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

["Robert Cohen" <Robert.Cohen@po-box.esu.edu>]

OK, I'll restate my question.
 
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?
 
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?
 
Or, if you prefer a nuclear example, consider nucleons in a less stable configuration moving to a more stable configuration.

________________________________

From: phys-l-bounces@carnot.physics.buffalo.edu on behalf of John Denker
Sent: Fri 5/1/2009 4:58 PM
To: Forum for Physics Educators
Subject: Re: [Phys-l] Conservation of energy in nuclear reactions



On 05/01/2009 11:29 AM, Robert Cohen wrote:

> > .... the rest energy mc^2.
>
> Still, the question is whether this form of energy is *in addition to*
> the energies we considered previously (like the gravitational energy or
> whatever) or *instead of*.  That is what I am confused about and how I
> could simply explain this to non-scientists.
>
> Consider an astronaut dropping a rock on the moon.  While the rock is
> falling, how does the kinetic energy of the moon/rock/astronaut change?
> How does the rest energy of the moon/rock/astronaut change?  Are there
> any other forms of energy that are changing?

First of all, let us take note of some unhelpful background:  In
some circles it is taught that there is gravitational potential
energy "in" the rock.  However, the law of universal gravitation
gives us an energy term G M m / r which is symmetric w.r.t interchange
of M and m.  So really the energy is a property of the rock/moon
system, not either one separately.  In the limit where M >> m we
can approximate the moon as stationary and the local gravitational
field as unchanging, so that the system energy "effectively" depends
only on what m is doing ... but we need to keep things in perspective
and recognize these approximations as approximations.  They can
cause problems when we are dealing with epistemological issues.

In electrostatics, when we have one electron interacting with another,
it is conventional and almost entirely satisfactory to consider the
interaction energy as residing in the electric field, not in either
electron. 

When dealing with gravitational fields, it is not entirely conventional
to consider the interaction energy as residing in the field.  You can
do that if you want, or you can just say the energy is a property of
the rock/moon _system_ and leave it at that.  I don't care.  Hypotheses
non fingo.

Secondly, we are reminded that "conserved" is not synonymous with "constant".
Local conservation of energy says that the energy in any region is constant
_except_ insofar as energy flows across the boundary of that region.  We
really must insist on this point.
  http://www.av8n.com/physics/conservative-flow.htm
Our main region of interest contains the rock and nothing else.

Now we can address the question that was asked.  We consider the energy
that is strictly inside the region of interest, i.e. inside the rock; 
this does not include the gravitational interaction energy.

 -- Initially the energy is just E = m c^2 i.e. the rest energy.

 -- After it has dropped a ways, the energy is larger.  The rock has some
  kinetic energy in addition to the rest energy it started with.  The
  energy of the rock is conserved but not constant.  Energy has flowed
  into the rock from the gravitational field, in accordance with the
  work/KE theorem.

It is highly amusing to show, by considering the invariant length of the
4-momentum vector, that to lowest order the kinetic energy of the rock
must be .5 (ps)^2 / (2m) where ps is the 3-momentum, i.e. the spatial
part of the 4-momentum vector p.  Relativity tells us what the non-
relativistic result must be.
  http://www.av8n.com/physics/mass.htm#main-KE


If that doesn't answer the question, please re-ask.

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