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Re: [Phys-l] Relativity Question about spring



On Tuesday, May 08, 2007 9:31 AM, William Maddox wrote

"The more general problem of
what to do with potential energy is often
ignored. One way this question is dealt with is
to say that if the system is chosen to be a
single object then E = (moc2 + sum of PEs) / (1-v2/c2)1/2"

This is not correct expression for the energy.
ALL kinds of energy of an object in its rest frame
are included into what we call the rest mass of this object.


"If the object is at rest in the chosen reference
frame this becomes E = moc2 + sum of PEs."

The same. The second term on the right is counted
twice here. It is already contained in the rest mass.

"If the reference frame is chosen so PE(grav) = mgh = 0,..."

This is a rather subtle point. In non-relativistic mechanics,
the potential energy is defined up to an additive constant.
If I am lifted 4 floors up from the ground floor, I can still
claim that I have no potential energy, if I explicitly indicate
the floor of my office, rather than the ground floor, as a
reference level. In other words, it is potential difference
between the levels that matters.
In relativity, since the potential energy as measure of
interaction between two parts of an object contributes to the
objects rest mass, the freedom of choice is lost. The potential
energy must be chosen so that the reference level corresponds to
zero interaction. (This is what is actually used in Newton's or
Coulomb's law for potential energy). In our case this would
correspond to Earth and an item infineiely far apart. If the item
now happens to get any closer and fall freely, the potential energy
decreases, which brings in the corresponding negative term into
the rest mass of the system "Earth + the item". But this does not
result in any change of the system's rest mass since it is exactly
compensated for by the corresponding positive contribution from
the increasing kinetic energy.
This argument ignores the very small loss in the form
of gravitational radiation caused by the fall. It is totally
negligible in our case, but may become prominent if the same item
falls into a black hole. Generally, the rest mass will decrease by
the amound of energy radiated away, and this would make the
gravitational mass defect. In nuclear physics such defect is usually
due to gamma radiation which may accompany formation of a compound
system.
In case of fast motions, the notion of potential energy becomes
less meaningful because of retardation effects. In this case we
must explicitly include into the rest mass of a system also the
energy of all the interaction fields in terms of the field theory,
e.g., E(EM)= 1/2(E^2 + B^2) for the EM field (stupid constants dropped).


"...then E = moc2 + 1/2kx2."

This is also wrong for the same reasons as above, if we are
to read E as the rest energy.


Moses Fayngold,
NJIT