined
  by what the rock is doing.

Compare the complexity of analysis (2) with the simplicity of analysis (1).

> Here is an alternative scenario, if you'd prefer.
>  
> Consider an electron and proton separated by an finite amount. While
> the two particles approach, how does the kinetic energy of the
> proton/electron system change? How does the rest energy of the
> proton/electron system change? Are there any other forms of energy
> that are changing?

Same question, same answer.

> Or, if you prefer a nuclear example, consider nucleons in a less
> stable configuration moving to a more stable configuration.

Same question, same answer.  The only difference here is that we cannot
separate the nucleon/nucleon system into a "small" subsystem (the rock) 
and a "big" subsystem (the moon).  But the analysis I gave above does not '
depend on this separation, so the conclusions are unchanged.  (There are 
additional observations that one could make based on such separation, 
but I leave those for the interested reader to work out.)